Anybody interested in Dirac's equation?

[AnssiH]

It is no more than the definition of “K” which yields “c” as the velocity of light. The same issue came up in the deduction of Schrödinger's equation (go back and look at it once).

 

The “2” on the right arose from squaring the alpha and beta operators (which yielded 1/2). Later, in order to obtain exactly Schrödinger's equation, I have to define [imath]c=\frac{1}{K\sqrt{2}}[/imath]. It doesn't really make any difference as, in my analysis of relativity, I show that the actual velocity, v_? is immaterial; we have to define a clock before we can define velocities.

 

The whole thing is really a simple consequence of my definition of the alpha and beta operators. I should perhaps define them to be in alignment with the Pauli spin matrices and Dirac's work as it would be less confusing, but I defined them for a different purpose and had been using my original definition long before (probably something like ten years before) I managed to pull down my first approximate solution. It never felt right to go back and change things after the fact.

 

Ooooohhhh, actually, yeah, what was throwing me off there was that the definition for energy was different in Schrödinger's thread, but it dawns on me that there's the C there all over the place now, so is the issue simply that we are now talking about relativistic energy... Different thing than what we were talking about in Schrödinger's...?

 

There is no meaning to it at all since they simply commute with one another. (I can prove that if you need to have it proved.)

 

No I don't think that's necessary at this point.

 

The only reason I changed the order here was so I could explicitly show the time derivative of [imath]\vec{\Psi}_2[/imath] by itself. The given derivative of [imath]\vec{\Psi}_1\vec{\Psi}_2[/imath] is,

[math]\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2 = \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1,[/math]

 

Okay, now I must say this whole step is little bit tricky to me... After a lot of head scratching, I think I'm starting to see what is going on here, but I need to move with little tiny steps to make sure I pick it up correctly (I've gone through very many interpretations in my head by now :)

 

So here's what I gather now... What we had before this step was:

 

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2+c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2= \frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

[/math]

 

Where the tau components are separated into their own terms.

 

Now this move:

 

[math]\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2 = \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1[/math]

 

Should I take it that, that separation means exactly that the first term (on the right side) is the time derivative for the first element, and thus would allow us to write:

 

[math]

\left\{ c\vec{\alpha}_1 \cdot \vec{p}_1 + \alpha_{1\tau} m_1c^2 \right\} \vec{\Psi}_1\vec{\Psi}_2= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2

[/math]

 

And likewise, the second term is the time derivative for the second element, and thus would allow us to write:

 

[math]

\left\{ c\vec{\alpha}_2 \cdot \vec{p}_2 + \alpha_{2\tau} m_2c^2 \right\} \vec{\Psi}_1\vec{\Psi}_2= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

Or, putting the tau component back into [imath]\vec{p}_2[/imath]:

 

[math]

c\vec{\alpha}_2 \cdot \vec{p}_2 \vec{\Psi}_1\vec{\Psi}_2= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

Which is essentially what you have in the OP.

 

Even then, I get confused by your comment;

 

and I am interested in getting rid of the second term: i.e., via [imath]c\vec{\alpha}_2\cdot\vec{p}_2\vec{\Psi}_2-i\hbar\frac{1}{2}\frac{\partial}{\partial t}\vec{\Psi}_2 =0[/imath] where [imath]\vec{p}_2[/imath] is taken to be the four dimensional version of the momentum in my Euclidean space. This is just a quick and dirty way of getting rid of some terms.

 

I would understand it if it said [math]c\vec{\alpha}_2\cdot\vec{p}_2\vec{\Psi}_2-i\hbar\frac{1}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2 =0[/math] (either a typo or I am overlooking something), except that even then I do not know what do you mean by "getting rid of the second term" (which term in which equation?) via that fact...

 

At any rate;

 

I suspect you are not understanding my comment, “(which can be deduced from the fact that the dot product is the component of alpha in the direction of the momentum times the magnitude of that momentum)”.

 

That was indeed the problem. I was trying to think of what do I know about the component of the alpha in the direction of the momentum, and things started to seem very complicated :)

 

Remember, we are working in a Euclidean coordinate system and that fact yields a number of vector relationships which allow us to simplify vector product calculations (take a quick look at the wikipedia entry on dot products; there is a lot of valuable information there) . If we take the momentum in the tau direction to be “mc” then

[math]

\vec{p}_2 = p_x \hat{x}+p_y \hat{y}+p_z \hat{z}+p_\tau \hat{\tau}= p_x \hat{x}+p_y \hat{y}+p_z \hat{z}+mc \hat{\tau}

[/math]

 

and [imath]\vec{\alpha}_2\cdot\vec{p}_2[/imath] (the left hand side of the expression of interest) is a simple dot product. In a Euclidean coordinate system, the dot product of two vectors is invariant under rotation ([imath]\vec{a}\cdot\vec{b}=|a|\;|b| cos \theta[/imath] where theta is the angle between the two vectors). It can thus be seen as the magnitude of one vector multiplied by the component of the other in the direction of the first.

 

Fundamentally, from a physics perspective, the expression we are working with is a simple statement that the energy (as defined through the momentum) is the same as the energy (as defined through the time derivative): i.e., that fact has nothing to do with the coordinates used to represent the phenomena under examination, it is a characteristic of the phenomena itself. Thus we can always rotate our coordinate system to a state where [imath]\vec{p}_2[/imath] is parallel to one of the four orthogonal axes (such a rotation will not change the result of the dot product). In that case, the dot product is the component of [imath]\vec{\alpha}[/imath] along that axis (that single component) times the magnitude of [imath]\vec{p}_2[/imath] or [imath]\sqrt{\frac{1}{2}}|\vec{p}_2|[/imath] (we have already shown earlier that [imath]\alpha_i^2=1/2[/imath] so [imath]|\alpha_i|[/imath] must be [imath]\sqrt{\frac{1}{2}}[/imath]).

 

I spent a lot of time scratching my head on the above too, and I see the point of dot product being invariant upon rotation, and that consequently we are able to choose a convenient coordinate system to see the result, and I remember the anti-commuting elements squaring to [imath]\frac{1}{2}[/imath].

 

And overall that all looks valid to me now, but the only stone in my shoe is that I do not understand at all why do you point out that bit of "If we take the momentum in the tau direction to be “mc” then..."

 

If it's important for the validity of the rest of the paragraph, then I suspect I am misunderstanding something... :I

 

Another way to see the same result is to view [imath]\vec{\alpha}_2\cdot\vec{p}_2= i\hbar\sqrt{\frac{1}{2}}\frac{\partial}{\partial t}[/imath] as an operator identity when operating on [imath]\vec{\Psi}_2[/imath]. If we operate twice we end up with [imath]\frac{1}{2}|p_2|^2\vec{\Psi}_2=-\hbar^2\frac{1}{2}\frac{\partial^2}{\partial t^2}\vec{\Psi}_2[/imath] which is exactly the square of our first result again implying that [imath]\vec{\alpha}_2\cdot\vec{p}_2=\sqrt{\frac{1}{2}}|\vec{p}_2|[/imath] when operating on [imath]\vec{\Psi}_2[/imath].

 

That all looks valid to me too, except I do not know where you got that first expression exactly... :I

 

-Anssi

[Doctordick]

Ooooohhhh, actually, yeah, what was throwing me off there was that the definition for energy was different in Schrödinger's thread,

From the Schrödinger's thread:

First, I will define ”the Energy Operator” as [imath]i\hbar\frac{\partial}{\partial t}[/imath] (and thus, the conserved quantity required by the fact of shift symmetry in the t index becomes “energy”: i.e., energy is conserved by definition). A second definition totally consistent with what has already been presented is to define the expectation value of “energy” to be given by

[math]E=i\hbar\int\vec{\Psi}^\dagger\cdot\frac{\partial}{\partial t}\vec{\Psi}dV.[/math]

and, from the Dirac thread (this thread)

The partial derivative [imath]i\hbar\frac{\partial}{\partial t}[/imath] is exactly the same energy operator defined through my deduction of Schrödinger's equation.

So there is absolutely no difference in the definition of Energy. You are confusing the “wave equation” (which has the constant K which is directly related to the wave velocity) with the energy equation. Remember, I multiplied through by [imath]-\hbar c[/imath] as one of the steps in the Schrödinger derivation. That multiplication (together with the definition [imath] c=\frac{1}{K\sqrt{2}}[/imath]) essentially moved the c over to the momentum and potential terms and out of what becomes the Energy term. (Note that the energy of a photon is c times the magnitude of its momentum and, if the momentum in the tau direction is mc, that term becomes [imath]mc^2[/imath]). The c is not “all over the place” it turns up in very specific positions.

... but it dawns on me that there's the C there all over the place now, so is the issue simply that we are now talking about relativistic energy... Different thing than what we were talking about in Schrödinger's...?

No, energy is exactly the same thing in both places.

So here's what I gather now... What we had before this step was:

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2+c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2= \frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

[/math]

 

Where the tau components are separated into their own terms.

Correct!

Now this move:

 

[math]\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2 = \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1[/math]

Also correct! But at this point you begin to go astray. The next step is to write down

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2+c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2[/math]

 

[math]= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

which can be rewritten as

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2[/math]

 

[math] +\left[\{c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2\}\vec{\Psi}_2-\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\right]\vec{\Psi}_1= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2

[/math]

 

essentially moving the time derivative of [imath]\vec{\Psi}_2[/imath] to the left hand side of the equation. It is the resulting term in square brackets which I wish to set to zero. You make the error of presuming the same thing can be done with the other element. If what I have just said is true, then

And likewise, the second term is the time derivative for the second element, and thus would allow us to write:

[math]

\left\{ c\vec{\alpha}_2 \cdot \vec{p}_2 + \alpha_{2\tau} m_2c^2 \right\} \vec{\Psi}_1\vec{\Psi}_2= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

is false. What you are missing is the fact that the factorization into [imath]\vec{\Psi}_1[/imath] and [imath]\vec{\Psi}_2[/imath] can presume one is a function only of its position but not both. Which ever one is chosen to not depend upon the probability of finding the other, sets the other as possibly depending upon the firsts position (if they are totally independent elements then the second need not depend upon the first; however, in this case, they are not independent elements). Essentially, it is the term [imath]-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)[/imath] which provides the connection between them. If we say the probability of finding element two can be established (not knowing where element one is; which is exactly what we have essentially done here) then the equation which controls the probability of finding element one can depend upon the probability distribution of element two and the connection is provided by [imath]-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)[/imath].

I would understand it if it said [math]c\vec{\alpha}_2\cdot\vec{p}_2\vec{\Psi}_2-i\hbar\frac{1}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2 =0[/math] (either a typo or I am overlooking something), except that even then I do not know what do you mean by "getting rid of the second term" (which term in which equation?) via that fact...

It is indeed a typo (I do a lot of those don't I? :doh: ) By “getting rid of the second term”, I mean the second differential with respect to t and the momentum terms (and mass) of the second element. I expressly said, “where [imath]\vec{p}_2[/imath] is taken to be the four dimensional version of the momentum in my Euclidean space” because I wanted to include the mass term (the momentum in the tau direction) which I had earlier defined not to be there: i.e., I had specifically defined [imath]\vec{p}[/imath] to be Dirac's three dimensional momentum and, for this particular series of steps, I wanted to temporarily go back to my four dimensional notation. I was trying to write things out in a simple way and managed to make it confusing; I apologize.

That was indeed the problem. I was trying to think of what do I know about the component of the alpha in the direction of the momentum, and things started to seem very complicated :)

Remember, we are doing the rotation for the sole purpose of evaluating [imath]\vec{\alpha}_2\cdot\vec{p}_2[/imath] (which does not depend upon that orientation). If we use the coordinate system where [imath]\vec{p}_2[/imath] lies along the x axis then [imath]p_x = |\vec{p}_2|[/imath] and all the other components vanish. In that case, the only non-zero part of the dot product results from [imath]\alpha_x[/imath] which provides that [imath]\sqrt{\frac{1}{2}}[/imath] term. I could just as well have let [imath]\vec{p}_2[/imath] lie along the tau axis (in that case, we would have been looking at element 2 in its own rest frame: i.e., the three dimensional version of momentum would then vanish. I didn't do that because I didn't want to imply that the evaluation depended upon the frame of reference chosen. Some things are just easier to do in one frame rather than another.

... but the only stone in my shoe is that I do not understand at all why do you point out that bit of "If we take the momentum in the tau direction to be “mc” then..."

Purely because I have asserted that [imath]\vec{p}[/imath] is to be Dirac's three dimensional momentum and I wanted to point out my intentions explicitly whenever I was violating that assertion. Again, I apologize for the confusion.

That all looks valid to me too, except I do not know where you got that first expression exactly... :I

It is essentially the relationship in square brackets above; that collection of terms I am trying to get rid of. This is just another way of looking at the problem of evaluating [imath]\vec{\alpha}_2\cdot\vec{p}_2[/imath].

 

You are doing good and I hope I have cleared up some of your confusion.

 

Have fun -- Dick

[AnssiH]

From the Schrödinger's thread:

and, from the Dirac thread (this thread)

So there is absolutely no difference in the definition of Energy. You are confusing the “wave equation” (which has the constant K which is directly related to the wave velocity) with the energy equation. Remember, I multiplied through by [imath]-\hbar c[/imath] as one of the steps in the Schrödinger derivation. That multiplication (together with the definition [imath] c=\frac{1}{K\sqrt{2}}[/imath]) essentially moved the c over to the momentum and potential terms and out of what becomes the Energy term. (Note that the energy of a photon is c times the magnitude of its momentum and, if the momentum in the tau direction is mc, that term becomes [imath]mc^2[/imath]).

 

Oh okay, I think I see what you mean. What was throwing me off was simply the statement in the OP:

 

If the vector [imath]\vec{p}_2[/imath] is taken to be a four dimensional vector momentum in my [imath]x,y,z,\tau[/imath] space we can write the energy as

[math]\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1 =c\vec{\alpha}_2\cdot\vec{p}_2\vec{\Psi}_2\vec{\Psi}_1=\sqrt{\frac{1}{2}}|cp_2|\vec{\Psi}_2\vec{\Psi}_1[/math]

 

i.e. when I saw "we can write the energy as", my mathematically unsophisticated mind went looking for an explicit statement of energy (the exact form of the definition) and couldn't find it anywhere :)

 

But at this point you begin to go astray. The next step is to write down

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2+c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2[/math]

 

[math]= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

which can be rewritten as

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2[/math]

 

[math] +\left[c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2\vec{\Psi}_2-\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\right]\vec{\Psi}_1= \left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2

[/math]

 

Ah of course... Tiny typo there too btw, [imath]\vec{\Psi}_2[/imath] is not connected to the first term in the square brackets like it should...

 

essentially moving the time derivative of [imath]\vec{\Psi}_2[/imath] to the left hand side of the equation. It is the resulting term in square brackets which I wish to set to zero.

 

Yup....

 

You make the error of presuming the same thing can be done with the other element. If what I have just said is true, then

is false. What you are missing is the fact that the factorization into [imath]\vec{\Psi}_1[/imath] and [imath]\vec{\Psi}_2[/imath] can presume one is a function only of its position but not both. Which ever one is chosen to not depend upon the probability of finding the other, sets the other as possibly depending upon the firsts position (if they are totally independent elements then the second need not depend upon the first; however, in this case, they are not independent elements). Essentially, it is the term [imath]-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)[/imath] which provides the connection between them. If we say the probability of finding element two can be established (not knowing where element one is; which is exactly what we have essentially done here) then the equation which controls the probability of finding element one can depend upon the probability distribution of element two and the connection is provided by [imath]-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)[/imath].

 

Ah, correct. I am so unfamiliar with this stuff that even though I still remembered [imath]\vec{\Psi}_2[/imath] was the function chosen to not depend on the probability distribution of [imath]x_1[/imath], I was not sure how to apply that information here.

 

It is indeed a typo (I do a lot of those don't I? :doh: ) By “getting rid of the second term”, I mean the second differential with respect to t and the momentum terms (and mass) of the second element.

 

Haha, I made a very unfortunate misinterpretation there originally; I didn't realize you were talking about the terms of [imath]x_2[/imath] for the simple reason of getting rid of them in the next step in the OP... I was very puzzled as what I was looking at was "here's how we get rid of [imath]x_2[/imath]" and I just saw an equation which was stripped of [imath]x_1[/imath] instead :D

 

At any rate, with that, I understand the OP all the way to:

 

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau}m_1c^2 -2i\hbar c \beta\delta(\vec{x}_1-\vec{x}_2\}\vec{\Psi}_1\vec{\Psi}_2 = \sqrt{\frac{1}{2}}\vec{\Psi}_2 i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

i.e. I understand what you've done on the right side there.

 

Onto the next step:

If one now defines a new operator [imath]\vec{\gamma}=\vec{\alpha}_1\beta[/imath], since [imath]\vec{\alpha}_1\cdot\vec{\alpha}_1=2[/imath] (it consists of four components each being 1/2) it should be clear that [imath]\vec{\alpha}_1\cdot\vec{\gamma}=2\beta[/imath]. Making that substitution in the above we have

[math]\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau}m_0c^2 -i\hbar c \vec{\alpha}_1\cdot\vec{\gamma}\delta(\vec{x}_1-\vec{x}_2\}\vec{\Psi}_1\vec{\Psi}_2 = \sqrt{\frac{1}{2}}\vec{\Psi}_2 i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1.[/math]

 

Well I'm (still) not exactly fluent with dot products, but since you state [imath]\vec{\alpha}_1\cdot\vec{\gamma}=2\beta[/imath], i.e. [imath]\vec{\alpha}_1\cdot\vec{\alpha}_1\beta=2\beta[/imath], I take it that [imath](\vec{\alpha}_1\cdot\vec{\alpha}_1)\beta = \vec{\alpha}_1\cdot(\vec{\alpha}_1\beta)[/imath]. I.e. the order of the operations doesn't matter... Sounds reasonable to me anyway and I suspect this is well defined property of dot product.

 

And yes in that case that substitution is trivially true (and I didn't even spot any typos ;) )

 

Thank you; I'll continue from here soon...

 

-Anssi

[Doctordick]

i.e. when I saw "we can write the energy as", my mathematically unsophisticated mind went looking for an explicit statement of energy (the exact form of the definition) and couldn't find it anywhere :)

Yes, I should have said, “the energy relationship as”; the [imath]\sqrt{\frac{1}{2}}[/imath] arises from my definition of alpha. (I think I will change the wording.)

Ah of course... Tiny typo there too btw, [imath]\vec{\Psi}_2[/imath] is not connected to the first term in the square brackets like it should...

Thanks, I have already fixed it.

Ah, correct. I am so unfamiliar with this stuff that even though I still remembered [imath]\vec{\Psi}_2[/imath] was the function chosen to not depend on the probability distribution of [imath]x_1[/imath], I was not sure how to apply that information here.

It's my fault for not making the issue clear. If [imath]\vec{\Psi}_2[/imath] is the probability not dependent upon the other element, then the other element could be anywhere (including on the far side of the galaxy) so the interaction term can be presumed to be zero insofar as it comes to establishing [imath]\vec{\Psi}_2[/imath].

Well I'm (still) not exactly fluent with dot products, but since you state [imath]\vec{\alpha}_1\cdot\vec{\gamma}=2\beta[/imath], i.e. [imath]\vec{\alpha}_1\cdot\vec{\alpha}_1\beta=2\beta[/imath], I take it that [imath](\vec{\alpha}_1\cdot\vec{\alpha}_1)\beta = \vec{\alpha}_1\cdot(\vec{\alpha}_1\beta)[/imath]. I.e. the order of the operations doesn't matter... Sounds reasonable to me anyway and I suspect this is well defined property of dot product.

The important issue there is that beta is not a vector operator and, as such, is a factorable term. The dot product is essentially identical to ordinary multiplication except for the need to take into account the fact that the dot product between unit vectors (the things with the “[imath]\hat{ }[/imath]”) is zero if they are orthogonal and one if they are same unit vectors.

And yes in that case that substitution is trivially true (and I didn't even spot any typos ;) )

I guess I was lucky!

 

Very good -- Dick

[AnssiH]

I haven't been able to concentrate on this during the weekend quite the way I wished I would, so here's just a small step...

 

I guess I was lucky!

 

Actually there was a typo there, it was just too tiny even for me to spot at first (even when I did look quite closely)! :D Nothing fatal though; the Dirac delta function is just missing the closing parenthesis in those two last expressions I've been through.

 

The last step in this process is to left multiply by [imath]\vec{\Psi}_2^\dagger\cdot[/imath] and integrate over [imath]\vec{x}_2[/imath]. One will obtain a unit coefficient in every term except for the term containing the Dirac delta function (I will refer to this as the interaction term as it is the only term containing [imath]\vec{x}_2[/imath], the coordinate of the second element). Integration over the interaction term will introduce a coefficient consisting of [imath]\vec{\Psi}_2^\dagger(\vec{x}_1,t)\cdot\vec{\Psi}_2^(\vec{x}_1,t)[/imath] (integration over a delta function yields a spike at the point [imath]\vec{x}_1=\vec{x}_2[/imath]: i.e., we obtain exactly the explicit probability amplitude that element #2 will be at exactly the same point as element #1). After this integration (and multiplying everything by [imath]\sqrt{2}[/imath] the fundamental equation has the form,

[math]\sqrt{2}\left\{c\vec{\alpha}_1\cdot \vec{p}_1 +\alpha_\tau m_1c^2 - \vec{\alpha}_1 \cdot i\hbar c\left[\vec{\gamma}\vec{\Psi}_2^\dagger(x_1,t)\cdot\vec{\Psi}_2(x_1,t)\right]\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1.[/math]

 

Yup.

 

It is almost trivial to identify this with Dirac's equation: one need only recognize that Dirac's vector representation of his anti-commuting alpha matrix (i.e., in their mathematical consequences) is exactly equivalent to [imath]\sqrt{2}[/imath] times the first three components of my vector alpha operator and his anti-commuting beta matrix corresponds exactly to [imath]\sqrt{2}\alpha_\tau[/imath].

 

I'm still supposed to take a good look at your explanation of how Dirac equation and his matrix notation works exactly, but right now I just have to take the above on faith.

 

In addition, his equation makes the assumption that there exists a four component complex solution: i.e., his function [imath]\Psi(\vec{x},t)[/imath] can certainly be seen as a simplified approximation to my more general [imath]\vec{\Psi}(\vec{x},t)[/imath] (which, of course, includes many much more complex possibilities).

 

That part I think I understand.

 

(Note further that [imath]\vec{x}_2[/imath] has vanished from the equation so the subscript is no longer necessary). Making these substitutions (and I sincerely apologize for having used the same Greek letters for my operators as Dirac used for his as I understand the confusion it can lead to), we then have

[math]\left\{c\vec{\alpha}\cdot \left(\vec{p} -i\hbar c\sqrt{2}\left[\vec{\gamma} \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right)+\beta mc^2 -i\hbar c\sqrt{2}\left[ \gamma_\tau \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right\}\Psi= i\hbar \frac{\partial}{\partial t}\Psi[/math]

 

I am not able to find my way to that result... :(

 

What I started with is:

 

[math]

\left\{ \sqrt{2} c\vec{\alpha}_1\cdot \vec{p}_1 + \sqrt{2} \alpha_\tau m_1c^2 - \sqrt{2} \vec{\alpha}_1 \cdot i\hbar c\left[\vec{\gamma}\vec{\Psi}_2^\dagger(x_1,t)\cdot\vec{\Psi}_2(x_1,t)\right]\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

Then, since you said "Dirac's vector representation of his anti-commuting alpha matrix is exactly equivalent to [imath]\sqrt{2}[/imath] times the first three components of my vector alpha operator", I suppose it means I'm ought to multiply the alpha operators by [imath]\sqrt{2}[/imath] (ending up with just 2 times the alpha operator since those terms already contained [imath]\sqrt{2}[/imath]).

 

And since you said "his anti-commuting beta matrix corresponds exactly to [imath]\sqrt{2}\alpha_\tau[/imath]", I substituted that in there. I noticed the last term contains the factor [imath]2c\vec{\alpha}_1[/imath] just like the first term, so I just moved them together in an attempt to get as close to your result as possible, ending up with:

 

[math]

\left\{ 2 c\vec{\alpha}_1\cdot \left( \vec{p}_1 - i\hbar \left[\vec{\gamma}\vec{\Psi}_2^\dagger(x_1,t)\cdot\vec{\Psi}_2(x_1,t)\right]\right) + \beta m_1c^2 \right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

...but now I can't find any route to your result. I was a bit uncomfortable with so many moves that I did, and so I think I may have well made a mistake somewhere... Better if I ask help...

 

-Anssi

[Doctordick]

Actually there was a typo there, it was just too tiny even for me to spot at first (even when I did look quite closely)! :D Nothing fatal though; the Dirac delta function is just missing the closing parenthesis in those two last expressions I've been through.

Fixed it; thanks.

I'm still supposed to take a good look at your explanation of how Dirac equation and his matrix notation works exactly, but right now I just have to take the above on faith.

Yeah, we can get onto that later.

I am not able to find my way to that result... :(

 

What I started with is:

 

[math]

\left\{ \sqrt{2} c\vec{\alpha}_1\cdot \vec{p}_1 + \sqrt{2} \alpha_\tau m_1c^2 - \sqrt{2} \vec{\alpha}_1 \cdot i\hbar c\left[\vec{\gamma}\vec{\Psi}_2^\dagger(x_1,t)\cdot\vec{\Psi}_2(x_1,t)\right]\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

Then, since you said "Dirac's vector representation of his anti-commuting alpha matrix is exactly equivalent to [imath]\sqrt{2}[/imath] times the first three components of my vector alpha operator", I suppose it means I'm ought to multiply the alpha operators by [imath]\sqrt{2}[/imath] (ending up with just 2 times the alpha operator since those terms already contained [imath]\sqrt{2}[/imath]).

You are just not thinking carefully. The above expression contains exactly the factor [imath]\sqrt{2}\vec{\alpha}_1[/imath] and since Dirac's [imath]\vec{\alpha}[/imath] is exactly equivalent to [imath]\sqrt{2}[/imath] times mine (for the x,y,z components), all you need do is substitute his [imath]\vec{\alpha}[/imath] (just as you properly did when you substituted his [imath]\beta[/imath] for my [imath]\sqrt{2}\alpha_\tau[/imath]). What you did essentially was to multiply by [imath]\sqrt{2}[/imath] twice (once when you got rid of the [imath]\sqrt{2}[/imath] in the denominator of the time derivative and again when you erroneously multiplied a second time to covert to Dirac's [imath]\vec{\alpha}[/imath]: i.e., that factor “2” should not be there. With that correction (and converting to Dirac's alpha and beta operators plus his [imath]\vec{p}[/imath]) you should have obtained:

[math]\left\{c\vec{\alpha}\cdot \left(\vec{p} -i\hbar c\sqrt{2}\left[\vec{\gamma} \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right)+\beta mc^2 -i\hbar c\sqrt{2}\left[ \gamma_\tau \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right\}\Psi= i\hbar \frac{\partial}{\partial t}\Psi[/math]

 

The other important issue is that, since Dirac's [imath]\vec{\alpha}[/imath] and [imath]\vec{p}[/imath] have only three components (x,y,z), [imath]\vec{\alpha}\cdot\vec{\gamma}[/imath] has only those same three components. In order to be complete, you have to stick in the tau results which yield [imath]\beta mc^2[/imath] (which you have included: his beta is my [imath]\sqrt{2}\alpha_\tau[/imath]) and another term proportional to [imath]\gamma_\tau[/imath] which you omitted.

 

See if that makes a little sense to you.

 

Have fun – Dick

[AnssiH]

While thinking about this step, I noticed little detail in the previous step that I do not understand.

 

I noticed that in your expression...

 

[math]

\sqrt{2}\left\{c\vec{\alpha}_1\cdot \vec{p}_1 +\alpha_\tau m_1c^2 - \vec{\alpha}_1 \cdot i\hbar c\left[\vec{\gamma}\vec{\Psi}_2^\dagger(x_1,t)\cdot\vec{\Psi}_2(x_1,t)\right]\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

...you have placed the gamma inside the square brackets. Is there some meaning to placing it in there?

 

You are just not thinking carefully. The above expression contains exactly the factor [imath]\sqrt{2}\vec{\alpha}_1[/imath] and since Dirac's [imath]\vec{\alpha}[/imath] is exactly equivalent to [imath]\sqrt{2}[/imath] times mine (for the x,y,z components), all you need do is substitute his [imath]\vec{\alpha}[/imath] (just as you properly did when you substituted his [imath]\beta[/imath] for my [imath]\sqrt{2}\alpha_\tau[/imath]). What you did essentially was to multiply by [imath]\sqrt{2}[/imath] twice (once when you got rid of the [imath]\sqrt{2}[/imath] in the denominator of the time derivative and again when you erroneously multiplied a second time to covert to Dirac's [imath]\vec{\alpha}[/imath]: i.e., that factor “2” should not be there.

 

Well "doh!" :D

 

The other important issue is that, since Dirac's [imath]\vec{\alpha}[/imath] and [imath]\vec{p}[/imath] have only three components (x,y,z), [imath]\vec{\alpha}\cdot\vec{\gamma}[/imath] has only those same three components. In order to be complete, you have to stick in the tau results which yield [imath]\beta mc^2[/imath] (which you have included: his beta is my [imath]\sqrt{2}\alpha_\tau[/imath]) and another term proportional to [imath]\gamma_\tau[/imath] which you omitted.

 

Hmmm yeah that whole issue with 4 vs 3 dimensional representation can lead to some confusion if one is not careful, so let me just make sure I understanding something correctly; the operator called [imath]\vec{p}_1[/imath] has thus far ALWAYS been conceived as 3 dimensional in the OP? It was just the [imath]\vec{p}_2[/imath] that had the [imath]\tau[/imath] component plugged in for a while just for the purpose of getting rid of it. (Looking at the equations, I think this must be correct)

 

Okay, so after substituting to Dirac's alpha and beta, I'm at:

 

[math]

\left\{ c\vec{\alpha} \cdot \vec{p} + \beta mc^2 - \vec{\alpha} \cdot i\hbar c\left[\vec{\gamma}\vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

...except that it's still not quite Dirac's version as the [imath]\vec{\alpha}[/imath] still contains [imath]\tau[/imath]... So in other words, right now in the above expression, the inner products are essentially [imath]\vec{\alpha}_{xyz\tau} \cdot \vec{p}_{xyz}[/imath] and [imath]\vec{\alpha}_{xyz\tau} \cdot \vec{\gamma}_{xyz\tau}[/imath]. (Stating this for future reference; let me know if I'm missing something)

 

And to convert into 3 dimensional [imath]\vec{\alpha}[/imath], in the first inner product, the [imath]\tau[/imath] component yields 0 so that's easy; we just forget it.

 

In the second one, the separation must be:

 

[math]

\vec{\alpha}_{xyz\tau} \cdot \vec{\gamma}_{xyz\tau} i\hbar c\left[\vec{\Psi}_2^\dagger\cdot\vec{\Psi}_2\right] = \vec{\alpha}_{xyz} \cdot \vec{\gamma}_{xyz} i\hbar c\left[\vec{\Psi}_2^\dagger\cdot\vec{\Psi}_2\right] +\vec{\alpha}_{\tau} \cdot \vec{\gamma}_{\tau} i\hbar c\left[\vec{\Psi}_2^\dagger\cdot\vec{\Psi}_2\right]

[/math]

 

I think...

 

So overall now I'm at:

 

[math]

\left\{ c\vec{\alpha}_{xyz} \cdot \vec{p}_{xyz} + \beta mc^2 - \vec{\alpha}_{xyz} \cdot \vec{\gamma}_{xyz} i\hbar c\left[\vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]

- \vec{\alpha}_{\tau} \cdot \vec{\gamma}_{\tau} i\hbar c\left[\vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]

\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

Or to tidy it up a bit towards how you express it:

 

[math]

\left\{ c\vec{\alpha}_{xyz} \cdot \left ( \vec{p}_{xyz} - \vec{\gamma}_{xyz} i\hbar \left[\vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right] \right )

+ \beta mc^2 - \vec{\alpha}_{\tau} \cdot \vec{\gamma}_{\tau} i\hbar c\left[\vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]

\right\}\vec{\Psi}_1= i\hbar \frac{\partial}{\partial t}\vec{\Psi}_1

[/math]

 

I'm still not quite finding my way all the way to your result:

 

[math]

\left\{c\vec{\alpha}\cdot \left(\vec{p} -i\hbar c\sqrt{2}\left[\vec{\gamma} \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right)+\beta mc^2 -i\hbar c\sqrt{2}\left[ \gamma_\tau \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right]\right\}\Psi= i\hbar \frac{\partial}{\partial t}\Psi

[/math]

 

1. I do not have the [imath]c[/imath] that you have in the term [math]- i\hbar c\sqrt{2}\left[\vec{\gamma} \vec{\Psi}_2^\dagger(x,t)\cdot\vec{\Psi}_2(x,t)\right][/math] anymore, as I moved it outside the parenthesis (either you've made a typo or I'm still missing something)

 

2. I do not know where the [imath]\sqrt{2}[/imath]'s are coming from.

 

3. I have no idea what to do with [imath]\vec{\alpha}_{\tau}[/imath]... Somehow you've gotten rid of it.

 

(I've got the haunting feeling the last two problems are related :) )

 

-Anssi

[Doctordick]

Hi Anssi, I hate to say it but modest and Erasmus are just “pulling your chain”. All they are doing is plucking supposed facts out of thin air which they can no more prove than the man in the moon for the simple purpose of getting you to run around in circles. They can no more capable of accepting the possibility that I am right than a religionist can consider the possibility that God doesn't exist. Even if it is possibly true, they would rather not believe it so they have no interest in understanding what we are talking about.

 

Before I get into clearing up your mathematical problems, let me first answer your first question.

..you have placed the gamma inside the square brackets. Is there some meaning to placing it in there?

From a mathematical perspective, no; from a “connection” perspective, yes. I put it inside the square brackets for the simple reason that what is in the square brackets is exactly what should be defined to be “the expectation value of [imath]\vec{\gamma}[/imath]”, I defined what I meant by “the expectation value” way back in my derivation of Schrödinger's equation

A second definition totally consistent with what has already been presented is to define the expectation value of “energy” to be given by

[math]E=i\hbar\int\vec{\Psi}^\dagger\cdot\frac{\partial}{\partial t}\vec{\Psi}dV.[/math]

Essentially I am defining the expectation values of operators here. The concept is actually quite analogous to the idea of setting [imath]\vec{\Psi}^\dagger \cdot \vec{\Psi}dV[/imath] to be the expectation value of the arguments of [imath]\vec{\Psi}[/imath]: i.e., the fundamental definition of [imath]\vec{\Psi}[/imath].

 

But we need to get to the major problems you are having in this thread. It seems to me that I need to rephrase the issues you are having difficulties with. They arise totally out of the confusion generated by the difference between Dirac's three dimensional notation and my four dimensional notation together with the fact that we both are using the Greek letters alpha and beta to stand for subtly different things. Let us do this whole thing over, first entirely in my four dimensional notation, and then make the conversion over to Dirac's notation. I will presume you understood my removal of the terms related to [imath]\vec{\Psi}_2[/imath] (element #2): the removal of those terms via the expression

[math]\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1 =c\vec{\alpha}_2\cdot\vec{p}_2\vec{\Psi}_2\vec{\Psi}_1=\sqrt{\frac{1}{2}}|cp_2|\vec{\Psi}_2\vec{\Psi}_1[/math]

 

where I explicitly state that [imath]\vec{p}_2[/imath] is taken to be my four dimensional momentum: i.e., [imath]-i\hbar\vec{\nabla}\vec

{\Psi}_2[/imath]. I have assumed you were not confused by that step in view of your comment:

It was just the [imath]\vec{p}_2[/imath] that had the [imath]\tau[/imath] component plugged in for a while just for the purpose of getting rid of it. (Looking at the equations, I think this must be correct)

After removal of those terms, my fundamental equation will appear essentially as follows (purely in my four dimensional notation):

[math]\{\vec{\alpha}_1\cdot\vec{\nabla}_1+\beta_{12}\delta(\vec{x_1}-\vec{x_2})+\beta_{21}\delta(\vec{x_2}-\vec{x_1})\} \vec{\Psi}_1\vec{\Psi}_2=\vec{\Psi}_2\left\{K\frac{\partial}{\partial t}\vec{\Psi}_1\right\}[/math]

 

If we left multiply by [imath]\vec{\Psi}_2[/imath] and integrate over [imath]\vec{x}_2[/imath] we will get unity in every term except for the Dirac delta functions where we will obtain the function [imath](\beta_{12}+\beta_{21})\vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)[/imath]. The resultant equation is

[math]\left\{\vec{\alpha}_1\cdot\vec{\nabla}_1+\left[(\beta_{12}+\beta_{21})\vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=K\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

At this point, we want to transform the representation into a form we can compare to Dirac's equation. The first step is to get the right hand side to be a representation of the associated representation of the energy operator which is defined to be [imath]i\hbar\frac{\partial}{\partial t}[/imath]. We can get to that result by first multiplying the equation through by [imath]-i\hbar c[/imath] and setting [imath]K=-\frac{1}{c\sqrt{2}}[/imath]. At which point we will have

[math]\left\{-i\hbar c\vec{\alpha}_1\cdot\vec{\nabla}_1-i\hbar c\left[(\beta_{12}+\beta_{21})\vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1[/math].

 

Multiplying that through by [imath]\sqrt{2}[/imath] and we obtain a result where the right hand side is exactly the desired energy term.

[math]\left\{-i\hbar c\sqrt{2}\vec{\alpha}_1\cdot\vec{\nabla}_1-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

The next step is to convert to Dirac's three dimensional notation and his definitions of [imath]\vec{\alpha}[/imath] and [imath]\beta[/imath]. His [imath]\vec{\alpha}=\sqrt{2}\left [\alpha_{1x}\hat{x}+ \alpha_{1y}\hat{y}+\alpha_{1z}\hat{z}\right][/imath] and his [imath]\beta=\sqrt{2}\alpha_{1\tau}[/imath]. Inserting these definitions into the equation (plus picking up your notation for the three dimensional [imath]\vec{\nabla}[/imath] as [imath]\vec{\nabla}_{xyz}[/imath] and making a slight rearrangement in the order of terms), we have,

[math]\left\{c\vec{\alpha}\cdot (-i\hbar)\vec{\nabla}_{xyz}+\beta c (-i\hbar)\frac{\partial}{\partial \tau}-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

Using my definition of the “mass” operator

Likewise, this leads to a second definition: the expectation value of “mass” to be given by

[math]m=-i\frac{\hbar}{c}\int\vec{\Psi}^\dagger\cdot\frac{\partial}{\partial \tau}\vec{\Psi}dV.[/math]

and Newton's definition of the three dimensional momentum, [imath]\vec{p}_{xyz} =-i\hbar\vec{\nabla}_{xyz}[/imath], the above can be written

[math]\left\{c\vec{\alpha}\cdot \vec{p}_{xyz}+\beta mc^2-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

At this point it is valuable to take a look at Dirac's equation in order to understand the difference between my expression and his. One should remember that, in the deduction of Schrödinger's equation, the potential functions arose out of the contribution of the Dirac delta function interactions. That interaction, in this case, has the form [imath]-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/imath].

 

Dirac's equation is

[math]\left\{c\vec{\alpha}\cdot\left(\vec{p} -\frac{e}{c}\vec{A}\right)+\beta mc^2 +e\Phi \right\}\Psi=i\hbar\frac{\partial \Psi}{\partial t}[/math].

 

where [imath]\vec{A}[/imath] is the magnetic vector potential and [imath]\Phi[/imath] is the electric potential. Notice that the magnetic potential is inside the same parenthesis as the momentum operator (remember, only moving charges create magnetic fields) and its contribution to the energy is a consequence of a dot product with the same [imath]\vec{\alpha}[/imath] as is the momentum whereas the electric potential is a scalar term which stands together with the mass term. It is this fact which led me to define my [imath]\vec{\gamma}[/imath] operator as [imath]\vec{\gamma}= \vec{\alpha}_1\frac{1}{2}(\beta_{12}+\beta_{21})[/imath]. (We really don't have to go through that intermediate step: i.e., defining another [imath]\beta=\frac{1}{2}(\beta_{12}+\beta_{21})[/imath] as we already have too many things called [imath]\beta[/imath] and there is plenty of confusion without it.)

 

When I define that gamma (which is four dimensional by the way as it is defined via my four dimensional operators), I can express the interaction term in a manner which allows me to factor out Dirac's [imath]\vec{\alpha}_{xyz}[/imath]: i.e., in terms of my [imath]\vec{\alpha}_{1xyz\tau}[/imath] the expression

[math]\vec{\alpha}_1\cdot \vec{\gamma}\equiv \left[\vec{\alpha}_1\cdot \vec{\alpha}_1\right]\frac{1}{2}(\beta_{12}+\beta_{21}) =\left[\alpha_{1x}^2+\alpha_{1y}^2+\alpha_{1z}^2+\alpha_{1\tau}^2\right]\frac{1}{2}(\beta_{12}+\beta_{21})[/math]

 

[math]=\left[\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right]\frac{1}{2}

(\beta_{12}+\beta_{21})=\beta_{12}+\beta_{21}[/math]

 

(Note that this result is equivalent to the “[imath]2\beta[/imath] I obtained in the OP.)

 

It follows directly from that result that the interaction term can be written in terms of [imath]\vec{\gamma}[/imath] as

[math]-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]=\vec{\alpha}_{1xyz\tau}\cdot(-i\hbar c\sqrt{2})\left[ \vec{\gamma} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math]

 

 

There is only one problem with that expression. It is expressed in terms of my four dimensional operators: i.e., element being factored out is [imath]\vec{\alpha}_{1xyz\tau}[/imath] (a four dimensional vector operator) and what I wanted to be able to factor was Dirac's [imath]\vec{\alpha}_{xyz}[/imath] (a three dimensional vector operator with a magnitude [imath]\sqrt{2}[/imath] times mine). The magnitude is easy to handle as all I need to do is bring in that [imath]\sqrt{2}[/imath] which is already present in the “interaction term” expressed above. The problem that Dirac's [imath]\vec{\alpha}_{xyz}[/imath] is three dimensional is also pretty straight forward:

[math]\vec{\alpha}_{1xyz\tau}\cdot \vec{\gamma}\left[-i\hbar c\sqrt{2}\right]=\left\{\vec{\alpha}_{xyz} +\sqrt{2}\alpha_{1\tau}\hat{\tau}\right\}\cdot \vec{\gamma} \left[-i\hbar c\right][/math].

 

In other words, the interaction term will end up being two terms, one from which Dirac's [imath]\vec{\alpha}[/imath] is a factor and one which must stand alone: i.e., Dirac's [imath]\vec{\alpha}[/imath] is not a factor. At this point I need to point out that [imath]\alpha_{1\tau}\hat{\tau} \cdot \vec{\gamma}=|\alpha_{1\tau}|\gamma_\tau=\sqrt{\frac{1}{2}}\gamma_\tau[/imath]. Thus it is that the final result of all this algebra is the fact that the interaction term may be written in the form

[math]\vec{\alpha}\cdot(-i\hbar c)\left[ \vec{\gamma} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]+(-i\hbar c)\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math].

 

Note that, although [imath]\vec{\gamma}[/imath] is defined to be a four dimensional operator, there is no need for that fact to be explicitly expressed in the above equation as the dot product between Dirac's [imath]\vec{\alpha}[/imath] and [imath]\vec{\gamma}[/imath] will not pick up that tau component as there is no tau component in Dirac's [imath]\vec{\alpha}[/imath]. The final result is the fact that my fundamental equation can be written

 

[math]\left\{c\vec{\alpha}\cdot\left(\vec{p} -i\hbar\left[ \vec{\gamma} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]

\right)+\beta mc^2 -i\hbar c\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]

\right\}\Psi=i\hbar\frac{\partial \Psi}{\partial t}[/math].

 

All that is left is to identify the expectation values of [imath] \vec{\gamma}[/imath] with the electromagnetic potentials

[math]\vec{A}=\frac{i\hbar c}{e}\left[ \vec{\gamma}_{xyz} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math]

 

and

[math]\Phi= -\frac{i\hbar c}{e}\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math]

 

and the result is exactly Dirac's equation.

 

***PROBLEM***Note that these are not exactly the results I obtained in the original post. This may be the root two problem you were having. I have traced the difference down and found it in my failure to properly handle the magnitude of my [imath]\vec{\alpha}_{1xyz\tau}[/imath] in the OP. Not only that, but I apparently made the same error in my original work thirty years ago. I have checked over those notes and found exactly the same error. Sorry about that. What it goes to show is that proof reading is a very important issue in any paper. I have edited the OP to reflect this correction but I haven't tracked down all the consequences. I will look at again sometime later. If you spot any problems, let me know.

 

That was a real ***** to write out! I hope I haven't made any typos. I am sure, if I have, you will find them. I thank you in advance.

Hmmm yeah that whole issue with 4 vs 3 dimensional representation can lead to some confusion if one is not careful, so let me just make sure I understanding something correctly; the operator called [imath]\vec{p}_1[/imath] has thus far ALWAYS been conceived as 3 dimensional in the OP? It was just the [imath]\vec{p}_2[/imath] that had the [imath]\tau[/imath] component plugged in for a while just for the purpose of getting rid of it. (Looking at the equations, I think this must be correct)

You are absolutely right; however, I have tried to be a little less caviler in this post.

(either you've made a typo or I'm still missing something)

I may have made some typos in that post; I have not checked it. I re-did the algebra as I have because your attack was somewhat askew of what I would call reasonable algebraic work. Don't worry about it, just let me know if this presentation is a little more coherent.

3. I have no idea what to do with [imath]\vec{\alpha}_{\tau}[/imath]... Somehow you've gotten rid of it.

I hope it is a little clearer in this post. The point is that the indicated dot product essentially yields the magnitude of [imath]\vec{\alpha}_{\tau}[/imath] which is [imath]\sqrt{\frac{1}{2}}[/imath]. This is probably also the source of your problem with [imath]\sqrt{2}[/imath] factors coming and going: i.e., you are right, the two problems are related.

 

If you can find all my errors, we will be up to where you left off with your last post.

 

Good luck -- Dick

[AnssiH]

Hi Anssi, I hate to say it but modest and Erasmus are just “pulling your chain”. All they are doing is plucking supposed facts out of thin air which they can no more prove than the man in the moon for the simple purpose of getting you to run around in circles. They can no more capable of accepting the possibility that I am right than a religionist can consider the possibility that God doesn't exist. Even if it is possibly true, they would rather not believe it so they have no interest in understanding what we are talking about.

 

I get that feeling myself too. It's not particularly hard for me to see where their intuitive thoughts lead them astray, but it's a bit of a shame to not be able to communicate that issue... Well, I think (and hope) there are people out there who are little bit more analytical in their thinking, and at any rate, trying to explain the issues to Erasmus and Modest is kind of secondary importance to me right now. Kind of a little challenge to give a stab at if I have time :) (I once managed to convince a creationist that evolution is a valid explanation to organisms! Think about that :D)

 

Btw, no need to worry I would just drop out of the conversation. I can certainly see the point of this sort of analysis from quite many angles.

 

Before I get into clearing up your mathematical problems, let me first answer your first question.

From a mathematical perspective, no; from a “connection” perspective, yes. I put it inside the square brackets for the simple reason that what is in the square brackets is exactly what should be defined to be “the expectation value of [imath]\vec{\gamma}[/imath]”, I defined what I meant by “the expectation value” way back in my derivation of Schrödinger's equation

Essentially I am defining the expectation values of operators here. The concept is actually quite analogous to the idea of setting [imath]\vec{\Psi}^\dagger \cdot \vec{\Psi}dV[/imath] to be the expectation value of the arguments of [imath]\vec{\Psi}[/imath]: i.e., the fundamental definition of [imath]\vec{\Psi}[/imath].

 

Ah, okay, I think I understand that.

 

But we need to get to the major problems you are having in this thread. It seems to me that I need to rephrase the issues you are having difficulties with. They arise totally out of the confusion generated by the difference between Dirac's three dimensional notation and my four dimensional notation together with the fact that we both are using the Greek letters alpha and beta to stand for subtly different things. Let us do this whole thing over, first entirely in my four dimensional notation, and then make the conversion over to Dirac's notation. I will presume you understood my removal of the terms related to [imath]\vec{\Psi}_2[/imath] (element #2): the removal of those terms via the expression

[math]\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1 =c\vec{\alpha}_2\cdot\vec{p}_2\vec{\Psi}_2\vec{\Psi}_1=\sqrt{\frac{1}{2}}|cp_2|\vec{\Psi}_2\vec{\Psi}_1[/math]

 

where I explicitly state that [imath]\vec{p}_2[/imath] is taken to be my four dimensional momentum: i.e., [imath]-i\hbar\vec{\nabla}\vec

{\Psi}_2[/imath]. I have assumed you were not confused by that step in view of your comment:

After removal of those terms, my fundamental equation will appear essentially as follows (purely in my four dimensional notation):

[math]\{\vec{\alpha}_1\cdot\vec{\nabla}_1+\beta_{12}\delta(\vec{x_1}-\vec{x_2})+\beta_{21}\delta(\vec{x_2}-\vec{x_1})\} \vec{\Psi}_1\vec{\Psi}_2=\vec{\Psi}_2\left\{K\frac{\partial}{\partial t}\vec{\Psi}_1\right\}[/math]

 

(Took the liberty to fix the above equation; you left in a lone right parenthesis)

 

Okay, so let me step back to;

 

[math]

\{\vec{\alpha}_1\cdot\vec{\nabla}_1+\vec{\alpha}_2\cdot\vec{\nabla}_2 +\beta_{12}\delta(\vec{x_1}-\vec{x_2})+\beta_{21}\delta(\vec{x_2}-\vec{x_1})\} \vec{\Psi}_1\vec{\Psi}_2 =K\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

[/math]

 

And now to perform the move to remove element 2, it's:

 

[math]

\vec{\alpha}_2\cdot\vec{\nabla}_2 \vec{\Psi}_1\vec{\Psi}_2 = \vec{\Psi}_1 \left \{K\frac{\partial}{\partial t}\vec{\Psi}_2\right \}

[/math]

 

that allows us to remove them from the equation and arrive at:

 

[math]

\{\vec{\alpha}_1\cdot\vec{\nabla}_1+\beta_{12}\delta(\vec{x_1}-\vec{x_2})+\beta_{21}\delta(\vec{x_2}-\vec{x_1})\} \vec{\Psi}_1\vec{\Psi}_2 =\vec{\Psi}_2 \left \{K\frac{\partial}{\partial t}\vec{\Psi}_1\right \}

[/math]

 

Great, I think I understand that move.

 

If we left multiply by [imath]\vec{\Psi}_2[/imath] and integrate over [imath]\vec{x}_2[/imath] we will get unity in every term except for the Dirac delta functions where we will obtain the function [imath](\beta_{12}+\beta_{21})\vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)[/imath]. The resultant equation is

[math]\left\{\vec{\alpha}_1\cdot\vec{\nabla}_1+\left[(\beta_{12}+\beta_{21})\vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=K\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

And that move.

 

At this point, we want to transform the representation into a form we can compare to Dirac's equation. The first step is to get the right hand side to be a representation of the associated representation of the energy operator which is defined to be [imath]i\hbar\frac{\partial}{\partial t}[/imath]. We can get to that result by first multiplying the equation through by [imath]-i\hbar c[/imath] and setting [imath]K=-\frac{1}{c\sqrt{2}}[/imath]. At which point we will have

[math]\left\{-i\hbar c\vec{\alpha}_1\cdot\vec{\nabla}_1-i\hbar c\left[(\beta_{12}+\beta_{21})\vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1[/math].

 

Yes I got that result too.

 

Multiplying that through by [imath]\sqrt{2}[/imath] and we obtain a result where the right hand side is exactly the desired energy term.

[math]\left\{-i\hbar c\sqrt{2}\vec{\alpha}_1\cdot\vec{\nabla}_1-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

Yup.

 

The next step is to convert to Dirac's three dimensional notation and his definitions of [imath]\vec{\alpha}[/imath] and [imath]\beta[/imath]. His [imath]\vec{\alpha}=\sqrt{2}\left [\alpha_{1x}\hat{x}+ \alpha_{1y}\hat{y}+\alpha_{1z}\hat{z}\right][/imath] and his [imath]\beta=\sqrt{2}\alpha_{1\tau}[/imath]. Inserting these definitions into the equation (plus picking up your notation for the three dimensional [imath]\vec{\nabla}[/imath] as [imath]\vec{\nabla}_{xyz}[/imath] and making a slight rearrangement in the order of terms), we have,

[math]\left\{c\vec{\alpha}\cdot (-i\hbar)\vec{\nabla}_{xyz}+\beta c (-i\hbar)\frac{\partial}{\partial \tau}-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

I was able to walk through that completely as well, to arrive exactly at your result.

 

Using my definition of the “mass” operator

and Newton's definition of the three dimensional momentum, [imath]\vec{p}_{xyz} =-i\hbar\vec{\nabla}_{xyz}[/imath], the above can be written

[math]\left\{c\vec{\alpha}\cdot \vec{p}_{xyz}+\beta mc^2-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]\right\} \vec{\Psi}_1=i\hbar\frac{\partial}{\partial t}\vec{\Psi}_1[/math]

 

Yes.

 

At this point it is valuable to take a look at Dirac's equation in order to understand the difference between my expression and his. One should remember that, in the deduction of Schrödinger's equation, the potential functions arose out of the contribution of the Dirac delta function interactions. That interaction, in this case, has the form [imath]-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/imath].

 

Dirac's equation is

[math]\left\{c\vec{\alpha}\cdot\left(\vec{p} -\frac{e}{c}\vec{A}\right)+\beta mc^2 +e\Phi \right\}\Psi=i\hbar\frac{\partial \Psi}{\partial t}[/math].

 

where [imath]\vec{A}[/imath] is the magnetic vector potential and [imath]\Phi[/imath] is the electric potential. Notice that the magnetic potential is inside the same parenthesis as the momentum operator (remember, only moving charges create magnetic fields) and its contribution to the energy is a consequence of a dot product with the same [imath]\vec{\alpha}[/imath] as is the momentum whereas the electric potential is a scalar term which stands together with the mass term. It is this fact which led me to define my [imath]\vec{\gamma}[/imath] operator as [imath]\vec{\gamma}= \vec{\alpha}_1\frac{1}{2}(\beta_{12}+\beta_{21})[/imath]. (We really don't have to go through that intermediate step: i.e., defining another [imath]\beta=\frac{1}{2}(\beta_{12}+\beta_{21})[/imath] as we already have too many things called [imath]\beta[/imath] and there is plenty of confusion without it.)

 

When I define that gamma (which is four dimensional by the way as it is defined via my four dimensional operators), I can express the interaction term in a manner which allows me to factor out Dirac's [imath]\vec{\alpha}_{xyz}[/imath]: i.e., in terms of my [imath]\vec{\alpha}_{1xyz\tau}[/imath] the expression

[math]\vec{\alpha}_1\cdot \vec{\gamma}\equiv \left[\vec{\alpha}_1\cdot \vec{\alpha}_1\right]\frac{1}{2}(\beta_{12}+\beta_{21}) =\left[\alpha_{1x}^2+\alpha_{1y}^2+\alpha_{1z}^2+\alpha_{1\tau}^2\right]\frac{1}{2}(\beta_{12}+\beta_{21})[/math]

 

[math]=\left[\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right]\frac{1}{2}

(\beta_{12}+\beta_{21})=\beta_{12}+\beta_{21}[/math]

 

(Note that this result is equivalent to the “[imath]2\beta[/imath] I obtained in the OP.)

 

Ahha, yes I followed that and everything looks valid.

 

It follows directly from that result that the interaction term can be written in terms of [imath]\vec{\gamma}[/imath] as

[math]-i\hbar c\sqrt{2} \left[(\beta_{12}+\beta_{21}) \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]=\vec{\alpha}_{1xyz\tau}\cdot(-i\hbar c\sqrt{2})\left[ \vec{\gamma} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math]

 

 

There is only one problem with that expression. It is expressed in terms of my four dimensional operators: i.e., element being factored out is [imath]\vec{\alpha}_{1xyz\tau}[/imath] (a four dimensional vector operator) and what I wanted to be able to factor was Dirac's [imath]\vec{\alpha}_{xyz}[/imath] (a three dimensional vector operator with a magnitude [imath]\sqrt{2}[/imath] times mine). The magnitude is easy to handle as all I need to do is bring in that [imath]\sqrt{2}[/imath] which is already present in the “interaction term” expressed above. The problem that Dirac's [imath]\vec{\alpha}_{xyz}[/imath] is three dimensional is also pretty straight forward:

[math]\vec{\alpha}_{1xyz\tau}\cdot \vec{\gamma}\left[-i\hbar c\sqrt{2}\right]=\left\{\vec{\alpha}_{xyz} +\sqrt{2}\alpha_{1\tau}\hat{\tau}\right\}\cdot \vec{\gamma} \left[-i\hbar c\right][/math].

 

In other words, the interaction term will end up being two terms, one from which Dirac's [imath]\vec{\alpha}[/imath] is a factor and one which must stand alone: i.e., Dirac's [imath]\vec{\alpha}[/imath] is not a factor.

 

Yup, all clear once again. (I must say it is very nice to be able to move through this so quickly :)

 

At this point I need to point out that [imath]\alpha_{1\tau}\hat{\tau} \cdot \vec{\gamma}=|\alpha_{1\tau}|\gamma_\tau=\sqrt{\frac{1}{2}}\gamma_\tau[/imath].

 

Ahha, yes, I remember that issue, and I spotted where that fact comes in handy this time; it'll cancel out that [imath]\sqrt{2}[/imath].

 

Thus it is that the final result of all this algebra is the fact that the interaction term may be written in the form

[math]\vec{\alpha}\cdot(-i\hbar c)\left[ \vec{\gamma} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]+(-i\hbar c)\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math].

 

Yes, I was able to follow that too.

 

Note that, although [imath]\vec{\gamma}[/imath] is defined to be a four dimensional operator, there is no need for that fact to be explicitly expressed in the above equation as the dot product between Dirac's [imath]\vec{\alpha}[/imath] and [imath]\vec{\gamma}[/imath] will not pick up that tau component as there is no tau component in Dirac's [imath]\vec{\alpha}[/imath].

 

Yes, the only little discomfort that I get from that is that, I started thinking whether it would be possible to end up doing algebraic steps that would move it in position with another four dimensional operator, and end up with an error that way. After little bit of thinking I thought I don't want to spend much time trying to figure that out right now :D

 

The final result is the fact that my fundamental equation can be written

 

[math]\left\{c\vec{\alpha}\cdot\left(\vec{p} -i\hbar\left[ \vec{\gamma} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]

\right)+\beta mc^2 -i\hbar c\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right]

\right\}\Psi=i\hbar\frac{\partial \Psi}{\partial t}[/math].

 

Yes, I got that exact result now myself too.

 

Phew! That certainly was much easier this time around, it only took me one 3 hour sitting to walk through the whole thing quite carefully (and I didn't even spot any typos apart from that one parenthesis; I guess you are not completely senile yet :)

 

All that is left is to identify the expectation values of [imath] \vec{\gamma}[/imath] with the electromagnetic potentials

[math]\vec{A}=\frac{i\hbar c}{e}\left[ \vec{\gamma}_{xyz} \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math]

 

and

[math]\Phi= -\frac{i\hbar}{e}\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x}_1,t) \cdot \vec{\Psi}_2(\vec{x}_1,t)\right][/math]

 

Oh, just when I thought it was typo free, I think the definition for [imath]\Phi[/imath] is missing a c... It seems to be correct in the OP.

 

And also I notice that your definition of [imath]\vec{A}[/imath] is different here than in the OP, which seems to trace down to somewhere earlier in the derivation. Is that what you are referring to here:

 

***PROBLEM***Note that these are not exactly the results I obtained in the original post. This may be the root two problem you were having. I have traced the difference down and found it in my failure to properly handle the magnitude of my [imath]\vec{\alpha}_{1xyz\tau}[/imath] in the OP. Not only that, but I apparently made the same error in my original work thirty years ago. I have checked over those notes and found exactly the same error. Sorry about that. What it goes to show is that proof reading is a very important issue in any paper. I have edited the OP to reflect this correction but I haven't tracked down all the consequences. I will look at again sometime later. If you spot any problems, let me know.

 

I'm not sure where the difference occurs, but it's kind of nice that you have uncovered such an old error :)

 

That was a real ***** to write out! I hope I haven't made any typos. I am sure, if I have, you will find them. I thank you in advance.

 

Thank you for putting it down in such explicit small steps, it was quite easy to follow. And I was a bit surprised myself that I found only those 2 little typos; either I am getting senile too, or your brain just needed bit of a dusting :)

 

I may have made some typos in that post; I have not checked it. I re-did the algebra as I have because your attack was somewhat askew of what I would call reasonable algebraic work. Don't worry about it, just let me know if this presentation is a little more coherent.

 

I would say this was definitely clearer.

 

I hope it is a little clearer in this post. The point is that the indicated dot product essentially yields the magnitude of [imath]\vec{\alpha}_{\tau}[/imath] which is [imath]\sqrt{\frac{1}{2}}[/imath]. This is probably also the source of your problem with [imath]\sqrt{2}[/imath] factors coming and going: i.e., you are right, the two problems are related.

 

Was very clear.

 

If you can get though that post

 

Third typo! ;)

 

-Anssi

[Doctordick]

(I once managed to convince a creationist that evolution is a valid explanation to organisms! Think about that :D)

I am impressed! Maybe you can reach modest and Erasmus.

(Took the liberty to fix the above equation; you left in a lone right parenthesis)

Thank you, I have fixed the post.

Yes, the only little discomfort that I get from that is that, I started thinking whether it would be possible to end up doing algebraic steps that would move it in position with another four dimensional operator, and end up with an error that way. After little bit of thinking I thought I don't want to spend much time trying to figure that out right now :D

Hopefully whoever moves it to such a position is aware of the problem and handles it properly. Such a thing can certainly be done as division by zero (in an algebraic expression where it isn't obvious) is part of a common joke proof that 1=2.

Oh, just when I thought it was typo free, I think the definition for [imath]\Phi[/imath] is missing a c... It seems to be correct in the OP.

You are correct; the [imath]\Phi[/imath] is indeed missing a c. But you are wrong about the OP. In the OP I apparently overlooked the fact that one c was factored out with the [imath]\vec{\alpha}[/imath] and had a [imath]c^2[/imath] in my resulting definition of [imath]\vec{A}[/imath]. I have made the correction. I checked my old notes and that error was also in my original work. That may explain my problems with that OP; I was having a lot of difficulty obtaining the same thing I had in my notes and had trouble finding all my errors. Apparently some of the errors were actually in the original stuff.

And also I notice that your definition of [imath]\vec{A}[/imath] is different here than in the OP, which seems to trace down to somewhere earlier in the derivation. Is that what you are referring to here:

No it isn't; I had already fixed all the errors I had found before I posted that latest post (most of them were [imath]\sqrt{2}[/imath] factors coming and going). I just hadn't noticed that problem with the extra c in [imath]\vec{A}[/imath].

Thank you for putting it down in such explicit small steps, it was quite easy to follow. And I was a bit surprised myself that I found only those 2 little typos; either I am getting senile too, or your brain just needed bit of a dusting :)

Putting it down in such explicit small steps helped me too. Most of the errors I found were not in my latest post but in the OP which I guess was done poorly.

 

I also corrected that third typo you found. :lol:

 

Thank you very much for your kind attention to detail -- Dick

[AnssiH]

I am impressed! Maybe you can reach modest and Erasmus.

 

Yeah maybe, certainly they would be capable of following it if they were motivated to concentrate on it. But yeah, it's not that important to get their attention, and I really think there are people out there who can see what's going on here with this analysis. Just need to get their attention. So first I'm focusing mostly on getting through the thing myself.

 

You are correct; the [imath]\Phi[/imath] is indeed missing a c. But you are wrong about the OP. In the OP I apparently overlooked the fact that one c was factored out with the [imath]\vec{\alpha}[/imath] and had a [imath]c^2[/imath] in my resulting definition of [imath]\vec{A}[/imath].

 

Ah okay, so that was indeed a real error (it's that exact c that was giving me the trouble in post #24). When I wrote the previous reply, I realized it was because of that c that you had the [imath]c^2[/imath] in the definition, but had forgotten I'd already suspected it to be a mistake :D

 

Another tiny thing that I forgot to mention, in the OP, inside your definition of [imath]\Phi[/imath] you are not using the subscript 1 with [imath]\vec{x}[/imath], but you are with the definition of [imath]\vec{A}[/imath]. In the previous step they were removed though. So just to avoid confusion, should probably do something about that, just to maximise clarity.

 

It follows that Dirac's equation can be seen as an approximation to my fundamental equation under some very specific conditions. It is important that these conditions be examined carefully. First, this is an approximate solution to my equation if we have an interaction between two events in total isolation from the rest of the universe. Except for the "two" events part, that is exactly the common approximation made when using Dirac's equation. In the Dirac's equation, the electromagnetic field potentials are certainly not analogous to what I have put forth as a fundamental element: i.e., a “point” entity interacting with the rest of the universe via a Dirac delta function.

 

However, the photon (which does bear a close resemblance to what I have put forth as a fundamental element) is quite often described as being the consequence of quantizing the electromagnetic field. Anyone who has followed the derivation of my fundamental equation knows that, under that derivation, two lone fermions cannot interact: i.e., the Dirac delta function vanishes via the Pauli exclusion principle. Since the electron (the particle element number one is to represent) is a fermion, the second element has to be a boson. If that is the case, the second element in the above deduction must obey Bose-Einstein statistics: i.e., an unlimited number of particles may occupy the same state at the same time.

 

That being the case, under the assumption that interaction between photons is negligible and that any number of bosons may occupy the same state (the same function [imath]\psi(\vec{x},t)[/imath]) we may clearly include as many photons (the name I have used to refer to as element number two in the above deduction) as we wish. Thus it is that “electromagnetic field potentials” controlling the behavior of the electron could be defined by the expressions

[math]\Phi(\vec{x},t)=-i\frac{\hbar c}{e}\sum_{i=2}^\infty \left[\gamma_\tau \vec{\Psi}_i^\dagger(\vec{x},t) \cdot \vec{\Psi}_i(\vec{x},t)\right][/math]

 

and

 

[math]\vec{A}(\vec{x},t)=i\frac{\hbar c}{e}\sum_{i=2}^\infty \left[\vec{\gamma}\vec{\Psi}_i^\dagger(\vec{x},t) \cdot \vec{\Psi}_i(\vec{x},t)\right][/math].

 

This is clearly a representation which yields the electromagnetic field potentials as the collective result of a collection of photons. This picture is essentially quite equivalent to seeing photons as quantized electromagnetic fields except from exactly the opposite direction. The philosophic question is, of course, which is the more fundamental perspective. Modern physics pretty well takes the field picture as more fundamental: i.e., it is their presumption that the quantized elements (what they call particles) are to be discovered by “quantizing field solutions to their physical problems”. For example, right now, the big issue in physics is the attempt to “quantize Einstein's general relativistic field equations” in order to obtain the characteristics of the “graviton”. I will show explicitly that it is more rational to hold the quantized elements as fundamental and the fields as deduced consequences. That view removes a lot of subtle difficulties in the modern physics perspective that field theory is fundamental.

 

Now I must wonder if those subtle difficulties are exactly what led to the Standard Model modeling electromagnetism via force exchange particles. Am I right to presume that your attack is quite equivalent to that view (is that exactly what you are referring to when you say "equivalent to seeing photons as quantized electromagnetic fields"?)

 

If so, I would think it may not be accurate assessment anymore these days that fields are seen as more fundamental, at least not in everybody's mind?

 

At any rate, indeed the Wikipedia entry mentions that "force exchange particles are bosons" and "photons mediate the electromagnetic force"...

 

And most of all this is quite interesting from epistemological perspective, certainly there's no reason to take one or another presentation (fields or discreet particles) as more fundamental in ontological sense, but it's interesting to think about whether the necessity to model reality in discreetly defined portions leads to entirely discreet descriptions becoming more valid (in the sense of avoiding subtle difficulties in the connection of discreet and continuous aspects of the view). Hmmm...

 

Laying that issue aside for the moment, there is still a subtle difficulty with the result I have achieved. I have shown that my fundamental equation yields a result quite analogous to Dirac's equation: i.e., Dirac's equation presumes the electromagnetic potentials are given and, in my deduction of my fundamental equation, I proved there always exists a potential function which will yield the observed behavior. Actually all this proves is that my expression above “could be” a valid expression of the electromagnetic potentials. That really isn't sufficient to identify my result with Dirac's equation as the electromagnetic potentials are specifically defined.

 

The problem here is that, from the perspective of my fundamental equation (and the work above), the electromagnetic potential is a many body problem and, as such, is a problem we cannot solve. That being the case, let me take the same attack to discover the form of [imath]\vec{\Psi}_2[/imath] which I used to discover the form of [imath]\vec{\Psi}_1[/imath]: i.e., presume a solution for [imath]\vec{\Psi}_1[/imath] and examine (via the interaction term) the kind of equation which the expectation value of [imath]\vec{\gamma}[/imath] must obey.

[math]<\vec{\gamma}>=\left[\vec{\gamma}\vec{\Psi}_2^\dagger(\vec{x}_2,t)\cdot \vec{\Psi}_2(\vec{x}_2,t)\right][/math].

 

I have to ask to be sure; what is the meaning of the left side of the equation?

 

as derived from

[math]\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2+c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2-2i\hbar c\beta \delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1\vec{\Psi}_2= \frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2[/math].

 

Um, okay, I'm not really sure what is happening here... So, how the above equation would be after definition of [imath]\vec{\gamma}=\vec{\alpha}_1\beta[/imath] and left multiply by [imath]\vec{\Psi}_2^\dagger\cdot[/imath]:

 

[math]

\{c\vec{\alpha}_1\cdot\vec{p}_1+\alpha_{1\tau} m_1c^2+c\vec{\alpha}_2\cdot\vec{p}_2+\alpha_{2 \tau}m_2c^2 - i\hbar c \vec{\alpha}_1\cdot\vec{\gamma}\delta(\vec{x}_1-\vec{x}_2)\}\vec{\Psi}_1 \vec{\Psi}_2^\dagger\cdot \vec{\Psi}_2 = \frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1 \vec{\Psi}_2^\dagger\cdot \vec{\Psi}_2

[/math]

 

Not really sure where to move from there, but I suppose what we are interested of is the impact of that [imath]\gamma[/imath] operator on [imath]\vec{\Psi}_2^\dagger\cdot \vec{\Psi}_2[/imath]

 

(Please note that I have returned to my alpha and beta operators, [imath]\alpha[/imath] and [imath]\beta[/imath] are not Dirac's matrices.) In this case, I will simply approximate the solution for element number one (pulling off the dependence on [imath]\tau[/imath] in terms of the rest mass of the electron) as

[math]\vec{\Psi}_1(\vec{x},t)=\psi(\vec{x},t)e^{-i\frac{m_e c}{\hbar}\tau}[/math]

 

where

 

[math]\psi^\dagger(\vec{x},t)\psi(\vec{x},t)\approx a\delta(\vec{x}-\vec{v}t)[/math]

 

where a is some constant setting the "amplitude" of the effective delta function.

 

I don't really know what is happening here either... I'll try to substitute that in and see if I can work something out, but I kind of feel like I'm stabbing in the dark right now so maybe you can help a bit (smaller steps are always helpful if you can lay some down)... And perhaps you can explain little bit what the meaning of the latter equation is?

 

-Anssi

[Doctordick]

Ah okay, so that was indeed a real error (it's that exact c that was giving me the trouble in post #24). When I wrote the previous reply, I realized it was because of that c that you had the [imath]c^2[/imath] in the definition, but had forgotten I'd already suspected it to be a mistake :D

Yeah, with complex algebra it is easy to find an error and then forget what the error was; particularly with the behavior of the LaTex implementation on this forum. When I find an error in a formula, I then have to find the LaTex formula in the source. By the time I find the actual LaTex formula, which you can only find by remembering the text immediately before or after the formula, I have quite often forgotten what the error was. Once I do find and fix the error, and again preview the thing (which goes back to the beginning), the source also pops back into the original position: i.e., it goes back to displaying the first line. If the thing just stayed where it was, at least I would know where I was in the post. Instead, if I find another error I have to go through he whole stupid routine again. Sooner or later, it gets mind boggling.

Another tiny thing that I forgot to mention, in the OP, inside your definition of [imath]\Phi[/imath] you are not using the subscript 1 with [imath]\vec{x}[/imath], but you are with the definition of [imath]\vec{A}[/imath]. In the previous step they were removed though. So just to avoid confusion, should probably do something about that, just to maximize clarity.

I have just removed the subscript from the definition of [imath]\vec{A}[/imath]. I think I had found this error earlier but fixed it only in the second case; the one with the sum over i.

Now I must wonder if those subtle difficulties are exactly what led to the Standard Model modeling electromagnetism via force exchange particles. Am I right to presume that your attack is quite equivalent to that view (is that exactly what you are referring to when you say "equivalent to seeing photons as quantized electromagnetic fields"?)

Essentially, yes.

If so, I would think it may not be accurate assessment anymore these days that fields are seen as more fundamental, at least not in everybody's mind?

Perhaps, as a graduate student, I sort of saw the two as different ways of explaining forces. I was always kind of astonished by the way “believers in Einstein's General Relativity” tended to want to avoid thinking of exchange forces as a fundamental issue. (by the way, exchange forces amount to a fascinating subject all on their own.)

 

But back to the way physicists tend to view the situation. Most all the physicists I know have a driving urge to see fields as fundamental and “exchange forces” being due to quantized energy exchange via these fields. If you go back and read Qfwfq's position on this issue, you will discover that he is quite confident that the correct approach is to see fields as fundamental (that is probably why his avatar is

 

By the way, I have never figured out your avatar is that some kind of beast sitting on your shoulders or what?

At any rate, indeed the Wikipedia entry mentions that "force exchange particles are bosons" and "photons mediate the electromagnetic force"...

 

And most of all this is quite interesting from epistemological perspective, certainly there's no reason to take one or another presentation (fields or discreet particles) as more fundamental in ontological sense, but it's interesting to think about whether the necessity to model reality in discreetly defined portions leads to entirely discreet descriptions becoming more valid (in the sense of avoiding subtle difficulties in the connection of discreet and continuous aspects of the view). Hmmm...

Well, fermions can not serve the purposes of exchange forces (particularly between fermions) because of “Fermi statistics” which essentially state that you cannot have two identical fermions in the same state. (That is easy to prove; if you are interested, I will show it to you.) They could cause an exchange force between bosons but it would have to be weak for the same reason.

 

But the fundamental issue is a bit worse than the simple differences in point of view; that issue will be quite clear by the time we finish this discussion of the Dirac equation.

I have to ask to be sure; what is the meaning of the left side of the equation?

I just threw that in to see what your reaction would be. Actually it is a somewhat common shorthand (pseudo Dirac notation) for the expression on the right. Under Dirac notation the state vectors are represented by bras, <”state”|, and kets, |”state”>, and the operator to be evaluated. When you see something bracketed by <...> you know they are talking about some specific “expectation” value (not generally a universal expectation: i.e., not integrated over some range). Putting a name in the middle (in this case [imath]\vec{\gamma}[/imath]) it is taken to mean a specific expectation value of [imath]\vec{\gamma}[/imath] (at least by most people I worked with forty years ago). I suspect you jumped to the correct interpretation. :lol:

Um, okay, I'm not really sure what is happening here...

Oh God, there is that [imath]\beta[/imath] again. Instead of [imath]2\beta[/imath] I should have stayed with [imath]\beta_{12}+\beta_{21}[/imath]. I will leave the OP as it is for the moment but I really think I should stick with the new notation we are supposedly using.

So, how the above equation would be after definition of [imath]\vec{\gamma}=\vec{\alpha}_1\beta[/imath] and left multiply by [imath]\vec{\Psi}_2^\dagger\cdot[/imath]

I think you are jumping the gun here. We have come to the conclusion that the potential fields we are interested in here (the potential fields of interest generated through the Dirac delta function) are given by the expectation value of [imath]\vec{\gamma}[/imath]. We accomplished this without ever actually determining (or coming up with a way of determining) [imath]\vec{\Psi}_2[/imath]. What we had developed was an equation for determining [imath]\vec{\Psi}_1[/imath]: i.e., the equation Dirac's electron was to obey; the so called Dirac's equation. Dirac' equation takes the electromagnetic potentials as given information.

 

In order for us to come up with those “given” electromagnetic potentials, we need to know what the function [imath]\vec{\Psi}_2[/imath] looks like. Without a way of actually determining the expectation value of [imath]\vec{\gamma}[/imath] the fact that it will be “the potential of interest” is a rather useless piece of information. So our problem resolves down to finding a method of determining [imath]\vec{\Psi}_2[/imath]. That is the reason why it is necessary to go all the way back to the representation of my two elements interacting via that Dirac delta function. Whereas, in the first case I removed [imath]\vec{\Psi}_2[/imath] from the equation so that I could obtain an equation which would define [imath]\vec{\Psi}_1[/imath], in this case I want to remove [imath]\vec{\Psi}_1[/imath] from the equation so that I can obtain an equation which will define [imath]\vec{\Psi}_2[/imath].

 

I use exactly the same procedure to remove [imath]\vec{\Psi}_1[/imath] as I used to remove [imath]\vec{\Psi}_2[/imath] earlier. I use the fact that (for the analysis I am now performing: that is, at this particular point in my derivation) I can take [imath]\vec{\Psi}_1[/imath] as the function determining the probability of finding element one. In that case [imath]\vec{\Psi}_2[/imath] is the probability of finding element two “given the probability that we already have the solution for element number one”. Now it is clear that this is not exactly what defined our original [imath]\vec{\Psi}_1[/imath] (where we obtained Dirac's equation) as that probability depended upon the probability of finding element number two; but, in this case we will deal only with an approximation to [imath]\vec{\Psi}_1[/imath] so that the implied “error” here is somewhat ameliorated.

 

Another way to look at the issue is that Dirac's equation is to provide us with the correct [imath]\vec{\Psi}_1[/imath]: i.e., that result is what is to be expected for a little object (an electron) moving around through the coordinate system. Whatever it turns out to be, it is reasonable to approximate [imath]\vec{\Psi}_1[/imath] with

[math]\vec{\Psi}_1(x,y,z,\tau,t)=\vec{\sigma}\psi(x,y,z,t)e^{-i\frac{m_e c}{\hbar}\tau}[/math]

 

where [imath]\vec{\sigma}[/imath] supplies the vector component and

 

[math]\psi^\dagger(x,y,z,t)\psi(x,y,z,t)\approx a\delta(\vec{x}-\vec{v}t)[/math]

 

and “a” is some constant setting the "amplitude" of the effective delta function. You should understand that this is entirely consistent with asserting [imath]\vec{x}=\vec{f}(t)[/imath] where [imath]\vec{v}=\frac{d}{dt}\vec{f}(t)[/imath]: i.e., an infinite collection of points (a line) where [imath]\vec{x}=\vec{f}(t)[/imath] is the explicit path of our electron (at least approximately).

 

All that is saying is that the solution [imath]\vec{\Psi}_1[/imath] will yield a probability of finding our electron as being zero everywhere except in the vicinity of of the position [imath]\vec{x}_1[/imath] at time t: i.e., P(\vec{x}_1,t) (the probability of finding our electron) is roughly very analogous to a point object moving through space and thus representable by a Dirac delta function. We are not solving for its position; we are instead, taking its position as a given information exactly analogous to the to the way the Dirac equation took the electromagnetic potentials to be given information. What we are assuming is that, whatever the correct [imath]\vec{\Psi}_1[/imath] is, it will be interpreted as representing a point object moving through space. How it actually moves will depend upon the electromagnetic potentials: i.e., we have still essentially left the question of how it behaves to be an open issue. Thus it is that we really haven't made any assumptions at all concerning the equation [imath]\vec{\Psi}_1[/imath] has to obey. I hope that is at least somewhat clear.

 

At any rate, we will use

[math]\left\{i\hbar c\vec{\alpha}_1\cdot \left( \vec{\nabla}_{1xyz\tau}\right) -i\hbar \sqrt{\frac{1}{2}}\frac{\partial}{\partial t}\right\}\vec{\Psi}_1=0[/math]

 

to remove all the terms depending upon the state of element one from our fundamental equation. Substituting that expression for the specified terms related to element one yields the interaction term as the only term with any dependence upon [imath]\vec{x}_1[/imath] and the resultant equation is as follows:

[math]\left\{-i\hbar c\vec{\alpha}_2\cdot \vec{\nabla}_{2xyz\tau} -i\hbar c (\beta_{12}+\beta_{21}) \delta (\vec{x}_1-\vec{x}_2)\right\}\vec{\Psi}_1\vec{\Psi}_2=\frac{i\hbar}{\sqrt{2}}\vec{\Psi}_1\frac{\partial}{\partial t}\vec{\Psi}_2[/math].

 

If we now multiply through by [imath]\vec{\Psi}_1^\dagger\cdot[/imath] and integrate over [imath]\vec{x}_1[/imath], we get unity as the contribution from every term except the interaction term. The interaction term once again spikes when [imath]\vec{x}_1=\vec{x}_2[/imath] and the result is the amplitude of the probability that our electron is in the position referred to as [imath]\vec{x}_2[/imath]: i.e., it is exactly given by [imath]\vec{\Psi}_1^\dagger(\vec{x}_2,t)\cdot\vec{\Psi}_1(\vec{x}_2,t) = \psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)[/imath] with the function [imath]\psi[/imath] defining the position of the electron as defined above. The resultant equation can thus be rewritten as

[math]\left\{-i\hbar c\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -\left[i\hbar c (\beta_{12}+\beta_{21}) \psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right] \right\}\vec{\Psi}_2=\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2[/math].

 

(If I haven't made any typos.)

Not really sure where to move from there, but I suppose what we are interested of is the impact of that [imath]\gamma[/imath] operator on [imath]\vec{\Psi}_2^\dagger\cdot \vec{\Psi}_2[/imath]

Again you are somewhat ahead of the game here. What we want is to know here are the differential equations [imath]\Phi(\vec{x},t)[/imath] and [imath]\vec{A}(\vec{x},t)[/imath] must obey. What I have just deduced is the equation [imath]\vec{\Psi}_2[/imath] must behave. If you go back and look at the deduction of Dirac's equation, you will see that, in order for that to be Dirac's equation, we need [imath]\Phi(\vec{x},t)[/imath] and [imath]\vec{A}(\vec{x},t)[/imath] to be given by,

[math]\Phi(\vec{x},t)=-i\frac{\hbar c}{e}\left[\gamma_\tau \vec{\Psi}_2^\dagger(\vec{x},t)\cdot \vec{\Psi}_2(\vec{x},t)\right][/math]

 

and

 

[math]\vec{A}(\vec{x},t)=i\frac{\hbar c}{e}\left[\vec{\gamma}\vec{\Psi}_2^\dagger(\vec{x},t)\cdot \vec{\Psi}_2(\vec{x},t)\right][/math].

 

Noting that, except for [imath]i\frac{\hbar c}{e}[/imath] which is a simple constant, both [imath]\Phi(\vec{x},t)[/imath] and [imath]\vec{A}(\vec{x},t)[/imath] are nothing more than expectation values of [imath]\vec{\gamma}[/imath] so what we really need is to convert that equation above (for [imath]\vec{\Psi}_2[/imath]) into an equation for [imath]\vec{\gamma}[/imath]. That is actually a pretty straight forward procedure very similar to what was done in my deduction of Schrödinger's equation.

 

Starting with

[math]\left\{-i\hbar c\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -\left[i\hbar c (\beta_{12}+\beta_{21}) \psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right] \right\}\vec{\Psi}_2=\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2[/math].

 

We can assert that, so long as we are operating on the correct [imath]\vec{\Psi}_2[/imath] the above (divided through by [imath]i\hbar c[/imath]) can be seen as implying the operator identity

[math]\left\{-\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -\left[(\beta_{12}+\beta_{21}) \psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right] \right\}=\frac{1}{c\sqrt{2}}\frac{\partial}{\partial t}[/math].

 

Thus it is that we can operate on [imath]\vec{\Psi}_2[/imath] with that operator twice and obtain

[math]\left\{-\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -\left[(\beta_{12}+\beta_{21}) \psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right] \right\}\left\{-\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -\left[(\beta_{12}+\beta_{21}) \psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right] \right\}\vec{\Psi}_2[/math]

 

[math]=\frac{1}{c\sqrt{2}}\frac{\partial}{\partial t}\frac{1}{c\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2=\frac{1}{2c^2}\frac{\partial^2}{\partial t^2}\vec{\Psi}_2[/math].

 

Now, if we perform the indicated multiplication ([imath]\{\cdots\}\{\cdots\}[/imath]),noting the fact that because of the commutation properties of the alpha and beta operators (which appear in every term making up those two factors) the cross terms of those operators all vanish and the direct terms all yield exactly [imath]\frac{1}{2}[/imath] we will get,

[math]\left\{\frac{1}{2} \nabla_{2xyz\tau}^2 +\left[(\frac{1}{2}+\frac{1}{2}) \left(\psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right)^2\right] \right\}\vec{\Psi}_2=\frac{1}{2c^2}\frac{\partial^2}{\partial t^2}\vec{\Psi}_2[/math].

 

Finally, multiplying through by two we have a final differential equation [imath]\vec{\Psi}_2[/imath] must obey

[math]\left\{\nabla_{2xyz\tau}^2 +2 \left(\psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right)^2 \right\}\vec{\Psi}_2=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec{\Psi}_2[/math].

 

If we left multiply this by [imath]\vec{\gamma}\vec{\Psi}_2^\dagger\cdot[/imath] we obtain the equation

[math]\left\{\vec{\gamma}\vec{\Psi}_2^\dagger\cdot\nabla_{2xyz\tau}^2 +\vec{\gamma}\vec{\Psi}_2^\dagger\cdot2 \left(\psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right)^2 \right\}\vec{\Psi}_2=\vec{\gamma}\vec{\Psi}_2^\dagger\cdot\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec{\Psi}_2[/math].

 

Which can be rewritten (keeping in mind that none of the differential operators operate on [imath]\vec{\gamma}[/imath]),

[math]\vec{\gamma}\vec{\Psi}_2^\dagger\cdot\nabla_{2xyz\tau}^2\vec{\Psi}_2 +2 \left(\psi^\dagger(\vec{x}_2,t)\psi(\vec{x}_2,t)\right)^2\vec{\gamma}\vec{\Psi}_2^\dagger\cdot\vec{\Psi}_2=\vec{\gamma}\vec{\Psi}_2^\dagger\cdot\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec{\Psi}_2[/math].

 

I think this presentation is considerably clearer than the original (hopefully without typos). If you can follow that, we are down to the point where I make a change in the differential operators.

At this point, I want to bring up an interesting mathematical relationship,

[math]\frac{\partial^2}{\partial x^2}\left\{\vec{\Psi}^\dagger(\vec{x},t)\cdot\vec{\Psi}(\vec{x},t)\right\}=\frac{\partial}{\partial x}\left\{\left( \frac{\partial}{\partial x}\vec{\Psi}^\dagger(\vec{x},t)\right)\cdot\vec{\Psi}(\vec{x},t) + \vec{\Psi}^\dagger(\vec{x},t)\cdot\left(\frac{\partial}{\partial x}\vec{\Psi}(\vec{x},t)\right)\right\}[/math]

 

[math]=\left\{\left( \frac{\partial^2}{\partial x^2}\vec{\Psi}^\dagger(\vec{x},t)\right)\cdot\vec{\Psi}(\vec{x},t) + 2 \left(\frac{\partial}{\partial x}\vec{\Psi}^\dagger(\vec{x},t)\right)\cdot\left(\frac{\partial}{\partial x}\vec{\Psi}(\vec{x},t)\right)+ \vec{\Psi}^\dagger(\vec{x},t)\cdot\left(\frac{\partial^2}{\partial x^2}\vec{\Psi}(\vec{x},t)\right)\right\}.[/math]

 

If you get this far we can continue with rest of the original post.

 

Sorry about the complications -- Dick

[Rade]

.... What we are assuming is that, whatever the correct [imath]\vec{\Psi}_1[/imath] is, it will be interpreted as representing a point object moving through space....

I have a few questions about your above assumption. Does your comment above mean then that [imath]\vec{\Psi}_1[/imath] will be interpreted as representing a point object moving through time also--i.e., moving through space-time ? Another question, I see you use the term "point object"--how would this differ from a "non-point object" so moving, could you please give an example ? My interest is in the concept of how "objects" as you call them in your above comment (point, non-point, or both), when they move, are related to my concepts of "time", which I define = that which is intermediate between two moments, and "space", which is defined = that which is intermediate between two existents. Thus I find interest in your concept of the [imath]\vec{\Psi}_1[/imath] which appears to be what my definitions above would refer to as being a "moment of an existent" moving through "space-time"

Edit: To help me with my understanding (or lack of) I refer to this comment on the nature of [imath]\vec{\Psi}_1[/imath]:

I will divide my [math]\vec{\Psi}[/math] into three components. [math]\vec{\Psi}=\vec{\Psi}_1\vec{\Psi}_2\vec{\Psi}_0[/math]

 

where [math]\vec{\Psi}_0[/math] represents the entire rest of the universe, taken to be essentially independent of [math]\vec{\Psi}_1 and \vec{\Psi}_2. \vec{\Psi}_1[/math] is the function we are looking for (the function which yields the expectations for the Dirac particle) and [math]\vec{\Psi}_2[/math] is to yield the known expectations for the electromagnetic potential arising from a single photon

 

Thus, if as you say, the [imath]\vec{\Psi}_1[/imath] is a "point object" moving through "space" (i.e., a single photon for example), and [math]\vec{\Psi}_2[/math] yields the known expectations for the electromagnetic potential arising from [math]\vec{\Psi}_1[/math], would it not then be required that "space", as you call it above in relation to [math]\vec{\Psi}_1[/math] must be included within [math]\vec{\Psi}_0[/math] and if so, how can [math]\vec{\Psi}_0[/math] be "essentially independent of [math]\vec{\Psi}_1[/math]" since you "assume" that [math]\vec{\Psi}_1[/math] "moves though space" ? If an "object" moves through something such as "space" then it does not seem logical to say that the object is "independent" of space (my confusion ?). Now, if "space" is not included in [math]\vec{\Psi}_0[/math], where exactly is it included ? --that is, exactly where in your Fundamental Equation do we find the "space" that the [math]\vec{\Psi}_1[/math] must move through, and what exactly is the mathematical relationship of "space" to the [imath]\vec{\Psi}_1[/imath] so moved ?

[AnssiH]

Sorry about the delays, I've been too busy for the last two weeks to concentrate on this issue.... ...weekends included :P

 

But back to the way physicists tend to view the situation. Most all the physicists I know have a driving urge to see fields as fundamental and “exchange forces” being due to quantized energy exchange via these fields. If you go back and read Qfwfq's position on this issue, you will discover that he is quite confident that the correct approach is to see fields as fundamental (that is probably why his avatar is

 

Why, what does that equation mean?

 

By the way, I have never figured out your avatar is that some kind of beast sitting on your shoulders or what?

 

It's a sphynx cat, it looks like it is holding a cane or something, but it is actually licking its tail :D

It's not my cat, it's just a random picture from the internet.

 

Well, fermions can not serve the purposes of exchange forces (particularly between fermions) because of “Fermi statistics” which essentially state that you cannot have two identical fermions in the same state. (That is easy to prove; if you are interested, I will show it to you.)

 

Yes please :)

(I also went back and quickly reviewed exchange symmetry from the "What can we know of reality" thread)

 

They could cause an exchange force between bosons but it would have to be weak for the same reason.

 

Oh right, because you could only have one fermion in that interaction at a time...?

 

But the fundamental issue is a bit worse than the simple differences in point of view; that issue will be quite clear by the time we finish this discussion of the Dirac equation.

 

Okay...

 

I just threw that in to see what your reaction would be. Actually it is a somewhat common shorthand (pseudo Dirac notation) for the expression on the right. Under Dirac notation the state vectors are represented by bras, <”state”|, and kets, |”state”>, and the operator to be evaluated. When you see something bracketed by <...> you know they are talking about some specific “expectation” value (not generally a universal expectation: i.e., not integrated over some range). Putting a name in the middle (in this case [imath]\vec{\gamma}[/imath]) it is taken to mean a specific expectation value of [imath]\vec{\gamma}[/imath] (at least by most people I worked with forty years ago). I suspect you jumped to the correct interpretation. :lol:

 

Well I guess. Since you said "...examine the kind of equation which the expectation value of [imath]\vec{\gamma}[/imath] must obey", I thought it probably meant just that; "expectation value of [imath]\vec{\gamma}[/imath]", but thought it could mean something more... And I guess it did mean something more in that it's a "specific" expectation value :) At any rate, let it be said that at some point, I really need to get back to the standard Dirac Equation explanation at some point (at the earlier parts of this thread), to get a better grasp of it.

 

Oh God, there is that [imath]\beta[/imath] again. Instead of [imath]2\beta[/imath] I should have stayed with [imath]\beta_{12}+\beta_{21}[/imath]. I will leave the OP as it is for the moment but I really think I should stick with the new notation we are supposedly using.

 

Yup, I think at some point it might be valuable to rework the OP into clearer form, but let's see if we can find more errors first.

 

(Especially I think using the subscripts "xyz" and [imath]\tau[/imath] when those components are separated should make things clearer)

 

I think you are jumping the gun here. We have come to the conclusion that the potential fields we are interested in here (the potential fields of interest generated through the Dirac delta function) are given by the expectation value of [imath]\vec{\gamma}[/imath]. We accomplished this without ever actually determining (or coming up with a way of determining) [imath]\vec{\Psi}_2[/imath]. What we had developed was an equation for determining [imath]\vec{\Psi}_1[/imath]: i.e., the equation Dirac's electron was to obey; the so called Dirac's equation. Dirac' equation takes the electromagnetic potentials as given information.

 

In order for us to come up with those “given” electromagnetic potentials, we need to know what the function [imath]\vec{\Psi}_2[/imath] looks like. Without a way of actually determining the expectation value of [imath]\vec{\gamma}[/imath] the fact that it will be “the potential of interest” is a rather useless piece of information. So our problem resolves down to finding a method of determining [imath]\vec{\Psi}_2[/imath]. That is the reason why it is necessary to go all the way back to the representation of my two elements interacting via that Dirac delta function. Whereas, in the first case I removed [imath]\vec{\Psi}_2[/imath] from the equation so that I could obtain an equation which would define [imath]\vec{\Psi}_1[/imath], in this case I want to remove [imath]\vec{\Psi}_1[/imath] from the equation so that I can obtain an equation which will define [imath]\vec{\Psi}_2[/imath].

 

I use exactly the same procedure to remove [imath]\vec{\Psi}_1[/imath] as I used to remove [imath]\vec{\Psi}_2[/imath] earlier. I use the fact that (for the analysis I am now performing: that is, at this particular point in my derivation) I can take [imath]\vec{\Psi}_1[/imath] as the function determining the probability of finding element one. In that case [imath]\vec{\Psi}_2[/imath] is the probability of finding element two “given the probability that we already have the solution for element number one”. Now it is clear that this is not exactly what defined our original [imath]\vec{\Psi}_1[/imath] (where we obtained Dirac's equation) as that probability depended upon the probability of finding element number two; but, in this case we will deal only with an approximation to [imath]\vec{\Psi}_1[/imath] so that the implied “error” here is somewhat ameliorated.

 

Right... It is unfortunate that we have to change that definition a bit :(

 

Another way to look at the issue is that Dirac's equation is to provide us with the correct [imath]\vec{\Psi}_1[/imath]: i.e., that result is what is to be expected for a little object (an electron) moving around through the coordinate system. Whatever it turns out to be, it is reasonable to approximate [imath]\vec{\Psi}_1[/imath] with

[math]\vec{\Psi}_1(\vec{x},t)=\psi(\vec{x},t)e^{-i\frac{m_e c}{\hbar}\tau}[/math]

 

where

 

[math]\psi^\dagger(\vec{x},t)\psi(\vec{x},t)\approx a\delta(\vec{x}-\vec{v}t)[/math]

 

and “a” is some constant setting the "amplitude" of the effective delta function. You should understand that this is entirely consistent with asserting [imath]\vec{x}=\vec{f}(t)[/imath] where [imath]\vec{v}=\frac{d}{dt}\vec{f}(t)[/imath]: i.e., an infinite collection of points (a line) where [imath]\vec{x}=\vec{f}(t)[/imath] is the explicit path of our electron (at least approximately).

 

Um, hmm. I think I understand what this is all about, but I would like to understand that first equation correctly, and I'm struggling a bit.

 

That first step is I guess what you refer to as "pulling off the dependence on [imath]\tau[/imath] in terms of the rest mass of the electron" in the OP. I remember we did these sorts of moves to remove some partial derivatives from differential equations. But, refereshing my memory on how it worked back then, all I can figure out is that:

 

[math]

\frac{\partial}{\partial \tau}\vec{\Psi}_1 = \frac{\partial}{\partial \tau}\vec\psi e^{-i\frac{m_e c}{\hbar}\tau} = \vec\psi -i\frac{m_e c}{\hbar}e^{-i\frac{m_e c}{\hbar}\tau}

[/math]

 

but I don't know where and how to apply that exactly in this case... I also do not know what [imath]m_e[/imath] is.

 

The [imath]\psi^\dagger(\vec{x},t)\psi(\vec{x},t)\approx a\delta(\vec{x}-\vec{v}t)[/imath] I think I figured out. The value of [imath]\vec{x}[/imath] is essentially constrained to be given by [imath]\vec{v}t[/imath]... ...at least approximately.

 

All that is saying is that the solution [imath]\vec{\Psi}_1[/imath] will yield a probability of finding our electron as being zero everywhere except in the vicinity of of the position [imath]\vec{x}_1[/imath] at time t: i.e., P(\vec{x}_1,t) (the probability of finding our electron) is roughly very analogous to a point object moving through space and thus representable by a Dirac delta function.

 

Yup.

 

We are not solving for its position; we are instead, taking its position as a given information exactly analogous to the to the way the Dirac equation took the electromagnetic potentials to be given information.

 

Yup.

 

What we are assuming is that, whatever the correct [imath]\vec{\Psi}_1[/imath] is, it will be interpreted as representing a point object moving through space. How it actually moves will depend upon the electromagnetic potentials: i.e., we have still essentially left the question of how it behaves to be an open issue. Thus it is that we really haven't made any assumptions at all concerning the equation [imath]\vec{\Psi}_1[/imath] has to obey. I hope that is at least somewhat clear.

 

I think I'm starting to get the idea... (at least approximately ;) )

 

At any rate, we will use

[math]\left\{i\hbar c\vec{\alpha}\cdot \left( \vec{\nabla}_{1xyz\tau}\right) -i\hbar c\sqrt{\frac{1}{2}}\frac{\partial}{\partial t}\right\}\vec{\Psi}_1=0[/math]

 

Okay, getting to that expression clearly requires understanding of the stuff I asked about above, so I hope you can clear it out for me... In the meantime, I'll take the validity of the above expression on faith and move onwards.

 

(The alpha there must be the alpha for element #1, should it have the subscript? And likewise the alphas of the consequent expressions should have the subscript 2?)

 

to remove all the terms depending upon the state of element one from our fundamental equation. Substituting that expression for the specified terms related to element one yields the interaction term as the only term with any dependence upon [imath]\vec{x}_1[/imath] and the resultant equation is as follows:

[math]\left\{-i\hbar c\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -i\hbar c (\beta_{12}+\beta_{21}) \delta (\vec{x}_1-\vec{x}_2)\right\}\vec{\Psi}_1\vec{\Psi}_2=\frac{i\hbar}{\sqrt{2}}\vec{\Psi}_1\frac{\partial}{\partial t}\vec{\Psi}_2[/math].

 

Okay, I went backwards in this thread to the beginning of #12, where I still have the terms of elements #1 and #2 in there, and where I've just multiplied the entire equation by [imath]-ic\hbar[/imath]:

 

[math]

-ic\hbar \{\vec{\alpha}_1\cdot\vec{\nabla}_1+\vec{\alpha}_2\cdot\vec{\nabla}_2 +\beta_{12}\delta(\vec{x_1}-\vec{x_2})+\beta_{21}\delta(\vec{x_2}-\vec{x_1})\} \vec{\Psi}_1\vec{\Psi}_2 = -ic\hbar K\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

[/math]

 

Where all the components are still 4-dimensional.

 

So let's see... Some re-ordering first:

 

[math]

\left\{ -ic\hbar \vec{\alpha}_1 \cdot \vec{\nabla}_1 \right\} \vec{\Psi}_1\vec{\Psi}_2

+

(-ic\hbar)

\left\{ \vec{\alpha}_2 \cdot \vec{\nabla}_2 + (\beta_{12} + \beta_{21}) \delta(\vec{x_1}-\vec{x_2}) \right\} \vec{\Psi}_1\vec{\Psi}_2

=

-ic\hbar K\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

[/math]

 

With the definition of [imath]c=-\frac{1}{K\sqrt{2}}[/imath], the right hand side is:

 

[math]

-ic\hbar K\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

=

\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\vec{\Psi}_2

=

\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

So we're at:

 

[math]

\left\{ -ic\hbar \vec{\alpha}_1 \cdot \vec{\nabla}_1 \right\} \vec{\Psi}_1\vec{\Psi}_2

+

(-ic\hbar)

\left\{ \vec{\alpha}_2 \cdot \vec{\nabla}_2 + (\beta_{12} + \beta_{21}) \delta(\vec{x_1}-\vec{x_2}) \right\} \vec{\Psi}_1\vec{\Psi}_2

=

\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_1\right\}\vec{\Psi}_2+\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

Trying to move the time derivative term of [imath]\vec{\Psi}_1[/imath] to the left side;

 

[math]

\left [\left\{ -ic\hbar \vec{\alpha}_1 \cdot \vec{\nabla}_1 \right\}\vec{\Psi}_1

-i\hbar \sqrt{\frac{1}{2}}

\left (\frac{\partial}{\partial t}\vec{\Psi}_1 \right )\right ] \vec{\Psi}_2

+

(-ic\hbar)

\left\{ \vec{\alpha}_2 \cdot \vec{\nabla}_2 + (\beta_{12} + \beta_{21}) \delta(\vec{x_1}-\vec{x_2}) \right\} \vec{\Psi}_1\vec{\Psi}_2

=

\left\{\frac{i\hbar}{\sqrt{2}}\frac{\partial}{\partial t}\vec{\Psi}_2\right\}\vec{\Psi}_1

[/math]

 

So now, it is easy to see that the term in square brackets is almost the same as:

 

[math]\left\{i\hbar c\vec{\alpha}\cdot \left( \vec{\nabla}_{1xyz\tau}\right) -i\hbar c\sqrt{\frac{1}{2}}\frac{\partial}{\partial t}\right\}\vec{\Psi}_1=0[/math]

 

The only problem is that you have that one mysterious "c" in there. Since I'm not sure how you got to that expression, I wouldn't want to start guessing as to who went wrong and where, but I did notice you don't have that "c" there in the OP, so I suspect you have made a typo somewhere :)

 

btw, in that same expression in OP, the [imath]\vec{\Psi}[/imath] is missing the subscript 1, and that second term inside the curly brackets is positive whereas here it is negative.

 

At any rate, if I just manage to set that square bracket as "0", I can trivially see the rest of the steps to:

 

[math]

\left\{-i\hbar c\vec{\alpha}\cdot \vec{\nabla}_{2xyz\tau} -i\hbar c (\beta_{12}+\beta_{21}) \delta (\vec{x}_1-\vec{x}_2)\right\}\vec{\Psi}_1\vec{\Psi}_2=\frac{i\hbar}{\sqrt{2}}\vec{\Psi}_1\frac{\partial}{\partial t}\vec{\Psi}_2

[/math]

 

Phew, I think I'll pause here for a bit to avoid more confusion, since I've been guessing way too much by now already. Scanning the rest of the post, it looks quite a bit easier though, if I can just get the above issues cleared...

 

-Anssi

[Doctordick]

Sorry about the delays, I've been too busy for the last two weeks to concentrate on this issue.... ...weekends included :P

It is good to hear from you. In your absence I have been reading some of the other posts one the “philosophy” forum. The absolute absence of thought is just plain astonishing; take the “what is time” thread for example..., 764 posts totally without an accepted definition of what they are talking about.

I'm afraid it's the kinds of arguments that we are seeing in this thread that is not furthering the dialogue on this subject, because it is just impossible to even understand what does someone mean when they say "time" or "spacetime" or use all kinds of poetic ways to explain what time is "in their opinion". It really does require quite a bit more analytical approach.

You are dead on the money. The phrase, “what does someone mean when they say...”, is asking for a definition as without definition, everyone is just talking past everyone else. This fact seems to be beyond their comprehension. Give me a definition of what you mean; I don't care what your definition is as that makes no difference. What one means is the central issue; without a definition, intelligent discussion is impossible.

Why, what does that equation mean?

Rather little actually. You know initiates into supposedly profound knowledge like to have symbols of their supposed knowledge which can be understood only by other members of their station; the symbols, which actually provide utterly no information, bring “power” to their position of authority. That is all that really what stands behind that equation. Many kinds of shorthand notation often rise to such uses; such as the use of Feynman diagrams to imply that one understands the mathematics of QED. QED is really a quite complex perturbation attack where the “Feynman diagrams” are actually no more than a shorthand notation used to refer to specific terms in an infinite sum. Just for the fun of it, you might take a look at the wikipedia entry for QED (not that I would expect you to understand any of it but rather it is an excellent example of esoteric shorthand notation only someone intimately involved would understand). The following quote, taken directly from that entry, kind of tickles my fancy as my worthless Ph.D. thesis was actually directly related to actually calculating such perturbation results.

A brilliant argument by Freeman Dyson shows that the radius of convergence of the perturbation series in QED is zero⁵. The basic argument goes as follows: if the coupling constant were negative, this would be equivalent to the Coulomb force being negative. This would "reverse" the electromagnetic interaction so that like charges would attract and unlike charges would repel. This would render the vacuum unstable against decay into a cluster of electrons on one side of the universe and a cluster of positrons on the other side of the universe. Because the theory is sick for any negative value of the coupling constant, the series do not converge, but are an asymptotic series. This can be taken as a need for a new theory, a problem with perturbation theory, or ignored by taking a "shut-up-and-calculate" approach.

(Twenty years after I got my degree, I happened to stop by the physics department theoretical alcoves where I worked when I was a student and there was a copy of my thesis on someone's desk. I asked why it was there and a student said, “Oh the appendices are a great summary of how to actually do perturbation calculations!” so I guess it wasn't totally worthless. :lol:)

 

But back to Qfwfq's avatar; I could be wrong but I presume the [imath]\partial[/imath] with the slash through it is a shorthand notation for the differential operator in Minkowski geometry; the partial with respect to space being momentum and the partial with respect to time being energy (both of which are multiplied by [imath]\sqrt{-1}[/imath], the common definition of “i”). The [imath]\Psi[/imath] clearly refers to the “wave function” defined by the equation and “m” is the mass of the quantum entity associated with the “field” being defined by the function [imath]\Psi[/imath]. In essence it is a shorthand notation for what is commonly called the [????] equation. I am definitely going senile here as the name of the equation is on the tip of my tongue but totally eluding me; it is essentially the square root of the Klein-Gordon equation and is usually brought up as defining the fundamental relativistic field which is to be quantized in order to get a general relativistic quantum theory. For calculational purposes, it is a pretty worthless expression (that is my personal opinion only) though you may recognize a distinct similarity to my fundamental equation (all you have to do is ignore the interaction terms which should point out the difficulty with the Klein-Gordon equation and its decendents).

It's a sphynx cat, it looks like it is holding a cane or something, but it is actually licking its tail :D

It's not my cat, it's just a random picture from the internet.

To me it looks like some demon sitting on a guy's head holding on to his chin (the dark shadow is the guy's head; all you can actually see of the fellow is his shoulders). I know that's a pretty imaginative image but it's exactly what pops into my head whenever I see it.

Yes please :)

Well first, you need to have two identical elements (call them element #1 and element #2). The wave function which yields the probability of finding them at positions [imath]\vec{x}_1[/imath] and [imath]\vec{x}_2[/imath] would be written as [imath]\Psi(\vec{x}_1,\vec{x}_2,t)[/imath]: i.e., the actual probability would be given by [imath]\Psi^\dagger(\vec{x}_1,\vec{x}_2,t)\Psi(\vec{x}_1,\vec{x}_2,t) dV[/imath] ([imath]dV=(dx_1dy_1dz_1)(dx_2dy_2dz_2)[/imath] being the actual differential volume associated with the positions of interest). Notice that, since the probability of finding them somewhere in the universe must be unity, the probability of finding them at exactly a specific defined point must be zero: i.e., there must be some volume associated with the probability [imath](dxdydz)\neq 0[/imath].

 

Now, since the probability is obtained by squaring the magnitude of [imath]\Psi[/imath], a change is sign is of no consequence as [imath](-1)^2=1^2[/imath]. Given that we have two identical elements, element #1 found at [imath]\vec{x}_1[/imath] and element #2 found at [imath]\vec{x}_2[/imath], we can ask the question, what is the probability that element #2 is found at [imath]\vec{x}_1[/imath] while element #1 is found at [imath]\vec{x}_2[/imath]. Clearly, since the elements are identical, that probability must be the same. On the other hand, through the definition of [imath]\Psi[/imath], that probability must also be exactly given by [imath]\Psi^\dagger(\vec{x}_2,\vec{x}_1,t)\Psi(\vec{x}_2,\vec{x}_1,t) dV[/imath]. Thus it is that the issue referred to as “exchange” comes up.

 

Of interest is, how [imath]\Psi(\vec{x}_1,\vec{x}_2,t)[/imath] is related to [imath]\Psi(\vec{x}_2,\vec{x}_1,t)[/imath]. Since the elements are identical (by definition) the resultant probability cannot change. The only possible change that can occur is that the sign either changes or it does not (since such a sign change has no consequences in the resultant probability). It follows that all possibilities can be divided into two categories: either the sign changes under “exchange” or the sign does not change. These two categories are defined as “fermions” (the sign changes) and “bosons” (the sign does not change).

 

No sign change is essentially “no change at all” so we can see boson exchange as having no consequences of interest; however, a sign change has a subtle but very important consequence. Suppose we are talking about “fermions”: i.e., there is a sign change [imath]\Psi(\vec{x}_1,\vec{x}_2,t)=-\Psi(\vec{x}_2,\vec{x}_1,t)[/imath]. We can then ask the question, what is the probability that both elements are in the same position: i.e., what happens if [imath] \vec{x}_1=\vec{x}_2[/imath]. In that case, since exchanging the arguments can have no effect (they are identical) the resultant [imath]\Psi[/imath] must be identical. It follows that it must be identical and that it must change sign! There is only one mathematical element which is identical to its negative; that element is zero! Thus it is that the probability two identical fermions are in exactly the same place must be exactly zero.

 

There is one minor cavil here that any competent physicist will probably assert. That is the fact that positions are not the whole story; one should really point out that their “state” must be the same. This is actually little more than “defining” what is meant by “identical”. The spin on an electron allows us to put two electrons in the same position so long as their spins sum to zero. I call it a minor cavil because, if their spins are not in the same direction they are clearly not identical elements (if they are identical, their spins they cannot sum to zero).

Oh right, because you could only have one fermion in that interaction at a time...?

Absolutely correct. With bosons, you can have as many elements in a specified state as you want.

At any rate, let it be said that at some point, I really need to get back to the standard Dirac Equation explanation at some point (at the earlier parts of this thread), to get a better grasp of it.

Dirac's equation and Dirac's notation are quite different issues. Except for his introduction of his alpha and beta operators (quite analogous to Pauli's spin matrices) an issue which can also be handled by using a vector notation for [imath]\Psi[/imath], Dirac's equation can be written in Schrödinger's notation (which is what I have with inclusion of a little matrix notation). What is important here is the algebraic relationships involved; which are much easier to see (at least for me) in the Schrödinger notation. We can get back to the matrix and Dirac notation later.

Yup, I think at some point it might be valuable to rework the OP into clearer form, but let's see if we can find more errors first.

I agree with you.

Right... It is unfortunate that we have to change that definition a bit :(

Actually, it is exactly what is done in ordinary conventional physics. What I have referred to as “compartmentalization” elsewhere. Dirac's equation, by design, is an equation which yields the correct relativistic behavior of an electron when one takes the electromagnetic fields as given information. And Maxwell's equations yield the correct behavior of the electro-magnetic fields when one takes the positions and motions of electrons as given information. Thus what I am doing is no more than equivalent to the standard compartmentalization of everyday electrodynamics.

Um, hmm. I think I understand what this is all about, but I would like to understand that first equation correctly, and I'm struggling a bit.

Again it is my somewhat ambiguous notation which is confusing you. The expression, [imath]\vec{\Psi}_1(\vec{x},t)[/imath] is in my four dimensional Euclidean space [imath](x.y,z,\tau)[/imath] whereas the [imath]\psi_1(\vec{x},t)[/imath] is in the standard three dimensional space of conventional physics (x,y,z). You were expected to pick up on that fact by noticing the tau dependence had been pulled out. I should have written the thing

[math]\vec{\Psi}_1(x,y,z,\tau,t)=\vec{\sigma}\psi(x,y,z,t)e^{-i\frac{m_ec}{\hbar}\tau}[/math]

 

where [imath]\vec{\sigma}[/imath] is an abstract unit vector supplying the vector attributes of [imath]\vec{\Psi}_1[/imath].

 

That first step is I guess what you refer to as "pulling off the dependence on [imath]\tau[/imath] in terms of the rest mass of the electron" in the OP. I remember we did these sorts of moves to remove some partial derivatives from differential equations. But, refereshing my memory on how it worked back then, all I can figure out is that:

 

[math]

\frac{\partial}{\partial \tau}\vec{\Psi}_1 = \frac{\partial}{\partial \tau}\vec\psi e^{-i\frac{m_e c}{\hbar}\tau} = \vec\psi -i\frac{m_e c}{\hbar}e^{-i\frac{m_e c}{\hbar}\tau}

[/math]

 

but I don't know where and how to apply that exactly in this case... I also do not know what [imath]m_e[/imath] is.

First of all, [imath]m_e[/imath] is the mass of an electron (that is what the subscript “e” denotes). Second, since [imath]\psi[/imath] is not a function of tau, [imath]\frac{\partial}{\partial \tau}\psi =0[/imath] and finally, you have misrepresented the differential of a product. Starting with the second term of your expression, the rest of the line should have been written as

[math]

\frac{\partial}{\partial \tau}\psi e^{-i\frac{m_e c}{\hbar}\tau}= \left\{\frac{\partial}{\partial \tau}\psi \right\}e^{-i\frac{m_e c}{\hbar}\tau} +\psi\left\{\frac{\partial}{\partial \tau} e^{-i\frac{m_e c}{\hbar}\tau}\right\}= -i\frac{m_e c}{\hbar}\psi e^{-i\frac{m_e c}{\hbar}\tau}

[/math]

 

which essentially states that the given function is an eigenfunction of the operator [imath]i\hbar\frac{\partial}{\partial \tau}[/imath] (multiply the above equation through by [imath]i\hbar[/imath]) with an eigenvalue of m_ec or c times the rest mass of the electron. There is one other thing you missed and that is the disappearance of the vector notation in going from [imath]\vec{\Psi}[/imath] to [imath]\psi[/imath]. I dropped out that notation because I was only concerned with the position of the electron and the vector notation was actually only there to handle the spin properties of the electron wave function; a factor which I really haven't gone into here because its impact upon the behavior of Maxwell's equations is a slightly more complex then what is being presented here. It's important, but what I am deducing can easily be seen as a raw deduction of the form of Maxwell's equations. Bringing in spin effects is a subtle extension which I am trying to avoid for the moment. Life is complex enough without it.

 

As an aside, Maxwell's equations deal with the electromagnetic fields one obtains from moving electric charges. Adding in spin yields quantum mechanical augmentation of those fields due to fundamental charge motion implied by consistency with the spin characteristics of the electron. What I am fundamentally deducing here are the simple consequences of moving charges.

The [imath]\psi^\dagger(\vec{x},t)\psi(\vec{x},t)\approx a\delta(\vec{x}-\vec{v}t)[/imath] I think I figured out. The value of [imath]\vec{x}[/imath] is essentially constrained to be given by [imath]\vec{v}t[/imath]... ...at least approximately.

Yes, sometimes approximate results are quite useful at least when we are discussing implications and not the details of those implications.

(The alpha there must be the alpha for element #1, should it have the subscript? And likewise the alphas of the consequent expressions should have the subscript 2?)

You caught it! I have edited the thing and fixed it.

Okay, getting to that expression clearly requires understanding of the stuff I asked about above, so I hope you can clear it out for me... In the meantime, I'll take the validity of the above expression on faith and move onwards.

See, every time you take what I say on faith, you are making a mistake. That expression is WRONG! It happens to be exactly where the problem with “c” arises. When I do things in my head, I almost always make one error or another. Note that I have multiplied through by [imath] -ic\hbar[/imath] and removed the constant “K” but have failed to remove the “c” which should have disappeared because [imath]K=-\frac{1}{c\sqrt{2}}[/imath]. Sloppy sloppy sloppy!

With the definition of [imath]c=-\frac{1}{K\sqrt{2}}[/imath], the right hand side is:

But you did it right!

The only problem is that you have that one mysterious "c" in there. Since I'm not sure how you got to that expression, I wouldn't want to start guessing as to who went wrong and where, but I did notice you don't have that "c" there in the OP, so I suspect you have made a typo somewhere :)

 

btw, in that same expression in OP, the [imath]\vec{\Psi}[/imath] is missing the subscript 1, and that second term inside the curly brackets is positive whereas here it is negative.

I think it is my failure to bring in that negative sign in the definition of K in terms of c. I have a suspicion that I have been very sloppy in all my conversions from K to c. Sorry about that. I am going to change the OP as your algebra seems to be error free.

Phew, I think I'll pause here for a bit to avoid more confusion, since I've been guessing way too much by now already. Scanning the rest of the post, it looks quite a bit easier though, if I can just get the above issues cleared...

I don't think you haven't been guessing at all. You have just been much more careful than I and I greatly appreciate it. Thank you for your great effort. Between the two of us, we just might be able to get the OP to and error free state.

 

Sorry I have been so slow, I was distracted by an article on “Thinking Machines” in my new discovery magazine. I sent you an e-mail about it which I presume you got. I don't expect anything to come of it but anything is possible.

 

Thank you again -- Dick

[AnssiH]

Just a quick reply before I go to sleep....

 

It is good to hear from you. In your absence I have been reading some of the other posts one the “philosophy” forum. The absolute absence of thought is just plain astonishing; take the “what is time” thread for example..., 764 posts totally without an accepted definition of what they are talking about.

 

Yeah they should change the title of that thread into "what is waste of time?", then at least they would have an answer right there.

 

QED is really a quite complex perturbation attack where the “Feynman diagrams” are actually no more than a shorthand notation used to refer to specific terms in an infinite sum. Just for the fun of it, you might take a look at the wikipedia entry for QED (not that I would expect you to understand any of it but rather it is an excellent example of esoteric shorthand notation only someone intimately involved would understand).

 

Yeah, didn't really open up to me :)

 

The following quote, taken directly from that entry, kind of tickles my fancy as my worthless Ph.D. thesis was actually directly related to actually calculating such perturbation results.

 

Didn't really understand what that was about either.

 

To me it looks like some demon sitting on a guy's head holding on to his chin (the dark shadow is the guy's head; all you can actually see of the fellow is his shoulders). I know that's a pretty imaginative image but it's exactly what pops into my head whenever I see it.

 

Oh, now I see it :) Hehe..

 

Well first, you need to have two identical elements (call them element #1 and element #2)...

 

Okay so that was essentially the same proof as what you discussed earlier in the "What can we know of reality", although it wasn't quite this succint back then :)

 

There is one minor cavil here that any competent physicist will probably assert. That is the fact that positions are not the whole story; one should really point out that their “state” must be the same.

 

I thought of that too. But then...

 

This is actually little more than “defining” what is meant by “identical”. The spin on an electron allows us to put two electrons in the same position so long as their spins sum to zero. I call it a minor cavil because, if their spins are not in the same direction they are clearly not identical elements (if they are identical, their spins they cannot sum to zero).

 

...I immediately thought of that also.

 

Dirac's equation and Dirac's notation are quite different issues.... ...We can get back to the matrix and Dirac notation later.

 

Ah, of course.. And yup...

 

Actually, it is exactly what is done in ordinary conventional physics. What I have referred to as “compartmentalization” elsewhere. Dirac's equation, by design, is an equation which yields the correct relativistic behavior of an electron when one takes the electromagnetic fields as given information. And Maxwell's equations yield the correct behavior of the electro-magnetic fields when one takes the positions and motions of electrons as given information. Thus what I am doing is no more than equivalent to the standard compartmentalization of everyday electrodynamics.

 

Hmmm, yeah, true.

 

Again it is my somewhat ambiguous notation which is confusing you. The expression, [imath]\vec{\Psi}_1(\vec{x},t)[/imath] is in my four dimensional Euclidean space [imath](x.y,z,\tau)[/imath] whereas the [imath]\psi_1(\vec{x},t)[/imath] is in the standard three dimensional space of conventional physics (x,y,z). You were expected to pick up on that fact by noticing the tau dependence had been pulled out. I should have written the thing

[math]\vec{\Psi}_1(x,y,z,\tau,t)=\vec{\sigma}\psi(x,y,z,t)e^{-i\frac{m_ec}{\hbar}\tau}[/math]

 

Ah, right... I recognized that move as removing the tau derivative from the equation, which of course is the same thing as removing tau dependence from the function... I'm just too unfamiliar with this stuff still to even draw that connection :P

 

where [imath]\vec{\sigma}[/imath] is an abstract unit vector supplying the vector attributes of [imath]\vec{\Psi}_1[/imath].

 

Okay

 

First of all, [imath]m_e[/imath] is the mass of an electron (that is what the subscript “e” denotes).

 

Okay, I see, that's why you said "...pulling off the dependence on [imath]\tau[/imath] in terms of the rest mass of the electron".

 

Second, since [imath]\psi[/imath] is not a function of tau, [imath]\frac{\partial}{\partial \tau}\psi =0[/imath] and finally, you have misrepresented the differential of a product. Starting with the second term of your expression, the rest of the line should have been written as

[math]

\frac{\partial}{\partial \tau}\psi e^{-i\frac{m_e c}{\hbar}\tau}= \left\{\frac{\partial}{\partial \tau}\psi \right\}e^{-i\frac{m_e c}{\hbar}\tau} +\psi\left\{\frac{\partial}{\partial \tau} e^{-i\frac{m_e c}{\hbar}\tau}\right\}= -i\frac{m_e c}{\hbar}\psi e^{-i\frac{m_e c}{\hbar}\tau}

[/math]

 

Oooops! I meant to say exactly what your result is, didn't look at that negative sign too carefully :D

 

Have to try and be careful with those things.

 

which essentially states that the given function is an eigenfunction of the operator [imath]i\hbar\frac{\partial}{\partial \tau}[/imath] (multiply the above equation through by [imath]i\hbar[/imath]) with an eigenvalue of m_ec or c times the rest mass of the electron. There is one other thing you missed and that is the disappearance of the vector notation in going from [imath]\vec{\Psi}[/imath] to [imath]\psi[/imath]. I dropped out that notation because I was only concerned with the position of the electron and the vector notation was actually only there to handle the spin properties of the electron wave function; a factor which I really haven't gone into here because its impact upon the behavior of Maxwell's equations is a slightly more complex then what is being presented here. It's important, but what I am deducing can easily be seen as a raw deduction of the form of Maxwell's equations. Bringing in spin effects is a subtle extension which I am trying to avoid for the moment. Life is complex enough without it.

 

Yes, we can get to that later then.

 

As an aside, Maxwell's equations deal with the electromagnetic fields one obtains from moving electric charges. Adding in spin yields quantum mechanical augmentation of those fields due to fundamental charge motion implied by consistency with the spin characteristics of the electron. What I am fundamentally deducing here are the simple consequences of moving charges.

 

Okay.

 

See, every time you take what I say on faith, you are making a mistake.

 

Seems like it :)

 

That expression is WRONG! It happens to be exactly where the problem with “c” arises. When I do things in my head, I almost always make one error or another.

 

Well then I must say I have the advantage that I can't do these things in my head yet :D

 

Note that I have multiplied through by [imath] -ic\hbar[/imath] and removed the constant “K” but have failed to remove the “c” which should have disappeared because [imath]K=-\frac{1}{c\sqrt{2}}[/imath]. Sloppy sloppy sloppy!

 

Yup.

 

I don't think you haven't been guessing at all. You have just been much more careful than I and I greatly appreciate it. Thank you for your great effort. Between the two of us, we just might be able to get the OP to and error free state.

 

Yes I hope so. I should have time over the weekend to move onwards with this.

 

Sorry I have been so slow, I was distracted by an article on “Thinking Machines” in my new discovery magazine. I sent you an e-mail about it which I presume you got. I don't expect anything to come of it but anything is possible.

 

No I didn't actually receive an e-mail, maybe try re-sending it, it must have gotten buried underneath all the spam... :(

 

-Anssi

[Doctordick]

Didn't really understand what that was about either.

Mostly it was the, ignore or take a "shut-up-and-calculate" approach. That is essentially the -"direction"- my thesis advisor gave me some forty years ago.

Yes, we can get to that later then.

Since it relates to something Qfwfq commented about two years ago, I decided to answer his post. I don't think the issue really needs to be addressed by me as I am sure there are many physicists much more qualified to work out those issues. Read my new post to the “What can we know of reality?” thread.

No I didn't actually receive an e-mail, maybe try re-sending it, it must have gotten buried underneath all the spam... :shrug:

You are probably right. I forget that Yahoo is well known as a “spam source”. A lot of people don't get my e-mails unless I am replying to one of their notes to me.

 

Such is life, I probably shouldn't have written the thing anyway. Lot's of times I make comments I later wish I hadn't.

 

Have fun -- Dick