Require A Mathematical Help On The Function Of The Derivative On An Operator

[Dubbelosix]

The equation I used was:

 

 

[math]\frac{\ddot{R}}{R} \Theta \psi = -i \frac{\hbar}{MR^2}\Theta \frac{d\psi}{dt} [/math]

 

 

Now, keep in mind, within literature, the quotients of scale factor, radius and particle number can be seen in light of:

 

[math]\frac{\dot{a}}{a} = \frac{\dot{R}}{R} = \frac{\dot{N}}{N}[/math]

 

[math]\equiv (\frac{a}{a} = \frac{R}{R} = \frac{N}{N}) \Theta [/math]

 

(Theta here is just a time derivative called  a fluid expansion)

 

 

Direct substitution for the particle number identity gives

 

 

[math]\frac{\ddot{N}}{N}\Theta \psi = -i \frac{\hbar}{MR^2}\Theta \frac{d\psi}{dt} [/math]

 

 

Taking the operator form of [math]\frac{\ddot{N}}{N}[/math] has raised a question about the application of the mathematics. Without too much explanation, the operator form will lead to ~

 

[math]\Sigma_k \frac{aa^{\dagger}}{( a^{\dagger}a + 1)}[/math]

 

Which is a configuration of creation and annihilation operators, but there are two time derivatives [supposed] to be in there!

 

Though the double time derivative was in ''the particle number quotient'' I have had to end up writing [math]\Theta^3[/math] for the last equation, because I've never seen a creation or annihilation operator with a time derivative, let alone two of them.

 

 

[math]\Sigma_k \frac{aa^{\dagger}}{( a^{\dagger}a + 1)} \Theta^3 \psi = -i \frac{\hbar}{MR^2}\Theta \frac{d\psi}{dt} [/math]

 

 

But I feel I've moved the time derivative, may not change the physics but it feels smudged. Maybe I am just being hard on myself, but if anyone knows how to carry the quantization onto the creation and annihilation operators (if possible) I would appreciate it --- so really for the mathematicians, and mathematical physicists that may be about, because I am bit stumped myself. 

Edited April 6, 2017 by Dubbelosix