Light Wave

[Little Bang]

If I have a detector for light, say green light, and I connect the output to a scope. Can anyone explain what the amplitude of that frequency means?

[Erasmus00]

If I have a detector for light, say green light, and I connect the output to a scope. Can anyone explain what the amplitude of that frequency means?

 

The amplitude is related to number of photons per second hitting the detector.

-Will

[Little Bang]

Will, I'm not saying that your wrong, but it seems to me that one cycle of the wave should be related to one photon.

[infamous]

Will, I'm not saying that your wrong, but it seems to me that one cycle of the wave should be related to one photon.

I was just thinking the same thing Little Bang, maybe Will can explain this misunderstanding for us?

[Erasmus00]

Will, I'm not saying that your wrong, but it seems to me that one cycle of the wave should be related to one photon.

 

Why? First, you can't really detect one photon as a wave pattern. If you are getting a wave pattern it means that you have lots of photons hitting your detector building up the pattern. Now, classicly, the amplitude of the wave is related to the energy, and since at the same frequency we have the same energy/photon, a higher intensity wave must have more photons per second.

 

Maybe a good way to see this is as follows: Notice that one photon has some given amplitude and wavelength. Draw the wave. What happens if we have two photons? We add two of these waves together. If they add in phase, we get one wave with twice the height (amplitude).

-Will

[Little Bang]

You may be right, but there are ten gigahertz, single shot scopes now that should be able to detect a single photon. if that is true what does the amplitude mean?

[Erasmus00]

You may be right, but there are ten gigahertz, single shot scopes now that should be able to detect a single photon. if that is true what does the amplitude mean?

 

It wouldn't look like a wave though. Just a brief voltage blip. You need a whole lot of particles to build a wave up.

-Will

[Little Bang]

Single shot means that I can build a waveform from a single event. If my detector sees one photon it can display the waveform that describes the photon.

[EWright]

Why? First, you can't really detect one photon as a wave pattern. If you are getting a wave pattern it means that you have lots of photons hitting your detector building up the pattern. Now, classicly, the amplitude of the wave is related to the energy, and since at the same frequency we have the same energy/photon, a higher intensity wave must have more photons per second.

 

Maybe a good way to see this is as follows: Notice that one photon has some given amplitude and wavelength. Draw the wave. What happens if we have two photons? We add two of these waves together. If they add in phase, we get one wave with twice the height (amplitude).

-Will

Hi Will, I"m certainly not challenging your knowledge here - just trying to understand. It sounds as if you're saying higher energy waves are not a matter of intensity per photon, but rather, the number of photons given off by the source within a given time, ie more photon output?

 

I'm interpreting it this way because when you say "more photons per second" this obviously cannot mean at a *faster rate* because all photons are required to travel at the same constant speed.

 

Your description also leads me to ask if then single photons do actually travel as a wave, or if that is just the interpretation of a mass of photons?

[Carlos]

Talking about photomultipliers? :cup:

[Erasmus00]

Single shot means that I can build a waveform from a single event. If my detector sees one photon it can display the waveform that describes the photon.

 

Right, but a single photon will just exhibit some voltage in the detector at some time, not a wave. The minute it hits the detector its waveform collapses. All you get is that single blip. To build a probability wave, you need lots of photons.

-Will

[Erasmus00]

Hi Will, I"m certainly not challenging your knowledge here - just trying to understand. It sounds as if you're saying higher energy waves are not a matter of intensity per photon, but rather, the number of photons given off by the source within a given time, ie more photon output?

 

It could be either. Turning up the frequency and keeping the number of photons emmited per second the same will result in more energy. We could also keep the frequency the same, and increase the number of photons. Either way we increase the intensity of the wave.

 

Your description also leads me to ask if then single photons do actually travel as a wave, or if that is just the interpretation of a mass of photons?

 

The standard interpretation is that the wave represents the probability of finding a photon at some point in space and time. The photon itself is a point particle. When you puts lots of photons together, the system becomes classical, and we see a wave built out of lots of individual detection events.

-Will

[CraigD]

I'm interpreting it this way because when you [Will] say "more photons per second" this obviously cannot mean at a *faster rate* because all photons are required to travel at the same constant speed.

Photons are bosons. Unlike the other major class of fundamental particles, fermions, bosons don’t obey the Pauli exclusion principle, which means that there’s no inherent limit to how many of them you can have in a single place at the same time. So, regardless of the speed of light, there’s no practical upper limit to the intensity of light.

Your [Will’s] description also leads me to ask if then single photons do actually travel as a wave, or if that is just the interpretation of a mass of photons?

Single photons do actually travel as a wave.

 

This can (and has) been verified by a simple (and famous) experiment:

1. Construct a dual-slit aperture, such that shining a light source (eg: laser, neon tube, glowing metal) through it onto a photographic film on the other produces a clear interference pattern (after a short exposure);
   
2. Interpose so many photographic filters (eg: smoked glass) between the light source and the apparatus that the film show no change after even a long exposure;
   
3. Remove filters one at a time, until a image again appears on the film;
   
4. Calculate and take a long enough exposure that the same pattern appears as did in step test 1

It can be shown that such an arrangement rarely contains more than 1 photon at a time, yet produces the same interference pattern as it did when many photons were present at the same time. Therefore, each photon exhibited interference, a wave phenomena..

[CraigD]

… a single photon will just exhibit some voltage in the detector at some time, not a wave. The minute it hits the detector its waveform collapses. All you get is that single blip. To build a probability wave, you need lots of photons.

You can measure the frequency (and thus wavelength) of a single photon. One way is to carefully select the type and state of the atoms your detector is made of, so that only specific photons are reflected or generate a measurable (eg: photoelectric) effect. Just as very intense light below the frequency given by a photoelectric material’s work function won’t cause it to produce even a single electron of current, bombarding such a detector with oddly timed streams of photons won’t cause it to relate these timings to the light’s frequency.

[Erasmus00]

You can measure the frequency (and thus wavelength) of a single photon. One way is to carefully select the type and state of the atoms your detector is made of, so that only specific photons are reflected or generate a measurable (eg: photoelectric) effect. .

 

This method allows you to carefully select the energies of the photons you are detecting. However, you aren't "seeing" the wavelength of the photon in the sense to which Little Bang was referring, but extrapolating from its energy.

-Will

[Little Bang]

Ok Will and I am honestly trying to learn so bear with me, I have a detector that is detecting billions of photons. My scope produces a pure sinewave with a peak positive voltage and a peak negative voltage. What do these two peaks tell me about the photons that I'm detecting?

[Erasmus00]

Ok Will and I am honestly trying to learn so bear with me, I have a detector that is detecting billions of photons. My scope produces a pure sinewave with a peak positive voltage and a peak negative voltage. What do these two peaks tell me about the photons that I'm detecting?

 

The frequency represents the energy/photon. Classicaly, the amplitude is related to the energy in the whole wave(or more precisely, the energy is proportional to the amplitude squared). So, the amplitude squared is representative of the number of photons per second impinging on your detector.

-Will