Square of absolute value of amplitude for a single photon

In summary: Maxwell's equations and classical electrodynamics before you'll be in able to take on the quantum electrodynamics description of light as photons - otherwise you'll just confuse yourself. (This shouldn't be surprising, as QED is basically quantizing the electromagnetic field, and you'd expect to have to understand something before you can understand how it's quantized).In summary, the probability being discussed in this conversation is related to the detection of a single photon on a surface where it is pumped by a laser. The probability is proportional to the expectation value of the energy-density of the radiation field, similar to classical electrodynamics. However, it is not possible to define a wave function for a single photon in the same sense as
  • #1
DavidReishi
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I understand that this determines a probability, but of what exactly for a single photon? The probability that the photon will be detected on a surface where the photon is pumped, e.g. where on the surface the laser is aimed?
 
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  • #2
DavidReishi said:
I understand that this determines a probability, but of what exactly for a single photon? The probability that the photon will be detected on a surface where the photon is pumped, e.g. where on the surface the laser is aimed?
Your intended meaning is unclear. Amplitude of what? Please reference your statements.
 
  • #3
Orodruin said:
Your intended meaning is unclear. Amplitude of what? Please reference your statements.

Sorry about that. I was under the impression that when a laser is aimed at a surface, for example, in emitting very low intensity light at a CCD plate, the amplitude of the light is known. So I figured that the square of the absolute value of that amplitude gives a probability having to do with where each photon will be detected on the plate. Thus my question was what exactly is this probably dealing with. Where the photon will be detected radially on the plate in relation to where the laser is aimed?
 
  • #4
It is not possible to define a wave function for a single photon in the same sense as for a massive particle in non-relativistic physics. The reason is that there is no position operator for a single massless quantum with spin ##\geq 1##.

The detection probability is proportional to the expectation value of the energy-density of the radiation field, similar to classical electrodynamics.

Further a laser never emits single photons but an electromagnetic field that is quantum-theoretically described as a socalled coherent state.
 
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  • #5
vanhees71 said:
It is not possible to define a wave function for a single photon in the same sense as for a massive particle in non-relativistic physics. The reason is that there is no position operator for a single massless quantum with spin ##\geq 1##.

The detection probability is proportional to the expectation value of the energy-density of the radiation field, similar to classical electrodynamics.

Further a laser never emits single photons but an electromagnetic field that is quantum-theoretically described as a socalled coherent state.

If we have the same set-up as the one-photon-at-a-time double slit experiment using a CCD surface as the back-plate, except we remove the front plate with the slits, each photon will by no means necessarily be detected by the CCD exactly where the light was aimed, right? Is there not a formula to determine the probability of where each photon will be detected, say, in terms of its distance on the CCD from where it was aimed?
 
  • #7
vanhees71 said:
It's proportional to the intensity as calculated by classical electrodynamics, and the intensity is proportional to the energy-density distribution.

Do you mean that, in my example, the probability of where each photon will be detected is proportional to the light's relative intensity on the plate if it were much more light than single photons being used?


I'm confused by these links. I thought diffraction comes into play only if we keep the front plate with the slits. Or is it a fact that light coming directly out a source follows the same principle? The opening from which the light is emitted being a "first slit," out of which it diffracts?
 
  • #8
DavidReishi said:
Do you mean that, in my example, the probability of where each photon will be detected is proportional to the light's relative intensity on the plate if it were much more light than single photons being used?
You may be better off thinking in terms of the expected number of photons detected at a given point per unit time. This will be directly related to the intensity of the light; if the light is brighter at a given point more photons will be detected there in any given time interval. (The problems with just looking at the "probability of a photon being detected" without thinking "per unit time" are: if you wait long enough, you're likely to find a photon just about anywhere; and the light doesn't need to be very bright at all before the probability of detecting at least one photon approached 100% and the interesting question becomes "how many?").
The opening from which the light is emitted being a "first slit," out of which it diffracts?
Yes, that's what's going on. You will have to learn the classical model of light as electromagnetic waves governed by Maxwell's equations and classical electrodynamics before you'll be in able to take on the quantum electrodynamics description of light as photons - otherwise you'll just confuse yourself. (This shouldn't be surprising, as QED is basically quantizing the electromagnetic field, and you'd expect to have to understand something before you can understand how it's quantized).
 
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  • #9
Nugatory said:
You may be better off thinking in terms of the expected number of photons detected at a given point per unit time. This will be directly related to the intensity of the light; if the light is brighter at a given point more photons will be detected there in any given time interval. (The problems with just looking at the "probability of a photon being detected" without thinking "per unit time" are: if you wait long enough, you're likely to find a photon just about anywhere; and the light doesn't need to be very bright at all before the probability of detecting at least one photon approached 100% and the interesting question becomes "how many?").

Thanks, that's helpful. I was actually visualizing photons per unit of time, but still one at a time. I just couldn't get it outta my head when I read about the modern double-slit experiment, in which classes of undergraduates get to watch an interference pattern build up a single photon at a time. Lol, or did I read that wrong?

Yes, that's what's going on.You will have to learn the classical model of light as electromagnetic waves governed by Maxwell's equations and classical electrodynamics before you'll be in able to take on the quantum electrodynamics description of light as photons - otherwise you'll just confuse yourself. (This shouldn't be surprising, as QED is basically quantizing the electromagnetic field, and you'd expect to have to understand something before you can understand how it's quantized).

Funny you mention that. Not even half an hour ago I sprang outta my seat to see if I had Maxwell's book. I don't, but I do have a facsimile of his Matter and Motion. Thanks for the advice about studying electrodynamics before quantum electrodynamics. Though I'd guess one could argue the exact opposite, that it's better to learn about the unit before taking on the combination of those units. Or, perhaps even better, that it's best to learn about both simultaneously.
 
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  • #10
DavidReishi said:
I just couldn't get it outta my head when I read about the modern double-slit experiment, in which classes of undergraduates get to watch an interference pattern build up a single photon at a time. Lol, or did I read that wrong?
That image is just fine, but you're still best off thinking in terms of dots appearing per unit time at each point on the film. And if you aren't prepared to do the full quantum electrodynamic treatment, that appearance rate is calculated from the intensity of the light at that point, and the intensity is calculated classically.
Thanks for the advice about studying electrodynamics before quantum electrodynamics. Though I'd guess that one could argue the exact opposite, that it's better to learn about the unit before taking on the combination of those units. Or, perhaps even better, that it's best to learn about both simultaneously.
That makes about as much sense as suggesting that it's best to learn elementary algebra and elementary integral calculus at the same time. There's a reason why it takes four years or more to get from classical E&M to quantum field theories, and it's the same reason that it usually takes about four years to get from elementary algebra to calculus - there's a lot to learn in between.
 
  • #11
Nugatory said:
That image is just fine, but you're still best off thinking in terms of dots appearing per unit time at each point on the film. And if you aren't prepared to do the full quantum electrodynamic treatment, that appearance rate is calculated from the intensity of the light at that point, and the intensity is calculated classically.

You mean the inverse square law? What would the full quantum electrodynamic treatment of the question look like?
 
  • #12
DavidReishi said:
You mean the inverse square law?
No, not the inverse square law (although the inverse square law does fall out of the classical solution for the intensity of radiation emitted from a point source). I was talking about the appearance of interference patterns in the double slit experiment which you mentioned above, and the appearance of diffraction patterns that vanhees71 had mentioned. The common thread across these examples is that you can start with the classical wave solution including diffraction and interference, calculate the intensity at each point, and get from that the photon arrival rate.
What would the full quantum electrodynamic treatment of the question look like?
I can give you two answers:
1) Try Feynman's "QED: The strange theory of light and matter". I've already recommended this in another of your threads and it's about as good as it gets for a non-specialist.
2) To see what a full quantum electrodynamic treatment would look like... Mark Srednicki's QFT textbook is available online; the section on spin-one particles is what you're looking for. Note that I am not recommending Srednicki's textbook as a good start for self-study (for that Blundell and Lancaster's "Quantum field Theory for the Gifted Amateur" would be better), just giving an example of what the real thing looks like.
 
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  • #13
DavidReishi said:
Funny you mention that. Not even half an hour ago I sprang outta my seat to see if I had Maxwell's book.

Maxwell's treatise is famous for being more or less unreadable; it sold quite well when it was published; but anecdotally there were only about three people in the world who understood it:rolleyes:
What we now call Maxwell's theory was formulated quite a bit later using more modern notation etc by a group of other scientists.

Hence, don't try to learn physics by reading anything he wrote. Pick up a modern book instead, I believe Griffith's book is supposed to be quite good.
 
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Related to Square of absolute value of amplitude for a single photon

1. What is the square of absolute value of amplitude for a single photon?

The square of absolute value of amplitude for a single photon refers to the square of the magnitude of the amplitude of a single photon in a quantum system. It is a measurement of the intensity or strength of the photon's wave function.

2. How is the square of absolute value of amplitude for a single photon calculated?

The square of absolute value of amplitude for a single photon is calculated by squaring the absolute value of the amplitude of the photon's wave function. This can be represented by the equation |A|², where A is the amplitude.

3. What does the square of absolute value of amplitude for a single photon tell us about the photon?

The square of absolute value of amplitude for a single photon provides information about the probability of the photon's position or momentum in a quantum system. It is also related to the photon's energy and wavelength.

4. How does the square of absolute value of amplitude for a single photon relate to the uncertainty principle?

The square of absolute value of amplitude for a single photon is related to the uncertainty principle, which states that the more accurately we know the position of a particle, the less we know about its momentum, and vice versa. The square of absolute value of amplitude for a single photon represents the probability of its position or momentum, thus contributing to the uncertainty principle.

5. Can the square of absolute value of amplitude for a single photon be greater than 1?

No, the square of absolute value of amplitude for a single photon cannot be greater than 1. This is because the probability of the photon's position or momentum cannot exceed 100%. If the square of absolute value of amplitude for a single photon is equal to 1, it means that the photon's position or momentum is certain, and if it is less than 1, there is a probability associated with it.

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