Is there a difference between EM waves and photon wavefunctions?

[A. Neumaier]

There are CQED methods that allow you to "sense" the number of photons in a Fock state dispersivly.

How do they respond to dim coherent laser light?

 

[vanhees71]

And in the case of prepared 1-photon states one knows the presence of the single photon only because of the preparation, not because of the detection! The detection only informs one about the presence of a local burst of the energy density (sufficient to fill the cup, in the classical analogy). Together with the knowledge about the preparation one infers that the photon arrived.

Exactly. Now we completely agree :-)).

 

[A. Neumaier]

Blais, Alexandre, et al. "Cavity quantum electrodynamics for superconducting electrical circuits: An architecture for quantum computation." Physical Review A 69.6 (2004): 062320.

These systems can be used to create and manipulate just about any photon state you can imagine, including obviously simple Fock states.

I was talking about photons in a beam extending over a laboratory desk, and their detection. Your setting is quite different.
How do you get the photons from a photon beam into the cavity to be detected by your solid state devices?

 

[A. Neumaier]

Note also that is is possible to perform tomography of photon states using "classical" detectors by using methods borrowed from radio engineering (IQ mixing) . It is is of course an indirect method since you can't "see" individual events, but as long as you are able to average over many events you can re-construct the state of the photons generated by your source.

This can be done for the state of any source - one doesn't need any assumptions on whether or not it contains ghostlike particles.

The density matrix of an $n$-level system has $n^2-1$ independent degrees of freedom, and it is not difficult to find this many independent observables such that, from the measurement of their mean (by repeated observation of individual systems emitted by the source), the density matrix can be reconstructed by solving a linear system of equations.

 

[Elemental]

Sciencejournalist00,

On your original question,

Or is the electromagnetic wave considered to be the wavefunction of the photon?

consider the following. One performs a "box quantization" of the electromagnetic field by treating each cavity mode of the electromagnetic field as a harmonic oscillator. Here though you are actually starting with cavity modes of the electromagnetic field rather than modes of the Schrödinger wave function but it goes the same way. Singling out a particular mode, from Maxwell's equations it is found one can treat the electric field as the generalized coordinate and the magnetic field as the conjugate momentum. (Of course other approaches are possible including treating the magnetic field as the coordinate or more commonly quantizing the vector potential, but this approach illustrates the relation of wave function to EM field.)

You write the Hamiltonian for the EM field energy, then re-express it in terms of annihilation/creation operators exactly as is done for the harmonic oscillator. Presto, you obtain EXACTLY the same Schrödinger equation and equations of motion as you get for the harmonic oscillator. This has a wave function whose allowed modes have the same frequency as those of the electromagnetic field you started with! One can then define coherent states like you get from a laser, Fock (number) states, squeezed states, etc.

So the modes of the wave function sure look a lot like electromagnetic field modes. But what about its interpretation? For the harmonic oscillator, the magnitude squared of the wave function gives the probability of the position (coordinate) of the particle. For the quantized electromagnetic field as derived here, it does not give position probability, it gives the probability that the electric field will have a particular value. Not the same thing. However I've noticed in many quantum optics papers that the difference in interpretation between the Schrödinger wave function and the electromagnetic filed is glossed over.

 

[vanhees71]

One should stress that there is no consistent interpretation of relativistic wave equations in the sense of the "first quantization formalism", except in certain low-energy cases, where the physics is close to non-relativistic dynamics, i.e., only for massive quanta (e.g., bound states, where the binding energy is small compared to the masses of the bound quanta).

For massless quanta, and photons are with amazing accuracy observed as massless quanta, no such "wave-function interpretation" makes any sense. Since photons are described as massless spin-1 fields, they do not admit the definition of a position operator, and thus the position of a photon is not describable as an observable. So the localization of a photon is not even definable.

Another reason is that photons are very easily produced and destroyed. This leads to the description of relativistic quanta in terms of a quantum many-body, where quanta can produced and destroyed, and thus the number of quanta is not conserved. This is precisely what quantum field theory does in a very natural way and as explained in the previous posting #40.

The "box quantization" described in this posting is not only an interesting real physical situation in "nano physics", where one investigates single photons or states with a small number of photons in cavities ("cavity QED"), but also a convenient mathematical technique to quantize the (completely gauge fixed, using the radiation gauge) electormagnetic field canonically. In this case, you rather use periodic than rigid boundary conditions, because then you have a well defined momentum operator for the photons in the box-regularized version. At the end of the calculation you can take the infinite-volume limit, leading to the usual continuous momenta of photons in free space.

For a very detailed explanation, including a very good and intuitive introduction to the necessity for the field-quantization approach for relativistic particles, of this approach to photons, see Landau&Lifshitz, Course of Theoretical Physics, vol. IV.

 

[tade]

This is not correct. it is well-known that photon location cannot be described by a probability distribution

What about the double-slit experiment, and the interference fringe pattern?

Aren't the bright fringes bright because photons have a higher probability of being there, and the dark fringes are dark because the photons have a low probability of being there?

What about double-slit experiments where they reduce the intensity to one photon at a time, yet there is still an interference pattern?

 

[A. Neumaier]

What about the double-slit experiment, and the interference fringe pattern?

The point where a beam hits a screen determines only the transversal position, not the longitudinal one. Transversal position is reasonably well-determined in the paraxial approximation.

 

[tade]

The point where a beam hits a scree determines only the transversal position, not the longitudinal one. Transversal position is reasonably well-determined in the paraxial approximation.

I'm sorry, I don't understand this terminology.

How does this relate to probability waves, or that photons cannot be described by a probability distribution?

 

[A. Neumaier]

I'm sorry, I don't understand this terminology.

If a beam goes into the z-direction, the transversal position are the x,y coordinates only, while z is the longitudinal position. The paraxial approximation defines what a beam is in term of a field.

How does this relate to probability waves, or that photons cannot be described by a probability distribution?

There are commuting operators for transversal position but not for longitudinal position.

 

[tade]

If a beam goes into the z-direction, the transversal position are the x,y coordinates only, while z is the longitudinal position. The paraxial approximation defines what a beam is in term of a field.There are commuting operators for transversal position but not for longitudinal position.

Ok, is that considered a transversal probability distribution?

My understanding so far is described by this Wikipedia passage:

Sending particles through a double-slit apparatus one at a time results in single particles appearing on the screen, as expected. Remarkably, however, an interference pattern emerges when these particles are allowed to build up one by one (see the image to the right).

This demonstrates the wave-particle duality, which states that all matter exhibits both wave and particle properties: the particle is measured as a single pulse at a single position, while the wave describes the probability of absorbing the particle at a specific place of the detector. This phenomenon has been shown to occur with photons, electrons, atoms and even some molecules, including buckyballs.

This is the electron build up over time. What's the fundamental difference between electrons and photons in this scenario?

I posed a question about wavefunction and EM waves in my thread: https://www.physicsforums.com/threa...ld-and-probability-amplitude-of-waves.869530/
which is what led me to this thread.

 

[Nugatory]

Aren't the bright fringes bright because photons have a higher probability of being there, and the dark fringes are dark because the photons have a low probability of being there?

No, and the problem is in the words "being there" - wording it that way is already assuming that the photon has a position and hence a probability distribution for that position. There's no substitute for actually learning and using quantum electrodynamics, but if you want a non-rigorous and intuitive sort of understanding, you could say that the bright fringes are bright because there is a higher probability that the electromagnetic field will deliver energy to those areas, or you could say "appearing there" instead of "being there".

 

[tade]

No, and the problem is in the words "being there" - wording it that way is already assuming that the photon has a position and hence a probability distribution for that position. There's no substitute for actually learning and using quantum electrodynamics, but if you want a non-rigorous and intuitive sort of understanding, you could say that the bright fringes are bright because there is a higher probability that the electromagnetic field will deliver energy to those areas, or you could say "appearing there" instead of "being there".

I see.
Am I right to say that a photon cannot be described by a scalar wavefunction? (a wavefunction similar to the wavefunctions of the Schrödinger equation) That only a vector field can describe it?

 

[A. Neumaier]

I see.
Am I right to say that a photon cannot be described a scalar wavefunction?

The wave function is given by the Silberstein vector field.
You may wish to browse my theoretical physics FAQ.

 

[tade]

The wave function is given by the Silberstein vector field.

Ahhh, thank you.

 

[A. Neumaier]

Ahhh, thank you.

See http://arnold-neumaier.at/ms/optslides.pdf

 

[tade]

See http://arnold-neumaier.at/ms/optslides.pdf

Thanks.

By the way, can you help with me these problems?

https://www.physicsforums.com/threads/what-is-the-theoretical-basis-for-kramers-law.869641/

https://www.physicsforums.com/threads/plancks-1st-derivation-of-plancks-law.869497/