I Photon interference and beamforming

[vanhees71]

I have a feeling that this demonstrates trying to teach students to run before they can walk. If elementary wave theory is properly sorted out first then the probability concept is more of a shoe in.
The problem is, of course, that we are dealing with an ever expanding canon of topics and brains are no bigger than they used to be.

It would of course help a lot, if you'd not teach them anymore completely outdated subjects like "photons are little particles" or Bohr's model of the hydrogen atom. It would help to learn the right thing first, avoiding the effort to unlearn models that give a wrong intuition and are outdated by almost 100 years! So the canon doesn't need to expand that much!

 

[antonantal]

What does "two different photons" actually mean? You may have a two-photon state, in which case the photons are indistinguishable. That system may have superposition that constructively or destructively interferes with itself.

Let me check if I understand correctly what you mean.
If two photons are emitted at exactly the same time from two sources that are in perfect phase sync, at the screen there is no way to tell which photon is which, so they are indistinguishable.
Then, we don't have two single-photon wave functions that are interfering each with itself and with the other resulting in the incorrect 8-fold increase in intensity (calculated in post #14, point 3), but instead we have a single two-photons wave function which is only interfering with itself and follows the usual probability density (calculated in post #14, point 2): $$P(p) = |\tilde \psi(p)|^2 = \frac 1 {2\pi} \left[1 + cos\left(\frac {p(x_1-x_2)} {\hbar}\right)\right]$$
But this time, it represents the probability of finding 2 photons (since it is the wave function of a two-photon state) at a certain angle on the screen. This way we get a factor of 2 (because 2 photons) compared to the double-slit/single-photon case, resulting in the correct factor of 4 compared to the single-slit/single-photon case.

 

[vanhees71]

Photons have no wave function. A general two-photon state is given by
$$|\Psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 p_1 \int_{\mathbb{R}^3} \mathrm{d}^3 p_2 \sum_{\lambda_1,\lambda_2 \in \{-1,1\}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1) \hat{a}^{\dagger}(\vec{p}_2,\lambda_2) \Psi(\vec{p}_1,\lambda_1;\vec{p}_2,\lambda_2) |\Omega \rangle,$$
where $\Psi(\vec{p}_1,\lambda_1;\vec{p}_2,\lambda_2)$ is a square-integrable function, $\hat{a}^{\dagger}(\vec{p},\lambda)$ are creation operators for photons with momentum $\vec{p}$ and helicity $\lambda$, and $|\Omega \rangle$ is the vacuum state.

 

[antonantal]

Photons have no wave function. A general two-photon state is given by
$$|\Psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 p_1 \int_{\mathbb{R}^3} \mathrm{d}^3 p_2 \sum_{\lambda_1,\lambda_2 \in \{-1,1\}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1) \hat{a}^{\dagger}(\vec{p}_2,\lambda_2) \Psi(\vec{p}_1,\lambda_1;\vec{p}_2,\lambda_2) |\Omega \rangle,$$
where $\Psi(\vec{p}_1,\lambda_1;\vec{p}_2,\lambda_2)$ is a square-integrable function, $\hat{a}^{\dagger}(\vec{p},\lambda)$ are creation operators for photons with momentum $\vec{p}$ and helicity $\lambda$, and $|\Omega \rangle$ is the vacuum state.

This is beyond my basic QM knowledge, but since the integrals are with respect to momentum, it means $|\Psi \rangle$ is a function of helicity $\lambda$ (don't know what that is). So, can it be interpret as a superposition of photons with different helicities? And does helicity identify in any way the source of the photon, same as position would?

 

[vanhees71]

This is a two-photon state, i.e., if you make a two-photon measurement, i.e., you put a detector registering photons, you'll find two photons. Helicity is the equivalent of spin for massless particles. It's the component of the total angular momentum in direction of the photon's momentum, and it can take values $\pm 1$. The corresponding em. waves are left-circular or right-circular polarized plane waves.

 

[antonantal]

I would expect that the state vector $|\Psi \rangle$ is the sum of some eigenvectors weighted by probability amplitudes. Since these can not be momentum eigenvectors (because $\vec p$ vanishes after integration), I expect they are position eigenvectors. But since there is no position-related variable in the formula I am a bit confused. What is the basis in which $|\Psi \rangle$ is represented?

 

[vanhees71]

It's just the state ket in representation-free form.

 

[sophiecentaur]

It would help to learn the right thing first,

Do you mean to learn the current thing first? Diffraction and interference are good old classical phenomena which students get presented with at an appropriate time.
I'd suggest that teaching QM comes in later than basic wave theory. If you kick off their QM education by deliberately ignoring waves then many (especially the bright ones) will spot the parallels. Neither approach is necessarily right or wrong. Perhaps there should be a caveat about this at the start of every chapter and lesson. The last thing we want is for the "Newton was wrong" brigade to get a look in.

 

[vanhees71]

Of course, you should learn non-relativistic QM first. I think the old-fashioned way via wave mechanics is the best intro, but you should get as quickly as possible to Dirac's representation-free bra-ket formulation.

Relativistic QT should nowadays be taught only as relativistic QFT. Relativistic quantum mechanics is completely outdated and has not much merit in teaching it. Particularly if it comes to photons any classical-particle picture is misleading.

 

[sophiecentaur]

Perhaps it would be best not to use the term "Interference" at all in this context. If the results of interaction between small numbers of photons are subtly different then 'something' needs to be bolted onto the straightforward wave approach Perhaps Quinterference could describe what happens.
Or is it possible that the actual experiments are flawed in some strange way when they are claiming to identify the parameters of the photons they are examining.

 

[Charles Link]

I am led to believe that there are still some missing pieces to the puzzle. In the last 40 years there have been a couple experiments, where as I recall, they showed interference with a beamsplitter (or beamsplitters) and a single photon. The researchers didn't know before the experiment what the outcome would be. Perhaps there is a standard textbook that covers these topics in detail. @vanhees71 Can you furnish us with a good reference that has most of the latest on the subject, and hopefully is also somewhat easy reading?

Edit: I did a little searching of my own: See https://www.physicsforums.com/threads/interference-of-a-single-photon-in-an-interferometer.920022/ I see this topic has been discussed on Physics Forums in the last couple of years, and the discussion is interesting.

 

[antonantal]

Edit: I did a little searching of my own: See https://www.physicsforums.com/threads/interference-of-a-single-photon-in-an-interferometer.920022/ I see this topic has been discussed on Physics Forums in the last couple of years, and the discussion is interesting.

Post #27 from that topic cites the following statement:
"Interference can occur if two or more different ways to produce the same result cannot be distinguished with the apparatus."

Now let's apply it to the case of two photons that are emitted at exactly the same time from two sources that are in perfect phase sync. At the screen there is no way to tell which photon is which, so they cannot be distinguished and interference will occur.
Does my post #52 describe this correctly?
In other words, do we have a two-photon state which is a superposition of "both photons emitted by first source" and "both photons emitted by second source"?

 

[sophiecentaur]

Now let's apply it to the case of two photons that are emitted at exactly the same time from two sources that are in perfect phase sync.

Isn't this a very unreal scenario? How would you be able to tell that they are emitted at exactly the same time? Not being like little bullets, they cannot be put past a start line with a stop watch. Only when you consider a single photon can you discuss anything "exactly" the same between the two possible paths by the two 'alternative' versions of the photon which interfere with each other.

It's a bit late to be going over the basics of this thread but, just to make sure I have appreciated the principles of the experiments, it seems to me that individual photons are 'gated' through a shutter and then delay adjusted to be equal so that they pass through a slit each. Then, when everything is arranged correctly, a 'conventional' two slit pattern is observed. Is this in any way along the right lines? (I have to apologise if my idea is too much like the RF equivalent.)

 

[antonantal]

@sophiecentaur , you are correct, what I meant by "emitted at the same time" is just that "they are emitted in such a way that they could interfere at the screen".

 

[sophiecentaur]

@sophiecentaur , you are correct, what I meant by "emitted at the same time" is just that "they are emitted in such a way that they could interfere at the screen".

It worries me that it’s such a big IF.
But, once they have shown interference in space then why not also in time? (ie all the other interference phenomena) it all seems to hang on coherence - as ever.