Photon interference and beamforming

[PeroK]

Why would it be? The basic idea is the same: you add amplitudes, not probabilities. That's ultimately where all quantum interference effects come from.

If we add two amplitudes and get a number of magnitude squared greater than 1, then it can no longer represent a probability.

 

[vanhees71]

Of course you always have to normalize your distributions appropriately to get probabilities.

 

[Charles Link]

That simply follows from the Hamiltonian for the cavity photons,
$$\hat{H}=\sum_k \hat{N}_k \omega_k,$$
where the sum runs over all single-photon cavity states. There's nothing mysterious in this. It's just one of the first exercises in QFT to deal with free particles/photons.

Maybe a single photon can't be treated as a state with a sinusoidal electric field. If you do go this route and try to sum photons in the same mode, it becomes apparent that the phases can't all be the same, because the energy density must follow the classical result that it is proportional to the second power of the electric field amplitude. Perhaps this is something that there isn't a definitive answer to, but in any case I find it of interest. If there is such a thing as a phase for a single photon, a phasor diagram with random phases (for multiple photons) would result in a random walk for the resultant electric field amplitude, with a result that on the average the resultant amplitude would be proportional to the square root of $N$, just as it needs to be for energy conservation.

See also https://core.ac.uk/download/25351228.pdf for a discussion of the phase operator. I find it interesting reading, but it doesn't appear that they are able to yet give a conclusive answer.

 

[PeterDonis]

If there is such a thing as a phase for a single photon

What the reference you gave is telling you is that no, there isn't. The best you can do is more like "sine of the phase" and "cosine of the phase".

 

[vanhees71]

Maybe a single photon can't be treated as a state with a sinusoidal electric field. If you do go this route and try to sum photons in the same mode, it becomes apparent that the phases can't all be the same, because the energy density must follow the classical result that it is proportional to the second power of the electric field amplitude. Perhaps this is something that there isn't a definitive answer to, but in any case I find it of interest. If there is such a thing as a phase for a single photon, a phasor diagram with random phases (for multiple photons) would result in a random walk for the resultant electric field amplitude, with a result that on the average the resultant amplitude would be proportional to the square root of $N$, just as it needs to be for energy conservation.

See also https://core.ac.uk/download/25351228.pdf for a discussion of the phase operator. I find it interesting reading, but it doesn't appear that they are able to yet give a conclusive answer.

For cavity photons the "sinusoidal electric fields" are indeed true states (in contradistinction to the case in free space, where of course the monochromatic plane wave is no true state, and you have to integrate over some square-integrable test function (wave packet); that's not much different from classical optics).

Only now I understand, what's discussed here: The question whether there is a observable representing the phase. As written in the nice paper you linked that's the question, whether there is a canonically conjugate operator to the number operator $\hat{N}$.

The problem is, as with angles as observables in quantum mechanics, that you can define Hermitean but not self-adjoint angle operators. For the angle it's very easy to give an intuitive insight to the problem. Let's consider a particle in a plane and work in polar coordinates. Now consider the angular-momentum operator, $\hat{L}=\hat{x}_1 \hat{p}_2 - \hat{x}_2 \hat{p}_1$, which has the representation in polar coordinates of the position representation,
$$\hat{L} \psi(R,\varphi) = -\mathrm{i} \partial_{\varphi} \psi(R,\varphi).$$
The eigenfunctions with eigenvalues $m$ obviously are
$$u_m(\varphi)=\exp(\mathrm{i} m \varphi)/\sqrt{2 \pi}$$
with $m \in \mathbb{Z}$. That there are only integer eigenvalues follows from using the ladder operators of the symmetric harmonic oscillator in 2D,
$$\hat{a}_j = \sqrt{\frac{m \omega}{2}} \hat{x}_j + \frac{\mathrm{i}}{\sqrt{2 m \omega}} \hat{p}_j, \quad j \in \{1,2 \},$$
fulfilling the commutator relations
$$[\hat{a}_j,\hat{a}_k]=0, \quad [\hat{a}_j,\hat{a}_{k}^{\dagger}]=1.$$
Then also
$$\hat{b}=\frac{1}{\sqrt{2}}(\hat{a}_1+\mathrm{i} \hat{a}_2)$$
is an annihilation operator,
$$[\hat{b},\hat{b}^{\dagger}]=0.$$
The corresponding "number operators"
$$\hat{N}_j=\hat{a}_j^{\dagger} \hat{a}_j, \quad \hat{N}_b=\hat{b}^{\dagger} \hat{b}$$
are self-adjoint and have eigenvalues $n_1,n_2,n_b \in \mathbb{N}_0=\{0,1,2,\ldots \}$. Now
$$\hat{L}=\hat{N}_1+\hat{N}_2-2\hat{N}_b,$$
which thus has only integer eigenvalues. Note that
$$[\hat{N}_1+\hat{N}_2,\hat{N}_b]=0.$$
This also implies that the wave functions must be square integrable over $R \in \mathbb{R}_+$ and $\varphi \in (0,2 \pi)$ and being $2 \pi$-periodic in $\varphi$, because all wave functions can be written in terms of eigenfunction of $\hat{L}$,
$$\psi(R,\varphi)=\sum_{m=-\infty}^{\infty} \psi_m(R) u_m(\varphi).$$
Since the $m$ are integer that's a $2\pi$-periodic function in $\varphi$.

The angle operator then should be $\hat{\phi}=\varphi$. That's a hermitian operator and canonically conjugated to $\hat{L}$,
$$\frac{1}{\mathrm{i}} [\hat{\phi},\hat{L}]=1.$$
The problem, however, is that it is obviously is not self-adjoint, because for all $m \in \mathbb{Z}$
$$\hat{\phi} u_{m}(\varphi)=\varphi \exp(\mathrm{i} m \varphi)/\sqrt{2 \pi},$$
which is not $2 \pi$-periodic. So there is no angle observable in quantum mechanics, at least not in this direct sense.

Here the way out is that one can use $\cos \hat{\phi}$ and $\sin \hat{\phi}$, which are self-adjoint and commute. Their common generalized eigenfunctions are obviously $u_{\varphi_0}(\varphi)=\delta(\varphi-\varphi_0)$, where the eigenvalues are in $[0,2 \pi)$. In this sense we can define the angle as an observable by measuring simultaneously $\cos \hat{\phi}$ and $\sin \hat{\phi}$.

For the phase operator for the 1D harmonic oscillator the math starts similarly, but here the would-be phase observable were the canonical conjugate of the number operator. Again you can try to define operators $\hat{C}$ and $\hat{S}$ akin to $\cos \hat{\phi}$ and $\sin \hat{\phi}$, but here $\hat{N}$ has not $\mathbb{Z}$ but only $\mathbb{N}_0$ as its spectrum, and this makes $\hat{C}$ and $\hat{S}$ non-commuting. For the detailed analysis, see Ref. [19] in the above quoted paper:

https://doi.org/10.1103/RevModPhys.40.411

 

[sophiecentaur]

The key point to look at is whether your experiment has one light source or multiple light sources.

If there are two phase locked oscillators then how can their outputs be distinguished from a single oscillator with a 'power splitter'? Any source will have a bandwidth due to random fluctuations an doesn't this imply that two photons from a given source will still not be 'identical'? Perhaps I should take off my RF Engineering hat but I'd still expect some commonality.

Maybe a single photon can't be treated as a state with a sinusoidal electric field.

This is a problem I have. Unless the photon has infinite length of existence, it will not be a simple sinusoid. It will have an envelope / bandwidth. Two photons will need to be produced within some sort of time window (coherence length?) to allow interference. Won't that hold for either for one or two sources.

 

[Cthugha]

One should never mix up 'classical' interference effects with – so to speak – 'non-classical' interference effects. Dirac refers to classical interference effects which must in fact be understood as ensemble outcomes of individual entities where each of which is in a superposed state, i.e., Dirac refers to expectation values of single-photon observables. Collective many-photon interference effects should be understood in terms of many-photon observables.

Kirk T. McDonald remarks, for example, in “Bunching of Photons When Two Beams Pass Through a Beam Splitter”:

“Dirac has written [1] Each photon then interferes only with itself. Interference between two different photons never occurs. Indeed, a practical definition is that “classical” optics consists of phenomena due to the interference of photons only with themselves. However, photons obey Bose statistics which implies a “nonclassical” tendency for them to ‘bunch’.”

The problem is that this distinction is not tenable. Higher-order interference is not necessarily non-classical. The HBT-effect is a good example for that. Considering single photons you need Ugo Fanos explanation that utilizes Bose statistics. However, for a classical thermal light field you also observe photon bunching which can be modeled extremely simply. Just take a huge amount of harmonic oscillators where each of them is subject to random phase jumps. Add up all the amplitudes of the individual oscillators with their relative phase. You will get Bose-Einstein photon number distributions and photon bunching out of it. Classical noise with the right shape is enough to create bunching. That is actually quite easy to see. The second order correlation function for delay zero is given by:
$$g^{(2)}(0)=\frac{\langle: n^2 :\rangle}{\langle n \rangle^2}$$,
where the double stops denote normal ordering of the underlying field operators which ensure that the dection of the first photon reduces the photon number by one, so you get:
$$g^{(2)}(0)=\frac{\langle: n (n-1)\rangle}{\langle n \rangle^2}$$.
You can now replace the instantaneous photon number n by its mean value and some fluctuations around it:
$$g^{(2)}(0)=\frac{\langle (\langle n\rangle +\delta n)(\langle n\rangle +\delta n-1) \rangle}{\langle n \rangle^2}$$.
Few terms survive the averaging:
$$g^{(2)}(0)=\frac{\langle: (\langle n\rangle^2 -\langle n \rangle +\langle \delta n^2 \rangle}{\langle n \rangle^2}=1-\frac{1}{\langle n \rangle}+\frac{\langle \delta n^2 \rangle}{\langle n \rangle^2}$$.
If you find a photon number distribution, where the fluctuation term (which is more or less the variance of the photon number distribution) amounts to the square of the photon number plus the photon number, you get bunching identified by a second-order correlation function that equals 2. This is exactly the case for Bose-Einstein statistics. However, any other very noisy distribution will also result in bunching.

Many-photon observables identify non-classicality for antibunching, but bunching for thermal light does not fulfill any criteria of nonclassicality. One also does not need to do a HBT experiment using two detectors to investigate bunching. One can get that as well from the statistics of the detected quadratures in homodyne detection without any need to do coincidence counting. This has been pioneered by Mike Raymer already in the nineties (Phys Rev A 55, R1609(R)).

 

[Cthugha]

If there are two phase locked oscillators then how can their outputs be distinguished from a single oscillator with a 'power splitter'? Any source will have a bandwidth due to random fluctuations an doesn't this imply that two photons from a given source will still not be 'identical'? Perhaps I should take off my RF Engineering hat but I'd still expect some commonality.This is a problem I have. Unless the photon has infinite length of existence, it will not be a simple sinusoid. It will have an envelope / bandwidth. Two photons will need to be produced within some sort of time window (coherence length?) to allow interference. Won't that hold for either for one or two sources.

You cannot distinguish these situations. Two perfectly phase locked LOs with an identical spectrum are effectively a single light source.

And: yes, single photons are never sinusoidal in practice. They do not have to. All you need to do to have a Fock state is to have a photon number of 1 in some well defined mode structure. This mode structure is typically some weighted sum over sinusoidal modes. Actually you can turn whatever mode with any arbitrary spectral decomposition into a single photon mode by finding an emitter which emits exactly into this mode spectrum.

If one really is interested in the fields associated with modes of the light field, the continuous variable picture of the Wigner function is much better suited for describing the physics. A reasonably low-level introduction can be found, e.g., here: http://gerdbreitenbach.de/gallery/ - which is an extended and commented version of Breitenbach's Nature paper on the Wigner function of squeezed light (http://gerdbreitenbach.de/publications/nature1997.pdf). In the "one single photon" section, one can play also with experimental data and create field measurement data and Wigner functions for different types of light fields when clicking on "animation of their data".

 

[Cthugha]

In classical physics, the reason why double-slit has 4 times peak intensity of single-slit is this:
- at the single-slit, the electric field vector has magnitude $E_0$ and the wave has intensity $I_0 \sim E_0^2$
- at each slit of the double-slit, the electric field vector has the same magnitude $E_0$, and if they are in phase at the screen, the vector sum will be $2E_0$ and intensity $I \sim (2E_0)^2 = 4I_0$
The fact that the field is $E_0$ at each of the 2 slits in the double-slit and in the single-slit tells me that the intensity ($\sim E_0^2$) is the same at each of the 3 slits, which means that each slit passes the same amount of light. In other words double-slit uses 2 times more light than single-slit, so it is correct to compare double-slit/two-photons to single-slit/single-photon and expect to get the factor of 4.
Is this not true?

Well, it is misleading. You get a factor of two from interference and you get another factor of two from doubling the incoming intensity. Actually, I have never seen the "factor of 4 thing" being taught to students because it obscures things. The reasonable comparison is to illuminate the left slit and the right slit seperately and to compare the added spectrum to the double slit spectrum - which gives a factor of two in peak intensity for having the same total intensity arriving. If you now change the illumination in the noninterfering case, e.g., by just having one slit open, you of course change the ratio accordingly, but there is no meaningful physics involved here. The ratio you will get is just the factor two from interference vs. no interference times the ratio of the intensities which arrive at the slits in the interfering/non-interfering cases, respectively.

 

[Charles Link]

Well, it is misleading. You get a factor of two from interference and you get another factor of two from doubling the incoming intensity. Actually, I have never seen the "factor of 4 thing" being taught to students because it obscures things. The reasonable comparison is to illuminate the left slit and the right slit seperately and to compare the added spectrum to the double slit spectrum - which gives a factor of two in peak intensity for having the same total intensity arriving. If you now change the illumination in the noninterfering case, e.g., by just having one slit open, you of course change the ratio accordingly, but there is no meaningful physics involved here. The ratio you will get is just the factor two from interference vs. no interference times the ratio of the intensities which arrive at the slits in the interfering/non-interfering cases, respectively.

See my post 30. Energy is conserved in the multiple slit interference, with the necessary result that the peaks become narrower as $N$ increases. The $N$ slit interference problem is very important in understanding diffraction grating based spectroscopy. The formula for the intensity $I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)}$ where $\phi=\frac{2 \pi d \sin{\theta}}{\lambda}$ is very useful, and with it, the widths of each of the primary maxima can be readily computed.

(These principal maxima occur when $\phi/2=m \pi$, and the width is found by locating the distance to the first adjacent zero in the numerator in the expression for $I(\theta)$.
When $\phi/2=m \pi$, both numerator and denominator are zero, and when using this formula for these principal maxima, it is taken as the limit as $\phi/2 \rightarrow m \pi$. The result is $I(\theta_{max})=N^2 I_o$, with the width $\Delta \phi=2 \pi/N \approx 2 \pi d \Delta \theta/\lambda$, so that $\Delta \theta=\lambda/(Nd)$, and notice the width $\Delta \theta$ is proportional to $1/N$. Larger $N$ makes for increased resolving power with a diffraction grating spectrometer because the spectral lines (the primary maxima) are narrower ).

 

[Cthugha]

(These principal maxima occur when $\phi/2=m \pi$, and the width is found by locating the distance to the first adjacent zero in the numerator in the expression for $I(\theta)$.
When $\phi/2=m \pi$, both numerator and denominator are zero, and when using this formula for these principal maxima, it is taken as the limit as $\phi/2 \rightarrow m \pi$. The result is $I(\theta_{max})=N^2 I_o$, with the width $\Delta \phi=2 \pi/N \approx 2 \pi d \Delta \theta/\lambda$, so that $\Delta \theta=\lambda/(Nd)$, and notice the width $\Delta \theta$ is proportional to $1/N$. Larger $N$ makes for increased resolving power with a diffraction grating spectrometer because the spectral lines (the primary maxima) are narrower ).

Yes, I fully agree with that. I just think that we are talking about fundamentally different topics within different physics courses. I would introduce the physics of going from 1 slit to n slits in a course on classical optics or maybe in a course on spectroscopy, while the "double slit vs. single slit with photons"-topic arises in an introduction to quantum mechanics. I would not bother students in classical optics with single photons - introducing photons does not help at all at this point - and I would also not bother with going to more than two slits in an introductory qm course. This is also likely to create confusion.

 

[sophiecentaur]

Actually, I have never seen the "factor of 4 thing" being taught to students because it obscures things.

The pattern is produced by the vector (E) addition of the fields. The intensity (power) is proportional to the square of the field (E²). Two squared = Four.
I can't imagine why that wouldn't have already been taught and sorted out for all students.

 

[Cthugha]

The pattern is produced by the vector (E) addition of the fields. The intensity (power) is proportional to the square of the field (E²). Two squared = Four.
I can't imagine why that wouldn't have already been taught and sorted out for all students.

Let me emphasize it again: This includes a setup where you compare scenarios where the total power arriving at the setup is different. You compare the field squared to twice the field squared which yields the factor of 4. You have twice the total intensity making it through the slits in the two-slit case.

In the qm textbook scenario, you start with a single electron (or a single photon) arriving at the double slit and you assume that there is a 100% probability of the photon making it through the double slit towards the detector. If you now replace this by a single slit with the same diameter as one of the double slits, you will reduce the input intensity by 50%. At this point the students have no idea about photons/electrons and are usually not yet well versed in the probabilistic interpretation of qm. So typically they will be worried about what it might mean to have "half a photon arriving at the slit" and are typically quite distracted from what the lecture is supposed to get across.

Sure, you can formulate all of this in terms of probability amplitudes and can describe all this with count rates in the statistical ensemble and will find an enhancement by a factor of 4. But doing so would distract students completely from the main topic: interference happens on the field level also for individual particles.

Of course at some point students are taught what that factor of 4 means, but there is a good reason why the double slit with individual particles is usually introduced as "both slits open" compared to "sum of only left slit open and only right slit open". It helps in teaching and also in discussions to minimize the possible ways to be sidetracked. And for the question we are discussing here, it makes sense to distinguish between the changes introduced by interference and the changes introduced by comparing different input intensities.

This is also what Feynman did in his lectures: He introduced the incoherent sum of two single slit patterns and compares this to the double slit pattern - first with water then with individual electrons. He keeps the incoming flux constant - and for a good reason.

 

[sophiecentaur]

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.

 

[vanhees71]

See my post 30. Energy is conserved in the multiple slit interference, with the necessary result that the peaks become narrower as $N$ increases. The $N$ slit interference problem is very important in understanding diffraction grating based spectroscopy. The formula for the intensity $I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)}$ where $\phi=\frac{2 \pi d \sin{\theta}}{\lambda}$ is very useful, and with it, the widths of each of the primary maxima can be readily computed.

(These principal maxima occur when $\phi/2=m \pi$, and the width is found by locating the distance to the first adjacent zero in the numerator in the expression for $I(\theta)$.
When $\phi/2=m \pi$, both numerator and denominator are zero, and when using this formula for these principal maxima, it is taken as the limit as $\phi/2 \rightarrow m \pi$. The result is $I(\theta_{max})=N^2 I_o$, with the width $\Delta \phi=2 \pi/N \approx 2 \pi d \Delta \theta/\lambda$, so that $\Delta \theta=\lambda/(Nd)$, and notice the width $\Delta \theta$ is proportional to $1/N$. Larger $N$ makes for increased resolving power with a diffraction grating spectrometer because the spectral lines (the primary maxima) are narrower ).

An given that $I_0$ is the intensity of the single slit at the zeroeth maximum, you get, for the $N$-slit grating at $\phi=0$
$$I(0)=I_0 \lim_{\phi \rightarrow 0} \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)} = \lim_{\phi
rightarrow 0} \frac{N^2 \phi^2/4 + \mathcal{O}(\phi^4)}{\phi^2/4 + \mathcal{O}(\phi^4)} = I_0 N^2,$$
i.e., $N^2$ times the intensity of a single slit. The reason is that all Huygens's partial waves interfer constructively at this place and thus you get $N$ times the amplitude of the single slit. Since the intensity is the amplitude squared you get this factor $N^2$.

 

[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.