Interference between two photons, tensor product of individual wave functions?

I have learned that the wave function cannot be visualized as a real physical wave like for example the EM field, because for multi-particle systems, it is not a wave in $\mathbb{R}^3$ but in $\mathbb{R}^{3N}$. See this question if I haven't expressed myself clearly.

If I understand things correctly, this is a consequence of the QM fact that the combination of two individual quantum systems with Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ is described by the tensor product $\mathcal{H}_1\otimes \mathcal{H}_2$. The wave function of two free particles is not a wave in $\mathbb{R}^3$, but in $\mathbb{R}^3\otimes \mathbb{R}^3 = \mathbb{R}^6$.

I'm wondering how that fits into the following modified double-slit experiment:

Imagine two light sources emitting photons that are as coherent and equal as possible (same frequency, same polarization, etc). Each light source is placed before one of the slits, and there is a wall between the light sources, so if hole 1 is closed, only photons from source 2 can pass to the detector.

Let's define $\mathcal{H}_1$ as the Hilbert space of the quantum system where source 2 is off and only source 1 emits photons:

Similarly, $\mathcal{H}_2$ describes photons from source 2 and no photons from source 1:

In both systems, there is no interference pattern on the screen. In $\mathcal{H}_1$, we detect most of photons around the position A on the detector, and in $\mathcal{H}_2$ most of the photons land at position B.

Now, for the system $\mathcal{H}_c$ of the combined system, where both sources emit photons simultaneously, I assume that $\mathcal{H}_c = \mathcal{H}_1 \otimes \mathcal{H}_2$.

Is that correct so far?

Now, as far as I know, in $\mathcal{H}_c$ there should be an interference pattern of photons at the position C of the detector. The photons from source 1 interfere with those from source 2.

How can this interference be explained with respect to the tensor product? The wave functions of the photons from source 1 do not "live in" the same $\mathbb{R}^3$ as those from source 2, so how can they interfere?

EDIT: Continuation moved to other question.

• $\mathbb{R}^3\otimes\mathbb{R}^3\neq \mathbb{R}^6$, in general (although they can be made isomorphic). However, when considering tensor product spaces, the notation $\psi$ is to be intended as $\psi_1\otimes 1_2$, therefore as element of the tensor product space obtained by tensoring with the identity on the other side; if you carry this notation on you will find that everything is consistent eventually. – gented Oct 23 '15 at 9:13
• @GennaroTedesco in the wave picture, the Identity of the Hilbert space is the vector space zero $\psi_0 \equiv 0$. So for a photon from $\mathcal{H}_1$ and one from $\mathcal{H}_2$ we have $\psi_1 \otimes \psi_2$ as the combined state. I cannot see where I need $\mathbf{1}_i$ here. – Bass Oct 23 '15 at 9:32
• If you are considering the system of the two particles then yes, $\psi_{12} = \psi_1\otimes\psi_2$ and consequently the wave function is $\psi(1,2) = (\langle x_1|\otimes\langle x_2 |)\otimes(|\psi_1\rangle\otimes |\psi_2\rangle) = \psi_1(x_1)\cdot\psi_2(x_2)$ by definition of tensor product. – gented Oct 23 '15 at 9:56
• @GennaroTedesco: I think that's exactly my problem: How can there be destructive interference in $\psi_1(x_1) \cdot \psi_2(x_2)$? If both amplitudes at a certain point on the detector are non-zero, their product cannot vanish too. – Bass Oct 23 '15 at 10:01
• The tensor product of $\mathbb{R}^3$ with itself is $\mathbb{R}^9$, not $\mathbb{R}^6$. We have $\mathbb{R}^n\otimes\mathbb{R}^m = \mathbb{R}^{n\cdot m}$, while the sum is for the direct product $\mathbb{R}^n\oplus\mathbb{R}^m = \mathbb{R}^{n+m}$. – ACuriousMind Oct 23 '15 at 14:41

3 Answers

This answers the title

Interference between two photons, tensor product of individual wave functions?

Photons are quantum mechanical entities, described by wavefunctions which are solutions of quantum mechanical equations on which the boundary conditions are imposed.

Photons have extremely weak interactions between them. Photon photon interactions are box diagrams and the probability of a photon interacting with another photon is minuscule. . The classical interference pattern in the Young experiment showed the wave nature of light within classical electrodynamics. The observed single photon at a time interference demonstrates the quantum mechanical nature of photons. Classical light beams are emergent from an under-layer of innumerable photons.

Though a tensor product of individual wave functions may be written down , it has no interference type information (phases) unless the wavefunctions come from the solution of the quantum mechanical equations with the same boundary conditions, i.e. are entangled. This can only happen in dimensions commensurate to h_bar, quantum mechanical dimensions, not the macroscopic dimensions shown in the figure.

In addition , the velocity of light is enormous, there is no way that two independent light sources can be made synchronous in releasing individual photons.

You need to distinguish functions from $\mathbb R^{3N}$ and $\mathbb R^{3N}$ itself.

For instance if you have eigenstates of the harmonic oscillator, they are functions like $\psi_1:\mathbb R^3\rightarrow \mathbb C$ and the the tensor product of a space spanned by such functions is itself spanned by $\psi:\mathbb R^6\rightarrow \mathbb C.$

You can think of functions like $\psi(x_1,y_1,z_1,x_2,y_2,z_2)=\psi_1(x_1,y_1,z_1)\psi_2(x_2,y_2,z_2)$ as like a basis and they are functions from $\mathbb R^6$ and they are factorizable as the product of two functions but their linear combinations ate not factorizable like that but instead is a more general function from $\mathbb R^6.$

And as long as we are correcting misconceptions. In general, the function isn't into the complex numbers, it is into a spin spin state and so the multiparticle version isn't into the complex numbers but is into the joint spin state (which is a tensor product of the single particle spin states).

Now for some further problems. The Schrödinger equation is for non relativistic particles and doesn't handle particle creation. So when you have a source that creates photons you have the problem of particle creation and the problem of a massless particle.

But you could always ask the corresponding question about electron beams and ignore the creation process.

So you might imagine a function from $\mathbb R^6$ but you can't actually have just any function. In particular your function that sends $(x_1,y_1,z_1,x_2,y_2,z_2)\mapsto \psi(x_1,y_1,z_1,x_2,y_2,z_2)$ in facts has to have the symmetry $\psi(a,b,c,d,e,f)=-\psi(d,e,f,a,b,c)$ (and if they Z bosons instead you would have to have $\psi(a,b,c,d,e,f)=+\psi(d,e,f,a,b,c)$).

Which means if there is a nonzero probability of one electron going through the slit there is a nonzero probability of the other electrons going through the same slit.

Which means there are two totally different senses of labelling the electrons. One corresponds to whether you are referring to $(x_1,y_1,z_1)$ or $(x,2,y_2,z_2)$ (which is the sense everyone else uses) and another is which slit is goes through (the sense you are using) so you have to watch out for that.

You can definitely talk about waves corresponding to one electron heading (probability current pointing) towards the slit on the right and the other one heading (probability current pointing) in a completely different direction, a direction orthogonal to the slit.

If those waves evolve to have just one central peak that has some nonzero values at C then you can ask how the state evolves with a state of both beams heading towards the slits. But know because of the symmetry those basic eaves we build things out of aren't just $\psi(x_1,y_1,z_1,x_2,y_2,z_2)=\psi_1(x_1,y_1,z_1)\psi_2(x_2,y_2,z_2)$ they are waves like $\psi(x_1,y_1,z_1,x_2,y_2,z_2)=\psi_1(x_1,y_1,z_1)\psi_2(x_2,y_2,z_2)-\psi_2(x_1,y_1,z_1)\psi_1(x_2,y_2,z_2)$ and so it is not possible to say something like one particle is going through one slit and the other is going through the other.

One think you can do however is slow the rate of particle such that the time to go from slit to screen is so much less than the times between hits that one one particle is going through any slit at a single moment. Hence just one particle is going through any slits at any moment.

But that's the usual single particle case. There is a wave at both slits and a single particle going through. All that has happened now is you have been honest about the effects on and of the other particles that haven't gone through yet.

Apparently, I got this wrong from the beginning. Two photons from different sources can interfere (if the experiment is set up appropriately, but it's very hard to set that up).

However, they don't interfere at the quantum mechanical wave function level, but at the level of the electromagnetic field.

According to @ACuriousMind (who explained this to me) this could be solved in quantum optics, and most likely in QED too.

• I think your answers is itself still a misunderstanding. I don't think the focus on photons and creation and distinct sources helps to understand interference and how it is related (or not related) to multiparticle effects such as tensor products. And the electromagnetic field is an emergent phenomena that happens from a buildup of information from a multiparticle quantum system. So all the physics happens at that deeper quantum level to cause the interference of the classical field, so you've got the causality completely backwards. – Timaeus Oct 25 '15 at 16:49
• @Timaeus: That might well be. The reason I'm asking it this way is the order I've learned things: 1. First I've learned that light's a wave, it shows interference effects. 2. Then I've learned that light is quantized, has particle-like behaviour, and that even a single photon shows interference effects. This can be hard to swallow if you're accustomed to classical physics. 3. Once I accepted that, I wondered: How can the classical, "macroscopic" interference be seen in the tensor product? Now I realized these interferences happen at a different level. – Bass Oct 25 '15 at 20:26
• @Timaeus: I suppose that the "deep quantum level" you're talking about becomes apparent in QFT, where quantum mechanical effects like photon self-interference are coupled with field effects like macroscopic (many-particle) interference effects. I'm an absolute beginner in QFT, so at the moment I'm happy to know that the classical interference does not happen at the wave-function level. – Bass Oct 25 '15 at 20:30
• Don't be happy to be wrong. The only reason classical interference isn't happening at the quantum level is that a classical wave is like temperature, it doesn't exist for a single particle its a property of specific kinds of multiparticle states. The true interference is happening at the level of the quantum wave and that is what ends up causing the classical interference, but you'd have to learn how to see a classical field be built out of many photons before you could see how a classical field has interference. – Timaeus Oct 25 '15 at 21:19
• @Timaeus "Build a classical field out of many photons", is that usually taught in QM or QFT courses? – Bass Oct 26 '15 at 7:37