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I'm trying to calculate the interference of two non-entangled photons, like in a double-slit experiment with two photon sources, one behind each slit (follow-up on this question).

The individual particles have the wave function $\psi_i\in\mathcal{H}_i$ where $i\in\{1,2\}$. The tensor product of their wave function is

$$ \Psi(1,2) = c_1\Psi_1 + c_2\Psi_2 = c_1(\psi_1\otimes 1_2) + c_2(1_1\otimes\psi_2) $$

This is what I tried:

$$ \begin{align} |\Psi(1,2)|^2 & = \langle c_1(\psi_1\otimes\mathbf{1}) + c_2(\mathbf{1}\otimes\psi_2) | c_1(\psi_1\otimes\mathbf{1}) + c_2(\mathbf{1}\otimes\psi_2) \rangle \\ & = \sum_i|c_i|^2|\psi_i|^2 + \overline{c_1}c_2\langle\psi_1\otimes\mathbf{1} | \mathbf{1}\otimes\psi_2\rangle + \overline{c_2}c_1\langle\mathbf{1}\otimes\psi_2|\psi_1\otimes\mathbf{1}\rangle \\ & = \sum_i|c_i|^2|\psi_i|^2 + 2\,\mathfrak{Re}\left(\overline{c_1}c_2\langle\psi_1\otimes\mathbf{1} | \mathbf{1}\otimes\psi_2\rangle\right) \\ & = \sum_i|c_i|^2|\psi_i|^2 + 2\,\mathfrak{Re}\left(\overline{c_1}c_2\langle\psi_1|\mathbf{1}\rangle\langle\mathbf{1}|\psi_2\rangle\right) \end{align} $$

Is that correct so far? How to continue to show the interference?

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  • $\begingroup$ Wait. What is $1_2$ supposed to be? Those are states, not operators. The state of the two photons is simply $\psi_1\otimes\psi_2$. $\endgroup$ – ACuriousMind Oct 23 '15 at 14:37
  • $\begingroup$ @ACuriousMind I have taken this notation from the only answer to my other question: physics.stackexchange.com/a/214147/86781 I thought it was the (non-normalizable) "wave function" $\mathbf{1}_2(\mathbf{x}) \equiv 1$, but maybe I didn't understand this correctly. $\endgroup$ – Bass Oct 23 '15 at 14:41
  • $\begingroup$ That's not a wavefunction. I have left a comment on that answer as well. $\endgroup$ – ACuriousMind Oct 23 '15 at 14:42
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Photons are quantum mechanical entities. This means that their interactions can be described by Feynman diagrams, which are reinterpreted as creation and annihilation operators operating on ground states with propagators in between.

It has little meaning to talk of two individual photons "interfeering" without writing down the Feynman diagrams. Photon photon interactions are four vertex at lowest order. This means a factor of ~10^-8, as the electromagnetic coupling constant is of order 10^-2.

twophoton

A Feynman diagram (box diagram) for photon–photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

The four propagators for energies below gamma energies are highly off mass shell and thus diminish the probabilities even further.

So it is very improbable that two photons will have a measurable interference.

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  • $\begingroup$ So when Thomas Young discovered light interference (and thus stated that light behaves wave-like), the interference came mostly from photon self-interference, and not from photon-photon interference? $\endgroup$ – Bass Oct 23 '15 at 21:33
  • $\begingroup$ Also, are we talking about the same thing? I'm talking about light interference, and I've never heard that this means photon-photon scattering. $\endgroup$ – Bass Oct 23 '15 at 21:35
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    $\begingroup$ You are talking about particle interference and the particles that make up light are the photons. Here is how zillions of photons add up to classical light motls.blogspot.com/2011/11/… . Classical wave equations for light explain light interference. It is when single photon experiments are performed youtube.com/watch?v=GzbKb59my3U and still interference is observed that the quantum mechancical framework has to be used for describing the phenomenon. $\endgroup$ – anna v Oct 24 '15 at 3:21
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    $\begingroup$ The photons in a light beam do not interfere with each other, the built up wave does. In a sense it is as if the photons are the medium on which the light wave is built. They keep the phases to each other they have been generated with, until the beam hits something. Going through the double slits, the individual photons follow the probabilities, and the probabilities makeup the interference pattern consistently, between classical and quantum mechanical under layer. $\endgroup$ – anna v Oct 24 '15 at 3:24
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I'm trying to calculate the interference of two non-entangled photons,

That's like saying you are trying to compute the wave function of a unicorn.

The individual particles have the wave function $\psi_i\in\mathcal{H}_i$ where $i\in\{1,2\}$.

That's not quite right. There is a space $\mathcal{H}_i$ of single particle states. And the space of states of a multiparticle space is a subset of the tensor product of the single particle states. And the tensor product is $\mathcal{H}_1\otimes\mathcal{H}_2$ and what subset it is depends on whether the single particles are fermions or bosons and whether the two particles are identical or different.

Now why did I object so early? Because a general state of the multiple system is not of the form $\psi_1\otimes \psi_2$ with $\psi_i\in\mathcal{H}_i$ even when they are distinct. A state like $\psi_1\otimes \psi_2$ is a special multiparticle state, one where there is a single particle state for each particle (namely $\psi_1$ and $\psi_2$). And in general this doesn't happen. In general these states are like the x axis and y axis in $\mathbb R^2$ they are possible, but not general. The general states are linear combinations of these basis states and hardly ever are these states.

The tensor product of their wave function is

$$ \Psi(1,2) = c_1\Psi_1 + c_2\Psi_2 = c_1(\psi_1\otimes 1_2) + c_2(1_1\otimes\psi_2) $$

The state you just wrote out as $c_1(\psi_1\otimes 1_2) + c_2(1_1\otimes\psi_2)$ is not the tensor product of $\psi_1$ and $\psi_2$ the tensor product of $\psi_1$ and $\psi_2$ is $\psi_1\otimes\psi_2$ which is different. And it is, for instance, factorizable. Whereas the one you have is in general not.

Which is another problem, you said you wanted a non entangled state. But that means you want a factorizable state. And in general that isn't possible. Sure you could have bosons in a state that is identical i.e. if $\psi_1=\psi_2$ then $\psi_1\otimes\psi_2$ is possible but in general only states like $c\psi_1\otimes\psi_2+c\psi_2\otimes\psi_1$ Are possible which means you have to learn to distinguish between states on the left of the tensor symbol and states on the right rather than letting the numbers do the distinguishing for you. I mean you could let $a=\psi_1$ and let $b=\psi_2$ and then talk about a state like $a\otimes b +b\otimes a $ but guess what? That state is entangled. Entanglement isn't just common, like being not on the x axis or the y axis is common entanglement is almost required.

And when it isn't required the states are the same, so it is hard to have a state aimed at just one slit and then not be entangled.

Hence you are asking for the wave function of a unicorn. Except evolution could have produced a unicorn and scientists could try to someday make one. Whereas these symmetry rules come from the statistics required to observationally get the thermodynamics we observe. So there are really good reasons to think that it is literally impossible to have what you want.

So tour whole goal of a non entangled state with each particle aimed at just one slit is misguided and impossible. And you didn't write down the tensor product correctly, and you wrongly assumed a multiparticle state allows there to be single particle states instead of being made out of linear combinations of products of single particle states.

But what if it were? Then, if you compute $|\psi_1(x_1,y_1,z_1)\otimes\psi_2(x_2,y_2,z_2)|^2$ you get $|\psi_1(x_1,y_1,z_1)|^2|\psi_2(x_2,y_2,z_2)|^2$. And check my work questions are off topic which is why I spent so much time addressing more fundamental conceptual errors. And I can't answer your question about how to proceed towards showing interference for non entangled photons where each one is heading towards just one slit since that doesn't sound physically possible to have such a thing even before it went through the slits.

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