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Consider two photons emitted in two different modes $l$ and $l'$ (for instance by the annihilation of an electron and a positron), such that the initial state of the system is $\left|\psi\right\rangle =\left|1_{l}\right\rangle \otimes\left|1_{l'}\right\rangle$ .

Now at some point, I reverse the second photon and put it in the mode $l$ (for instance using a reflection on a mirror to change the direction and polarization of the photon).

What is the good description of the final state ?

  • Is it a two photons state $\left|2_{l}\right\rangle$ ? But shouldn't the phase between both photons be taken into account so that the interference could be positive or destructive ?

  • Is it twice a one photon state $\left(1+e^{i\varphi}\right)\left|1_{l}\right\rangle$ ?

  • Is it something else ?

  • Does the question make sense ?

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    $\begingroup$ The Hilbert space would surely still be $|l\rangle \otimes |l^\prime\rangle$, it might be $|2\rangle \otimes |0\rangle$, though, I suppose. $\endgroup$
    – innisfree
    Commented Feb 28, 2017 at 6:13
  • $\begingroup$ Then, how is the phase accumulated before the reflection taken into account ? $\endgroup$
    – Pen
    Commented Feb 28, 2017 at 6:55
  • $\begingroup$ I would say that $|1_{\ell}\rangle \otimes |1_{\ell'}\rangle \to |1_{\ell}\rangle \otimes |1_{\ell}\rangle$, since they are still uncorelated. The two-photon number state $|2_{\ell}\rangle $ cannot be factorized to two one-photon states. $\endgroup$
    – kuzand
    Commented Mar 8, 2017 at 9:22

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You cannot merge two photons from distinguishable modes into the same mode with unitary operations, see for example "Is it possible to construct a lens which focuses all the light rays from an extended object in one point?".

What you can do is to combine the two photons on a beamsplitter as sketched below:

The beamsplitter translates photons in modes $l$ and $l'$ into photons in modes $k$ and $k'$: $$| 1_l \rangle \otimes | 1_{l'} \rangle \to \left( | 2_k \rangle \otimes | 0_{k'} \rangle + | 0_k \rangle \otimes | 2_{k'} \rangle \middle) \middle/ \sqrt{2} \right.$$ The result is a superposition of two photons leaving the system in mode $k$ and two photons leaving the system in mode $k'$. This is known as the Hong-Ou-Mandel effect. If you were to ignore mode $k'$ (mathematically speaking, if you were to perform a partial trace over the second entry of the state vector), you would see a two-photon state $| 2_k \rangle$ in $50 \, \%$ of the cases and a zero-photon state $| 0_k \rangle$ in the other $50 \, \%$ of the cases.

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