# Two photons or twice one photon?

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 ?

• The Hilbert space would surely still be $|l\rangle \otimes |l^\prime\rangle$, it might be $|2\rangle \otimes |0\rangle$, though, I suppose. – innisfree Feb 28 '17 at 6:13
• Then, how is the phase accumulated before the reflection taken into account ? – Pen Feb 28 '17 at 6:55
• 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. – AndyK Mar 8 '17 at 9:22