Can a twin photon pair entagled in polarization get disentangled after one of the photons meets a PBS (polarization beam splitter). I suppose after the PBS the photon will go one way or another depending what path it chooses. It can be in superposition of vert/horiz polarization (accordongly left/rigth path) but if a supervision (measurement) is made of left and absense of photon is observed than the photon should be in right channel and has horizontal polarization. Then its twin far away will get a vertical polarization and both are not anymore entangled?

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    $\begingroup$ A global unitary operation can take any state to any state, so it can take an entangled state to a separable one. Local unitaries cannot change the entanglement properties $\endgroup$ Commented Aug 17, 2022 at 17:23
  • $\begingroup$ @QuantumMechanic can you reformulate this in more experimental terms. After measuring that photon 1 of the twin pair is not in left channel the state must become $\left| h\rangle _{1}\right| v\rangle _{2}$ from $\left| h\rangle _{1}\right| v\rangle _{2}+\left| v\rangle _{1}\right| 1\rangle _{2}$. How is $\left| h\rangle _{1}\right| v\rangle _{2}$ entangled? $\endgroup$
    – Mercury
    Commented Aug 17, 2022 at 20:26
  • $\begingroup$ @QuantumMechanic Here is a measurement taken. It is non-unitary. So your comment seems strange to me. $\endgroup$
    – Mercury
    Commented Aug 18, 2022 at 18:30
  • $\begingroup$ I was only talking about unitary operations. Non-unitary operations can do anything, but if you want to do something without measurement or wavefunction collapse or anything like that then you'll want unitary operations. Now, whether you can easily implement a particular unitary is the question $\endgroup$ Commented Aug 18, 2022 at 19:45

1 Answer 1


Yes! It's easiest to talk about this in Fock space, so that we can account for the other input port of the polarizing beam splitter (PBS) that is empty. If we want to be really explicit and label the modes by their polarization $H$ or $V$ and their paths $a$ or $b$, we can write the entangled state as proportional to $$|1\rangle_{a,H}\otimes|0\rangle_{a,V}\otimes |0\rangle_{b,H}\otimes|1\rangle_{b,V}+|0\rangle_{a,H}\otimes|1\rangle_{a,V}\otimes |1\rangle_{b,H}\otimes|0\rangle_{b,V}.$$ (There are other possible entangled states but this one will suffice for our consideration.) The photon in path $b$ then goes through a PBS, so we have to consider this whole system including some other input path $c$ to the beam splitter that is initially empty: $$\left(|1\rangle_{a,H}\otimes|0\rangle_{a,V}\otimes |0\rangle_{b,H}\otimes|1\rangle_{b,V}+|0\rangle_{a,H}\otimes|1\rangle_{a,V}\otimes |1\rangle_{b,H}\otimes|0\rangle_{b,V}\right)\otimes|0\rangle_{c,H}\otimes|0\rangle_{c,V}.$$ Let's say the horizontal photons remain in path $b$ and the vertical photons exit in path $c$. Then the state after the PBS is $$|1\rangle_{a,H}\otimes|0\rangle_{a,V}\otimes |0\rangle_{b,H}\otimes|0\rangle_{b,V}\otimes|0\rangle_{c,H}\otimes|1\rangle_{c,V}+|0\rangle_{a,H}\otimes|1\rangle_{a,V}\otimes |1\rangle_{b,H}\otimes|0\rangle_{b,V}\otimes|0\rangle_{c,H}\otimes|0\rangle_{c,V}.$$ In a first-quantized notation, this would look like $$|H\rangle_a\otimes |V\rangle_c+|V\rangle_a\otimes |H\rangle_b,$$ but that's not very easy to deal with because it can't account for the lack of a photon in path $b$ or $c$. Finally, when we measure path $c$ and find no photon there, we have collapsed the state to $$|0\rangle_{a,H}\otimes|1\rangle_{a,V}\otimes |1\rangle_{b,H}\otimes|0\rangle_{b,V}\otimes|0\rangle_{c,H}\otimes|0\rangle_{c,V}.$$ This is indeed a separable state, of the form $|V\rangle_a|H\rangle_b$.

  • $\begingroup$ Now I can apply rotation and get two completely indistintiguishable photons. Absolutely identical. $\endgroup$
    – Mercury
    Commented Aug 18, 2022 at 20:31
  • $\begingroup$ In my mind this is the only scheme to get two identical particles. $\endgroup$
    – Mercury
    Commented Aug 18, 2022 at 20:34
  • $\begingroup$ @Mercury ahh not quite; if you apply a rotation you'll get something like $|H\rangle_a |H\rangle_b$; the two photons will be distinguishable in the path degree of freedom. If you want to make them completely indistinguishable you'll have to do something else that isn't unitary $\endgroup$ Commented Aug 19, 2022 at 13:03
  • $\begingroup$ I am not sure it must be unitary because I can send them in the two ports of BS like in Hong Ou Mandel scheme and they both will exit together on one path in one of the two exits but it is not necessary to do a measurement which collapses the WF. $\endgroup$
    – Mercury
    Commented Aug 21, 2022 at 19:33
  • $\begingroup$ @Mercury indeed you can do that and guarantee that the photons bunch together, but you cannot predict which path they will exit. The state will still be a superposition. I guess you'd like to measure one port and post-select on finding no photons - then yes the other arm will have two indistinguishable photons. This is a non-unitary process (we did a measurement) but it does work $\endgroup$ Commented Aug 22, 2022 at 14:34

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