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A single photon, passing through a beamsplitter (BS), ends up in the state $\frac{1}{\sqrt2}(a_1^\dagger + a_2^\dagger)\lvert\operatorname{vac}\rangle$, which is a coherent superposition of the two possible spatial modes. Let us consider its momentum.

When the photon goes straight its momentum doesn't change. However, when the photon gets deflected, some amount of momentum must have been transferred to the BS, which therefore acquires some (small) amount of momentum in the opposite direction of the photon (see image below). This new tiny momentum will correspond to some phononic excitation of the BS crystalline structure, or something like that.

enter image description here

Because the photon ends up in a superposition of the two output modes, the complete output state must be something of the form: $$ \frac{1}{\sqrt2} \Big(\lvert\text{photon up}\rangle\lvert\text{BS steady}\rangle+\lvert\text{photon right}\rangle\lvert\text{BS momentum changed}\rangle \Big).$$ However, we do not care about the final state of the BS, so that the final state of the photon will be obtained by partial tracing over the BS degree of freedom.

And here comes the catch: a naive calculation would now lead to the conclusion that, after partial tracing, the final state of the photon is $$ \frac{1}{2} \Big( \lvert\text{photon up}\rangle\langle\text{photon up}\rvert + \lvert\text{photon right}\rangle\langle\text{photon right}\rvert \Big). $$ This is a completely mixed state, which means that the photon is not in a coherent superposition of the output modes.

Now, we know very well that this is not actually the case, as the fact that the photon does end up in a coherent superposition is easily tested and well enstablished. So what did we get wrong? The only logical conclusion (that I can see) is that $\lvert\text{BS steady}\rangle$ and $\lvert\text{BS momentum changed}\rangle$ are not orthogonal states, so that partial tracing does not lead to a totally mixed state for the photon.

So here comes the question: why are the two final states of the BS not orthogonal? Is there a clean way to see this?

Moreover, given that these two states are also necessarily not the same, this means that partial tracing will always lead to some small amount of mixedness in the final photon state. Does this mean that there is a fundamental limit in the purity of the final photon state?

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  • $\begingroup$ Outline of an answer: compute how much momentum must be transferred to the beam splitter, then compare that introduced uncertainty to the line width of a typical source. $\endgroup$
    – rob
    Dec 15, 2017 at 16:34
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    $\begingroup$ I asked a very similar question here. $\endgroup$
    – knzhou
    Dec 15, 2017 at 16:37
  • $\begingroup$ @knzhou ...which led me to this answer to another question of which this one is basically a duplicate. Thanks, I don't know why I failed to find these so similar questions. $\endgroup$
    – glS
    Dec 15, 2017 at 16:45
  • $\begingroup$ A very good question is asked. Have you had a "physical" answer now? Right now I have exactly the same question. The answers to the related question are not so satisfied. @glS $\endgroup$
    – hengyue li
    Mar 30, 2022 at 1:10
  • $\begingroup$ @hengyueli I found the answer in the linked post reasonably satisfying myself. The very rough intuition is that the transfer of momentum does not cause the beamsplitter to shift into a state orthogonal to its original one. It will only slightly alter it, but being the BS relatively massive, this will only cause a negligible "mixedness" into the result state of the photon. If you have a more specific issue with it you can open a new post to ask about it (remember to link that post if you do!) $\endgroup$
    – glS
    Mar 30, 2022 at 8:55

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