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?

  • $\begingroup$ 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 '17 at 16:34
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    $\begingroup$ I asked a very similar question here. $\endgroup$ – knzhou Dec 15 '17 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 '17 at 16:45