$\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\ket}[1]{\left|#1\right\rangle}$ Suppose I have a beam splitter that will either reflect a photon by 45 degrees, or will allow the photon to pass directly through.

If I send a single photon state through, that is, $\ket{1}_{initial}$, we have:

$\ket{1}_{initial} = \frac{1}{\sqrt{2}}(\ket{0}_a\ket{1}_b + i\ket{1}_a\ket{0}_b)$

where $a$ and $b$ represent the paths that pass straight through, and the reflecting paths respectively.

Now suppose instead of sending through a single photon, the initial state is two photons, that is, $\ket{2}_{initial}$. Classically I would suspect the state to look something like this:

$\ket{2}_{initial} \propto (\ket{0}_a\ket{2}_b + \ket{2}_a\ket{0}_b + 2\ket{1}_a\ket{1}_b$)

I have the factor of 2 in front of the $\ket{1}_a\ket{1}_b$ state because classically you could have the first photon passing through and the second photon reflecting, or vice-versa. Is this still correct for quantum mechanics, where photons are indistinguishable?

Also, could you please show how to derive result that I have simply guessed for two photons (given the 1 photon case in the first equation)?


1 Answer 1


Your first guess is more or less correct, you can easily derive it (using your conventions) via the unitary transformations: $$ a \rightarrow \frac{1}{\sqrt{2}}(a - i b) $$ $$ b \rightarrow \frac{1}{\sqrt{2}}(a + i b),$$ where $a$ is the annihilation operator pertaining both to the input mode in which you put two incoming photons and also the mode where those photons are transmitted, $b$ pertains to both the orthogonal input mode and output mode where your input photons are reflected. Hopefully there should be no confusion with this notation, let me know if it's not clear. Plugging in the transformations, you get $$ (a^{\dagger})^2|0\rangle \rightarrow [(a^{\dagger})^2 + (b^{\dagger})^2 +2i a^{\dagger}b^{\dagger}]|0\rangle. $$ This is not normalised, by the way, because I'm lazy and it is completely irrelevant to the physics.

Note that this is not the Hong-Ou-Mandel (HOM) effect. In the HOM effect, one has photons incoming in both modes. The portion of the $a$ mode that is transmitted is out of phase by $\pi$ with respect to the portion of the $b$ mode that is reflected, and vice versa. Therefore the destructive interference is complete, and the photons bunch. Here however, either of the indistinguishable photons may be reflected, but there is no incoming photon from the other side for the reflected photon to interfere with.

(NB. If you are thinking of the time-reversed HOM effect, you would need to have a coherent superposition of two incoming photons in each mode to implement that. Then by unitarity you can only have one photon in each outgoing mode, and the amplitudes for two outgoing photons vanish.)


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