# Output of a beamsplitter with photon number (Fock) state inputs

Given a beamsplitter drawn below, where $\hat{a}$ and $\hat{b}$ are input modal annihilation operators, transmissivity is $\tau\in[0,1]$, and output modal annihilation operators are $\hat{c}=\sqrt{\tau}\hat{a}+\sqrt{1-\tau}\hat{b}$ and $\hat{d}=\sqrt{1-\tau}\hat{a}+\sqrt{\tau}\hat{b}$, suppose the inputs $\hat{a}$ and $\hat{b}$ are in photon number (Fock) states $|m\rangle$ and $|n\rangle$, respectively. What are the states of the outputs $\hat{c}$ and $\hat{d}$? I understand that if one of the inputs is a vacuum state $|0\rangle$, then the output states are binomial mixtures of photon number states, with "probability of success" parameter being either $\tau$ or $1-\tau$ and the "number of trials" parameter being the photon number $n$ of the non-vacuum input (so, if $|0\rangle$ was input on mode $\hat{a}$ and $|n\rangle$ on mode $\hat{b}$, then mode $\hat{c}$ is in the state $\sum_{k=0}^n\binom{n}{k}(1-\tau)^k\tau^{n-k}|k\rangle$ and mode $\hat{d}$ is in the state $\sum_{k=0}^n\binom{n}{k}\tau^k(1-\tau)^{n-k}|k\rangle$). I am wondering how this generalizes to both input modes being in the non-vacuum states.

• And if you put in one photon at each input for a 50-50 beam splitter, two come out of one output and none the other output. – Peter Shor Jul 27 '14 at 22:15
• @PeterShor What makes you say that ? – ticster Jul 27 '14 at 22:35
• It's the Hong-Ou-Mandel effect. – Peter Shor Jul 27 '14 at 22:42
• @M.B.M.: using the same idea as in the HOM effect, it's easy to calculate for one photon at each input, and an arbitrary beam splitter: 2 out of one output with probability $2\tau(1-\tau)$; one from each output with probability $(1-2\tau)^2$. It gets more complicated when you have larger input photon numbers. See this paper. Also this one. Available here with no pay wall. – Peter Shor Jul 28 '14 at 15:12
• I think you should have a minus sign somewhere because your matrix ($\hat c,\hat d$function of $\hat a, \hat b$) is not unitary. – Trimok Jul 28 '14 at 15:49

The transformation equations you specify are not correct since they do not respect unitarity. The condition of unitarity (or energy conservation) for the action of the beam-splitter gives the following transformations:

$\hat{c}=\sqrt{\tau}\hat{a}+\sqrt{1-\tau}\hat{b}$

$\hat{d}=\sqrt{1-\tau}\hat{a}-\sqrt{\tau}\hat{b}$

The minus sign in the second equation ensures that unitarity is respected.

For reasons that will become clear soon, let us invert these equations to get the input mode operators $\hat{a}$ and $\hat{b}$ in terms of the output mode operators $\hat{c}$ and $\hat{d}$. As expected from arguments of reversibility, we get:

$\hat{a}=\sqrt{\tau}\hat{c}+\sqrt{1-\tau}\hat{d}$

$\hat{b}=\sqrt{1-\tau}\hat{c}-\sqrt{\tau}\hat{d}$

It is useful to look at this problem in the Heisenberg picture where the action of the beam-splitter is entirely on the mode creation and annihilation operators with initial field state being assumed as vacuum.

Since the input states being considered are the Fock states $|m\rangle_{a}$ and $|n\rangle_{b}$ the full intial field state can alternatively be written as:

${(a^{\dagger})^m(b^{\dagger})^n |0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}}$

Now we substitute the earlier expressions for $\hat{a}$ and $\hat{b}$ in terms of $\hat{c}$ and $\hat{d}$ given by the beam-splitter transformations. The field state after the mode transformations is,

$(\sqrt{\tau}\hat{c}^{\dagger}+\sqrt{1-\tau}\hat{d}^{\dagger})^m(\sqrt{1-\tau}\hat{c}^{\dagger}-\sqrt{\tau}\hat{d}^{\dagger})^n|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$

Thus the output states for a beam-splitter transformation on input Fock states have been obtained.

As Peter Shor correctly pointed out, a beautiful consequence of these transformations is the Hong-Ou-Mandel effect. It states that when single-photon states are incident at the same time on the input ports of the beam-splitter, both photons emerge from the same output port.

This may be verified easily from the equation we have obtained by putting $m=n=1$. Also for convenience let us put $\tau=0.5$ i.e the beam-splitter is $50:50$ ratio. The output field state is,

$\frac{1}{\sqrt{2}}(\hat{c}^{\dagger}+\hat{d}^{\dagger})\frac{1}{\sqrt{2}}(\hat{c}^{\dagger}-\hat{d}^{\dagger})|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$

$=\frac{1}{2}\big((\hat{c}^{\dagger})^2-(\hat{d}^{\dagger})^2\big)|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$

$=\frac{1}{\sqrt{2}}(|2\rangle_{c}|0\rangle_{d}-|0\rangle_{c}|2\rangle_{d})$

Thus, we clearly see that either both photons emerge from port $C$ or both emerge from port $D$. Such a state is referred to as a two-photon NOON state (the state looks like that when N=2) and this effect is of paramount importance in linear optical quantum computing schemes.