1:1 Beam-splitter and the Hong-Ou-Mandel effect

We can write the state of two photons in different modes: $$\tag{1} \hat{\alpha}^{\dagger} \hat{b}^{\dagger}|0,0\rangle_{a b}=|1,1\rangle_{a b}$$ According to the Wiki page on the Hong-Ou-Mandel effect (https://en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect), when the two modes are mixed in a $$1:1$$ beam splitter, they turn into modes $$c$$ and $$d$$, and the creation operators transform: $$\tag{2} \hat{a}^{\dagger} \rightarrow \frac{\hat{c}^{\dagger}+\hat{d}^{\dagger}}{\sqrt{2}} \text { and } \quad \hat{b}^{\dagger} \rightarrow \frac{\hat{c}^{\dagger}-\hat{d}^{\dagger}}{\sqrt{2}}$$ and when two photons enter the beam splitter, one on each side, the state becomes: $$\tag{3} |1,1\rangle_{a b}=\hat{a}^{\dagger} \hat{b}^{\dagger}|0,0\rangle_{a b} \rightarrow \frac{1}{2}\left(\hat{c}^{\dagger}+\hat{d}^{\dagger}\right)\left(\hat{c}^{\dagger}-\hat{d}^{\dagger}\right)|0,0\rangle_{c d}=\frac{1}{2}\left(\hat{c}^{\dagger 2}-\hat{d}^{\dagger 2}\right)|0,0\rangle_{c d}=\frac{|2,0\rangle_{c d}-|0,2\rangle_{c d}}{\sqrt{2}}$$ Now my question is: what would happen if we let $$n$$ photons enter from one side and $$m$$ photons on the other side? I would assume that we can write: $$\tag{4} \begin{array}{l} \left|n_{1}, m_{2}\right\rangle=\left(\alpha_{1}^{\dagger}\right)^{n}\left(\hat{\alpha}_{2}^{\dagger}\right)^{m}\left|0_{1}, 0_{2}\right\rangle \\ \Rightarrow\left(m ! n ! 2^{(n+m)}\right)^{-\frac{1}{2}}\left(\hat{a}_{3}^{\dagger}+\hat{\alpha}_{4}^{\dagger}\right)^{n}\left(\hat{a}_{3}^{\dagger}-\hat{\alpha}_{4}^{\dagger}\right)^{m}\left|0_{3}, 0_{4}\right\rangle \end{array}$$ Is the expression in eq. $$(4)$$ correct, and what would be the resulting quantum-state?

2 Answers

You are definitely on the right track - I can think of two ways to proceed from here. The answers have quite a few terms!

Exactly as you said, we start both answers with $$|n_1,m_2\rangle=\frac{\alpha_1^{\dagger n}\alpha_2^{\dagger m}}{\sqrt{n!m!}}|0_1,0_2\rangle\to \frac{\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle,$$ where I have dropped the hats on the operators to make life easier.

1. All we need is to apply binomial theorem a few times and then use the definition of creation operators.

Now, since $$a_3^\dagger$$ and $$a_4^\dagger$$ commute, we can swap their order as many times as we like, so we can directly apply binomial theorem to find $$\left(a_3^\dagger+a_4^\dagger\right)^{n}=\sum_{k=0}^n\binom{n}{k}a_3^{\dagger k}a_4^{\dagger n-k}$$ and $$\left(a_3^\dagger-a_4^\dagger\right)^{m}=\sum_{k=0}^m\binom{m}{k}a_3^{\dagger k}a_4^{\dagger m-k}(-1)^{m-k},$$ where we have used the binomial coefficients $$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$.

Next, we do some manipulation, but the key physical detail to keep in mind is that every term has the same total number of photons $$n+m$$. We will use the notation $$N=n+m$$ and $$K=k+l$$: \begin{align}\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m}&=\sum_{k=0}^n\binom{n}{k}a_3^{\dagger k}a_4^{\dagger n-k}\sum_{l=0}^m\binom{m}{l}a_3^{\dagger l}a_4^{\dagger m-l}(-1)^{m-l}\\ &=\sum_{k=0}^n\sum_{l=0}^m \binom{n}{k}\binom{m}{l}a_3^{\dagger (k+l)}a_4^{\dagger (n+m)-(k+l)}(-1)^{m-l}\\ &=\sum_{K=0}^N a_3^{\dagger K} a_4^{\dagger (N-K)}\sum_{l=0}^K \frac{n!m!(-1)^{m-l}}{(n-K+l)!(K-l)!l!(m-l)!}.\end{align} The last few factors are not so nice, but we always get a factor in the numerator of $$\sqrt{K!(N-K)!}$$ when acting on the vacuum and we need to incorporate the denominator from the original state: $$|n_1,m_2\rangle\to \sum_{K=0}^N \psi_K|K_3,(N-K)_4\rangle,$$ where we can compute the coefficients $$\psi_K=\sum_{l=0}^K \frac{\sqrt{n!m!K!(N-K)!}(-1)^{m-l}}{(n-K+l)!(K-l)!l!(m-l)!\sqrt{2^{n+m}}}.$$ I don't recall offhand whether there's a nice way to simplify the final formula...

1. Use some more manipulations once we establish whether (a) $$n, (b) $$n=m$$, or (c) $$n>m$$. Again, the key thing to notice is that all of the creation operators commute, so we can swap their order at will. We will use the difference of squares formula $$(a+b)(a-b)=(a^2-b^2)$$, which also holds for commuting operators $$a$$ and $$b$$.

(a) $$n: \begin{align}\frac{\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle&=\frac{\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m-n}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle\\ &=\frac{\left(a_3^{\dagger (2n)}-a_4^{\dagger (2n)}\right)\left(a_3^\dagger-a_4^\dagger\right)^{m-n}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle\\ &=\frac{\left(a_3^\dagger-a_4^\dagger\right)^{m-n}\left(a_3^{\dagger (2n)}-a_4^{\dagger (2n)}\right)}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle.\end{align} Now using the definitions of creation operators, we find $$\frac{\left(a_3^\dagger-a_4^\dagger\right)^{m-n}\left(a_3^{\dagger (2n)}-a_4^{\dagger (2n)}\right)}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle= \frac{\sqrt{(2n)!}\left(a_3^\dagger-a_4^\dagger\right)^{m-n}}{\sqrt{n!m!2^{n+m}}}\left(|2n_3,0_4\rangle-|0_3,2n_4\rangle\right).$$ From this expression we can use binomial theorem on the remaining operators to find $$\frac{\sqrt{(2n)!}\left(a_3^\dagger-a_4^\dagger\right)^{m-n}}{\sqrt{n!m!2^{n+m}}}\left(|2n_3,0_4\rangle-|0_3,2n_4\rangle\right)=\frac{\sqrt{(2n)!}}{\sqrt{n!m!2^{n+m}}}\sum_{k=0}^{m-n}\binom{m-n}{k}a_3^{\dagger k}a_4^{\dagger (m-n-k)}(-1)^{m-n-k}\left(|2n_3,0_4\rangle-|0_3,2n_4\rangle\right).$$ The expression has quite a few terms, again, which can all be found using rules such as $$a_3^{\dagger k}|2n_3\rangle=\sqrt{\frac{(2n+k)!}{(2n)!}}|(2n+k)_3\rangle$$.

(b) $$n=m$$: \begin{align}\frac{\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle&=\frac{\left(a_3^{\dagger (2n)}-a_4^{\dagger (2n)}\right)}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle\\ &=\frac{\sqrt{(2n)!}}{\sqrt{n!n!2^{2n}}}\left(|2n_3,0_4\rangle-|0_3,2n_4\rangle\right)\\ &=\frac{\Gamma(n+1/2)}{\sqrt{\pi}\Gamma(n+1)}\left(|2n_3,0_4\rangle-|0_3,2n_4\rangle\right),\end{align} where I snuck in an expression using the Gamma function $$\Gamma$$ for fun.

(c) $$n>m$$: same as (a) but with the role of $$3$$ and $$4$$ reversed.

EDIT 2) Can be made better if we put the operators in the opposite order!

a) $$n: \begin{align}\frac{\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle&=\frac{\left(a_3^\dagger+a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{n}\left(a_3^\dagger-a_4^\dagger\right)^{m-n}}{\sqrt{n!m!2^{n+m}}}|0_3,0_4\rangle\\ &=\frac{\left(a_3^{\dagger (2n)}-a_4^{\dagger (2n)}\right)}{\sqrt{n!m!2^{n+m}}}\sqrt{\frac{(m-n)!}{(m-n)!}}\sum_{k=0}^{m-n}\frac{(m-n)!}{k!(m-n-k)!}a_3^{\dagger k}a_4^{\dagger (m-n-k)}(-1)^{m-n-k}|0_3,0_4\rangle\\ &=\frac{\left(a_3^{\dagger (2n)}-a_4^{\dagger (2n)}\right)}{\sqrt{n!m!2^{n+m}}}\sqrt{(m-n)!}\sum_{k=0}^{m-n}\sqrt{\binom{m-n}{k}}(-1)^{m-n-k}|k_3,(m-n-k)_4\rangle.\end{align} I multiplied by $$\sqrt{\frac{(m-n)!}{(m-n)!}}$$ to give us a nice binomial coefficient in there. Then, we can act with the remaining creation operators to find a (perhaps more tractable) expression: $$\sqrt{\frac{(m-n)!}{{n!m!2^{n+m}}}}\sum_{k=0}^{m-n}\sqrt{\binom{m-n}{k}}(-1)^{m-n-k} \left( \sqrt{\frac{(k+2n)!}{k!}}|(k+2n)_3,(m-n-k)_4\rangle- \sqrt{\frac{(m+n-k)!}{(m-n-k)!}}|k_3,(m+n-k)_4\rangle \right) .$$ (b) works as before and (c) works by swapping the two modes.

• Very elaborate, thank you!
– Logi
Apr 27, 2021 at 11:43
1. Your eq. (4) is correct.
2. The state is certainly a superposition of many different Fock (photon number). As the other answer shows, there is no concise expression of the state. Your eq. (4) is good enough.
3. The problem has been extensively studied, in particular by the Boson sampling or generalized HOM effect studies initiated by Scott Aaronson.