# 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?

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