Hong-Ou-Mandel (HOM) effect - question on math of indistinguishability of photons

Question

I’m working on the transformations of multi-photon states through a beam splitter, and I encountered confusion in some normalization factors . Using beam splitter transformations:

$$ \hat{a}^\dagger \to t \hat{a}^\dagger - r \hat{b}^\dagger \qquad \hat{b}^\dagger \to t \hat{b}^\dagger + r \hat{a}^\dagger $$

I tried to write down the equation for transformation of state $|1,1\rangle$ in a two photon two input beam-splitter setup like in Hong-Ou-Mandel (HOM) effect.

$$ |1, 1\rangle \to (t \hat{a}^\dagger - r \hat{b}^\dagger)(t \hat{b}^\dagger + r \hat{a}^\dagger) = t^2 \hat{a}^\dagger \hat{b}^\dagger - r^2 \hat{a}^\dagger \hat{b}^\dagger + t r \hat{a}^\dagger \hat{a}^\dagger - r t \hat{b}^\dagger \hat{b}^\dagger $$

My question is why normalization is NOT done here? Because the resultant of this has $(t^2-r^2) |11\rangle$ instead of $\frac{(t^2-r^2)}{\sqrt(t^4+r^4)}|11\rangle$ ?

From what I understand that is the way to normalize due to the indistinguishability of photons - We find the total probability of the states which "were distinguishable" and divide by the square root of that. Thats what I think is done in the case of transformation of $|20\rangle$, the reason for having the $\sqrt2$ term: $$ |2, 0\rangle \to (t \hat{a}^\dagger - r \hat{b}^\dagger)^2 = t^2 |2, 0\rangle - \sqrt{2}tr |1, 1\rangle + r^2 |0, 2\rangle $$

So please point out where am I going wrong in my reasoning, and if possible, add a resource for me to study the correct method of doing this (already tried searching internet without luck). Thanks!

Cross-post link: https://quantumcomputing.stackexchange.com/q/39401/

Answer 1

Your state is perfectly normalized if you use that $(a^\dagger)^2|0\rangle = \sqrt{2}|2\rangle$ (where $|0\rangle$ and $|2\rangle$ are normalized states): Its norm is $$ (t^2-r^2)^2 + (\sqrt{2}tr)^2 + (\sqrt{2}tr)^2 = (t^2+r^2)^2 =1 \ . $$

Answer 2

The normalization of a state $|\psi\rangle$ is defined by applying the inner product of the state with itself: $\langle\psi|\psi\rangle=1$.

The beamsplitter is represented by a unitary operation. Therefore, if the input state is properly normalized, the output state would still be normalized. But you need to consider the complete state, not just a part of it.

For the transformation given by $$ \hat{a}^\dagger \to t \hat{a}^\dagger - r \hat{b}^\dagger \qquad \text{and} \qquad \hat{b}^\dagger \to t \hat{b}^\dagger + r \hat{a}^\dagger , $$ unitarity requires that $|t|^2+|r|^2=1$. For a 50:50 beamsplitter they would just be $$ r=t=\frac{1}{\sqrt{2}} . $$ Two of the terms would then cancel (due to the Hong-Ou-Mandel effect), leaving $$ |1,1\rangle \to \frac{1}{2} \hat{a}^\dagger \hat{a}^\dagger|\text{vac}\rangle - \frac{1}{2} \hat{b}^\dagger \hat{b}^\dagger|\text{vac}\rangle = \frac{1}{\sqrt{2}}|2,0\rangle - \frac{1}{\sqrt{2}} |0,2\rangle , $$ which is normalized.

In the case where $r\neq t$, you'll get $$ |1,1\rangle \to |\psi\rangle = t r \hat{a}^\dagger \hat{a}^\dagger |\text{vac}\rangle +(t^2 - r^2) \hat{a}^\dagger \hat{b}^\dagger|\text{vac}\rangle - r t \hat{b}^\dagger \hat{b}^\dagger|\text{vac}\rangle $$ $$ = |2,0\rangle \sqrt{2}tr + |1,1\rangle (t^2 - r^2) - |0,2\rangle \sqrt{2}tr. $$ Note that the three terms are all orthogonal. Therefore $$ \langle\psi|\psi\rangle = 2(tr)^2 + (t^2 - r^2)^2 + 2(tr)^2 = (t^2 + r^2)^2 = 1 . $$