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

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!

• Surely $|t|^2+|r|^2=1$ as it is any scattering process. I assume that they have chosen phases so that $t$ and $r$ are real.... Commented Aug 1 at 23:23

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 $$(t^2-r^2)^2 + (\sqrt{2}tr)^2 + (\sqrt{2}tr)^2 = (t^2+r^2)^2 =1 \ .$$

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 $$\langle\psi|\psi\rangle = 2(tr)^2 + (t^2 - r^2)^2 + 2(tr)^2 = (t^2 + r^2)^2 = 1 .$$

• actually your $\vert 1,1\rangle$ as written now is not normalized. $\hat a^\dagger \hat a^\dagger \vert \text{vac}\rangle=\sqrt{2}\vert 2,0\rangle$, and the $\sqrt{2}$ factor will correctly normalize your $\vert 1,1\rangle$. Let me know when you fix this and I can upvote your answer. Commented Aug 2 at 12:12
• @ZeroTheHero. Made the correction. Thank for pointing it out. Commented Aug 3 at 3:20
• @flippiefanus I actually understand well the cancelling of the terms etc. My question was specifically for the general case when the t,r coefficient are not equal. More specifically, I wanted to know how the normalization is done after adding the indistinguishable states. Commented Aug 3 at 23:46