# Hong Mandel 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-mandel 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!

• In case you aren't able to get answers here, you can try the physics stackexchange physics.stackexchange.com Commented Aug 1 at 12:30
• If it is not normalized, it is because you made a mistake. Commented Aug 4 at 11:21
• @NorbertSchuch it is a simple calculation which i believe is correct, and is very easy to verify. If you see an error, can you please point it out? Commented Aug 5 at 23:20
• Please cross-link your crossposts! Otherwise you cause double work to people, which is rather incosiderate! Commented Aug 6 at 10:14

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