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!
Cross-post link: https://physics.stackexchange.com/questions/823293/hong-mandel-effect-question-on-math-of-indistinguishability-of-photons
