I Terms canceling out in unitarily evolving state

[Swamp Thing]

The Hong-Ou-Mandel experiment is one of many examples where the amplitudes associated with two histories cancel out, leaving us with a reduced range of possible outcomes. Obviously, the total probability of those outcomes has to be unity.

My question relates to the fact that these processes are said to be unitary (i.e. conserve total probability). But when terms cancel as in HOM, we have to "manually" normalize the rest of the terms to get back to a total probability of 1. How then do we say that the transformation matrix itself is inherently unitary?

 

[tom.stoer]

In which mathematical formula do you expect any violation of unitarity?

 

[Swamp Thing]

I was thinking of, for example equations 10, 11 and 12 here: http://agatabranczyk.com/files/Branczyk2013 - Notes on Hong-Ou-Mandel Interference.pdf

It's not that I expect violation of unitarity, but the normalization doesn't seem to happen automatically, though the text says it does. If the middle two terms of (10) hadn't canceled out, we would have to have different coefficients for the two outer terms.

 

[vanhees71]

Maybe I don't understand your question, but for a two-mode system it's pretty obvious that (3) and (4) define a unitary transformation in the corresponding Hilbert space. Perhaps the author should have been remarked that the transformation law, defining this unitary transformation reads
$$\hat{a}_i^{\dagger} \rightarrow \hat{U}_{\text{BS}} \hat{a}_i^{\dagger} \hat{U}_{\text{BS}}^{\dagger}, \quad \hat{b}_j^{\dagger} \rightarrow \hat{U}_{\text{BS}} \hat{b}_j^{\dagger} \hat{U}_{\text{BS}}^{\dagger}$$
and, by taking the adjoint of these equations
$$\hat{U}_{\text{BS}} |0 \rangle=|0 \rangle.$$
Of course, this is obvious since if there are no photons present ("vacuum") then also a beam splitter won't produce anyone.

 

[Swamp Thing]

Thank you. I think my question was actually a lot more naive and uninformed than it may have appeared to you , so you gave me credit for better understanding than I had.

What I was thinking was something like this : if we sum the squares of all the terms before and after evolution, we should get the same answer, and we should not have to normalize "by hand" to get a total probability of one. It was looking to me that this was not happening. But now I realize that my confusion arises from the convention of writing out amplitudes only for the modes where the amplitude is non-zero. But in order to get "automatic" normalization we need to write out all the modes of interest including unoccupied ones (and the list of modes of interest should be the same before and after the transformation). -- When you have a moment, please confirm whether this is correct, of do I still need to think & learn more about this?

So thanks again, your answer helped me to get some valuable clarity.

Edit : So I think the root of my confusion is in not realizing this: If you write out amplitudes "by inspection" rather than 'shut up and calculate', then be prepared to have to manually scale back to a total squared amplitude of unity "by inspection".

Edit: Sorry, wait, there's still something wrong. I'll get back.

 

[vanhees71]

But $|\psi_{\text{out}} \rangle$ is normalized, if $|\psi_{\text{in}} \rangle$ is, because $\hat{U}_{\text{BS}}$ is indeed unitary. Note that the action on the creation operators implies its definition on the entire Fock space, as is immediately clear from using the occupation-number basis.