Use of the Beam Splitter Operator

[BeyondBelief96]

TL;DR Summary

Want to understand how to properly apply or decompose the beam splitter operator.

Hello, I am a senior undergrad doing research in quantum optics, and I am trying to work out at the moment the output state of sending a coherent state through one input port and a squeezed vacuum state through the other, just to see what happens tbh. The problem I have constantly been running into is how to properly decompose the beam splitter operator to apply it to the input states.

The beam splitter operator has the form: $\hat{B} = e^{\frac{\theta}{2}(\hat{a}^{\dagger}\hat{b}e^{i\phi} - \hat{a}\hat{b}^{\dagger}e^{-i\phi})}$

I have tried using the Baker-Campbell-Hausdorff Relation that says:

$e^{\hat{X} + \hat{Y}} = e^{\hat{X}}e^{\hat{Y}}e^{-\frac{1}{2}[\hat{X},\hat{Y}]}$

If and only if $[\hat{X}, \hat{Y}]$ also commutes with $\hat{X}$ and $\hat{Y}$

The way I have tried to decompose this operator is letting $\hat{X} = \frac{\theta}{2}e^{i\phi}\hat{a}^{\dagger}\hat{b}$

and $\hat{Y} = \frac{\theta}{2}e^{-i\phi}\hat{a}\hat{b}^{\dagger}$

however when doing so I find that the commutator $[\hat{X},\hat{Y}] = [ \frac{\theta}{2}e^{i\phi}\hat{a}^{\dagger}\hat{b}, \frac{\theta}{2}e^{-i\phi}\hat{a}\hat{b}^{\dagger}] = \frac{\theta^2}{4}(\hat{a}^{\dagger}\hat{a} - \hat{b}^{\dagger}\hat{b})$

which doesn't seem to commute with either of my original operators. So I dont' think this is the right way to go? Unless I have made a mistake. Any help would be appreciated. I am using a and b to denote the two different input ports. Also, I'm wanting to apply this beam splitter operator to the input state:

$\left|\Psi_I\right> = \hat{B} \left|\alpha\right>_a \left|\xi\right>_b = \hat{B}\hat{D}_a(\alpha)\hat{S}_b(\xi)\left|0 \right>_a \left|0\right>_{b}$

where $\hat{D}(\alpha)$ is the displacement operator for generating coherent states,and $\hat{S}(\xi)$ is the squeezing operator, and that I can express both of them acting on the vacuum state in terms of photon number states. Thank you

 

[Cthugha]

Are you really interested in the full beam splitter operator that might describe any possible beam splitter one might build? Otherwise, you might be better off to investigate the standard lossless 50/50 beam splitter first to roughly understand how things work and to consider the general operator afterwards.

For the lossless 50/50 beam splitter, $\phi=90^\circ$, so you get something like $\hat{B}=\exp({i\Theta (\hat{a}^\dagger \hat{b} + \hat{a} \hat{b}^\dagger)})$, where $\Theta=\frac{\pi}{4}$.

The usual approach would now be to expand the operator around unity. So you have an operator of the type $\hat{B}=\exp{i \Theta \hat{J}}$ and expand it as $\hat{B}=\hat{I}+i\Theta\hat{J}+...$ and check how many orders of the expansion you need. In many cases the first order in $\hat{J}$ is already completely sufficient.

 

[BeyondBelief96]

If you would like to see my work so far, I have a LaTeX file and I can send you the file in a PM if you would like to look carefully at my work.

 

[Cthugha]

Is there some post missing? The text of your post does not correspond to the one in my notification mail.

Anyway, the stuff you are interested in is discussed in reference 11 of the the manuscript you are interested in, which also makes use of the approximation I made (which is so standard that most prople do not even mention it).

 

[BeyondBelief96]

Is there some post missing? The text of your post does not correspond to the one in my notification mail.

Anyway, the stuff you are interested in is discussed in reference 11 of the the manuscript you are interested in, which also makes use of the approximation I made (which is so standard that most prople do not even mention it).

I had written a post but the format got messed up after posting it for some reason. So I had to delete it and since I am at work did not have time to rewrite the text.

I will look into the reference.