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As asked here, the beam splitter Hamiltonian has the form $\hat{H}=\frac{i}{\sqrt{2}}\left(\hat{a}_{1}^{\dagger} \hat{a}_{2}+\hat{a}_{2}^{\dagger} \hat{a}_{1}\right)$.

I would like to construct the standard quantum optics "unitary beam splitter operator" from this Hamiltonian using the time evolution operator $e^{-i Ht}$. In this case I get:

$\hat{U}=\exp \left[-\frac{i t}{\sqrt{2}}\left(\hat{a}_{1}^{\dagger} \hat{a}_{2}+\hat{a}_{2}^{\dagger} \hat{a}_{1}\right)\right]$.

How do I split up this exponential?

At first glance it seems like I can use the Baker-Campbell-Hausdorff (BCH) formula:

$$e^{A} e^{B}=e^{A+B+[A, B] / 2}$$

But this only works if [A, B] = c, where c is a constant. In my case, $A = -\frac{i t}{\sqrt{2}} \hat{a}_{1}^{\dagger} \hat{a}_{2}$ and $B = -\frac{i t}{\sqrt{2}} \hat{a}_{1} \hat{a}_{2}^{\dagger}$, and their commutation is nonzero (see calculation at end).

Any ideas how I can proceed?

Just to show my calculation that $[A, B] \neq c$:

$$ \frac{i t}{\sqrt{2}} \hat{a}_{1}^{\dagger} \hat{a}_{2} \frac{i t}{\sqrt{2}} \hat{a}_{1} \hat{a}_{2}^{\dagger}- \frac{i t}{\sqrt{2}} \hat{a}_{1} \hat{a}_{2}^{\dagger}\frac{i t}{\sqrt{2}} \hat{a}_{1}^{\dagger} \hat{a}_{2}$$

$$ \frac{t^2}{2}\left(-\hat{a}_{1}^{\dagger} \hat{a}_{2} \hat{a}_{1} \hat{a}_{2}^{\dagger}+ \hat{a}_{1} \hat{a}_{2}^{\dagger}\hat{a}_{1}^{\dagger} \hat{a}_{2} \right)$$

$$ \frac{t^2}{2}\left(-\hat{a}_{1}^{\dagger} \hat{a}_{1} \hat{a}_{2} \hat{a}_{2}^{\dagger}+ \hat{a}_{1} \hat{a}_{1}^{\dagger} \hat{a}_{2}^{\dagger} \hat{a}_{2} \right)$$

$$ \frac{t^2}{2}\left(\hat{a}_{1} \hat{a}_{1}^{\dagger} \hat{a}_{2}^{\dagger} \hat{a}_{2} -\hat{a}_{1}^{\dagger} \hat{a}_{1} \hat{a}_{2} \hat{a}_{2}^{\dagger} \right)$$

$$ [a, a^\dagger ] = 1$$ $$ a a^\dagger- a^\dagger a = 1$$ $$ a a^\dagger = 1 + a^\dagger a$$ $$ a_2 a_2^\dagger = 1 + a_2^\dagger a_2$$ $$ a_1 a_1^\dagger = 1 + a_1^\dagger a_1$$

$$ \frac{t^2}{2}\left(\left(1 + a_1^\dagger a_1 \right) \hat{a}_{2}^{\dagger} \hat{a}_{2} -\hat{a}_{1}^{\dagger} \hat{a}_{1} \left(1 + a_2^\dagger a_2 \right) \right)$$

$$ \frac{t^2}{2}\left(\hat{a}_{2}^{\dagger} \hat{a}_{2} -\hat{a}_{1}^{\dagger} \hat{a}_{1} \right)$$

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  • $\begingroup$ Life would definitely be much easier in the Heisenberg picture. (Probably even to derive $U$ in the Heisenberg picture and then map it back.) $\endgroup$
    – Norbert Schuch
    Commented Mar 30, 2021 at 18:45
  • $\begingroup$ @NorbertSchuch, yeah I've seen a solution you provided to a similar problem that did that. I am curious to see if it's possible to go in this direction, and if it was obvious how to do it without going to the Heisenberg picture. $\endgroup$
    – Steven Sagona
    Commented Mar 30, 2021 at 18:58
  • $\begingroup$ One option that seems to work is that if you have exactly 1 photon, you can work out your coefficients by representing the exponential matrix $e^{-i H t}$ as a Taylor series, and you end up with something like $cos(t/ \hbar)|0,1\rangle + i sin(t/ \hbar)|1,0\rangle$. $\endgroup$
    – Steven Sagona
    Commented Mar 30, 2021 at 19:01
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    $\begingroup$ @StevenSagona Maybe my reading comprehension skills are failing me, but I didn't find a unitary operator in the linked document. I saw some unitary transformations applied to specific states, but not a state-independent unitary operator. The question seems to be asking for a specific way of writing the latter. Does the linked document show an example of how you want $U$ to be expressed? $\endgroup$
    – Chiral Anomaly
    Commented Mar 31, 2021 at 2:30
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    $\begingroup$ @StevenSagona The formula you give is in the Heisenberg picture (and applies independent of the number of photons), and deriving it by starting from the Heisenberg equation for the operators is immediate. Does the reference you quote give a form of U in the Schroedinger picture? Otherwise, if you want to derive the form you quote, start from $\dot a = i[H,a]$ - this will be immediate. There's good reason people work in this representation. $\endgroup$
    – Norbert Schuch
    Commented Mar 31, 2021 at 10:29

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