2 photon state split by a beam-splitter (Issue with normalization)

Question

I'm having a simple issue in a calculation involving splitting a $|2\rangle$ state with a beamsplitter: How exactly do you calculate the probabilities of splitting a $|2\rangle$ state on a beam-splitter?

I see some very similar questions asked, that don't exactly answer my question. For example in one question asked here :

$\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\ket}[1]{\left|#1\right\rangle}$ Suppose I have a beam splitter that will either reflect a photon by 45 degrees, or will allow the photon to pass directly through.

If I send a single photon state through, that is, $\ket{1}_{initial}$, we have:

$\ket{1}_{initial} = \frac{1}{\sqrt{2}}(\ket{0}_a\ket{1}_b + i\ket{1}_a\ket{0}_b)$

where $a$ and $b$ represent the paths that pass straight through, and the reflecting paths respectively.

Now suppose instead of sending through a single photon, the initial state is two photons, that is, $\ket{2}_{initial}$. Classically I would suspect the state to look something like this:

$\ket{2}_{initial} \propto (\ket{0}_a\ket{2}_b + \ket{2}_a\ket{0}_b + 2\ket{1}_a\ket{1}_b$)

I have the factor of 2 in front of the $\ket{1}_a\ket{1}_b$ state because classically you could have the first photon passing through and the second photon reflecting, or vice-versa. Is this still correct for quantum mechanics, where photons are indistinguishable?

The accepted answerer ignores the normalization saying:

$$ (a^{\dagger})^2|0\rangle \rightarrow [(a^{\dagger})^2 + > (b^{\dagger})^2 +2i a^{\dagger}b^{\dagger}]|0\rangle. $$ This is not normalised, by the way, because I'm lazy and it is completely irrelevant to the physics.

But the normalization is the part that I'm having trouble with. Using the notation from this answer:

The condition of unitarity (or energy conservation) for the action of the beam-splitter gives the following transformations:

$\hat{c}=\sqrt{\tau}\hat{a}+\sqrt{1-\tau}\hat{b}$

$\hat{d}=\sqrt{1-\tau}\hat{a}-\sqrt{\tau}\hat{b}$

And taking the case when the outputs are equal:

\begin{align} \hat{c}=\frac{1}{\sqrt{2}}(\hat{a}+\hat{b}) \qquad \hat{d}=\frac{1}{\sqrt{2}}(\hat{a}-\hat{b}) \end{align}

which means that: \begin{align} \hat{a}=\frac{1}{\sqrt{2}}(\hat{c}+\hat{d})\, ,\qquad \hat{b}=\frac{1}{\sqrt{2}}(\hat{c}-\hat{d}) \end{align}

Then my output state looks like:

\begin{align} |2\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d} &= (\hat{a}^{\dagger})^2|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d} \tag{1}\\ &=(\frac{1}{\sqrt{2}}(\hat{c}^{\dagger}+\hat{d}^{\dagger}))^2|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}\, ,\\ &=(\frac{1}{2}(\hat{c}^{\dagger}\hat{c}^{\dagger}+\hat{d}^{\dagger}\hat{c}^{\dagger}+\hat{c}^{\dagger}\hat{d}^{\dagger}+\hat{d}^{\dagger}\hat{d}^{\dagger})|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}\, ,\\ &=|0\rangle_{a}|0\rangle_{b}\left(\frac{1}{2} (|2\rangle_{c}|0\rangle_{d}+2|1\rangle_{c}|1\rangle_{d}+|0\rangle_{c}|2\rangle_{d}) \right) \end{align}

Okay, so now I see that the probability of states $|2\rangle_{c}|0\rangle_{d}$ and $|2\rangle_{c}|0\rangle_{d}$ are as expected:

$(\frac{1}{2})^2 = 1/4$

But the probability of observing $|1\rangle_{c}|1\rangle_{d}$ is strangely not $\frac{1}{2}$ but is:

$(\frac{2}{2})^2 = 1!?$

Obviously I'm making some kind of mistake here..but it's really not obvious what the mistake is.

Now this is where I stopped. But I had a fear someone would tell me that all I have to do is renormalize it (which, I believe, there shouldn't be any reason why I should need to do this, considering everything was properly normalized to begin with). But if I go ahead and do this, I observe that I end up dividing by $1+\frac{1}{4}=\frac{5}{4}$, and end up with weird probabilities that are not $\frac{1}{4}$ and $\frac{1}{2}$

Any ideas what mistake I'm making?

Answer 1

You don't normalize your two-photon states properly. They are $|2\rangle = \tfrac{1}{\sqrt{2}}(a^\dagger)^2|0\rangle$ (and the same for the $c$ and $d$ modes). You are missing the $\tfrac{1}{\sqrt{2}}$.

Then, your output state is $\frac{1}{2}\tfrac{1}{\sqrt{2}}(\hat{c}^{\dagger}\hat{c}^{\dagger}+\hat{d}^{\dagger}\hat{c}^{\dagger}+\hat{c}^{\dagger}\hat{d}^{\dagger}+\hat{d}^{\dagger}\hat{d}^{\dagger})|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$ which has probability $\tfrac14$ for the two two-photon states and $\tfrac12$ for the $|1\rangle |1\rangle$ state.

Answer 2

Unless you are dealing with very complicated scenarios, a beam splitter is usually easily implemented in terms of ladder operators. I like to represent the transformation like this: $$ \hat{a}^{\dagger} = \hat{a}^{\dagger} C + i \hat{b}^{\dagger} S , $$ $$ \hat{b}^{\dagger} = i \hat{a}^{\dagger} S + \hat{b}^{\dagger} C , $$ where $|C|^2+|S|^2=1$. As such, the beamsplitter is represented by a unitary transformation that will maintain the normalization of the state.

The transformation for a single photon is straight forward. For a multi-photon state, the important part is to have the input state properly normalized when it is expressed in terms creation operators. An $n$-photon state is then given by $$ |n\rangle = \frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n|\text{vac}\rangle . $$ You can check this by contracting the state on itself and working through the commutations of the operators.

To see how this state transforms as it passes through the beam splitter, one substitutes the above transformation equation into the expression for the $n$-photon state and expand it. Then one can convert the terms back into number states for the two output ports, while making sure that the correct normalization factors are incorporated.

That should do the trick. As an example, I'll work through the 2-photon case.

For 2 photons, the input state is $$ |2\rangle = \frac{1}{\sqrt{2}} (\hat{a}^{\dagger})^2|\text{vac}\rangle . $$ after substituting the transformation in, we get \begin{align} |2\rangle \rightarrow |\psi\rangle = & \frac{1}{\sqrt{2}} (\hat{a}^{\dagger} C + i \hat{b}^{\dagger} S)^2|\text{vac}\rangle \\ = & \frac{1}{\sqrt{2}}\left[ (\hat{a}^{\dagger})^2 C^2 + i 2 \hat{a}^{\dagger} \hat{b}^{\dagger} SC - (\hat{b}^{\dagger})^2 S^2 \right] |\text{vac}\rangle \\ = & |2\rangle_a C^2 + i\sqrt{2}|1\rangle_a|1\rangle_b SC - |2\rangle_b S^2 . \end{align} One can now show that this state is properly normalized, but if that is not obvious one can check the case where $C=S=1/\sqrt{2}$. Then it becomes $$ |\psi\rangle=\frac{1}{2} |2\rangle_a + i\frac{1}{\sqrt{2}}|1\rangle_a|1\rangle_b - \frac{1}{2} |2\rangle_b . $$ So that $$ \langle\psi|\psi\rangle = \frac{1}{4}+\frac{1}{2}+\frac{1}{4} = 1 . $$