# Can we derive the classical beam-splitter theory from the quantum beam-splitter Fock state picture?

A classical beam splitter divides incoming light into two parts according to the reflection and transmission coefficients.

A quantum beam splitter on the other hand, can be modelled using Fock states, where the state can be written as $$|n\rangle|m\rangle$$. The two states correspond to the two input ports of the beam splitter. Then we quantize the electric field, transforming the creation-annihilation operators and getting the output Fock state.

Could we derive the classical beam splitter theory from the quantum beam splitter Fock state picture?

When does a quantum state of light start reaching the classical model?

• I believe the “classical model” is reached with input states being coherent states. – ZeroTheHero Oct 1 '18 at 0:50

The cleanest way to understand beam splitters in quantum optics is via their action on the creation and annihilation operators, which is the same as their action on the classical field amplitudes: if you have input beams with annihilation operators $$\hat a_1$$ and $$\hat a_2$$, then the output annihilation operators $$\hat b_1$$ and $$\hat b_2$$ will be given by $$\begin{pmatrix} \hat b_1 \\ \hat b_2 \end{pmatrix} = \begin{pmatrix} r & -t^*\\ t & r^* \end{pmatrix} \begin{pmatrix} \hat a_1 \\ \hat a_2 \end{pmatrix},$$ where the two sets are related with a unitary matrix whose entries are the (coherent) reflection and transmission coefficients.

With this in hand:

• You can calculate the action of the beam splitter on an initial Fock state as \begin{align} |m⟩|n⟩ & = \frac{(\hat a_1^\dagger)^m}{\sqrt{m!}} \frac{(\hat a_2^\dagger)^n}{\sqrt{n!}} |0⟩|0⟩ \\ & = \frac{(r \hat b_1^\dagger + t \hat b_2^\dagger)^m}{\sqrt{m!}} \frac{(-t^* \hat b_1^\dagger + r^*\hat b_2^\dagger)^n}{\sqrt{n!}} |0⟩|0⟩, \end{align} so it will come out to some large-ish product of binomials on different Fock states.

• For the classical limit, you simply put in a coherent state on each of the input ports, \begin{align} |\alpha_1⟩|\alpha_2⟩ & = e^{\alpha_1 \hat a_1^\dagger - \alpha_1^* \hat a_1}e^{\alpha_2 \hat a_2^\dagger - \alpha_2^* \hat a_2}|0⟩|0⟩ \\ & = \exp\left[ \begin{pmatrix}\alpha_1^* & \alpha_2^*\end{pmatrix}\begin{pmatrix}\hat a_1 \\ \hat a_2\end{pmatrix} - \begin{pmatrix}\alpha_1 & \alpha_2\end{pmatrix}\begin{pmatrix}\hat a_1^\dagger \\ \hat a_2^\dagger\end{pmatrix} \right]|0⟩|0⟩ \end{align} and again you re-express the state in terms of the $$\hat b_i$$ by putting in the right matrix (which will then transfer directly as the same matrix acting on the $$\alpha_i$$, as it should).

A quantum state of light typically behaves almost classically if the expectation of the number operator on it is very large. This is essentially the same as saying that the energy scale of the system is much larger than the energy of a single photon.

If the latter is true, the quantum state behaves like a classical probability distribution, with an error that converges to zero as the expectation of the number operator goes to infinity. Also, the quantum evolution is well approximated by the classical evolution (in this case, classical electromagnetism).

The classical probability corresponding to a given quantum state is explicitly characterizable, the one corresponding to a Fock vector $$\lvert n,m\rangle$$, with $$n,m\to \infty$$, $$\frac{n}{m}=\mathrm{const.}$$, is a bit complicated to write down (but nonetheless explicit).

I am no expert on the theory of beam splitters, so I am not able to say whether a semiclassical description of such system is easy to be formulated analytically. Nonetheless, one should expect that a quantum beam splitter with a state having a very large number of photons (i.e. a large expectation of the number operator) should behave essentially as a classical beam splitter, apart from very small quantum corrections.

• If your input state is pure and not the product of two coherent states, the output state will be entangled, even if it has a very large number of photons. So unless you have some noise (which one usually has), it is wrong to say the beamsplitter behave essentially as a classical beamsplitter – Frédéric Grosshans Oct 2 '18 at 16:49