General theory of the beam splitter

For a standard beam splitter with two independent interfering modes $a,b$, one can write the interaction Hamiltonian as $$H=\frac{i}{2}(e^{-i\phi}a^\dagger b + e^{i\phi}b^\dagger a)=-\frac{1}{2}(a^\dagger b+b^\dagger a) $$ where I have assumed a $\frac{\pi}{2}$ phase added upon reflection. This generates an evolution operator $U_\theta$ defined as $$U_\theta=\exp(-i\theta H), $$ applying which we can find the output modes $b'=U_\theta^\dagger bU_\theta$ and $a'=U_\theta^\dagger a U_\theta^\dagger$. The parameter is then $\theta$ is used to define both the transmittivity $T=\sin\theta$ and its complementary, the reflectivity $R=\sqrt{1-T^2}=\cos\theta$.

The problem

Consider now the following problem. We want to model propagation through empty state from a plane $P_0$ to an orthogonal plane $P_1$ at distance $d$. Instead of having only two input signals, there is now an infinite set of continuous modes $a^\dagger(r)$, each creating a photon at position $r$. These operators are not independent, but $\delta$-correlated as $$[a(r), a^\dagger(r')]=\delta_{rr'}$$ (note that I'm using a discretized picture for simplicity). Likewise, the output modes $b(r)$ also depend on position; furthermore, I will assume that the position vectors labeling both the input and output modes are $2$-dimensional, thus removing the necessity of a distinction between positions on different planes.

Let $\sqrt{\eta(r,r')}$ be the probability amplitude for a photon generated at position $r$ on the first plane to be later found at position $r'$ on the second plane. Then for any output mode $b^\dagger(r)$ we should be able to write $$b^\dagger(r)=\sqrt{\eta(r)}a^\dagger(r) + \sum_{r'\ne r} \sqrt{\eta(r',r)}a^\dagger(r'), $$ where the transmission (read no deviation) amplitude is given by $$\sqrt{\eta(r)}=\sqrt{1-\sum_{r\ne r'}\eta(r,r')}.$$

The question

What is the correct way to write the Hamiltonian $H$ that describes this evolution? I'd venture the guess $$H=-\frac{1}{2}\sum_{r,r'}\left(a^\dagger(r)a(r')+a^\dagger(r')a(r)\right), $$ but I haven't been able to find the expected expression from the output modes after evolution by $U_\theta$.

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    $\begingroup$ So $r$ is now both the reflectivity and the position vector. That is a bit confusing. $\endgroup$ Jun 14, 2023 at 9:57
  • $\begingroup$ @flippiefanus Fixed! $\endgroup$ Jun 14, 2023 at 9:58
  • $\begingroup$ It does not seem that what you are looking for has anything to do with a beam splitter. So I'm still a little confused. $\endgroup$ Jun 15, 2023 at 2:56

1 Answer 1


Not sure I understand correctly, but usually the beamsplitter represents an identity operator as far as spatiotemporal degrees of freedom are concerned. Or, stated differently, the beamsplitter does not care what the degrees of freedom (mode) of the input state is. Every photon has an equal probability to be transmitted or reflected. What this means is that if an input state in one of the input ports has a certain mode, then both output ports will produce that mode.

For such a beamsplitter, one simply integrated over all the degrees of freedom of the ladder operators. Usually ladder operators are represented in terms of three dimensional wave vectors to capture all the spatiotemporal degrees of freedom. So $$ H = \frac{1}{2} \int \hat{a}^{\dagger}(\mathbf{k}) \hat{b}(\mathbf{k})+\hat{b}^{\dagger}(\mathbf{k}) \hat{a}(\mathbf{k}) d^3\mathbf{k} . $$

The process of free space propagation is something completely different from what a beam splitter does. Where a beamsplitter represents a system with two input port and two output port, a free space propagation process represents a system with just one input port and one output port.

The is a whole bunch of physics that has been developed for free space propagation of classical optical fields. Since it is a linear process (no-interactions), the classical theory can also be applied to quantum states. So, it just comes down to formulating the process for free-space propagation in terms of quantum formalism. Obviously, to do that one must first understand the classical theory.

In short, what free space propagation does is to perform a propagation-distance-dependent phase modulation in the Fourier domain. Therefore, you first need to convert the everything to the Fourier domain. That is done by using the ladder operators in terms of wave vector, as I've shown above for the beamsplitter. The Hamiltonian would then a modified version of the number operator. Under the monochromatic and paraxial approximations it would have the form $$ H = \int \mathcal{N} \hat{a}^{\dagger}(\mathbf{k}_{\perp}) \frac{z|\mathbf{k}_{\perp}|^2}{k} \hat{a}(\mathbf{k}_{\perp}) d^2\mathbf{k}_{\perp} , $$ where $\mathbf{k}_{\perp}$ is the transverse part of the wave vector, $z$ is the propagation distance, and $k$ is the wave number. The constant $\mathcal{N}$ is added in case I forgot some constant factors. (I'm writing this from memory which fails my sometimes.)

  • $\begingroup$ Thank you, but I'm not 100% sure this is what I want. The objective is to model propagation through empty space from a plane to another orthogonal one as a 'beam splitter'. To do so, I need to relate the operators $b^\dagger(r')$ creating a photon at position $r'$ on the second plane to the operators $a^\dagger(r)$ creating a photon at position $r$ on the first plane. I'll probably edit the question for clarity. $\endgroup$ Jun 14, 2023 at 10:36
  • $\begingroup$ OK I'll wait for the clarification. $\endgroup$ Jun 14, 2023 at 12:26
  • $\begingroup$ Hopefully it's clearer now. $\endgroup$ Jun 14, 2023 at 13:52

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