If we have light of a particular phase that is incident on a beam splitter, I assume the transmitted beam undergoes no phase change. But I thought that the reflected beam would undergo a phase change of $\pi$. I have, however, read that it undergoes a phase change of $\pi/2$.

Which is it, and why?

  • $\begingroup$ Have you see this question and related answers? $\endgroup$
    – Floris
    Commented May 21, 2018 at 2:05
  • $\begingroup$ Refer to Chapter 6 of "Introductory Quantum Optics" by Gerry and Knight, it gives a more comprehensive analysis of beam splitters than any of the answers here. $\endgroup$ Commented Jun 22 at 18:01

5 Answers 5


It actually depends on what kind of beam spitter you have.

I'll give a general treatment and shows that the conclusions of both Emilio Pisanty and Steven Sagona are basically correct, corresponding to different specific beam splitters, which are all common in the laboratory. For simplicity we don't consider loss in this answer.

First of all, the definition of "phase shift" in this specific answer is chosen as the relative phase between reflected and transmitted light from the same port. More specifically, we let the transmission coefficient to be real number $t$, and the reflection coefficient will thus carries the information about relative phase shift $r e^{i\theta_\alpha}$, where $\alpha=1,2$ representing the light coming from beam splitter port 1 or 2 (see picture below). enter image description here

Instead of assuming symmetrical phase shift, we allow any possible phase shift, as we are discussing general beam splitter which is not necessarily symmetric.

Then we write the physics procedure happening in beam splitter into the following matrix form: \begin{eqnarray} \begin{pmatrix} E_3 \\ E_4 \end{pmatrix} = \underbrace{ \begin{pmatrix} t & r e^{i\theta_2} \\ r e^{i\theta_1} & t \end{pmatrix}}_{ M} \begin{pmatrix} E_1\\ E_2 \end{pmatrix} \end{eqnarray}

The conservation of energy requires $|E_3|^2 + |E_4|^2 = |E_1|^2 + |E_2|^2$, which is equivalent to the mathematical statement that the beam splitter matrix is unitary. That gives us \begin{eqnarray} M^\dagger M = \begin{pmatrix}r^2 + t^2 & rt(e^{i\theta_2}+e^{-i\theta_1}) \\ rt(e^{i\theta_1}+e^{-i\theta_2}) & r^2 + t^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{eqnarray}

It can be seen that $r^2 + t^2 =1$ is automatically true as we deal with lossless beam splitter here. The rest of the above matrix equation gives us \begin{eqnarray} e^{i\theta_2}+e^{-i\theta_1} \equiv 2 e^{i(\theta_2-\theta_1)/2} \cos \frac{\theta_1+\theta_2}{2} =0 \end{eqnarray} from which we conclude that \begin{equation} \theta_1 + \theta_2 = \pi \end{equation}

If we have a beam splitter with symmetric phase shifts, $\theta_1 =\theta_2 = \pi /2$, then \begin{equation} M= \begin{pmatrix} t & ir \\ ir & t \end{pmatrix} \end{equation} This is consistent with Steven Sagona's argument based on the same assumption.

If we have a beam splitter that is not only symmetric in phase shifts ($\theta_1 = \theta_2=\pi/2$), but also symmetric in reflection and transmission ($r=t=1/\sqrt 2$), then we have \begin{equation} M=\frac{1}{\sqrt 2} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \end{equation} Then we basically arrive at Emilio Pisanty's second equation.

If we have a beam splitter that is not symmetric in phase shifts ($\theta_1 = 0, \theta_2=\pi$), which is also very common in the lab, then we have \begin{equation} M=\frac{1}{\sqrt 2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \end{equation} Remember I have been putting the relative phase factor to reflection instead of transmission so that we have $\{t, r e^{i \theta_\alpha}\}$. Now, we can also put the phase factor to transmission so that we have $\{t e^{-i\theta_\alpha}, r\}$, without losing any physics. What we got after is Emilio Pisanty's first equation, namely \begin{equation} M=\frac{1}{\sqrt 2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}

  • $\begingroup$ Why can you assume transmission coefficient $t$ same in both directions? $\endgroup$ Commented Mar 30 at 13:25
  • $\begingroup$ The assumption is made based on the fact that, for the considered beam splitting layer, each transmission channel is made up of the same passive optical gradients, and have same geometric properties. $\endgroup$
    – G.C.
    Commented Mar 31 at 15:11
  • $\begingroup$ What is meant by passive optical gradients? Just a variation in optical densities? $\endgroup$ Commented Mar 31 at 16:12
  • $\begingroup$ Even then, is it obvious that it should be the same for both channels? (I am imagining some kind of dielectric mirror with non-trivial interference effects leading to its reflective properties). I know in practice it doesn't matter (as explained by Emilio) so this is just out of interest. $\endgroup$ Commented Mar 31 at 18:27

Your confusion stems basically from comparing results in different conventions.

Basically, there is always a phase difference of $\pi$ between the two output ports of a beam splitter, but this can only ever be 'morally' true, because that statement talks about the phase of the EM field at different points, and thus the phase will be different depending on where exactly you fix your measuring point on the two input and the two output ports. In situations where you have two waves co-propagating, then their relative phase is perfectly well-defined, but for the ports of the beam splitter, you're comparing phases at different beams in different positions and moving in different directions, so the whole thing is impossible without some artificial way to fix the convention.

Broadly speaking, there are two separate ways to fix the convention: one with an explicit $\pi$ phase shift, $$ \begin{pmatrix} a_{\mathrm{out},1} \\ a_{\mathrm{out},2} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} a_{\mathrm{in},1} \\ a_{\mathrm{in},2} \end{pmatrix}, $$ and one with several explicit $\pi/2$ phases: $$ \begin{pmatrix} a_{\mathrm{out},1} \\ a_{\mathrm{out},2} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \begin{pmatrix} a_{\mathrm{in},1} \\ a_{\mathrm{in},2} \end{pmatrix}. $$ Those two conventions are exactly equivalent, since they can be transformed by adding a $\pi/2$ phase to $a_{\mathrm{in},2}$ and $a_{\mathrm{out},2}$, $$ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix}, $$ and experimentally this is equivalent to adding a thin slab of glass on one input and one output port. More importantly, though, you didn't know the precise value of the optical path before the $(\mathrm{in}{,}2)$ port or after the $(\mathrm{out}{,}2)$ output, so quibbling about these phases is completely moot.

What you do need is for the matrix governing the coupling to be unitary, which comes from a strict requirement of energy conservation.

Now, this requirement of unitarity can indeed sound a bit intimidating and exotic, but it is important to note that the requirement of unitarity has nothing to do with quantum mechanics, and it is already present within the hamiltonian classical-mechanics description of the system.

More precisely, when we say that the beam splitter can be described by a matrix, we are making two core assertions about the electromagnetic fields we're considering:

  • First, we are asserting that the EM fields we are willing to consider need to be linear combinations of only two pre-specified modes, which basically look like this:


  • Second, we are recognizing that those fields can also be expressed as linear combinations of two modes which look like this,


    which are characterized by having all of the output energy on only one of the output ports.

Both of these sets of modes are bases for the same linear subspace of field modes, which means that each set can be expressed as a linear combination of the other set; in other words, that means that the amplitudes of each set of modes are related via some matrix.

More importantly, when we sit down to describe the (classical) field, we write either $$ \mathbf E(\mathbf r,t) = \mathrm{Re}\mathopen{}\left[ \sum_{j=1}^2 \alpha_{\mathrm{in},j}(t) \mathbf E_{\mathrm{in},j}(\mathbf r) \right] $$ or $$ \mathbf E(\mathbf r,t) = \mathrm{Re}\mathopen{}\left[ \sum_{j=1}^2 \alpha_{\mathrm{out},j}(t) \mathbf E_{\mathrm{out},j}(\mathbf r) \right], $$ where $\alpha_{\mathrm{in},j}(t)$ and $\alpha_{\mathrm{in},j}(t)$ are the complex-valued canonical variables that describe the dynamics of those modes, and which satisfy the dynamical equations $$ \frac{\mathrm d^2}{\mathrm dt^2}\alpha_{x,j}(t) = -\omega^2 \alpha_{x,j}(t). $$

The tricky part is the normalization: because $\alpha_{x,j}(t)$ and $\mathbf E_{x,j}(\mathbf r)$ only ever appear (thus far) in the product $\alpha_{x,j}(t) \mathbf E_{x,j}(\mathbf r)$, there is an ambiguity in a common complex factor that can be put on either side, including both normalization and phase.

  • For the normalization, there is an objective absolute standard that needs to be adhered to: namely, that for each of the modes, the total energy flow across the beam-splitter midline needs to be constant. This is the only way to correctly quantize the system.
  • For the phase, there is no objective or absolute standard - there is a complete phase ambiguity on all four of the $\mathbf E_{x,j}(\mathbf r)$, and correspondingly on the $\alpha_{x,j}(t)$. You're free to pick any phase that you find convenient, but you do need to choose one.

    And, moreover, the phase conventions that you choose for the $\mathrm{in}$ ports cannot be used to set those for the $\mathrm{out}$ ports, or vice versa, because they refer to completely different modes evaluated at different places. They're completely independent quantities.

Once you fix the total incoming energy flow per mode at $R$ (in joules per second), then the total energy flow can be shown to be both $$ F = R\sum_{j=1}^2 |\alpha_{\mathrm{in},j}(t)|^2 $$ and $$ F = R\sum_{j=1}^2 |\alpha_{\mathrm{out},j}(t)|^2, $$ and those two energy flows need to match, because of energy conservation. This means, therefore, that the expression of the output canonical variables as linear combinations of the input canonical variables, $$ \begin{pmatrix} \alpha_{\mathrm{out},1} \\ \alpha_{\mathrm{out},2} \end{pmatrix} = \begin{pmatrix} c & d \\ e & f \end{pmatrix} \begin{pmatrix} \alpha_{\mathrm{in},1} \\ \alpha_{\mathrm{in},2} \end{pmatrix} = M \begin{pmatrix} \alpha_{\mathrm{in},1} \\ \alpha_{\mathrm{in},2} \end{pmatrix}, $$ needs to preserve the norm $\sum_{j=1}^2 |\alpha_{x,j}(t)|^2$. Or, in other words, the matrix of the transformation needs to be unitary.

(Why unitary and not just having unit norm on each row or each column? Because the beam splitter needs to conserve $\sum_{j=1}^2 |\alpha_{x,j}(t)|^2$ both for cases where the input is on only one of the input ports, but also for cases where there's light on both. If the matrix isn't unitary, there will be a superposition of input beams that will have a different output energy than the sum of the inputs.)

All of the above is crucial for a correct hamiltonian classical-field-mechanical description of the beam splitter. Once you've done it correctly (but only after you've done the classical mechanics correctly), you're ready to move on to the quantum mechanics of the system, which is now very easy: since you've done the classical mechanics correctly, all you need to do is to replace the hamiltonian canonical variables with annihilation operators, $$ \alpha_{x,j}(t) \mapsto \hat{a}_{x,j}. $$ and you're done.

And, since you had a strict requirement of unitarity on the matrix that interlinked the hamiltonian canonical variables between the input and output sets, you have an identical requirement of unitarity on the matrix that links the input and output annihilation (and therefore creation) operators.

What you don't have, because you didn't have it on the classical fields, is any additional restriction on what the phases must be, either of $\pi/2$ or $\pi$ or anything, because (again) those are just doomed attempts at comparing phases on different places, which cannot be done to any absolute or objective standard.

  • $\begingroup$ Is my answer consistent or inconsistent with yours? I'm having trouble understanding some things. You say that the matrix needs to be unitary, but what exactly does that mean? (It doesn't look like any of these operators are equal to their hermitian conjugates). In my answer, I show that in order to make both "individual cases" and "superposition cases" - follow |T|^2+|R|^2 = 1 - that we must make $cos(i \theta) = 0$. It makes some sense to me that you can "rotate" matrices, but in this case our initial matrix is already not unitary (it's not equal to its hermitian conjugate). $\endgroup$ Commented Dec 17, 2018 at 20:05
  • $\begingroup$ Here you choose to transform your operator (by multiplying both sides by this transforming matrix) without applying a hermitian conjugate to one of the sides. $\endgroup$ Commented Dec 17, 2018 at 20:05
  • $\begingroup$ I'm familiar with both operators being used in quantum optics (and have heard that you can simply rotate to go from one to the other), but whenever words like "unitary" are used - they are very confusing to me in this context - and that's why I wrote my answer to try to resolve those questions. $\endgroup$ Commented Dec 17, 2018 at 20:07
  • $\begingroup$ @StevenSagona Your answer is somewhere between obfuscatory and dead wrong. But you're right that the unitarity requirement isn't always explained as well as it should be; see my expanded answer for the origins of the requirement. $\endgroup$ Commented Jan 1, 2019 at 1:11
  • $\begingroup$ As a technical comment - "in this case our initial matrix is already not unitary (it's not equal to its hermitian conjugate" - that's not what unitary means; you're confusing it with hermitian. Unitarity requires $U^\dagger = U^{-1}$, and both the initial and final matrix after the "rotation" are unitary. $\endgroup$ Commented Jan 1, 2019 at 1:18


The answer is $\pi/2$. Interestingly, it doesn't matter if it is plus or minus $\pi/2$ or if the phase change is in transmission or reflection in analysis of quantum or classical interferometry.

From the above referenced article:

"Quantum optics essentially provides black-box models of the beamsplitter. They all agree on the existence of a π/2 phase-shift, even if its sign and precise location (on the transmitted or reflected beams) are uncertain. Such inconsistencies, however, are not critical for what concerns the respect of energy conservation principle."


This is something that is often handwaved in quantum optics, and I think it's pretty confusing. Here is one explanation that is pretty simple but confusing. It says almost dogmatically that the beamsplitter operator has to be unitary. This is helpful to getting the "correct answer" but not helpful in developing an intuition for what is actually going on.

Here is what I think is an easy way to explain why it's $\pi/2$ and not $\pi$:

First let's write the outputs of the beamsplitter in the most general form: \begin{align} |\mathrm{out}_1\rangle & = Re^{i \theta_R}|\mathrm{input}_1\rangle + Te^{i \theta_T}|\mathrm{input}_2\rangle \\ |\mathrm{out}_2\rangle & = Te^{i \theta_T} |\mathrm{input}_1\rangle + Re^{i \theta_R}|\mathrm{input}_2\rangle \end{align}

An important point to note here is that I am saying that reflection from input 1 to output 1 is the same type of reflection as reflection from input to output 2. Typically in "intuitive answers" something is said about how one side of the beamsplitter has a different coating than the other, so one receives a phase shift and the other doesn't - but here we are NOT assuming that. We are assuming reflection on either side gives the same result.

\begin{align} |\mathrm{out}_1\rangle & = |R||\mathrm{input}_1\rangle + |T|e^{i( \theta_T - \theta_R)}|\mathrm{input}_2\rangle\\ |\mathrm{out}_2\rangle & = |T|e^{i( \theta_T - \theta_R)} |\mathrm{input}_1\rangle + |R||\mathrm{input}_2\rangle \end{align}

In a pretty standard way, we can remove a "global phase" of $e^i\theta_R$. I can explain this in more detail if anyone is confused by this step. Now let's find the sum of probabilities of both reflection and transmission:

$$|R|^2 + |T|^2 + 2|R||T|e^{i(\theta_T-\theta_R)} + 2|R||T|e^{-i(\theta_T-\theta_R)}.$$

Which, with a trig relation reduces to

$$|R|^2 + |T|^2 + |R||T| \cos(\theta_T-\theta_R) $$

Now here we have a problem. If you have the photon enter through just one of the two inputs, you will see that the total probability becomes $T^2 + R^2$. Since we know this probability should be $1$. So we are already constrained (by the single-input case) that $T^2 + R^2 = 1$. But when we send the photon as a superposition of both inputs, we identify that the total output probability becomes $$|R|^2 + |T|^2 + |R||T| \cos(\theta_T-\theta_R) = 1.$$ So now the only way this can happen is if $\cos(\theta_T-\theta_R) = 0$.

People throw around math equations and say unitarity a lot, but I think going through this example makes it much more clear what is going on. Our constraint of having probability = 1 for both individual and superposition cases require our phase to be $\pi/2$.

This is often why the beam splitter model is called "phenomenological", because it simply tries to fit the parameters to the data without reaching a mathematical contradiction. (As opposed to maybe trying to model the actual Hamiltonian of the material that is splitting the light)

  • $\begingroup$ Frankly, this answer is just a whole bunch of obfuscating algebra that does nothing but make it harder to understand what's going on. If you find yourself requiring quantum mechanics (instead of classical mechanics) or using explicit expressions for reflection and transmission coefficients, then you're already wrong. $\endgroup$ Commented Jan 1, 2019 at 1:01
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    $\begingroup$ And similarly, your conclusion ("it's pi/2 and not pi") is dead wrong - it's neither. If you want something concrete to pin the wrongness of your proof on, look to the phase ambiguity in the definition of your input and output kets. Those are arbitrary basis kets and their phases can be manipulated at will - which will directly impact your conclusions. $\endgroup$ Commented Jan 1, 2019 at 1:05
  • $\begingroup$ I believe the constraint I came up with is in terms of the relative phase ( $\theta_r - \theta_t$) and not the arbitrary individual phases, so I'm not sure how that resolved this difference. $\endgroup$ Commented Jan 3, 2019 at 12:14
  • $\begingroup$ As I said, your conclusion is dead wrong; it's on you to fix the arguments. I can point out the first mistake - it's the claim that your first equation is the most general form for that relationship. It isn't - there's no requirement that the phases on both lines be the same. Similarly, the step from eq. 1 to eq. 2 is wrong - you cannot redefine phases at will like that without explicit indication of what you're doing. By that point you've accumulated enough wrongness to completely sink your argument. $\endgroup$ Commented Jan 3, 2019 at 17:03
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    $\begingroup$ @ Steven You're splitting hairs, and you're continuing to miss the point. Your conclusion is wrong. You cannot fix the arguments that get you there because the goal is a false statement. $\endgroup$ Commented Jan 4, 2019 at 2:41

People get very excited when it turns about quantum. And generally one has the answer: "phase does not matter in quantum states", but it is not because that it does not matter that it doesn't exist! In the case of the beam splitter the phase is determined by the optical structure properties of the chosen BS. In the case of two dielectric layers of different indexes the Fresnel equations (esaily derived by Maxwell's equations) show (because of the index difference n1-n2 factoring the reflexion coefficient) thta reflected beams have opposite signs when incident on one side or the other of the BS. This is pure Optics and cannot be discarded.

  • $\begingroup$ The entirety of QC rests upon the manipulation of relative phase between qubit states... $\endgroup$ Commented Jun 22 at 18:00

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