# Mach-Zehnder interferometer and superposition

I've a doubt on interpretation of superposition in the interferometer

Is it correct to say that in the Mach-Zehnder interferometer, the photon exists in a state of quantum superposition of the two paths and this means that, until a measurement is made, the photon is not in one of the two paths definitively, but in a combination of both, so it’s in a superposition of the two paths? Is it correct to say that superposition is a state in which a particle, such as a photon, simultaneously exists in multiple possible states until a measurement is made?

My doubt is how to physically interpret the superposition because these two expressions, "the photon passes through both branches" and "the photon passes through one branch or the other," should both be wrong. So, where does the photon go?

Since QM is probabilistic, you can only know probability until a measurement is performed. Your doubt can be referred to a "1-input" beam-splitter since you're dealing only the branches, without analyzing the whole interferometer.

Now, if you trust this probabilistic interpretation, we can say avoiding any math,

• that a photon may pass through all the branches (with probability determined by the beam-splitter itself, that can be represented with a reflection, $$r$$ and a transmission coefficient, $$t$$, not independent);
• that a photon has passed through a determined branch after you have measured it.

Math model. Let's recap this interpretation without dealing with Mach-Zehnder interferometer but just focusing on a beam-splitter without any detail about the EM field and phase, since your doubt is about the presence of the photon in one or the other branch.

Being the branch taken by the output photon a measurable physical quantity, it can be represented by an Hermitian operator $$\hat{B}$$ whose (orthonormal) eigenstates $$\{|\psi_i\rangle\}_{i=1,2}$$ represent the state of the system if the photon takes the $$i^{th}$$ branch, while the eigenvalues $$b_i$$ are the labels of the branches. The state of the photon after the 1-input beam-splitter can be thus written as a superposition of these eigenstates,

$$|\psi\rangle = c_1 |\psi\rangle_1 + c_2 |\psi\rangle_2 \ ,$$

being the coefficients $$c_i$$ determined by the very nature of the beam splitter (e.g. 1: transmission $$c_1 = t$$, 2: reflection, $$c_2 = r$$) and determining the probability for the photon taking the $$i^{th}$$ branch, $$p_i= c_i^* c_i = |c_i|^2$$. Normalization condition reads

$$1 = \langle \psi | \psi \rangle = \langle c_1 \psi_1 + c_2 \psi_2 | c_1 \psi_1 + c_2 \psi_2 \rangle = |c_1|^2 + |c_2|^2 = t^2 + r^2 \ ,$$

showing that transmission and reflection are not independent.

As an example, numbering branches with natural numbers $$\{b_1, b_2\} = \{0,1\}$$ the expected value of the branch is evaluated as

$$\langle \hat{B} \rangle = \langle \psi | \hat{B} | \psi \rangle = \dots = |c_1|^2 b_1 + |c_2|^2 b_2 = r^2 \ ,$$

with the choice done.

Remarks. For a non-ideal beam-splitter some photons may be absorbed, so absorption need to be considered as well in the transition from input to output.

The answer people give to this question depends on what interpretation of quantum theory they use.

Advocates of the Copenhagen and statistical interpretations will say you can talk about what predictions makes but not about what is happening in reality to bring them about. So those theories would say there is no answer to your question. Since this means there is no standard by which to judge whether the experiment was set up properly this seems unsatisfactory for both practical and theoretical reasons.

Now it is worth looking at what you have to explain to give an account of single photon Mach Zehnder Interferometer (MZI) experiments. If you measure whether there is a photon in each branch you always find a photon in only one branch at a given time. Suppose you adjust the interferometer so that the photon comes out of one port with probability 1. If you block one of the arms with an opaque object, when a photon isn't blocked it will come out of either port with probability 1/2. If you put transparent material in one or both branches you will change the probability of the photon coming out of one or another port. If you adjust the length of either path or both paths you will change the probabilities. So there is something in each arm of the interferometer that acts like a photon except that you don't detect it. This raises a problem. If a photon you don't detect is blocked by a detector, then it must be interacting with something that acts like a detector but you don't see the detector go off so there is an undetected detector. This raises questions like "who put the undetected detector there?" "Is it connected to an undetected computer that records results you haven't detected?" and so on.

What does quantum theory without any modifications such as collapse actually say? The evolution of the photon's direction is described by a Heisenberg picture observable $$\hat{D}(t)$$ which has two possible values: one for each direction. That observable picks up relative phases for each arm depending on what happens to photons in those arms. The result of the experiment depends on those relative phases. If you put a detector in one arm or the other that is described by a coupling to another system that copies information from the photon to the other system causing decoherence and suppressing interference:

https://arxiv.org/abs/1911.06282

The result is that reality as described by quantum theory on a macroscopic scale looks like a collection of parallel universes to a good but not perfect approximation (this is commonly called the many worlds interpretation):

https://arxiv.org/abs/1111.2189

https://arxiv.org/abs/quant-ph/0104033

Where does the photon go? It goes through both branches and can undergo interference using information about what is happening in both branches as a result, but because of decoherence you will only ever directly detect one version of it. For a popular account of the single photon MZI specifically see "The Fabric of Reality" by David Deutsch Chapter 9.

Ever since Schrodinger derived the wave function for electrons academics have been trying to apply the theory to photons as well ... it leads to a lot of irrational conclusions. Dirac, Feynman and many other scientists said the photon determines its own path ... this is the most insight that science has really provided for the photon's behaviour. Unfortunately these great scientists got drawn into the nuclear age ... the photon was left to conjecture by many academics.

A common sense explanation is that the EM field is always active, i.e. all the electrons in the apparatus are constantly feeling the forces of each other at the speed of light c. An excited electron in the source is already interacting with the whole apparatus, the photon is dumb, it merely travels a path that all the electrons dictate. It is known that an excited atom/molecule/electron can stay in the excited state for a period of time.

The Feynman path integral provides additional insight, by summing the E field of Maxwell's photon equation as a function of path length the result is that a photon's most probable path(s) is one where the path lengths are integer multiples of the lights wavelength .... i.e the paths that are resonant are preferred.

So it is the EM field that sees both branches .... the photon just takes one path.