# What happens when a photon hits a beamsplitter?

Yesterday I read that we can affect the path and the 'form' (particle or wave) of a photon after the fact (Wheeler's delayed choice experiment). Part of what is puzzling me is the beam-splitter. Are the individual photons actually being split into two new photons of lesser energy?

This question implies that you cannot split a photon but it seems that beam splitters do exactly that.

The crucial word is "beam", in "beam splitter". Beam means an ensemble, in contrast to "photon" which is an individual particle.

A light beam is an ensemble of photons and if it is of a single frequency $\nu$, all photons have energy $E= h*\nu$. A light beam can be split in a beam spliter, i.e. the ensemble of photons can be split into two streams of photons: the intensity of the beam goes down, but the individual photons still have frequency $h*\nu$.

Now one can think of impinging photons one by one on a beam splitter. A photon is described by a wavefunction which when squared will give the probability of finding the photon in a particular (x,y,z). It will go either where one stream went or the other according to the probabilities, but it will be seen as a whole photon of energy $E=h*\nu$.

• Is this really true? I don't really study quantum light (we use classical models), but a good question is :Are the Fresnel equations valid for single photons? If so, where are the references for the proof? I always see these QM experiments where the assumption is made that both polarization optics and normal beam splitters work exactly the same for single photons...I'm doubtful of these claims, especially for polarization optics since the phenomena is a result of large scale wave interference (i.e. acting on the build up of coherent states, NOT single photons) – daaxix Mar 30 '14 at 7:14
• I'm not suggesting that a beam splitter will have no effect on single photons, I'm mainly questioning the probability assumptions going from the large scale (classical , build up of coherent states) regime to the single photon regime. – daaxix Mar 30 '14 at 7:23
• If you really want to see how the classical electromagnetic wave emerges from the ensemble of photons you need some background in quantum field theory and its mathematics. Here is a link that shows this motls.blogspot.com/2011/11/… the photons add up coherently to build up the phases . for visualization, think of a wave of people , as in a stadium. You could split the wave, but this would mean re-directing the phases of movement of individual people. People would not be split – anna v Mar 30 '14 at 8:35
• not a bad reference, thanks. I've been introduced to ladder operators, and coherent states (by no means well versed though). My point is, many optical elements are designed to work on the ensemble not on individual photons, i.e., the polarization and beam splitting effects are from the superposition of many coherent states, so is treating the polarization for a single photon in the same way valid? This is like saying that a single photons wavefunction is equivalent to a superposition of many photons wavefunctions...if it is true, do you have any references? – daaxix Mar 30 '14 at 17:39
• look a the answer by StickyCube . A photon is an elementary particle as much as an electron. It has a probability of going left or right in a splitter and a corresponding probability to line up in spin in such a way that the ensemble displays polarization; this for example is a reference I found in a similar question here of how one could have a wavefunction for a photon onlinelibrary.wiley.com/doi/10.1029/JB078i017p03356/… – anna v Mar 30 '14 at 19:19

A single photon is a quantised packet of Electromagnetic energy, the smallest indivisible unit imposed by boundary conditions according to quantum mechanics. In this regime I find it easier to think of the photon as a particle with a 'polarization' degree of freedom which can be horizontal $\left|H\right>$,vertical $\left|V\right>$ or in any linear superposition of the two.

When the photon meets a beam splitter it acts like a quantum particle and takes both paths with some probabilities. Much like the electron taking both paths in Young's famous double slit experiment.

It is not the actual photon being split into two new ones, only the 'particle-like' position wavefunction holds the information about where the photon is. Quantum mechanically, we can treat photons exactly like any other bosonic particle.

• "takes both paths with probabilities determined by the polarization" this is only true for a polarizing beamsplitter. A regular beamsplitter doesn't care about the polarization. – glS Dec 15 '17 at 15:36

Very short and "axiomatic" answer: You indeed can "split" one particle. In QM particles are treated as a "wave functions", maybe it will be more easy for you to imagine a splitting wave. However, only at the point when the photon is detected the particle is measured in one point of space. This is the very foundation of QM and I agree that it's hard to grasp the concept.

• I would define "splitting a particle" to mean that the energy of the particle was split into two pieces. When you fire a single photon at a beam splitter, there's no evidence that this sort of splitting happens. A beam splitter doesn't split an incident photon this way, but rather it splits the wavefunction giving two alternatives which can interfere with each other. – twistor59 May 30 '13 at 19:08
• Jan, yes this is a difficult one to wrap my head around. Let me ask a different but related question. If the individual photon is split and then later detected will it have the same energy as when it left the emitter? – CramerTV May 30 '13 at 19:42

Photon is quantum. Experiences with photons are quantum experiences. In quantum experiences, you are calculating transitions probabilities .

To calculate transitions probabilities $p$, you will have to use quantum complex amplitudes probabilities $A$, with $p$ = $|A|$

To calculate transition amplitudes $A$, you will have to consider all the paths that are available to the photon field, and sum other them :

$A$ = $\sum_{paths} A_{path} = \sum_{paths} e^{- i S_{path}}$,

where $S_{path}$ is the action of the photon on the considered path.

So, you will have to get rid of the classical view of the particle, localized in space-time, indivisible, and so on.

You are now in the quantum world, which has its own rules.

In short, no, the individual photons are not actually being split into two new photons of lesser energy. It is true that in the classical interpretation of light, an ensemble of many distinct waves of photons, will create destructive and constructive interference patterns ( e.g. Young's Double Slit Experiment ). It is even more intriguing that these constructive and destructive interference effects are present even in the case of a single photon.

As an aside, when we mention photons, we mean the abstract idea that includes the wave/particle features inherent to an unobserved/detected electromagnetic wave. This is to say that until we make a measurement ( think delayed-choice ), these photons can and will behave as both.

• Light of a given frequency (wavelength) still comes in discrete energy units of $E = hc/\lambda$ so the question is still interesting when you have a single discrete unit of light. I don't think your answer actually addresses the question. – Brandon Enright Nov 2 '14 at 20:37