If I pass individual photons through a M-Z interferometer with equal arms I will observe interference (eg only one detector will respond). As I increase the path length of one arm I will observe the two detectors responding alternately as I pass through each phase cycle. Eventually I suspect that at a certain point, the interference will disappear and the two detectors will respond with equal probability. What determines this point and what does this tell us about the 'length' of an individual photon. What does QM predict when the path difference is greater than this?

  • 2
    $\begingroup$ A single photon doesn't have a coherence length. How would you measure a length from a single data point? $\endgroup$ – CuriousOne Jun 1 '16 at 10:08
  • 3
    $\begingroup$ You can nevertheless have a source which consistently outputs single photons in the same state, so the experiment as posed is reasonable. The answer, however, is "it depends on the photon". Some sources will produce longer photons and some sources will produce shorter ones. $\endgroup$ – Emilio Pisanty Jun 1 '16 at 10:10
  • 3
    $\begingroup$ To make this clear once more: "a photon" is a single measurement on a quantum field. Photons don't have "lengths". They have an energy and a momentum, though, and if the energy distribution of all photons from a light source shows that they all have a very similar energy, then we have a nearly monochromatic light source. If they all have nearly the same momentum, then we have a nearly parallel light source. What "coherence" measures is the properties of these distributions. A distribution, however, can not be understood by looking at a single of its points. $\endgroup$ – CuriousOne Jun 1 '16 at 10:32
  • 1
    $\begingroup$ Photon number states (e.g. one photon) have completely indeterminate phase. In that sense, the coherence length is zero. $\endgroup$ – garyp Jun 1 '16 at 10:51
  • 4
    $\begingroup$ @CuriousOne It's similar to asking "what's the standard deviation of a single statistical sample" makes perfect sense at one level. But I'd argue that a pure quantum number state can have a spectral / frequency / spread and all the rest of it in quantum superposition. You just need to make many measurements to glean it. So you could talk about measuring photons of a particular pure number state, just as you could talk about a person from a population whose height is known to be normally distributed with a precise mean and standard deviation. $\endgroup$ – WetSavannaAnimal Jun 1 '16 at 12:22

There isn't a straight answer to this question, which sheds light onto the meaning of the subtle word coherence, because what we tend to call "decoherence" can have two main roots.

A practical experimental meaning of the word coherence is "ability to show interference", and there are two ways whereby observable interference can disappear: (1) (energy) spectral spread within the pure quantum states in question; (2) genuine quantum decoherence.

Before we address these mechanisms, we need to be very clear on dispelling the notion of a "photon" as a little ball with schizophrenia and I'd urge you to read Daniel Sank's Most Wonderful Description Here and my commentary on his answer here: the only actors in the scene we'll discuss are (1) THE electromagnetic field and (2) the measuring instruments in your laboratory. We're talking about the electromagnetic field in a "one photon state" - this simply means the electromagnetic field has been raised one "notch" above its ground state and is undergoing unitary evolution such that statistics of the potential measurement events vary with time. These events are individual detections of an idealized photomultiplier tube, which tells you it has interacted with the EM field, and at the same time the EM field drops back to its ground state.

1. Energy Spectral Spread with Pure Quantum States

If this mechanism is what is causing your inability to see fringes, then the answer to your question is very simple. Turn the laser up and remove all attenuators so that you can see interference fringes, where they fade out and other "coherence / incoherence" effects. Now turn your laser down so that there is "one photon in the kit at one time" and take zillions of measurements of single detection events at all places in space. The probability to detect at a given point is exactly proportional to the intensity of the high power field you saw before turning the light level down. Notwithstanding apparent "incoherence" effects cause the "fading" or lack of fringe visibility in the high power picture, you'll still see "lack of fringe visibility" show up extremely precisely in your probability densities even though we are dealing with pure quantum light states.

Let's try to understand the underlying physics more deeply.

In this scenario we have a fantastically ideal laser that always raises the electromagnetic field to the same pure quantum state. We have an idealized attenuator gating off your experimental kit, so that, at random, unpredictable times (that arise following a Poisson process) the field in the interferometer gets raised to the same, pure number state, but there are (almost) never any field excitations above those immediately above the EM field ground state (i.e. the rate $\lambda$ events per unit time of the Poisson process, inversely proportional to the setting on your attenuator, is small compared to the rate at which your kit can measure). This situation is often colloquially described as "there's only one photon in the kit at a time".

But a one photon number state is not a simple two dimensional state. The subspace of one photon states is itself infinite dimensional: there are two basis one photon number states, one for each polarization, for every wavevector $\vec{k}\in\mathbb{R}^3$. These are the energy / momentum eigenstates of the one-photon subspace. Their amplitudes evolve with time by taking on a phase of the form $\exp\left(i\,\hbar\,\frac{|\vec{k}|}{c}\,t\right)$.

A one photon pure state can, like any quantum state, be in quantum superposition of the basis states. Now it is not emphasized enough that this includes one photon number states with a spread in energy. It is indeed possible to have "white" (broadband) perfectly coherent light, in the sense of its being in a pure quantum state (although it is very difficult to achieve very broadband pure states in practice). We'd write down our pure one photon state as something like this:

$$\int_{\vec{k}} \left(\psi(\vec{k})_+\,\left|\left.1\right>\right._{\vec{k}\,+} +\psi(\vec{k})_-\,\left|\left.1\right>\right._{\vec{k}\,-}\right)\,\mathrm{d}^3 k$$

or a field theorist might write it down more like:

$$\int_{\vec{k}} \left(\psi_+(\vec{k})\,a^\dagger_+(\vec{k})+\psi_-(\vec{k})\,a^\dagger_-(\vec{k})\right)\,\mathrm{d}^3 k\,\left|\left.0\right>\right.$$

i.e. as a superposition of creation operators acting on the unique EM field ground state $\left|\left.0\right>\right.$. Here the $\pm$ stand for left and right circular polarization states. $\psi_\pm(\vec{k})$ are the complex superposition co-efficients. Such a state has a definite spread in "frequency", but it's still a pure quantum state. It's simply now that energy / momentum gleaned from any measurement is uncertain. Let's write $\left|\left.\Psi\right>\right.$ as a shorthand for the above monstrous superpostiions.

Now, it can be shown that, given a pure one photon number state, you can write down the following quantities

$$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\Psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{B}}^+(\vec{r},t)\left|\left.\Psi\right>\right. \end{array}$$

for any one photon state and that they fulfill Maxwell's equations exactly. Conversely, for any properly normalized solution of Maxwell's equations, one can calculate a corresponding one-photon state $\left|\left.\Psi\right>\right.$ (i.e. the superposition co-efficients $\psi_\pm(\vec{k})$). Don't worry too much about the details: the key point is that a solution of Maxwell's equations, properly normalized, is the equivalent information to a one photon number state. It gets better. The solutions can be interpreted as probability amplitudes to make a detection with an idealized photomultiplier tube, because the time/space detection probability density is:

$$p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,|\vec{\phi}_E|^2 + \frac{1}{2\,\mu_0}\,|\vec{\phi}_B|^2$$

So, what you have to do to calculate this probability density for your kit is the following:

  1. Solve Maxwell's equations for your kit
  2. Normalize them so that the integrated energy density over all space is unity
  3. The energy density $\frac{1}{2}\,\left(\epsilon_0\,|\vec{E}|^2 + \mu_0\,|\vec{H}|^2\right)$ of the normalized solution is the required probability density function of space and time.

So now, with this Maxwellian picture in mind, you can see what's going on. If the pure EM field state has a spread of energies, the interference fringes in the Maxwell equation solutions will be at different places for different frequencies, because they correspond to fields with slightly different wavelengths. So you can get fringes "washing each other out" and giving apparent "incoherence" behavior even though the underlying quantum state is perfectly pure or "coherent". Even an atomic transition in cryogenically cooled atoms (so there is no Doppler-induced uncertainty in the transition energy) gives rise to a spread in photon energies. This is because the atom is coupled to a very wide range of frequencies in the coupled one-photon EM field number states. It "tries" to relax into them all equally, but destructive interference as it does so means that the overall amplitude for excitation of a given frequency is very low unless the frequency matches that of the perfectly sharp transition well. A perfectly sharp transition thus gives a nonzero linewidth Lorentzian spectral shape: indeed doing calculations to model the broadband coupling just spoken of foretells exactly the Lorentzian lineshape for the pure quantum light state. The linewidth is inversely proportional to the coupling strength. Strong coupling means that the destructive interference I just spoke of acts more swiftly and strongly throughout the relaxation process.

2. Genuine Quantum Decoherence

In this case, before detection events, you can't think of the system as the evolution of the EM field alone. The quantum state space is now the tensor product $\mathscr{E}\otimes\mathscr{K}$ of the EM field one-photon state space $\mathscr{E}$ and that $\mathscr{K}$ of the experimental kit. Lasers' atoms get bumped around by their neighbors in thermalized systems. Cavities get vibrated. Optical tables get bumped by coffee cups being put down on them or shaken by the traffic on the freeway outside your lab. The light state becomes entangled with the hugely complicated state of the experimental kit.

In this case, if you try to look at the state of the light alone in $\mathscr{E}$, you're looking at a reduced quantum state and this appears as a classical mixture of pure quantum states. It is modelled by the density matrix formalism. (Conversely and interestingly, such a classical mixture can always be thought of as a reduced pure quantum state in a superset quantum state space: see the notion of quantum purification for more details). But the density matrix formalism is equivalent to the incoherent adding of the effects of the constituent pure states, i.e. you can do the Maxwellian analysis above for each pure state in the mixture and them sum the classical probability densities weighted by the probabilities to be in each state in the mixture.

This is genuine quantum decoherence (of the light state) and it is hard to tell what exactly the experimental outcome will be. If the whole experimental kit is genuinely time invariant, so that the probabilities of each pure state in the mixture are constant, then the answer to your question will be exactly the same as it was when we talked about energy spectral spread in pure quantum states: the probabilities of single detection events at low light levels will reflect the high light level intensities precisely. But in practice this time invariance is very difficult to achieve. You'll find that the notion of coherence length is very, very hard to measure rigorously, because the result will be highly dependent on the integration times, frequency responses, surface areas and so of in your detectors and processing electronics. You'll get a different (probably shorter) coherence length with increasing integration time, but this rule of thumb breaks down when you measure so fast that you begin to approach the "one photon in the kit at a time" régime.

  • $\begingroup$ Thank you for this detailed answer. I am not qualified to comment on the mathematics but would I be right in summarizing what you have said by saying that the point at which interference disappears is determined by the spread of photon energies which the laser produces dE and that this path difference will therefore be approximately equal to hc/dE? I understand that the experiment says nothing about the 'coherence length' of a single photon but it does say something about the coherence of the light source. $\endgroup$ – J Oliver Linton Jun 3 '16 at 10:41
  • $\begingroup$ @JOliverLinton Yes it is quite meaningful to talk of the coherence length of the light source AND it is quite meaningful to talk of the coherence length of the pure quantum state that has the spread of energies in the first place. But, as CuriousOne emphasized, one photon detection cannot tell you the parameters of that state - you need many for that. From an experimentalist standpoint a quantum state is a property of an ensemble, and you can only get confidence intervals on the parameters of that state that tighten as the number of measurements increases ...... $\endgroup$ – WetSavannaAnimal Jun 3 '16 at 11:34
  • $\begingroup$ @JOliverLinton .... $h\,c/\Delta E = \lambda^2/(\Delta\lambda)$ seems a good rough estimate for coherence length since it is the condition that this is the length $\Delta z$ needed for two waves of wavelengths $\lambda +\Delta\lambda/2$ and $\lambda -\Delta\lambda/2$ to get one cycle out of step - or so I've just worked out. This is how you've derived this formula, no? In practice I'd be looking more carefully at the laser spectrum and working out precise equations for what fringe visibility I ought to see where, which is why I don't tend to carry formulas like that one around in my head. $\endgroup$ – WetSavannaAnimal Jun 3 '16 at 11:38
  • $\begingroup$ @JOliverLinton The second part of my answer is about the light getting entangled with the kit - genuine quantum decoherence - and this is very hard to quantify experimentally: you find you'll get all kinds of nonrepeatabilities when you try to quantify things like a source's coherence length rigorously and it depends crucially on the exact experimental setup. $\endgroup$ – WetSavannaAnimal Jun 3 '16 at 11:40
  • $\begingroup$ I may not be able to understand the maths but I think I understand the physics a lot better now. If we were to construct laser sources with smaller and smaller dE we would be able to demonstrate intrference over longer and longer path lengths - but at the same time by the HUP we would have to wait longer and longer for the photons to pass through the apparatus. If we had a source with _dE_=0 we would have to wait an infinite time before detecting a single photon! Am I right? $\endgroup$ – J Oliver Linton Jun 3 '16 at 12:14

If you go to the wiki page for coherence length , the following is found as introduction:

In physics, coherence length is the propagation distance over which a coherent wave (e.g. an electromagnetic wave) maintains a specified degree of coherence. Wave interference is strong when the paths taken by all of the interfering waves differ by less than the coherence length. A wave with a longer coherence length is closer to a perfect sinusoidal wave. Coherence length is important in holography and telecommunications engineering.

This article focuses on the coherence of classical electromagnetic fields. In quantum mechanics, there is a mathematically analogous concept of the quantum coherence length of a wave function.

Now you ask:

Coherence length of a single photon

A photon as a quantum mechanical entity will fall into the quantum coherence length of the wavefunction which is a different concept than for classical waves, and is what the comments have been trying to elucidate.

The photon as measured in an experiment has one point of interaction, and the information it carries is only of its fourvector, energy and direction, and its spin which is either +1 or -1 . If it hits a grating, even one photon will fall on the appropriate "color" and its energy will be known, and one point on a screen or ccd will give its direction. If a magnetic field is used, the up or down of the spin may be also measured. That is all for an individual photon.

The wavefunction used for describing photons has sinusoidal properties, but the wavefunction complex conjugate squared is what can be measured, and it is a probability distribution.. By construction a probability distribution becomes manifest by many samples, not by one measurement. Any wavelength built in the probability distribution will only appear when a number of samples is taken, not from an individual measurement. Which is what the commenters were trying to clarify.

In conclusion, a single photon cannot have a coherence length. The probability of its manifesting has a coherence length and it is an attribute of collective behavior and can be measured by an ensemble of photons. The classical coherence length will also emerge in the behavior of the ensemble of photons, so, once measured, one can know how probable it is for a photon to manifest at that (x,y,z) in the experiment.

Photons in ensembles build up the classical electromagnetic wave which will have its coherence length . I have found this image of how polarization is built up by individual photons that just have +1 or -1 spin useful in getting an intuition of how the classical wave is built up.


Left and right handed circular polarization, and their associate angular momenta. A QED explanation of how individual photons build up a classical wave can be found here.

  • $\begingroup$ Thank you for these comments but can I point out that I did not ask 'what is the coherence length of a single photon?' but 'what determines the point at which interference disappears when carrying out an interference experiment with individual photons.' I suspect that the answer lies in the fact that no practical source of individual photons can produce multiple photons of precisely the same energy. Can anyone elucidate this? $\endgroup$ – J Oliver Linton Jun 2 '16 at 8:19
  • $\begingroup$ That point is the same as the classical point of disappearing interference. The fact that mathematically the classical beam emerges from the quantum state assures this. A quantum mechanical answer would be in terms of probability. The( secondary) probability of seeing interference in the probability distribution of a photon would be compatible to zero, i.e. the probability distribution would compute to flat beyond the classical coherence length. btw you should change the title. it is asking for "the coherence length of a single photon". $\endgroup$ – anna v Jun 2 '16 at 8:40

CuriousOne makes a good point in comments and I will elaborate on that. While you are asking about the coherence length of a single photon, in the experiment that you describe you will have to detect many photons to judge if they are detected in both arms with equal probability. Where do this photons come from?

If you assume that the photons are identical, then you would need some ideal light source with infinite coherence length. If you take a real light source, the difference between individual photons will define your decoherence length.

You can see, then, that what you are actually measuring with your experiment is the coherence length of the light source. A single photon does not have such property.

  • $\begingroup$ I think you are on to something here. So what determines the 'coherence length' of a light source? And if we establish (by my proposed experiment) that the 'coherence length' of a particular light source is,say 30 cm, why am I not allowed to say that the photons it produces have an average coherence length of 30 cm? $\endgroup$ – J Oliver Linton Jun 2 '16 at 8:31
  • $\begingroup$ Coherence length of a light source is determined by it's ability to emit photons of similar wavelength and phase. A light bulb will have short coherence length compared to a laser. However, one thing that coherence depend on is the solid viewing angle of the light source: the smaller it is, the larger coherent length you will observe, so star light appears much more coherent that light from the Sun. Solid viewing angle of a star can be determined by measuring it's coherence length with an interferometer. $\endgroup$ – gigacyan Jun 2 '16 at 11:06
  • $\begingroup$ If coherence length depends on the distance from the light source, then you cannot assign a constant coherence length to each emitted photon. $\endgroup$ – gigacyan Jun 2 '16 at 11:09
  • $\begingroup$ That's an interesting point about solid viewing angle - perhaps you could provide a reference? - but I am not sure how relevant it is to my hypothetical experiment. Since only single photons are involved surely the solid viewing angle is zero? $\endgroup$ – J Oliver Linton Jun 2 '16 at 11:19
  • $\begingroup$ But you need many photons to judge if there is coherence, and if they all come from a small solid angle, their wavefronts are more parallel. Here is one reference: books.google.ch/… $\endgroup$ – gigacyan Jun 2 '16 at 12:48

Now that discussion of this topic has ceased, may I offer my own simplified answer based on what I have learned?

Firstly, no source of light can produce identical photons with exactly the same energies and direction. If the spread of energies is ΔE then the spread of wavelengths Δλ = ΔE λ2/hc. The maximum number of wavelengths of path difference before interference effects disappear will be n = λ/Δλ and the maximum path difference will be = λ2/Δλ = hc/ΔE. For a sodium atom at modest temperatures I calculate this to be of the order of 30 cm which seems to be about right. Additional effects may conspire to reduce this apparent coherence length even further.

Secondly, it basically doesn't matter how weak the light source is; when you average the results over many photons you will see the same interference pattern. In other words, what we are measuring is the coherence length of the source of light, not the 'length' of the individual photons which it produces.


By length maybe you mean wavelength. A single photon traveling at the speed of light and oscillating at a certain frequency will oscillate through one cycle every wavelength or say 500 nm. As you increase the length of one arm of the experiment the interference will go in and out of phase every one half cycle or every 250 nm.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.