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I don't mean the wavelength, I mean the full length in the direction of propagation. What is the total extent of a single photon in space? Does it even have a length?

Edit to clarify: Let's say a have a polar molecule, e.g. water. A standing EM wave (in this case our photon) will cause the molecule to oscillate (this is how a microwave works). So with a single photon forming the standing wave, how many water molecules can I put in a line, such that the same photon will be affecting them all at the same time? I'm not sure if that even makes sense, but that's what I'm thinking of when I refer to length.

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    $\begingroup$ Depends on the photon. It’s like asking what the width of an ordinary particle’s wavefunction is. Sometimes it’s bigger and sometimes it’s smaller. $\endgroup$
    – knzhou
    Apr 18 '19 at 21:35
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    $\begingroup$ However, photons with a definite wavelength always have infinite spatial extent. $\endgroup$
    – knzhou
    Apr 18 '19 at 21:36
  • $\begingroup$ archive.briankoberlein.com/2015/04/14/… $\endgroup$
    – mavzolej
    Apr 18 '19 at 21:37
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    $\begingroup$ @safesphere Sure, that's another valid way to interpret the OP's question. But both ways are commonly used, so I wouldn't agree that I was confused. $\endgroup$
    – knzhou
    Apr 18 '19 at 22:21
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    $\begingroup$ @knzhou Sorry, a poor choice of words. I ment you were confusing (others), but not confused (yourself). I know well that you are one of a few top experts on this site who both knows and understands physics. $\endgroup$
    – safesphere
    Apr 19 '19 at 0:01
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As knzhou says in a comment:

Depends on the photon. It’s like asking what the width of an ordinary particle’s wavefunction is. Sometimes it’s bigger and sometimes it’s smaller.

Some sources, such as lasers, emit photons with very long coherence lengths. For instance, the laser in your CD player probably emits wavetrains with lengths of ~10 cm.

Other sources emit photons with much shorter wavetrains. For example, if you look at thin film interference patterns made by light from a sodium discharge tube, the patterns never have more than ~100 fringes, which is because the sodium atoms emit wavetrains with lengths of ~100 wavelengths. This is determined by the properties of the atom. For example, if the half-life of the transition is ~100 periods of the light wave, then the wavetrain will have a length of $\lesssim100$ wavelengths.

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    $\begingroup$ He is asking about a single photon, not a wavetrain. $\endgroup$ Apr 18 '19 at 21:58
  • $\begingroup$ Could the photon wave function for a 100 sodium wavetrain be broken down into 100 photon wave functions? Would each of the 100 wave functions have its own length from the emitting atom to the eventual receiving atom? I think the photon has no length because it is a detection but a photon wave function has a length? $\endgroup$ Apr 18 '19 at 22:40
  • $\begingroup$ @PhysicsDave for a discussion of the wavefunction of a photon see physics.stackexchange.com/questions/437/… $\endgroup$ Apr 19 '19 at 0:18
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    $\begingroup$ Yes it's a good article. We need point photons to explain energy/momentum transfer, wave photons to explain lasers and YDSE etc , and photon wave functions to explain quantum eraser and Mach-Zehnder like experiments. $\endgroup$ Apr 19 '19 at 18:13
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The answer by @BenCrowell is closest to correct. The OP clearly asks for the length of a photon in the direction of propagation. As @safesphere commented, a single photon detection has no spatial extent (at least under ordinary circumstances). However, the photon's wave nature is relatively easy to demonstrate, as in single-photon interference.

An analogous experiment can be set up to demonstrate the spatial extent of a photon, both in the direction of propagation and in the directions at right angles to the direction of propagation: single-photon interference in a Michaelson interferometer, for example. The pathlength difference at which fringe contrast drops to (and stays at) zero is equal to the coherence length of the photons.

A really graphic demonstration of the 3D extent of a photon can be done by setting up a "classical" single-photon 3D holography system, to make a hologram of a large 3D scene of known geometry, so that the path length difference for object and reference beam is known at a given point on the recording film, for every point in the scene. Looking at the scene through the developed hologram at the given point amounts to a map of the spatial extent of the (identical) photons that recorded the hologram.

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I want to extend knzhou's comment. In the quantum theory of the electromagnetic (EM) field the number of "photons" is the number of excitations of the EM field with a given wavevector $\mathbf k$ and a given polarization $\lambda$. In classical EM theory, a wave with definite $\mathbf k$ and $\lambda$ is a plane wave, which has infinite extent to begin with (so it would influence all the water molecules). Of course we may build a localized wave, but in order to do this we'll have to add many plane waves and therefore $\mathbf k$ will assume a range of values. So a standing wave does not have a definite $\mathbf k$ since it's a sum of two plane waves travelling in different directions! We already see that the notion of "extent" of a wave with definite values of $\mathbf k, \lambda$ is ill-defined.

Things carry out similarly in the quantum theory of the EM field, but some subtleties kick in. The first thing that changes is how we describe the field: we use an occupation number ket $|n_{\mathbf k, \lambda}\rangle$ that tells us how many photons (excitations) there are with a given $\mathbf k$ and $\lambda$ and the fields become field-operators that act on the ket. You may ask: "if the fields are operators now, how do we associate them with the classical fields (which are numbers)?". Well, we take the expectation value $$ \langle n_{\mathbf k, \lambda}| \mathbf E |n_{\mathbf k, \lambda}\rangle $$ which, if you carry out the calculations, will be zero for a state with a definite number of photons (but $\langle \mathbf E^2 \rangle$ will be infinite!). It'll be non-zero if our field-state is a superposition of different occupation numbers, for instance: $$ |\psi\rangle = c_0|0_{\mathbf k, \lambda}\rangle + c_1|1_{\mathbf k, \lambda}\rangle. $$ Indeed, the state that resembles a classical coherent wave is called a coherent state, which is a superposition of infinite photon number states. So even our notion of electric field depends on having an uncertainty in the number of photons.

So, in your example which water molecule would a single photon influence? The answer is still all of them, but only one will be excited by the photon! All molecules will feel the EM field of a single photon, but by conservation of energy only one can absorb the photon and excite. It's interesting to note that even in the absence of photons the field can influence the molecules and de-excite them, making them emit more photons.

In conclusion, both classical and quantum treatments of the EM field will not have a well-defined "length" of a wave/photon with definite $\mathbf k, \lambda$, and if you really want a "reasonable" length you'll have to give up the definiteness of $\mathbf k$, and hence have more than one photon.

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According to our currently accepted theory, the Standard Model, photons are elementary particles.

This means that they do not have a spatial extension or an internal structure. We call them point particles.

That is why we cannot talk about the size of a single photon.

We would need a Theory Of Everything where elementary particles are made up of strings to talk about such a size.

But in the currently accepted theory, photons are point particles.

Though, sometimes we talk about the fact that photons do have wave characteristics, and they do have certain wavelengths where we could talk about the minimal size like Compton. Or we can talk about the Planck length, as a minimal size that is important in QM.

But in reality, the photon is a point particle.

please see here:

Why doesn't De Broglie's wave equation work for photons?

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    $\begingroup$ This is wrong. The OP is asking about length in a sense that makes photons have a finite length. $\endgroup$
    – user4552
    Apr 18 '19 at 21:50
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    $\begingroup$ I think we really really need the OP to clarify the question. $\endgroup$
    – WillO
    Apr 19 '19 at 0:13
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    $\begingroup$ @ZeroTheHero A single photon doesn't exhibit wave properties. If you don't detect it, you have nothing. If you do, you have a point particle. Wave properties are statistical and show up only when you have many photons, even if one at a time. The statistics of one photon (or of one of anything) is undefined. E.g. you cannot predict where it hits the screen. I don't mean that a quantum wave is made up of many photons, as this would be a misconception, but wave properties of one single photon cannot be observed in principle. Perhaps a good answer should elaborate on the wave-particle dualism. $\endgroup$
    – safesphere
    Apr 19 '19 at 3:39
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    $\begingroup$ @safesphere given that a single photon can interfere with itself (a classic undergraduate experiment see www2.optics.rochester.edu/workgroups/lukishova/QuantumOpticsLab/…) it is difficult to see how one can argue a photon doesn’t exhibit wave properties. We must be talking at cross-purposes... $\endgroup$ Apr 19 '19 at 11:16
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    $\begingroup$ @ZeroTheHero The experiment in your link used many photons, not just one. Yes, the photon doesn't interfere with anything, but itself. However, ironically, to prove this, you need many photons. A single photon does not produce interference lines on the screen. It produces a single dot, because it is a point-like particle. What has a spatial extension is the probability of finding the photon. However, probability is a statistical concept not applicable to a single event. The physical observable of a single photon is a single point-like dot on the screen. Is this horse dead enough yet? :) $\endgroup$
    – safesphere
    Apr 19 '19 at 16:02
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If you are talking about the spatial extension of a photon it is zero. Photons are elementary particles which have no size, in an atom size is created be a relationship between the strong force and electromagnetic force. But elementary particles ie. Photons, quarks are sizeless.

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You can only answer this question if you find an isolated photon source that spontaneously generates exactly ONE photon and is not disturbed. One of the few examples is the famous hydrogen line with a wavelength of 21 cm. The remarkably low transition probability A21 = 2.85e-15 1/s results in an extreme frequency accuracy. With this value and with Heisenberg's uncertainty principle, a coherence length of very many kilometers can be calculated. Experiments with many photons or even lasers are unsuitable for determining the properties of individual photons.

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  • $\begingroup$ This doesn't really get at the point of the question - supposing one did have a (single) photon, what does its spatial distribution look like? $\endgroup$ Jan 3 at 15:20

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