What is the length of a photon?

Question

Some questions that look kind of similar have been asked before, and I find the answers quite confusing. I intend to ask this question in a way that clearly shows what I'm asking.

Imagine the following thought-experiment. (I don't know whether it's practical to actually do the experiment or not.)

Set up the usual double-slit experiment with a photon source that sends one photon at a time. But replace the double slits with a single pinhole. Over a period of time you should observe a diffraction pattern that looks something like this:

Now in front of the pinhole, put a metal disk with slots cut in it, that can spin.

When a slot is entirely in front of the pinhole, it should have very little effect on photons that go through the pinhole. Maybe some minimal diffraction.

When the solid disk is entirely in front of the pinhole, hardly any photons should get through. A few might take the long route around the solid metal disk and through a slot, or around the outside edge of the disk.

When the edge of a slot is close to the pinhole, it might create an observable diffraction pattern different from the open one. You can observe those effects by rotating the disk to various angles and waiting for the diffraction pattern to form.

When the disk spins, then sometimes photons can get through the pinhole and other times they can't. Most photons are independent of all of the others. (If our photon-emitter releases them at random times, very occasionally two might go through the pinhole at similar times.)

The faster the disk spins, the shorter the times that photons can get through. We can vary two things -- the rotation speed of the disk, and the "wavelength" of the photons.

My assumption from a naive reading, is that photons are point particles that travel at lightspeed. So no matter how fast the disk spins, it should have no effect on the diffraction pattern.

But classically, radiation came from an accelerated charge. A single charge with an oscillatory motion would produce a wave, and if it oscillated a thousand times then the wave it produced might continue for a thousand wavelengths.

What happens in reality? Does our single-photon source produce point-particle photons which are entirely unaffected by the spinning disk except when an edge is close to the pinhole? Or does it produce photons that have a length, that can be interrupted by the spinning disk? If they are each a thousand wavelengths long, then at some wavelength and at some rotation speed they will all be affected. If they are one wavelength long, similarly at some speed they will all be affected.

Maybe the reality simply does not match up to point-particle photons that have special properties which make them each behave statistically like a wave. And it doesn't match up to literal waves either. If these concepts are useful teaching tools that don't really fit, it would be interesting to get a clear idea what to replace them with.

Answer 1

A photon is not a point particle in the classical sense of the word (and neither is an electron, or any other fundamental 'particle'). Rather, it is a convenient word for describing some aspects of the electromagnetic field. It refers to the presence of energy in some particular 'mode' of the field. A mode can be thought of as an extended entity; it has a well-defined frequency and in free space it has a well-defined wavelength. It corresponds roughly to a plane wave.

In the type of diffraction /interference experiment in the question, the pattern observed on the screen is exactly as classical wave theory would predict, except that it appears as a set of dots rather than as completely continuous. However since the question concerns the pattern (i.e. the density distribution of the dots) we can use classical wave theory to answer the question.

The rings in the pattern occur at angles from the pinhole as given by diffraction theory. For a monochromatic source (without the chopping effect proposed in the question) the first minimum is at an angle given by $$ \theta \simeq 1.22 \lambda / a $$ where $\lambda$ is the wavelength and $a$ the radius of the pinhole.

If we now 'chop' the transmission through the pinhole, as proposed in the question, then the light emerging through the chopper is no longer monochromatic. It now has a range of frequencies whose spread $\Delta \nu$ is of the order of $$ \Delta \nu \simeq 1 / \Delta t $$ where $\Delta t$ is the time for which the pinhole is open. Consequently the light emerging from the chopped pinhole has a range of wavelengths. The range is given by $$ \Delta \lambda \simeq \Delta \nu \left| \frac{d\lambda}{d\nu} \right| = \frac{\lambda^2}{c} \Delta \nu = \frac{\lambda^2}{c \Delta t} $$ The effect of this range of wavelengths is to blur the interference pattern. The angle to the first minimum is proportional to the wavelength, so it gets blurred by $$ \Delta \theta = 1.22 \Delta \lambda / a . $$ It is useful to compare this blurring to the angle itself: $$ \frac{\Delta \theta}{\theta} = \frac{\Delta \lambda}{\lambda} = \frac{\lambda}{c \Delta t}. $$ Thus we find that when the pinhole is chopped so rapidly that it only allows through one wavelength at a time, then the diffraction pattern is completely blurred away.

To answer the overall question about "how long is a photon", the answer is that a truly monochromatic photon is infinitely long. That is, it is a way of referring to a state of excitation of a mode of infinite length and perfectly precise frequency. For more physically realistic cases the field excitation of not infinitely long; in this case one can imagine a pulse of light of some finite duration. More generally it is the coherence length that is the important quantity.

Answer 2

It's called the coherence length. It depends on how the photon was created. A photon from a short-lived excited state has a shorter coherence length than a photon (of the same wavelength) from a long lived state.

You can measure it: the easiest way to visualise this is through an interference/diffraction experiment a bit different from the one you describe. You take a standard Michelson interferometer with a non-laser source and set it up to show a fringe pattern. Now increase the length of one of the arms. You will still see fringes, according to whether the arriving waves are in phase or out of phase. But as you increase the length further the pattern fades and eventually disappears, at the point when the wave travelling the longer path arrives at the detector so late after the wave travelling the shorter path that the photon waves do not overlap. The extra path length when the pattern disappears is the length of the photon.

Answer 3

Consider the following:

There is no such thing as a photon 1000 wavelengths long, but there is such thing as an atom emitting 1000 photons, these are called quanta and are fundamental to photon/quantum theory, each one is unique and destined on its own path to be eventually absorbed by another atom/atoms/molecules. An atom takes time to emit a photon so a single atom can't overlap photons, but a laser with many excited atoms can have one "seed" photon create a cascade of many photons, these photons are separated spatially according to how the atoms are arranged.

Consider your wheel to have a pinhole instead of a slot and lets consider your wheel stopped but the pinholes slightly misaligned, obviously there will be attenuation but a few will make it thru both holes, the result is just another circular diffraction pattern as in your image but offset from your original one. If we average this over many revolutions of the wheel you are left with Gaussian (blob) distribution. You have effectively destroyed the single slit "interference" pattern! Your wheel is very similar to what has been attempted for the DSE, you are trying to determine path information ... which is just another way of saying that you're messing with the photons and altering their originally intended path.

Finally, if your wheel was really really fast you are NOT going to be able to change the photons or break them or alter their wavelength!, they will will just behave quantum mechanically or probabilistically .... some might get thru at speed V and even less at 2V but certainly none will get thru at speed c ....which is impossible for your wheel to go anyway.

Answer 4

Time and energy in quantum mechanics are conjugate parameters, linked by The Uncertainty Principle where: (delta t) x (delta E) is greater than or equal to Planck's Constant/4 Pi. My understanding is that...if you had a one photon 'at-a-time' source, the time location/coordinate of that one photon and its property of energy would be linked and bound by The Uncertainty Principle above. You can never be certain where that one photon is, or the exact energy that it has. So, I tend to think of a photon as yes a point particle, but with each photon somewhere 'Uncertain' within a 'packet' of say delta t in 'length', and with an amplitude energy of say delta E in the 3 spatial dimensions. One photon would illuminate one spot only on the screen. As the number of photons sent increases, The Uncertainty Principle tells us that they hit different places (coordinates) with different energies (properties). What we see then is a wave pattern of light and dark fringes. This actually shows a Probability Wave for where the photons are most likely to end up, or not. This is the meaning of 'wave-particle duality', for photons, or indeed for particles such as electrons in an electron diffraction tube where this duality may be easily demonstrated. The wave pattern and the intensity/energy of the fringes, diminishing from the centre outwards, mathematically is described by Schroedinger's second-order differential equation. Hope this is helpful.

Answer 5

Often, as in some answers here, the terms photon and electromagnetic wave are used as synonyms. Yet they mean very different things. An em wave has wavelength, frequency and coherence length and time. A photon just has energy, momentum and spin. It is a dimensionless point particle as far as we know. The wave describes the probability of finding photons with given energy, momentum and spin at at given place and time. It is similar to an electron wave function in this respect.

In your thought experiment the transmission is not only position dependent, as in the traditional slit, but also time dependent. Classically this means that your original monochromatic wave is amplitude modulated and now consists of a band of frequencies. In photon terms you will observe a distribution of photon energies, each following their own probability distribution or diffraction pattern. However, in practice such effects are unobservable since mechanical chopping cannot be done at optically relevant frequencies. You will just observe the same interference pattern as without chopping, but chopped.

Answer 6

We can determine photon duration given:

1. sunlight power flux of 1370 $W / m^{2}$
2. sunlight photon flux of $10^{18} / m^{2}$

Watts $\left( J/s \right)$ per photon of sunlight $= 1.37 \cdot 10^{15}$

Energy of green light photon (mid-point of sunlight spectrum, average energy) $= 4.3 \cdot 10^{-19} J \left( h \cdot f \right)$

Divide energy of photon by power of photon to determine time duration per photon $\left( J / W \right) = 3.13 \cdot 10^{-4} s$.

Multiply the time found above by $c$ to get length: 93,837.4 $m$ per photon.

This applies to all photons, not just visible light, since $h$ is in $J s/cyc$

Additionally, dividing $h$ by this time gives us the energy per cycle of field displacement