# 'size' of a photon

What's the smallest aperture a photon can pass through? I mean with no transmission at all.

I'm pretty sure I saw long ago an experiment when they were reducing the size of a hole in a gold film and at about a quarter or a third of the wavelength there was no transmission.

If so, can we say that the size of the aperture can be associated with the dimension of the photon?

• There is no such size. A smaller aperture will merely reduce the transmission probability, but there is no known cutoff. Indeed, there is currently a lot of interest in deep sub-wavelength imaging techniques. A photon, by the way, is not an object. It's a quantum, i.e. the discrete unit of change of a quantum field. It does not have a size any more than the color red does. – CuriousOne Sep 11 '15 at 21:41
• But I guess at some point ($d \ll \lambda$) the far field transmission will be exponentially suppressed (could be wrong – this is just my first intuition). – Sebastian Riese Sep 11 '15 at 23:13
• Sebastian Riese's intuition is almost right, the classical theory predicts a suppression by $(r/\lambda)^4$. With small holes in metals films that theory will break down, though, because surface plasmons are not satisfying the assumption of perfectly absorptive or reflective conditions. See e.g. "Light Transmission via Subwavelength Apertures in Metallic Thin Films" by V. A. G. Rivera, F. A. Ferri, O. B. Silva, F. W. A. Sobreira and E. Marega Jr. for an introduction to the reality of light scattering in this regime. – CuriousOne Sep 11 '15 at 23:19
• More on size of photon. – Qmechanic Sep 13 '15 at 12:22

I should like to capture the excellent comment by user CuriousOne (emphasis mine):

There is no such size. A smaller aperture will merely reduce the transmission probability, but there is no known cutoff. Indeed, there is currently a lot of interest in deep sub-wavelength imaging techniques. A photon, by the way, is not an object. It's a quantum, i.e. the discrete unit of change of a quantum field. It does not have a size any more than the color red does.

So the only precise version of the OP's question is "what is the transmission probability through an aperture" and the answer to that question is that once the aperture is subwavelength, then the transmission probability drops off exponentially with the thickness of the screen the aperture pierces. As CuriousOne says, techniques such as scanning nearfield optical microscopy (SNOM) can image arbitrarily small features by gathering light through abitrarily small apertures: the catch is the total thickness represented by the aperture "screen" and standoff must be much less than the size of feature imaged. The essential problem is that electromagnetic fields couple through subwavelength holes through evanescent wave coupling. If the screen is of thickness $t$ and the size of the aperture $d$, the transmission probability is estimated by:

$$\begin{array}{lcl}p(t,\,d) &\propto &\exp\left(-4\,\pi\,z\,\sqrt{\frac{1}{d^2}-\frac{1}{\lambda^2}}\right)\\&\approx&\exp\left(-4\,\pi\,\frac{z}{d}\right);\quad d\ll\lambda\end{array}$$

This is a horrifically fast dropoff. If your screen thickness is$1{\rm \mu m}$ and the aperture $50{\rm nm}\ll\lambda$, the probability is of the order of $10^{-110}$.

See my answer to question a related idea: the limits on subwavelength imaging with light for further information.

Light is made up of photons which are elementary particles. The standard model of particle physics fits the data very well with the hypothesis that all elementary particles are point particles.

In an experiment where the size of a slit is varied and the cross-section for a photon to pass through the slit is measured two consideration should enter when comparing the probability of a photon to go through the slit:

$\bullet$ The wavelength of the photon which defines its momentum $\dfrac{h}{\lambda}$,

$\bullet$ The size of the slit .

This, before solving the quantum mechanical problem "photon +slit" scattering, has the rule of thumb of the Heisenberg Uncertainty principle :

\begin{align}\color{red}{\Delta \mathbf{x}\Delta \mathbf{p} \gt \frac{\hbar}{2}}\\ \color{red}{\Delta \mathbf{E}\Delta \mathbf{t} \gt \frac{\hbar}{2}}\end{align}

I'm pretty sure I saw long ago an experiment when they were reducing the size of a hole in a gold film and at about a quarter or a third of the wavelength there were no transmission.

Without a link it is hard to accept this as a fact. A quarter or a third of the HUP constraints would allow a quarter or a third of the incoming beam photons to go through, if one ignores the special surface of gold. If it is a fact that means that one needs a solution of the quantum mechanical set-up " photon + gold surface +hole " scattering, which will increase the probability of the photon to interact with the fields of the gold surface. Googling I found this, which is the way to approach the problem.

The photon as an elementary particle is a quantum mechanical entity, i.e. depending on the boundary conditions its location is either classical particle like or expressed as a wave like probability amplitude. If your memory is correct the solution is that in this experiment one is observing the probability-wave-like nature which should be very small for a single photon to pass the hole.

• I've removed the pic & wrote the relation using mathjax. You wouldn't mind, would you? – user36790 Sep 12 '15 at 6:17
• You seem to be considering the slit as a region of space as opposed to an object. There are distinct questions, "If I form a 1 µm wide slit from 1mm thick tin, what chance does a photon stand of passing through?" and "If I point a light source directly at an ideal detector, how reliably can I determine that a photon passed through this 1µm wide region?" The answers are different in part because passing "through slit" can also involve going through the surface of the metal. Even with no hole, photons will still go through. A subwavelength Swiss cheese texture will help evanescent propagation. – Blackbody Blacklight Sep 12 '15 at 8:47
• @BlackbodyBlacklight look at the answer by Savanah where he considers the thickness of the slit. I was answering in terms of the gold foil of the question, i.e. a very thin layer of atomic structure. A photon will go through a transparent to its frequency medium, it means there was no energy level to capture it on the path. – anna v Sep 12 '15 at 9:44

If we use the model of an electromagnetic wave, then like theses, a wave with a wave length of lambda need at least a hole with lambda / 2 to escape a cavity. So, the size of a photon could be proposed for a deeper study and experiences as the wave length of the photon divide by 2. But.... just as clue to start ! In fact, what is a photon ! A smily ? ...

• By the way there is a big problem. A hole in the cavity is surround of atoms... We are talking of a hole with a diameter less than 700 nano meter for the visible light, so we are in the field of quantum physique. This means that the hole itself doesn't allow us to make a measure of the size of the photon because the space there is not our «ordinary» space. – Pierre Ghislain Jun 18 '17 at 6:49

The simplest answer would be 1 ℓP, Planck length. Less than that means no aperture available. Unfortunately, it is difficult to imagine what an "aperture" is at that level - a "hole" "through" "what".... It might be thought of as a single spatial transaction across (or along) time. Bearing in mind, spatial is not really "spatial", it is planar.

## protected by Qmechanic♦Oct 8 '16 at 15:21

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