# What is the spatial extent of single photon?

By the uncertainty principle particles cannot be specified in space and momentum simultaneously in the copenhagen interpretation of quantum mechanics.

If photons are moving with c in every posible frame what is the spatial extent of a photon? Does it mean it is everywhere in all the space?

Is that same as it exists forever even before is generated?

But in interference experiment photons are required to be at same time at same position to interfere.

• My apologies on my comment and close vote, which I have retracted – user108787 Sep 8 '16 at 18:13
• Maybe an answer based on QFT considerations would be helpful for the conversation here. – Constantine Black Sep 9 '16 at 8:49
• More on size of photon. – Qmechanic Sep 9 '16 at 14:14

Yes actually. A photon of a given energy isn't a particle in physical space, it is a particle in momentum space which means that it has a definite, exact momentum in the same way that a particle in physical space has a definite, exact position. The excitation for such a particle can be written as:

$$e^{i\mathbf{k} \cdot x -i\omega t}$$

where $\bf{k}$ is the wavevector and $\omega$ is the angular frequency. Such a photon has energy $\hbar\omega$ and momentum $\hbar \bf{k}$. It can be seen from inspection that the spatial extent of such a "particle" is all of space. When we do quantum field theory, we usually transform to a momentum representation because the fields mostly decouple in this representation and become much more mathematically manageable.

Since these excitations have to exist everywhere in the universe, they would have to be sourced by a process emitting for an infinite period of time. In reality this doesn't happen and all "photons" in our universe are wave-packets sourced by moving charges. Such a wave-packet can be written as the integral over a density of modes:

$$\int{\hat{\rho}(\mathbf{k})}e^{i\mathbf{k} \cdot x -i\omega t} ~\mathrm d\mathbf{k}$$

• The fourier transform and the wave packet implies k space distribution . Would that mean that single photon have energy distributions and are not just single energy let say 1eV ? – Anonymous Sep 8 '16 at 20:43
• However, a superposition of one-photon states with definite momentum is still a one-photon state, so we should be careful not to imply that a photon must have a definite momentum. – Brian Bi Sep 8 '16 at 21:30

A photon's momentum is calculated by $$p = \frac{h \nu}{c}$$ which is different to how you calculate momentum for particles with mass.

This means that a photon can have a finite spatial extent, it just must have an uncertainty associated with its frequency.

Your thought that a photon with no momentum uncertainty must have infinite position uncertainty is correct. However, you could also say that a photon with infinite frequency uncertainty has no position uncertainty.

But in interference experiment photons are required to be at same time at same position to interfere.

This is wrong.

There exist single photon interference experiments where an interference pattern appears in the accumulation of photons.

The photon is a quantum mechanical entity described by a wavefunction, and the probability of finding the photon when scattered by the two slits has the interference pattern, even one photon at a time , builds up the interference pattern of the light wave that appears from a confluence of innumerable photons. Their wavefunctions superpose and the classical beam and classical interference pattern appears.

• Anna, could you please explain your position on this? Sometimes you set others straight by explaining that the wave function and probabilities are just tools for calculating and should not be considered physical things. I agree with that but what are you saying up above? How would you physically describe the scattering of these photons? How do they physically make up this interference pattern? – Bill Alsept Sep 9 '16 at 4:13
• @BillAlsept they make it up by an accumulation of points from the probability distribution of the experimental set up. If you throw a dice, how does it make a probability distribution? – anna v Sep 9 '16 at 4:17
• That still does not answer what is physically happening. If I throw a dice a number of times it only makes probability distributions because I physically write down the data. In the same way a fringe pattern on a screen is there because something physically is happening. – Bill Alsept Sep 9 '16 at 4:25
• @BillAlsept That is quantum mechanics, one has to develop a new intuition and work only with probabilities. The physical is the measurement, the interaction. The in between we have found cannot be analyzed with classical tools. Only with the probabilistic method of quantum mechanics can we have predictive models for the microcosm. We can predict probabilities only – anna v Sep 9 '16 at 4:31
• So if I see a fringe pattern on the screen is that considered measuring it? Because it is there. The pattern was physically caused by something. I can easily describe how photons could physically form a pattern on a screen. Why is it that only probabilities are excepted and physical explanation are not excepted? – Bill Alsept Sep 9 '16 at 4:49

By the uncertainty principle particles cannot be specified in space and momentum simultaneously in the copenhagen interpretation of quantum mechanics.

If something can't be specified it does not mean that it has not exact values. A photon is a undivisible unit moving with c and it is following its geodesic path.

A bit more difficult is it if the photon is moving through an electric or an magnetic field. Since photons have an oszillating electric field component as well as an oszillating magnetic dipole moment a photon is under the influence if this fields. The photons field components are changing there sign periodically and the significance of changing the straight path of motion to a wiggling one is unremarkable for us. But in befringes materials (calcit) it leads to the deflection of light into two paths.

If photons are moving with c in every posible frame what is the spatial extent of a photon? Does it mean it is everywhere in all the space?

As I wrote on top photons are indivisible units. Of what? Of energy. Once emitted they take away from the emitting particle some amount of energy and momentum and can give this values to another particle during absorption.

In the theory of Quantum Mechanics photons are disturbtions of an overall existing electromagnetic field. This leads to same difficulties in the understanding, which you expressed in the question about the dislocation of photons all other the space. Let us see if it is possible to avoid a photon as a dislocation of all over the infinite space.

But in interference experiment photons are required to be at same time at same position to interfere.

The last question is the reason for your others? As anna v stated there are done experiments with single photons and after throwing photons one by one through a double slit the well know intensity distribution appears on an observer screen (an electronic devise or simple a photo plate). It has to be underlined that the result of some intensity distribution one get for single slits as well as for single edges!

As I told on top an bifringes material is able to split a light beam into two beams and this happens do to the photons electric and magnetic field components. But an interaction - as the expression stated - always has two players. The second player in our case are the surface electrons of the involved obstacle. So the intensity distributions behind edges are the result of the common and quantised field between the photons and the surface electrons of the edge(s). This point of view resolves a lot of difficulties.

For the question in the headline about the size of a photon see How does the size of the magnetic field vary with the wavelength of a photon?.

• Thank you. My question is what is the spatial extent of photon. It comes from if a laser emits one photon what is its pulse duration? – Anonymous Sep 9 '16 at 8:03
• Anonymous I read your quested after a answered this one. But I'm not sure I can answer this because of wrong ur unclear presumption about laser and one photon emission. Need more input :-) – HolgerFiedler Sep 9 '16 at 8:13