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I'm trying to understand what happens to photons when a beam of light is focused down to its waist. In the image attached, do photons take the path 1 or 2. That is, do the photons cross or just get 'deflected'.

Beam of light focused

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do photons take the path 1 or 2. That is, do the photons cross or just get 'deflected'.

They take path 1 and travel in straight lines.

However due to the finite effective size of the photon and Heisenberg Uncertainty Principle it is not possible for all of the lights straight lines to pass through the centre point.

hyperbole

See above image (ignore the circle and dot) for a series of rays passing through as close to the centre point as possible before diverging again.

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Each photon in a sense passes through every point in the region you have drawn. Imagine the following experiment. Imagine a gain medium with a population inversion of 100%, i.e. all the emitters are in their excited state. Also imagine that their spontaneous emission lifetime is very, very long. So they only relax when triggered by a photon. Now we focus an optical system so that the "waist" (as you have drawn) of the focus is in the medium, but we let one photon at a time go through the system. After we have let a huge number of them go through one by one, what is the distribution of relaxed emitters in space? What you would find is that the density of relaxed emitters would be precisely the intensity pattern in the field you have draw.

Each photon propagates precisely following Maxwell's equations, which can be written as a Schrödinger equation for the first quantised photon: Maxwell's equations are indeed:

$$i\,\hbar\,\partial_t \vec{F}_\pm = \pm \hbar\, c \,\nabla \wedge \vec{F}_\pm$$

where $\vec{F}_\pm$ is a vector encoding the photon's state, and the Hamiltonian is $\hat{H} = \pm \hbar\, c \,\nabla \wedge$. The $\pm$ components represent the left/right circular polarisations. If you write $\vec{F}_\pm = \vec{E} \pm i\,c\,\vec{B}$, you find that $\vec{E}, \vec{B}$ fulfil the same equations - Maxwell's equations - as macroscopic electric and magnetic fields do.

Each photon casts its "field of influence" into space. As each one propagates through the waist, the probability of its interacting with an emitter is proportional to $|\vec{F}_+|^2 + |\vec{F}_-|^2$.

I say a great deal more about these ideas in these answers:

"Electromagnetic radiation and quanta"

"How can we interpret polarization and frequency when we are dealing with one single photon?"

"Do Photons interfere when it passes through a slit (one)?"

and, on this mysterious thing $\vec{F}$:

"What does the complex electric field show?"

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In the far field (greater than a Rayleigh range from the waist) they look like they have taken path 1. Nearer to the waist it is harder to tell. This is the origin of the Gouy phase.

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I would think the wave character of light is the most important and waves can cross each other without changing direction, so I think path 1 is the more likely. As pointed out in the comments photons can annihilate each other, creating other particles, which in turn can create a new photon pair. Conservation of momentum and energy would require the photons to travel in the original direction but in the reference given I conclude this not to be the main event in crossing light rays. So I still opt for path 1

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  • $\begingroup$ Would the down voters please leave a comment, so I can learn something $\endgroup$
    – KvdLingen
    Commented Mar 6, 2014 at 7:21

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