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In the double-slit experiment, we emit a photon that is in a state of superposition (wave form) which travels through both slits to interfere with itself. When we measure which slit it went through, it "collapses" to a particle at that point removing the interference from the other wave.

From that point to the detector wall, does the photon now go back into a wave form to travel from the collapsed point to the detector wall? Is a photon always in a state of superposition while traveling through space?

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It's tempting to think of the light as a little ball (the photon), and since little balls have a definite position the little ball has to be in a superposition of a state where it goes through one slit and a state where it goes through the other. However this is not a good description of what actually happens.

The light is not a photon, and it's not a wave - it is an excitation in a quantum field. As a general rule it is a good approximation to model the light as a wave when studying it's propagation and a particle when studying its interactions, but these are only working approximations and should not be taken as literally true.

The light propagating in our double slit experiment does not have a position in the sense that macroscopic objects have a position. To ask the position of the photon is a meaningless question because there is no position. The light travels though both slits because the excitation in the quantum field spans both slits. When the light interacts with something, e.g. it hits the photographic plate or CCD, then the interaction happens at a point (though the position of the interaction is subject to Heisenberg's uncertainty principle). The interaction looks like a localised exchange of energy, so it looks like a photon.

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    $\begingroup$ thanks for the quick response John. i've got a few more questions: 1. is your explanation an explication of Quantum Field Theory? 2. how does the explanation of "excitation in a quantum field" fit in with de Broglie's hypothesis that all matter has a wave nature. an electron is a particle as is a bucky ball, and there have been experiments that show the wave property of these particles causing interference. thanks again! $\endgroup$ – brian kirkby Apr 17 '15 at 22:42
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    $\begingroup$ @briankirkby: quantum field theory is a huge and complex area and my summary of it here is just a charicature. However I think it captures the basic principles of what is going on. There is an elementary description of how QFT describes the Young's slits experiment here, though in this context elementary means non-nerds won't understand a word of it! Your questions 2 and 3 are huge questions in their own right, but see this search for some comments I've made on it here. $\endgroup$ – John Rennie Apr 18 '15 at 5:36
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As an experimental physicist, I go back to basics:

Here is a double slit experiment a single photon at a time:

singlephoton double slit1

sinlgephotonds2

singlephotonds3

singlephoton dbs4

The top panel shows dots from single photons coming in at the slits. The others the slow accumulation by which an interference pattern appears.

That is what the experiment shows, dots like one would expect from single particles, localized at a specific (x,y). From a single photon no wave nature can be guessed at. It is the statistical accumulation that shows the classical wave interference, similar to water waves and acoustic waves.

What we have experimentally is a probability distribution of where to find a photon thrown at these two specific slits.

One talks of the "wave nature" of photons and particles at the individual particle level referring to this probability distribution. The particle is not spread out all over the screen , it has a specific trajectory after the fact of passing through one or the other slit. Which slit it passes through depends on the boundary conditions of the solution of the quantum mechanical problem "photon and slits" which gives the probability distribution.

No matter how close to the slits in the z direction one registers the photons the trajectories will have this behavior, of a particle leaving the two slits. Except the behavior of the aggregate is not the classical behavior of little balls thrown at the slits.

So the quantum mechanical way of looking at it is that a particle is described by a wavefunction with a probability distribution for any interaction it will have . The particle interacting with the two slits has a probability distribution which displays its wave nature. The individual trajectory displays its particle nature.

There exist which way experiments that show how the boundary conditions affect the interference patterns after the slits, in this case with electrons:

Overall, the results suggest that the type of scattering an electron undergoes determines the mark it leaves on the back wall, and that a detector at one of the slits can change the type of scattering. The physicists concluded that, while elastically scattered electrons can cause an interference pattern, the inelastically scattered electrons do not contribute to the interference process.

So back to your question:

In the double-slit experiment, we emit a photon that is in a state of superposition (wave form) which travels through both slits to interfere with itself.

The photon ( or other particle) is described by a quantum mechanical wave function that gives the probability of finding it in a specific trajectory in space. Psi(x,y,z,t).

When we measure which slit it went through, it "collapses" to a particle at that point removing the interference from the other wave.

Collapse is a funny word. The wave function is not a balloon. A value is picked from the probability distribution that will give a particle trajectory for it.

From that point to the detector wall, does the photon now go back into a wave form to travel from the collapsed point to the detector wall?

Due to the boundary condition of two "slits" the wavefunction, and therefor the probability distribution, changes and carries phase information that the two slits imposed.

Is a photon always in a state of superposition while traveling through space?

The correct statement is that the photon/particle is described by a probability distribution in space and time.

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  • $\begingroup$ I have read your comments on this subject and was hoping to have some discussion. Above you say "The particle is not spread out all over the screen , it has a specific trajectory after the fact of passing through one or the other slit." and then you move on to boundary conditions of the solution etc. So it feels like you are leading up to a but. I believe that light can be described as a particle using less assumptions than waves. I describe this in my link at the top of my page. Above are you describing the propagation of a photon as a particle traveling from A to B? Thanks $\endgroup$ – Bill Alsept Apr 14 '16 at 8:04
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It basically boils down to this. Looking at a flying electron through a camera, there is no interference. Nothing special. But not trying to find which slit it went through and gradually observing the electrons to hit the detector there is interference pattern. In other words, when trying to find which slit the electron went through, wave function collapses (as they say) and electron acts like a particle. Not trying to find which slit it went through, the electron acts like a wave -> interference pattern. In other words, the electron acts weird when there is no observation. Observation -> everything is ok -> you see one electron flying. When not observing the electron, the electron is everywhere.

From that point to the detector wall, does the photon now go back into a wave form to travel from the collapsed point to the detector wall?

As I understand it, the detector collapses the electron on the sheet. But if there was no detector and not being observed the electron is in superposition of states.

Is a photon always in a state of superposition while traveling through space?

Only when it is not observed while traveling through space

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As in John Rennie's answer we cannot describe the photon as being in a superposition of "going through different slits" states.

But there is a sense wherein the answer to your question "Is a photon always in a state of superposition while traveling through space?" is almost always "yes". And the answer depends on a choice of co-ordinates.

To answer the question precisely, you need to define what the photon is a superposition of. In quantum mechanics we build general states from eigenstates, which are eigenvectors of operators called observables that model measurements. We can use any observable we like and impart co-ordinate transformations to the superposition weights to change our description to one where the basis of the quantum state space is the set of eigenstates of our chosen observable.

For photons, most often we use momentum eigenstates, which are plane waves. Any one photon state of the quantized electomagnetic field can be a superposition of these plane wave states. Notice that plane waves can have different frequencies. So a pure, one-photon quantum state is often in a superposition of energies: if we tried to measure the energy of such a state we'd instead get a probability distribution defined by the linewidth of the one-photon state.

I give more details in my answer here to the Physics SE question "Relation between Wave equation of light and photon wave function?".

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