Young's slit experiment with single photons

[DrClaude]

Since a laser emits a coherent state, how is the quoted statement a correct description of the laser emission process?

Whatever emitted the light (atoms, dye molecules, quantum wells, ...) emitted it one photon at a time. The quantum state of the emitters is of course compatible with the fact that the photon number in the light is indeterminate.

 

[PeterDonis]

Whatever emitted the light (atoms, dye molecules, quantum wells, ...) emitted it one photon at a time. The quantum state of the emitters is of course compatible with the fact that the photon number in the light is indeterminate.

I'm not sure how these two statements are consistent, although of course any attempt to describe what the actual math says in ordinary language is going to be imperfect. But the quantum state of the emitters, if the EM field is in a coherent state, is a superposition of the two energy levels involved. That means we cannot say that one photon has been emitted, because whether or not an emission has occurred from any particular emitter is indeterminate.

What we can say is that the frequency of the laser light is well-defined (at least in the idealized case where we ignore things like finite line width), since (at least as I understand it) a coherent state (at least in the idealized case) is an eigenstate of frequency. We just can't say that that frequency is a frequency of "photons" since the frequency and photon number operators do not commute.

As above, at least part of this might just be unavoidable imperfections in translating the actual physics and its mathematical description into ordinary language. I don't think we're in disagreement about the main points of the actual physics.

 

[Dale]

Even though a coherent state is not an eigenstate of the number operator it is still correct that

The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy E=hν.

If you prepare a thousand identical coherent states and measure the energy in each then you will find that it varies, but always by an integer multiple (lumps) of $h\nu$. So I agree that it is not an eigenstate, but I don’t think that fact invalidates the statement by @WernerQH

 

[PeterDonis]

Even though a coherent state is not an eigenstate of the number operator it is still correct that If you prepare a thousand identical coherent states and measure the energy in each then you will find that it varies, but always by an integer multiple (lumps) of $h\nu$.

That's because measuring the energy is equivalent to measuring the photon number; the energy operator is just $h \nu$ (more commonly written $\hbar \omega$ in the literature) times the photon number operator. So of course you're always going to get an eigenvalue of photon number times $h \nu$ as the measurement result.

As I said in response to @DrClaude, I think at least part of the issue here is that we are trying to describe physics in vague ordinary language instead of math, and there is no translation from the latter to the former that is fully accurate. My preference is to avoid having to do that as much as possible, and when it is unavoidable, to make the translation as direct as possible, by talking specifically about the EM field state (a coherent state) or the operator being measured (energy, aka $h \nu$ times photon number) instead of using vaguer terms like "occurs in lumps".

 

[PeterDonis]

That's because measuring the energy is equivalent to measuring the photon number; the energy operator is just $h \nu$ (more commonly written $\hbar \omega$ in the literature) times the photon number operator.

Actually, even this is not quite right, since the Hamiltonian also includes the zero point energy of $h \nu / 2$. @Dale might have been indirectly referring to this when he said the result of the energy measurement always "varies" by multiples of $h \nu$. But the fact that the zero point energy is included in the total energy means that the total energy is not a multiple of $h \nu$, and that makes it even less appropriate to say that the energy in the EM field "always occurs in lumps" of size $h \nu$, since that ignores the zero point energy. Giving the specific state and the specific operator being measured avoids issues like this.

 

[Marilyn67]

Thank you all very much for your very informative answers !
I can't find these answers elsewhere.

I know this is all complicated, and as PeterDonis points out, everyday language can cause confusion.

It is not easy to develop an experiment like this because it requires a lot of theoretical knowledge, and to put it into practice, I imagine the work of precision and adjustment that it represents, and also its cost.

I don't know who can do this alone in his cellar !

Then, apparently, you have to know how to interpret the results which can be positive biases or negative biases !

It's really not easy, and all the people who work and think in this area have merit !

I need to reread all your answers, to review this video again, to take notes and to make a synthesis to see more clearly.

Cordially,
Marilyn

 

[vanhees71]

I don't know what "real" single photons are. You always have to wait some time for a photon to be produced, although there are of course experiments where they are created very quickly. The photon concept applies to the absorption and emission of light, which occurs only in "lumps" having energy $E = h\nu$. The propagation of light, however, is well described as a wave phenomenon (Maxwell's equations).

A real single photon is a normalizable Fock state of exact photon number 1, i.e.,
$$|\Psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 k \sum_{\lambda=-1,1} A_{\lambda}(\vec{k}) \hat{a}^{\dagger}(\vec{k},\lambda) |\Omega \rangle,$$
where $\lambda \in \{-1,1 \}$ are helicities (polarizations) and $|\Omega \rangle$ is the photon vacuum state. It describes an (asymptotic) free photon, and $A_{\lambda}(\vec{k})$ are square-integrable functions such that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 k |A_{\lambda}(\vec{k})|^2=1.$$
Photons never behave like massless point particles but as a quantum state of the electromagnetic field, i.e., as a wave. Particularly all the discussion about coherence, coherence lengths and times, etc. also apply to these single-photon states, and the interference patterns that can be measured by using very many single photons are those of the corresponding classical waves, reflecting the probability distributions for registering a photon at a given time and position of the detector (or a photoplate).

It hasn't been easy to produce true single-photon states. Unfortunately that's reflected by the inaccurate statement that you just have to use a very dim light source like a laser, but that doesn't produce true single-photon states but coherent states that are not eigenstates of the photon number. A very "dim" coherent state consists mostly of the vacuum, and the distribution of photon numbers is Poissonian. Although you can make the average photon number as small as you like (also below 1!), you never have a true single-photon state but a superposition of all number states, and for such a "dim coherent state" it's most probable to find no photon at all but some small probability to find 1, and even smaller probability to find 2, etc. photons.

 

[Marilyn67]

Hello,

I just have one more question for you.

More precisely, for most light sources it is very difficult to set up an experiment that will detect meaningful interference phenomena.

I know that commercially available small cheap laser pointers have much lower beam quality than a laser like helium neon laser or argon laser for example.

I myself own a small green laser pointer, class IIIb, (<20mW), 532 Nm.

If nothing other than the source is changed in the experience previously described on the link,

https://www.physicsforums.com/threads/how-big-is-a-photon-interesting-video.1005059/

but by replacing the neon helium laser with this type of laser pointer, will interference be observed as easily ?
Is the coherence length sufficient to observe interferences in the experiment setup depicted in the video ?

In advance, thank you for your response.

Cordially,
Marilyn

 

[Marilyn67]

by replacing the neon helium laser with this type of laser pointer, will interference be observed as easily ?
Is the coherence length sufficient to observe interferences in the experiment setup depicted in the video ?

For me, yes

 

[PeterDonis]

by replacing the neon helium laser with this type of laser pointer, will interference be observed as easily ?

Probably not because the cheap laser pointer probably has a much shorter coherence length. It probably also has a wider frequency width in its beam (no laser is perfect so they all have some finite frequency width, but the narrower the width the easier it is to observe interference).

 

[Marilyn67]

Thank you @PeterDonis ,

Probably not because the cheap laser pointer probably has a much shorter coherence length. It probably also has a wider frequency width in its beam (no laser is perfect so they all have some finite frequency width, but the narrower the width the easier it is to observe interference).

Your answer is interesting.

I didn't expect laser pointers with such poor quality..!

It would be interesting to repeat the experiment by modulating the distance of the second path to have an order of magnitude of the distance from which the results change... (a few decimetres, a few centimeters perhaps...)

See you soon,
Marilyn

 

[PeterDonis]

I didn't expect laser pointers with such poor quality..!

"Poor quality" is relative. Cheap laser pointers aren't designed to help you run interference experiments.

 

[Marilyn67]

"Poor quality" is relative. Cheap laser pointers aren't designed to help you run interference experiments.

Yes, it's true
It's all the more interesting to try
(I've done this before with Young's slits, my laser pointer and a lens)

Cordially,
Marilyn

 

[tech99]

At school I did the experiment using a sodium flame, before the invention of the laser. We passed the light through a slit to create a degree of coherence.

 

[Marilyn67]

Hello @tech99 ,

At school I did the experiment using a sodium flame, before the invention of the laser. We passed the light through a slit to create a degree of coherence.

Tell you about an experience with Young's slots ?
From what I understood at the beginning of the discussion, it's very easy to get visible interference with Young's slits because the path differences are very small at the point of impact.

I guess in the video setup a sodium flame would have given a diffuse pattern, right ?

Cordially,
Marilyn

 

[tech99]

There is an experiment shown on this video using sunlight as the source.

The light is passed first through a slit to provide spatial coherence. It is just a question of making the geometry so that the path length differences from each edge of the first slit to the double slits are much less than a wavelength. The pattern is very similar to the laser result, but not quite as sharp. The frequency coherence issue causes the sunlight to give coloured fringes. We used sodium light, which is sufficiently monochromatic to give clear fringes.

 

[Marilyn67]

Thanks @tech99 for your response,

I ask myself 2 questions that you may find silly :

It is just a question of making the geometry so that the path length differences from each edge of the first slit to the double slits are much less than a wavelength.

1/ How to obtain such precision when we know that the dilation due to temperature changes and that the vibrations present in the environment are of an order of magnitude well above 1/2 micron ?

2/ Why such precision for Young's slit experiment ?
If the difference in path length from each edge of the first slit to the double slits is even a few percent, isn't the interference pattern just shifted across the screen to the right or left to compensate this difference ?

Cordially,
Marilyn

 

[tech99]

The long triangle formed by the rays from each edge ensures that close tolerances are unnecessary. For instance, the (height - hypotenuse) is easily made much less than a micron. We do not require great precision.

 

[Marilyn67]

Ok, I understand the "how" but not the "why".

Indeed, to have interfringes "i" visible to the naked eye on the screen, a large length "D" and a small distance "a" between the two slits (i = λ.D/a) are required and here, "D" >> "a".

But we can imagine a much shorter triangle, and observe smaller interfringes i thanks to the equivalent of a microscope (ok it's more complicated, but a path difference less than the wavelength is not "fundamentally necessary", or is there another reason ?)

Is there a fundamental reason, or is it just a convenience to have wide interfringes ?

 

[tech99]

Ok, I understand the "how" but not the "why".

Indeed, to have interfringes "i" visible to the naked eye on the screen, a large length "D" and a small distance "a" between the two slits (i = λ.D/a) are required and here, "D" >> "a".

But we can imagine a much shorter triangle, and observe smaller interfringes i thanks to the equivalent of a microscope (ok it's more complicated, but a path difference less than the wavelength is not "fundamentally necessary", or is there another reason ?)

Is there a fundamental reason, or is it just a convenience to have wide interfringes ?

When we used sodium light we used a traveling microscope (which we used to call a cathetometer) to observe the fringes. There is a trade-off between fringe spacing and brightness, and a laser is much brighter than a sodium source.

 

[hutchphd]

But we can imagine a much shorter triangle, and observe smaller interfringes i thanks to the equivalent of a microscope (ok it's more complicated, but a path difference less than the wavelength is not "fundamentally necessary", or is there another reason ?)

The real world is much more complicated than the idealized two slits in a plane. In fact there are interferences from different numbers of wavelengths ("orders" of diffraction) that are more and more difficult to capture because they are typically finer detail. In fact an arbitrary object when illuminated appropriately will produce a very very complicated pattern of interferences on all scales. This is a hologram !

 

[Marilyn67]

Thanks @tech99 for your very concise answer.

It's still instructive today to know how we did in the good old days and with hindsight, it gives good ideas for tomorrow !

Cordially,
Marilyn

 

[oknow]

If you are looking for something to help you visualize single-photon interference in the double slit experiment, consider the following model. Imagine each possible state of the experiment's photon exists in a different world. So, in one world the photon traces a certain path to the detector, while in other worlds it traces a different path. Next, apply the uncertainty principle, which tells us that prior to observation we cannot be sure exactly which path a given photon will take, or in this model, exactly which world that photon occupies. When the experiment is performed, such uncertainty means each individual photon is not confined to anyone particular world, and thus is free to interact with other photons, the ones in other states/worlds, thus causing each individual photon's landing location to indicate interference has occurred.

 

[PeroK]

If you are looking for something to help you visualize single-photon interference in the double slit experiment, consider the following model. Imagine each possible state of the experiment's photon exists in a different world. So, in one world the photon traces a certain path to the detector, while in other worlds it traces a different path. Next, apply the uncertainty principle, which tells us that prior to observation we cannot be sure exactly which path a given photon will take, or in this model, exactly which world that photon occupies. When the experiment is performed, such uncertainty means each individual photon is not confined to anyone particular world, and thus is free to interact with other photons, the ones in other states/worlds, thus causing each individual photon's landing location to indicate interference has occurred.

That sounds like hocus pocus to me!

 

[PeterDonis]

consider the following model

Do you have a reference for this model? Or is it just your personal speculation? Personal speculations are off limits here.

Imagine each possible state of the experiment's photon exists in a different world

This sounds somewhat like the Many Worlds Interpretation--which would, in itself, mean it is off limits in this particular thread, because discussions of particular QM interpretations belong in the interpretations subforum, not this one, and this thread is about a question in the context of basic QM without adopting any particular interpretation.

However, the resemblance is only "somewhat", since in the MWI, different "worlds" refers to different decoherent branches of the wave function. The parts of the wave function that refer to the photon going through different slits are not different decoherent branches, so they are not different "worlds" in the sense of the MWI. (The fact that the different parts of the wave function can "interact", combining to produce the final probability amplitudes at the detector, also means they aren't different "worlds" in the sense of the MWI, since different "worlds" in the MWI cannot interact.)

 

[vanhees71]

It's much simpler. A single photon is still a certain state of an electromagnetic field (one-photon Fock state) and thus describes a wave, and in the double-slit experiment there's interference between the partial waves moving through either slit, which makes the interference pattern.

It's, however, different from the classical electromagnetic waves (which are quantum-field-theoretically described by so-called coherent states, which are other states of the electromagnetic field which do not have a specified number of photons; the photon number in a single-mode coherent state is Poisson distributed). A single-photon state means that you can only detect strictly one photon by, e.g., using the photo effect in the detector material, leading to absorption of this one photon and kicking out an electron, which can be used as a signal to detect the photon at the place of the detector. The meaning of the electromagnetic wave for a single photon is the probability distribution for detecting a photon at a given place and is identical with the intensity of this em. wave (i.e., the energy density), properly normalized such that the probabilities sum to 1.

 

[oknow]

I was referring to the explanation of the miwoi model (a combination of Many Worlds and Uncertainty Principle). It's deceptively simple, yet is also consistent with other things quantum such as tunneling, Ehrenfest's theorem, zero-point energy, and more. The model's prediction that matter's gravitational field changes slightly upon decoherence subsequently gained support from the recent observation of such in colliding galaxies. This hints at a quantum role in dark matter/energy.

 

[HomesliceMMA]

This might not be responsive, but one thought after reading your first post @Marilyn67 - My understanding is that if there is no measurement, no wave collapse, then the way to think about it is the PROBABILITY WAVES of that single particle (not yet collapsed) go through both slits. It is those waves that interfere (or reinforce) each other as they interact on the other side of the slits.

Making one slit closer or further away will affect the interference pattern, but it will still be there (in absence of measurement).

 

[HomesliceMMA]

A different way to say the same thing - before collapse, the particle is still a wave, and that wave will take every possible path it can, as a wave. The wave taking those different paths is causing the interference.
Question for someone: if you remove one slit, so you have a single slit, will an interference pattern be shown? I could have sworn I read somewhere that it would still show up, but that would not make sense to me. How could there be interference where the wave could take only one possible path?

 

[jtbell]

Question for someone: if you remove one slit, so you have a single slit, will an interference pattern be shown?

Yes. The single-slit diffraction pattern is caused by interference of waves that pass through different parts of the slit, across its width.

[added] The general two-slit diffraction pattern combines the effects of interference between the two slits, and interference between the different parts of each slit.

As the slits become narrower, while keeping them the same distance apart, the central maximum of the single-slit pattern spreads out and eventually fills the "field of view" so that you see only the classic two-slit interference pattern.

Note the scales are different in the two two-slit patterns linked above. If the slits have the same separation in the two patterns (with only their width being different), the peaks in the two-slit interference pattern are the same width as the narrow peaks in the two-slit diffraction pattern.