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In a recent question I asked how double-slit diffraction conserves momentum and the answer was that each photon gets a "kick" from the slits. So my question now is, what does that look like at a microscopic detail level?

I have a candidate. The lowest order loop expansion I can come up with looks like so. (Apologies. I don't have a drawing package that does electrons and photons properly. I'd be glad to be pointed at one if it exists.)

So very schematically: The photon must split into two parts, which have to start as charged particles. Otherwise they could not be interacting with the photon. These must go one on each side, otherwise there can't be any interference from the two slits. The slits provide an elastic interaction photon that provides momentum, but no change in energy.

I am presuming that the interaction photon can arise from any point on the slits, and interact with any part of the loop.

Is this what is going on? There ought to be a "well known" experiment that sees that charged loop if it is there. Say by imposing an electric field parallel to the plane of the slits.

enter image description here

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  • $\begingroup$ Your view is not consistent with quantum mechanics. Photon can't split in half as well as any other elementary particle can't, unless it's a pair-production process from a quantum vacuum. But in double-slit experiment laser energies used are minuscule and not enough for a pair-production, so that is not the case here. In QM terms photon simply travels all possible paths at once. This can be best understood as a probability wave which self-interferes going through slits and bumping at the screen. $\endgroup$ Dec 15, 2021 at 9:17
  • $\begingroup$ See my comments below. The photon is best understood as a wave or a disturbance in the EM field for the question you are asking. You do not need to consider the double slit, even for a single slit we have diffraction. A photon interacts very much with electrons in atoms, they can scatter (Raleigh scattering), reflect, absorb, transmit. Not only do the photons interact but they also interact with all the surrounding materials and their respective EM fields. $\endgroup$ Dec 15, 2021 at 23:23
  • $\begingroup$ @AgniusVasiliauskas Vacuum polarization is a thing. Photons very much do split into charged pairs. I have personally reproduced the Lamb shift experiment. Lots of fun with a very special spectrograph. $\endgroup$
    – Dan
    Dec 16, 2021 at 18:35
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    $\begingroup$ Dan you need to clarify whether your question on "diffraction" includes both concepts of: 1) the spreading (scattering) of light and/or 2) the formation of the famous diffraction double slit pattern. THEY are 2 different questions/concepts if you want to get into the details. $\endgroup$ Dec 17, 2021 at 2:05
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    $\begingroup$ It has nothing to do with slits. It is the edges. You can get a diffraction pattern from a single edge. Photons diffract around and behind edges and photons scatter off of edges. A single slit has two edges that photons diffract around and two edges that photons scatter off of. These four sources overlap on the screen to form the pattern. A photon is a particle that has a frequency or an oscillation. When you truly consider photons as a particles you can derive any diffraction pattern. $\endgroup$ Dec 17, 2021 at 4:40

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There is no charged loop and no extra photon.

The one and unique photon in the two slits experiment acts as a wave. It does not just "split in two parts" to go through each slit. The wave "fills" the full widths of both two slits at the same time and gets "kicks" from all four ends of the slits at the same time.

How do we know it it so ? Because of the full diffraction picture. Sure, the fact that there are two slits is the cause of the interference pattern. But the width of the entire diffraction pattern is controlled by the width of each slit.

Remember, the width of the diffraction pattern in a single slit experiment in inversely proportional to the width of the slit.

If the slits are not narrow enough, there is no visible diffraction at all. Roughly, the width of the slits must be narrow compared to their separation, otherwise the diffraction pattern is narrowerer than the interference one and one cannot see interferences.

You have to think in terms of waves. Trying to understand quantum mechanics without accepting the notion of duality leads nowhere. One has to accept the weirdness of QM, and to think differently.

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    $\begingroup$ you are talking as with electromagnetic waves, not electrons and photons. The quantum mechanical waves are probability waves, not single photon representations. $\endgroup$
    – anna v
    Dec 15, 2021 at 7:04
  • $\begingroup$ The probability wave of a single photon behaves like the electromagnetic wave. The pattern obtained by allowing many, many times a single photon to propagate in an interferometer is exactly the one obtained with a "macroscopic" EM wave ! Didn't you see I started my post by mentioning a single photon ? $\endgroup$
    – Alfred
    Dec 15, 2021 at 8:44
  • $\begingroup$ If the probability wave of a single photon was not affected by diffraction because of the narrow width of each slit, the probability wave of going through each slit would describe photons hitting well-separated parts of the screen ! $\endgroup$
    – Alfred
    Dec 15, 2021 at 9:01
  • $\begingroup$ Yes I agree with this answer and would add the Feynman path integral approach. It basically says the photon considers all paths and chooses the probable ones. The probable ones are shown by the integral to have path lengths that are n multiples of the wavelength of the light. $\endgroup$ Dec 15, 2021 at 23:14
  • $\begingroup$ "The wave "fills" the full widths of both two slits" is good but this is likely due to the excited atom/electron where the electron generates fields (virtual photons) before and actual photon (energy transfer) occurs. $\endgroup$ Dec 15, 2021 at 23:17
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You do not state the level of your physics background in your profile, but from your question, I assume that you are not familiar with quantum mechanics.

There is no "charged loop " as you imagine in your drawing.

The material of the slits consists of atoms and molecules, which are neutral. Atoms have charged electrons around them and those are the ones photons will interact with. BUT electrons and photons are point particles, no extension in space that you show for the "charged" particle in your drawing. Their interactions are described mathematically by quantum field theory which gives the solutions in series terms of decreasing in magnitude order. The first order diagrams and calculations for electron photon scattering can be seen here

compt

In the first diagram (figure 7.7a), the incident photon ($k,\varepsilon$) is absorbed by the incident electron ($p_i,s_i$) and then the electron emits a photon ( $k^\prime,\varepsilon^\prime$) into the final state. In the second diagram (figure 7.7b), the incident electron ($p_i,s_i$)

This is for a free electron. It looks a bit like your diagram but it corresponds to mathematical formulas that calculate the probability of scattering in a given direction.

You can see that from momentum conservation the directions change.

When the electron is tied up in the atoms of the slit, it becomes more complicated mathematically , but momentum conservation is a strict rule. The atom on which the electron is bound takes up the momentum and transfers it to the lattice it belongs to.

For information here is how the double slit single photon at a time images are built up

snglephot

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The classical interference pattern appears after the accumulation of photons.

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  • $\begingroup$ The diagrams you posit cannot possibly produce a two-slit pattern. There is nothing to go through both slits. If the photon was absorbed then re-emitted it would be coming from a point source, not a slit, and certainly not two slits. The two-slit pattern is clearly produced by light coming from all of both slits. $\endgroup$
    – Dan
    Dec 16, 2021 at 18:31
  • $\begingroup$ @Dan The diagrams are used to calculate the probability for a photon to scatter from an electron . It is based on quantum mechanics. A single photon scattering off two slits, of given width thickness and separation has a quantum mechanical probability to pass through that in principle can be calculated using a sum of such diagrams. Lighti is composed of zillions of photons and obeys classical electrodynamics. I will include the single photon at a time double slit experiment to show how different single photons are in behavior to the accumulated light pattern. $\endgroup$
    – anna v
    Dec 16, 2021 at 19:25
  • $\begingroup$ Again, the diagrams you posit CANNOT produce the double slit pattern. Each atom involved would act as a point source. Since it has to be visible to both the source and screen, that effectively makes it a point source at the slit edge. CANNOT form a double slit pattern because nothing would go through both slits. $\endgroup$
    – Dan
    Dec 17, 2021 at 1:36
  • $\begingroup$ @Dan you need to clarify whether your question on "diffraction" includes both concepts of: 1) the spreading (scattering) of light and/or 2) the formation of the famous diffraction double slit pattern. THEY are 2 different questions/concepts if you want to get into the details. $\endgroup$ Dec 17, 2021 at 2:04
  • $\begingroup$ @Dan as the esperiment on single photon double slit shows single photons scatter and are recorded as points . A single photon hit does not spread out all over the screen. It is not a space wave. The accumulation towards the right shows the probability wave accumulating in the classical wave pattern. experiment is what defines how the photons behave and makes necessary quantum mechanics. $\endgroup$
    – anna v
    Dec 17, 2021 at 4:50
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The photons that don't interact with the screen are the ones that form the pattern. It's the ones that do interact that transfer the momentum to the slit system.

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  • $\begingroup$ The question I cited in my question here shows that the photons that go through must also interact with the slit to get a sideways elastic kick. Otherwise they could not conserve momentum. $\endgroup$
    – Dan
    Dec 17, 2021 at 1:39
  • $\begingroup$ @Dan Quantum mechanics does not allow such a conclusion. $\endgroup$
    – my2cts
    Dec 17, 2021 at 8:35
  • $\begingroup$ You better tell Eisntein and Bohr then. Sources in the cited question's answer. $\endgroup$
    – Dan
    Dec 18, 2021 at 3:56
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The diffraction pattern is not formed by photons splitting up into pairs of charged particles. Unless the light has a frequency that is high enough to produce electron-positron pairs, the contribution of those virtual loops that are shown in the diagram would be negligible.

To explain the conservation of momentum here, I need to use a bit more technical language. Please bare with me.

We need to say something about the state of the light. Let's start with a single photon state, (a number state with exactly one photon) which we'll denote as $|\psi_1\rangle$. It consists of a superposition (or spectrum) of plane waves. Crudely, $$ |\psi_1\rangle = |\mathbf{k}_a\rangle C_a + |\mathbf{k}_b\rangle C_b + |\mathbf{k}_c\rangle C_c + ... , $$ where the $\mathbf{k}$'s are the wave vectors of the plane waves and the $C$'s are complex coefficients. More accurately, we can represent it with an integral $$ |\psi_1\rangle = \int |\mathbf{k}\rangle C(\mathbf{k}) \text{d}\mathbf{k} . $$ We can represent the screen with the two slits as a transmission function $t(\mathbf{x})$. To get the state after the screen, we need to apply an operator $\hat{T}$ that imposes this transmission function on the state. Even if the state before the screen were just a single plane wave, the transmission function would cause the state after the screen to have a spectrum of plane waves. However, the transmission function causes a loss. Therefore, the operator does not maintain the normalization. So, we would need to normalize the state afterward.

To treat the loss correctly, we can model the screen as a beamsplitter that sends the part of the state that is blocked by the screen to a different output port where we can "trace them out." That would give us a mixed state if the original state contained $n>1$ photons (a number state with exactly $n$ photons). In such cases, the interference would be lost.

Fortunately, most experiments are done with laser light, which is represented by a coherent state, instead of a number state. Coherent states remain pure states when they suffer loss. Coherent states are parameterized by spectra, similar to the way we parameterize the single photon state, but in this case, the spectrum is not in general normalized. We'll assume that input state is given by a coherent state $|\alpha\rangle$ with a spectrum $\alpha(\mathbf{k})$.

Now we can use this picture to address the issue of the momentum. The part of the light that passes through the slits is given by $|\alpha'\rangle$ where the spectrum is $$ \alpha'(\mathbf{k}) = \int t(\mathbf{x}) \exp(i\mathbf{x}\cdot\mathbf{k}-i\mathbf{x}\cdot\mathbf{k}') \alpha(\mathbf{k}') \text{d}\mathbf{k}'\text{d}\mathbf{x} . $$ Note that even if $\alpha(\mathbf{k}')$ was a very narrow spectrum, the modulation with the transmission function will cause $\alpha'(\mathbf{k})$ to have a broader spectrum. On the other hand, the light that is blocked by the screen is given by $|\alpha''\rangle$ where $$ \alpha''(\mathbf{k}) = \int [1-t(\mathbf{x})] \exp(i\mathbf{x}\cdot\mathbf{k}-i\mathbf{x}\cdot\mathbf{k}') \alpha(\mathbf{k}') \text{d}\mathbf{k}'\text{d}\mathbf{x} . $$ Note that $$ \alpha'(\mathbf{k}) + \alpha''(\mathbf{k}) = \alpha(\mathbf{k}) . $$ In other words, the additional components that are present in $\alpha'(\mathbf{k})$ due to the modulation are removed by $\alpha''(\mathbf{k})$ to reproduce $\alpha(\mathbf{k})$. Moreover, for the transmission function we have $t^2(\mathbf{x})=t(\mathbf{x})$, which means that $$ \int \alpha''(\mathbf{k}) \alpha'(\mathbf{k}) \text{d}\mathbf{k} = 0 . $$

To determine the momentum, we can compute the expectation value of the wave vector. We can do this with an operator $$ \hat{P} = \hbar \int \mathbf{k} \hat{a}^{\dagger}(\mathbf{k}) \hat{a}(\mathbf{k}) \text{d}\mathbf{k} . $$ It then follows that $$ \langle\hat{P}\rangle = \langle\alpha|\hat{P}|\alpha\rangle = \hbar \int \mathbf{k} |\alpha(\mathbf{k})|^2 \text{d}\mathbf{k} , $$ because $$ \hat{a}(\mathbf{k})|\alpha\rangle = |\alpha\rangle \alpha(\mathbf{k}) . $$

For the light that passes through the slits, we get $\langle\hat{P}\rangle'$ and for the light that is block, we get $\langle\hat{P}\rangle''$, when we use $\alpha'(\mathbf{k})$ and $\alpha''(\mathbf{k})$ in the respective calculations. Based on the properties of these spectra, it then follows that $$ \langle\hat{P}\rangle' + \langle\hat{P}\rangle'' = \langle\hat{P}\rangle , $$ which shows that momentum is conserved. This happens because the screen receives the momentum of the part of the state that it absorbed.

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  • $\begingroup$ The photons do not scatter from the edges in a double slit experiment, they are concentrated along probable paths due to the tight geometric constraints of the experimental setup. The Feynman path integral approach best explains the phenomenon. $\endgroup$ Dec 15, 2021 at 23:11
  • $\begingroup$ Yes there are always some scattering from the edges in a physical setup, but its contribution is not the dominant effect here. The path integral is only one way to understand the effect. $\endgroup$ Dec 16, 2021 at 3:07
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    $\begingroup$ Sorry, but your answer just does not work. A single photon coming from the source can still diffract. And it starts with momentum very tightly bound forward. Nothing you mention deals with how it acquires side-ways momentum. Plus, you brring in a lot of irrelevant stuff about lasers and n-particle states. $\endgroup$
    – Dan
    Dec 16, 2021 at 18:46
  • $\begingroup$ It acquire the sideways momentum due to being modulated by the transmission function. The reason for the n-particle state is to justify doing the calculation with coherent states. $\endgroup$ Dec 17, 2021 at 3:45
  • $\begingroup$ See the edits for clarifications $\endgroup$ Dec 17, 2021 at 4:03

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