The two slit experiment

[Pinball1970]

So what am I seeing; the energy from the particle, the particle itself, or the wave cause by movement?

The wave has values in the EM field, excitations. The wave does not need to wave "through" any medium however, it can do this in a vacuum.
How it actually propagates through this space in terms of a path gets weird.
So the single slit, firing one photon, one wave packet at a time will make a flash on the detector.
All those flashes make up a blob in the middle which makes sense.
Two slits you get an interference pattern even one flash at a time which is weird.

 

[exchemist]

deleted

 

[exchemist]

Thank you for your kind words and link.


For the record, one more question.
I can understand a wave function spreading (dispersing) over an existing substance.

But in the case of a single photon, what is the substance the wave function is dispersed over?

Apologies, due to a missing comma in post 60, that post may seem to be self-contradictory. I had meant to write "No, dispersion occurs for an electron in free flight......"

But this is interesting: a wave-particle with mass, like an electron, behaves differently from a massless photon. The wave function of a free electron spreads out over time, making its position progressively more uncertain, as a result of dispersion, whereas the EM wave (note I do not call it a wave function*) of photons travelling in a vacuum does not spread out in this way.

The relevant paragraph from the link I previously quoted starts: "Note, from the previous analysis, that the rate of spreading of a wave-packet is ultimately governed by the second derivative of ω(k) with respect to k. " [I can't copy it as I get errors when I try to paste it on the forum]. The key point is that for photons this second derivative is zero so they do not spread out over time, whereas for free QM entities with mass, this second derivative is not zero so their wave function does spread out over time. Note also that this is for free QM entities. For electrons in an atom, which are in a bound state, this will not apply.

* Schrödinger's equation, which is non-relativistic, contains a mass term in the denominator, leading to an undefined result when mass = 0. So the EM waves associated with photons can't be regarded as wave functions, or at least not in the same sense as the wave functions of QM entities with mass. But this gets quite hairy, e.g. see this discussion of the issue: https://physics.stackexchange.com/q...describes-the-wavefunction-of-a-single-photon. Suffice it to say one should not mix up photons and electrons, as they do not necessarily behave in the same way.

But all this is at the very edge of my competence as a chemist. A real physicist may perhaps shed more light on it.

 

[Write4U]

Plenty grist for thought....

 

[exchemist]

Plenty grist for thought....View attachment 6298

Indeed. I am struggling to understand the significance of d²ω/dk² ≠ 0, as ω seems to denote angular frequency while k normally denotes angular wavenumber. That implies they are just the equivalent, in angular units, of frequency and reciprocal of wavelength, respectively. For a classical wave, the speed of the wave, s = νλ, so speed, s x 1/λ (i.e wavenumber) = ν, frequency i.e. they are proportional to one another, the proportionality factor being the speed of the wave. So one would, naïvely, expect this double derivative to be zero - as indeed it is for photons.

There are references in the article to further equations that cover all this, but the links don't seem to work so I am currently unable to follow the trail to understand what is going on. If I have time and interest enough over the weekend I may try again to find out where this comes from. (I can only cope with this sort of maths in the morning, 1st thing after breakfast.)

 

[James R]

exchemist:

The speed of a wave of a single wavelength or frequency can be written as
$$v=\omega/k$$
which is equivalent to v=f(lambda).

A dispersive wave is a superposition of simple sinusoidal waves of different frequencies, f, or equivalently different angular frequencies, omega. If waves with different wavelengths travel at different speeds (as is the case, for instance, with light travelling through a refractive medium), then the frequencies of the component waves can be expressed as some function of the wavelength lambda (or, equivalently, the wave number k). That function is omega(k).

The group velocity of a wave (including dispersive waves) is then v_g = d(omega)/dk.

A dispersive wave packet will tend to spread out over time. For example, think about a wave packet made up of a number of frequencies of light ranging from red to blue, travelling through a refractive medium like glass. The speed of blue light in the medium is smaller than the speed of red light, so the blue light will continuously fall further and further behind the red light as the light propagates through the medium. The effect of this will be to stretch the wave packet out more and more as the combined wave propagates. The "middle" of the wavepacket will appear to travel at a speed that is a bit faster than the speed of pure blue light in the medium and a bit slower than the speed of pure red light; that "middle" point travels at the group velocity.

It makes sense that the rate of spreading of the wave wavepacket will depend on the second derivative d^2(omega)/dk^2.

 

[exchemist]

exchemist:

The speed of a wave of a single wavelength or frequency can be written as
$$v=\omega/k$$
which is equivalent to v=f(lambda).

A dispersive wave is a superposition of simple sinusoidal waves of different frequencies, f, or equivalently different angular frequencies, omega. If waves with different wavelengths travel at different speeds (as is the case, for instance, with light travelling through a refractive medium), then the frequencies of the component waves can be expressed as some function of the wavelength lambda (or, equivalently, the wave number k). That function is omega(k).

The group velocity of a wave (including dispersive waves) is then v_g = d(omega)/dk.

A dispersive wave packet will tend to spread out over time. For example, think about a wave packet made up of a number of frequencies of light ranging from red to blue, travelling through a refractive medium like glass. The speed of blue light in the medium is smaller than the speed of red light, so the blue light will continuously fall further and further behind the red light as the light propagates through the medium. The effect of this will be to stretch the wave packet out more and more as the combined wave propagates. The "middle" of the wavepacket will appear to travel at a speed that is a bit faster than the speed of pure blue light in the medium and a bit slower than the speed of pure red light; that "middle" point travels at the group velocity.

It makes sense that the rate of spreading of the wave wavepacket will depend on the second derivative d^2(omega)/dk^2.

Ah yes, James, thanks. What was baffling me was that (angular) frequency, ω was said to be a function of (angular) wavenumber, k. Let me paraphrase your explanation back to you, to check I have understood.

The point seems to be that, if the velocities of the components of the Fourier sum are not all the same but depend on frequency, frequency and and wavenumber are no longer in the same ratio to one another for every component. For each component wave, the ratio of frequency to wavenumber will be different, because the velocity is different. That means that, considering all the components in the wavepacket, ω becomes a function of wavenumber: ω(k).

The key difference between photons and particles with mass such as electrons seems to be the nature of their momentum. This is a function of wavenumber by de Broglie's relation: p = h/λ (or alternatively p = hbar.k in angular terms). For light in vacuo, all components of the wavepacket travel at c irrespective of frequency or wavenumber. But for (non-relativistic) particles with mass, momentum is mv, so differences in wavenumber translate into differences in velocity. And so the wave function of a free electron spreads out with time, whereas a photon wavepacket in vacuo does not. (Though in a medium that couples to EM radiation, it may, as you say.)

Do I have that right?

By the way, looking at the source I referenced earlier I found a nice explanation of the relation between position and momentum in terms of Fourier transforms. They show the relation between a wavepacket in x-space (i.e position space) vs. the same wavepacket transformed into k space (effectively momentum space). Apparently the Gaussian is the only function that transforms into another Gaussian. This leads quite naturally to Heisenberg's Uncertainty Principle: https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/02:_Wave-Particle_Duality/2.10:_Wave-Packets

This is all very cool stuff, some of which I dimly remember, though these days I can only get my head round the maths 1st thing after breakfast, after a solid night's sleep.

P.S. I agree kudos to Write4U for raising an interesting issue. At this rate we might even get a few more physicists showing up.