How to collapse a water wave through a double slit into particle behaviour?

[Happiness]

Electrons passing through a double slit is in a superposition of passing through the left slit and the right slit, thereby producing an interference pattern on the screen. But when a detector is placed to detect which slit the electrons pass through, the interference pattern is destroyed.

How can we destroy the interference pattern of a water wave passing the double slit? Why is it that the wave function of the water wave doesn't collapse even though we are constantly looking at it and therefore constantly making measurement?

 

[PeterDonis]

How can we destroy the interference pattern of a water wave passing the double slit?

You can't. A water wave doesn't have the quantum coherence necessary to do this.

Why is it that the wave function of the water wave doesn't collapse

Why do you think it doesn't? "Collapse" is not the same as "destroy the interference pattern". In the double slit experiment with electrons, each electron collapses when it hits the detector screen and makes a spot.

 

[Happiness]

You can't. A water wave doesn't have the quantum coherence necessary to do this.

Why do you think it doesn't? "Collapse" is not the same as "destroy the interference pattern". In the double slit experiment with electrons, each electron collapses when it hits the detector screen and makes a spot.

Could you explain quantum coherence in simple terms?

A water wave of wavelength $\lambda$ should behave as a particle with momentum $p=\frac{h}{\lambda}$. How can we observe this particle? And how can we collapse the wave function of this particle so that it passes through either the left or the right slit, instead of being in a superposition of passing both slits? I was talking about this collapse to bring about the loss of the interference pattern, not the collapse occurring on the screen.

 

[Heikki Tuuri]

We can interpret a water wave as a huge collection of coherent phonons. A phonon is a quantum of vibration of matter.

What happens if we disturb perfectly still water so little that just an individual phonon is produced and it progresses toward the double slit?

If we were able to measure through which slit each individual phonon passes, then no interference pattern would form.

A realistic water wave is a huge collection of many phonons. When photons reflect from the surface of water, the photons measure something about that huge collection, but they do not measure the path of each individual phonon. Therefore, the interference pattern is only very slightly disturbed by the measurement.

An analogous experiment is when we direct a laser beam towards a double slit. Some light is diffracted from each slit, and we can observe with the naked eye that the beam hits both slits. Why that measurement does not destroy the interference pattern on the screen? Because we only get some crude information about a huge collection of coherent photons. We do not measure the path of each individual photon.

 

[PeterDonis]

Could you explain quantum coherence in simple terms?

It means that the entire quantum system can be treated as having a single phase of its wave function, or if subsystems have different phases, those phases have a fixed, known relationship.

A water wave of wavelength $\lambda$ should behave as a particle with momentum $\frac{h}{\lambda}$.

No, it shouldn't, because a single quantum particle has only 1 degree of freedom (if we're talking about motion in a single direction), whereas a water wave propagating as a plane wave in a single direction has something like $10^{25}$ degrees of freedom (since that's roughly how many water molecules are involved).

 

[PeterDonis]

We can interpret a water wave as a huge collection of coherent phonons. A phonon is a quantum of vibration of matter.

But for this interpretation to have any relevance at all, you would have to cool the water to the point where it was in its quantum ground state with respect to phonons (i.e., not a single vibration anywhere), and then excite just a single phonon and propagate it. But you can't do that, because long before you reached that point the water would freeze and would no longer propagate waves of the kind the OP is describing.

perfectly still water

Water that is "perfectly still" in the sense of not having any waves of the kind the OP is describing is nowhere even close to "perfectly still" in the sense of being in the ground state with respect to phonons. Thermal fluctuations in the water at any temperature where it is liquid will be constantly exciting phonon modes, and these excitations will be incoherent and will not exhibit any quantum properties; the water and the waves in it can be treated entirely classically. So the phonon degrees of freedom you are describing are simply not relevant for the situation the OP is describing.

 

[Heikki Tuuri]

Peter Donis is right that warm water is a mess of a very big number of phonons, or quantums of vibrations. It is probably impossible to measure the path of an individual phonon in a very small water wave.

However, we believe that warm water can in principle be described as a soup of quanta.

The laser beam experiment which I referred to is more clear cut: our eyes will observe a few diffracted photons. Most of the photons in the coherent collection will pass the slits unobserved, and they will form an interference pattern on the screen.

Almost all of our observations of nature are weak measurements in the sense of Yakir Aharonov: we measure something of a big collection of quanta - we do not measure the state of each quantum.

 

[PeterDonis]

we believe that warm water can in principle be described as a soup of quanta.

Sure, it can, but the soup will be incoherent and you won't be able to get any interference effects from it.

Almost all of our observations of nature are weak measurements in the sense of Yakir Aharonov: we measure something of a big collection of quanta - we do not measure the state of each quantum.

This is a valid point, but there is no such weak measurement for warm water and the phonon degrees of freedom. For the water waves described in the OP, the interference they exhibit can be described entirely classically and doesn't require any quantum coherence--but that also means it can't be destroyed by "measuring which slit the water goes through" the way you can with a single electron.

 

[Heikki Tuuri]

Peter Donis, you can get interference effects from incoherent quanta. Just shoot individual photons, in whatever phase, into a double slit. Each photon "passes through both slits", and the observed dots on the photographic screen will form an interference pattern once you have fired a large number of photons.

Classically, it is hard to describe a flux of monochromatic, incoherent, photons. If we just sum a large number of sine waves in various phases, then destructive interference would cancel most of the wave. In the particle interpretation of quantum mechanics, an incoherent flux makes sense.

 

[PeterDonis]

you can get interference effects from incoherent quanta

You can get a double slit interference pattern with incoherent photons, yes, but that's because "incoherent" in this case doesn't mean the same thing as "incoherent" means when it's used to describe why water waves behave classically.

For the photons, what is "incoherent" is the phases of the wave functions of different photons when compared with each other. But the interference pattern in the double slit doesn't depend on coherence between different photons. It only depends on each individual photon being, so to speak, "coherent with itself" (in the sense of the "pieces" of its wave function that go through both slits having a known phase relationship). This is why you can get the same interference pattern when you turn the intensity of the photon source down to the point where only one photon at a time is passing through the apparatus.

What the interference pattern does depend on is the photons being well collimated, i.e., having momentum that is sharply peaked around a single direction (from the source to the slits). This is what ensures that the phase of each photon's wave function is the same (or close enough to being the same) when it hits both slits.

In the case of water waves, "incoherent" means the wave is not made of a single particle, it's made of $10^{25}$ or so particles, and their phase relationships are random. So there's no way for a water wave to have a quantum wave function whose phase is well-defined and the same (or close enough to being the same) at both slits. The observed interference is just classical interference and has nothing to do with any quantum properties of the water.

 

[PeterDonis]

Classically, it is hard to describe a flux of monochromatic, incoherent, photons.

Classically you don't describe photons at all. You describe waves of the classical electromagnetic field. There is no particle interpretation of such waves.

 

[Happiness]

The observed interference is just classical interference and has nothing to do with any quantum properties of the water.

So is it just pure coincidence that the double-slit interference for light and electrons are described by the same equation as the classical water-wave interference?

 

[Vanadium 50]

So is it just pure coincidence

Whoa! Dial back the 'tude dude!

They are both described by the wave equation. But that is not the same as saying they are both described by quantum mechanics.

 

[PeterDonis]

So is it just pure coincidence that the double-slit interference for light and electrons are described by the same equation as the classical water-wave interference?

If "coincidence" is how you want to interpret "water waves are classical, photons and electrons are quantum", then sure, it's coincidence.

 

[Heikki Tuuri]

How to model incoherent light classically?

Let us work with just one spatial dimension.

If we build the flux from wave packets in various phases, then the Fourier decomposition of each packet contains many frequencies. The light cannot be monochromatic then.

The Fourier decomposition of monochromatic light contains just a single frequency. It is coherent then.

We conclude that, classically, monochromatic light is always coherent.

What about quantum mechanics? Is monochromatic light always coherent in quantum mechanics? If we prepare the photons as wave packets, then we can do the exact same analysis as in the classical case. Monochromatic light is always coherent.

We can model incoherent light classically. We just sum a bunch of wave packets in various phases. The light is not monochromatic, though. The analysis is the same in quantum mechanics.

The sum of the wave packets does not look like a nice wave. Rather, it looks like random noise.

I was wrong when I claimed that classically it is hard to model incoherent light. The analysis is the exact same in quantum mechanics.

Water waves are apparently produced by processes which create mostly coherent waves. Light, on the other hand, is in many cases thermal radiation, and is incoherent and contains many frequencies.

 

[PeterDonis]

How to model incoherent light classically?

This is off topic and a thread hijack. If you want to discuss this, please start a separate thread.

 

[PeterDonis]

Water waves are apparently produced by processes which create mostly coherent waves. Light, on the other hand, is in many cases thermal radiation, and is incoherent and contains many frequencies.

Again, you're conflating different meanings of "coherent" here.

 

[Demystifier]

Why is it that the wave function of the water wave doesn't collapse even though we are constantly looking at it and therefore constantly making measurement?

Because the measurement of water wave is not described as entanglement with another wave. In other words, the measurement of water wave is a classical measurement that does not change the wave itself.

 

[Happiness]

the measurement of water wave is a classical measurement that does not change the wave itself.

I would have thought all measurements, in one way or another, involve microscopic particles and are hence quantum in nature. If there are two kinds of measurements—one, quantum and the other, classical—then how do you tell them apart properly?

 

[PeterDonis]

I would have thought all measurements, in one way or another, involve microscopic particles and are hence quantum in nature.

Since the QM formalism makes "measurement" a separate thing from unitary evolution, and does not treat it as a quantum interaction, it's actually the opposite as far as the formalism is concerned: all "measurements" are classical. It's just that some such measurements can give results different from those that would be predicted by purely classical physics, while others can't. The double slit measurement is in the first category; measurements of water waves are in the second.

 

[DarMM]

I would have thought all measurements, in one way or another, involve microscopic particles and are hence quantum in nature. If there are two kinds of measurements—one, quantum and the other, classical—then how do you tell them apart properly?

How probabilities for different results mesh together. In classical measurements for two disjoint events $A$ and $B$ we have:
$$P\left(A\ or \ B\right) = P\left(A\right) + P\left(B\right)$$
In Quantum Theory this can fail.

Also note what @PeterDonis said.

 

[Demystifier]

I would have thought all measurements, in one way or another, involve microscopic particles and are hence quantum in nature. If there are two kinds of measurements—one, quantum and the other, classical—then how do you tell them apart properly?

All measurements are ultimately quantum (which doesn't mean that @PeterDonis is wrong, he is right but in a different sense). Quantum measurement always changes the wave function of the measured system. If that change doesn't affect those parts of the wave function that are relevant for the given measurement, we can say that the measurement is "classical". Macroscopic systems have a larger number of degrees of freedom, so it's easier to find some relevant degrees (e.g. the position of the center of the mass) that will not be affected by the measurement. That's why "classical" measurements are usually associated with macroscopic systems.