Why does wave amplitude indicate particle location in QM?

[kurt101]

TL;DR Summary

Why do we say that wave amplitude tells us where we are likely to find the particle versus where we are likely to find the wave from the particle?

Why do we say that wave amplitude tells us where we are likely to find the particle versus where we are likely to find the wave from the particle? Isn't the later a more accurate description of the QM math?

 

[PeterDonis]

Why do we say that wave amplitude tells us where we are likely to find the particle versus where we are likely to find the wave from the particle?

Because, first, there is no such thing as "the wave from the particle", and second, waves don't have a single spatial location anyway.

Isn't the later a more accurate description of the QM math?

No.

 

[Haborix]

Have a look at this video. I'm amazed every time I watch it. To me, the video shows why it is natural think of the wavefunction as giving the probability density for finding a particle at a location.

 

[kurt101]

Because, first, there is no such thing as "the wave from the particle"

If the wave amplitude does not come from particles, where does it come from? (in the QM mathematical sense)

, and second, waves don't have a single spatial location anyway.

I don't know what you mean by this.

 

[PeterDonis]

If the wave amplitude does not come from particles, where does it come from? (in the QM mathematical sense)

It doesn't "come from" anywhere. The wave function describes the probabilities for different possible measurement results. That's just what the wave function is.

I don't know what you mean by this.

If I ask you what the position of a wave is, what's your answer? There isn't one: the concept doesn't make sense. A wave doesn't have a single particular position the way a particle does. So it doesn't make sense to talk about the probability of finding a wave in a particular position, the way it makes sense to talk about such a probability for a particle.

 

[kurt101]

It doesn't "come from" anywhere. The wave function describes the probabilities for different possible measurement results. That's just what the wave function is.

Each particle contributes to the overall wave function of the system and so the wave function comes from the particles.

If I ask you what the position of a wave is, what's your answer? There isn't one: the concept doesn't make sense. A wave doesn't have a single particular position the way a particle does. So it doesn't make sense to talk about the probability of finding a wave in a particular position, the way it makes sense to talk about such a probability for a particle.

But it does make sense to talk about the contribution to the amplitude at a specific location from a wave source. If you have a particle in a system and then you add a second particle that second particle changes the probability amplitudes in the system. What could have changed the probability amplitudes in the system in this case other than a wave source?

 

[PeterDonis]

Each particle contributes to the overall wave function of the system

No, they don't. Each particle is a degree of freedom of the system, and the wave function describes probabilities for the system. There is no general sense in which the individual particles "contribute" to the wave function.

and so the wave function comes from the particles.

No, it doesn't.

it does make sense to talk about the contribution to the amplitude at a specific location from a wave source.

But the quantum wave function is not a wave with a source. It's not radiation. It's a different thing.

If you have a particle in a system and then you add a second particle that second particle changes the probability amplitudes in the system.

No. In ordinary non-relativistic QM you can't add particles to the system. You have to include all the particles at the start.

In relativistic QFT, particles can be created or destroyed, but "particles" aren't fundamental anyway; quantum fields are. "Adding a particle" is just a particular quantum field operation. Viewing it as "changing the probability amplitudes in the system" is not a useful way to view what is going on.

What could have changed the probability amplitudes in the system in this case other than a wave source?

Your whole conceptual scheme here seems to be wrong. This question isn't even answerable.

 

[vanhees71]

If the wave amplitude does not come from particles, where does it come from? (in the QM mathematical sense)I don't know what you mean by this.

In non-relativistic quantum mechanics you can describe some quantum states (the socalled pure states) of a particle by the wave function $\psi(t,\vec{x})$. It's physical meaning is that
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
is the probability density for the position of the electron when observed at time, $t$.

I've no clue what you mean by "the wave amplitude comes from pariticles".

 

[kurt101]

No, they don't. Each particle is a degree of freedom of the system, and the wave function describes probabilities for the system. There is no general sense in which the individual particles "contribute" to the wave function.

There are two choices. Either the wave function for a system of two particles is the same as the wave function for a system with one particle or it is different.

 

[vanhees71]

No, they don't. Each particle is a degree of freedom of the system, and the wave function describes probabilities for the system. There is no general sense in which the individual particles "contribute" to the wave function.

Particles are not degrees of freedom of the system. The particle is the system. If it's prepared in a pure state, in non-relativistic QM the particle can be described by a wave function $\psi(t,\vec{x})$, which obeys the Schrödinger equation,
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}),$$
where $\hat{H}$ is the Hamilton operator (in position representation). For a particle in an external potential it reads
$$\hat{H}=-\frac{1}{2m} \hat{\vec{p}}^2 + V(\hat{\vec{x}})=-\frac{\hbar^2}{2m} \Delta + V(\vec{x}).$$
The physical meaning is that
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
is the probability density for observing the particle at $\vec{x}$ at time, $t$.

As demonstrated by the nice Youtube movie (though it uses photons rather than particles, where the description with a wave function is impossible and QED must be used, but the qualitative picture is of course the same, i.e., you can for sure find movies of this kind, where massive particles, electrons in the non-relativistic regime are used), a single particle does not appear as some "smeared cloud" but manifests itself as making one dot when registered by the detector (here obviously a scintillation screen or something similar, unfortunately the Youtuber doesn't give a scientific reference/paper, where to find the details of the experiment). Only when accumulating a lot of equally prepared particles going through the slits the predicted probability distribution forms. It clearly shows and interference pattern as the Schrödinger equation predicts.

Of course, also the wave-particle duality is resolved since 1926 with Born's probability interpretation, but that's another story.

No, it doesn't.But the quantum wave function is not a wave with a source. It's not radiation. It's a different thing.

Indeed it's a representations of a single-particle pure quantum quantum state with the above given probabilistic meaning.

No. In ordinary non-relativistic QM you can't add particles to the system. You have to include all the particles at the start.

Here you simply have one particle at a time going through a double slit and being registered at the scintillation screen (and it's position stored).

In relativistic QFT, particles can be created or destroyed, but "particles" aren't fundamental anyway; quantum fields are. "Adding a particle" is just a particular quantum field operation. Viewing it as "changing the probability amplitudes in the system" is not a useful way to view what is going on.

Indeed, and photons cannot be in any sense interpreted as particles. They don't even admit the definition of a position observable in the strict sense. Nevertheless, what also QED provides in this case of the double-slit experiment is the probability for detecting a photon at a certain place on the screen (it's given by the energy density of the electromagnetic field), and this forms an interference pattern when many single photons are used. It's similar as for particles, although there's no "wave function" for photons.

Your whole conceptual scheme here seems to be wrong. This question isn't even answerable.

I think the standard formulation of QT perfectly answers the somewhat vague question in #1.

 

[kurt101]

I've no clue what you mean by "the wave amplitude comes from pariticles".

The wave function $\psi(t,\vec{x})$ in your own words "describe some quantum states of a particle". $\psi(t,\vec{x})$ is a function about waves and waves have amplitude. How can amplitude from the equation $\psi(t,\vec{x})$ come from anything but particles?

 

[PeterDonis]

There are two choices. Either the wave function for a system of two particles is the same as the wave function for a system with one particle or it is different.

Obviously it is different since the two wave functions are on two different Hilbert spaces (the one-particle Hilbert space and the two-particle Hilbert space).

 

[PeterDonis]

Particles are not degrees of freedom of the system.

I was a bit sloppy there, yes. A more precise statement would be: for each particle in the system (if we are using non-relativistic QM), there are three degrees of freedom corresponding to its position/momentum configuration space, plus however many more degrees of freedom it takes to describe their internal properties (like spin).

The particle is the system.

In non-relativistic QM, this is true as you state it for a one-particle system. For multi-particle systems, the particles can be viewed as subsystems.

 

[PeterDonis]

$\psi(t,\vec{x})$ is a function of waves

No, it isn't, it's a wave function. What the wave function is a function of is time $t$ and the particle coordinates $\vec{x}$. It's not a function of waves.

How can amplitude from the equation $\psi(t,\vec{x})$ come from anything but particles?

This has already been answered: the whole idea doesn't even make sense. Your conceptual scheme for this is wrong. You need to discard it and adopt a better one.

 

[kurt101]

No, it isn't, it's a wave function. What the wave function is a function of is time $t$ and the particle coordinates $\vec{x}$. It's not a function of waves.

I wasn't referring to the the parameters of the wave function. I made the correction.

 

[PeterDonis]

I wasn't referring to the the parameters of the wave function. I made the correction.

Your "correction", I presume, is this?

$\psi(t,\vec{x})$ is a function about waves

If so, it doesn't change anything. Your next question is:

waves have amplitude. How can amplitude from the equation $\psi(t,\vec{x})$ come from anything but particles?

The amplitude (squared) gives the probability of finding the system at time $t$ at the position described by $\vec{x}$. It doesn't "come from" anything. It's just a function giving the probability.

Again, your whole conceptual scheme about this appears to be wrong. Continuing to ask questions based on that wrong conceptual scheme will not change anything.

 

[PeterDonis]

The OP question has been answered and the thread is going around in circles. Thread closed.