# Why wavefunction is not seen as substance distribution?

1. Sep 24, 2015

### zhanhai

Why wavefunction (the square of its modulus) of an electron is not seen as a measure of substance/charge distribution of the electron?

2. Sep 24, 2015

### Staff: Mentor

That model only works if you're looking at the wave function of a single particle written in the position basis. In that case and only in that case will the wave function take the form $\psi(x,t)$ so that you can interpret it as the distribution of some substance at position $x$ and time $t$.

Thus, it's a dead end - you'll abandon it for the more general mathematical formalism before you're half-way through your first QM textbook.

3. Sep 29, 2015

### jk22

If you see it as a substance you get maybe the problem of measurement it should shrink to at least a peak when you measure position.
I suppose in copenhagen it is the knowledge about the position.

We have the same problem in classical mechanics : imagine you rotate your coordonate system then the coordinates of all objects change as far as infinity. No computer could do this instantaneously.

4. Sep 29, 2015

### stevendaryl

Staff Emeritus
The problem with thinking of the wavefunction as a physical quantity, or field, is that it doesn't exist in space, it exists in configuration space.

What I mean by that is this: Suppose we have two particles. The wave function for that pair is a function of the form:

$\Psi(x_1, y_1, z_1, x_2, y_2, z_2)$

which gives the probability amplitude for finding the first particle at $(x_1, y_1, z_1)$ and the second particle at $(x_2, y_2, z_2)$.

When you square it, you don't get the probability of finding anything at $(x_1, y_1, z_1)$, or of finding anything at $(x_2, y_2, z_2)$. You get the probability of simultaneously finding one particle at one location and the other particle at the other location.

5. Sep 30, 2015

### zhanhai

This makes very good sense. But it assumes that the wave functions (WFs) of the two particles cannot be separated. The general validity of this assumption should be by itself related to the subject question. On the other hand, when the two particles' wave functions can be separated, the overall wavefunction of the two, as a product of individual WFs (or a summation of such products), would be more or less artificial, and the proposed understanding of that WF of each particile be seen as the substance distribution of that particle can still make sense.

6. Sep 30, 2015

### stevendaryl

Staff Emeritus
Particle wave functions only separate if there is no interaction between them. So you can't think of the wave function of interacting particles as somehow giving the "density" of particle stuff. Quantum mechanics tells us how to compute probabilities for situations, where a situation may involve many different particles that are far apart.

If you try to force a particle density interpretation where that is possible (that is, when the wave functions are factorable), then how do you interpret it when it evolves into a wave function that is not factorable?

7. Sep 30, 2015

### Staff: Mentor

That assumption is built into the formalism of quantum mechanics. Any time you see two particles being treated as if their wave functions are separate entities, you are looking at an approximation (although if one of the particles is on earth and the other one is in the Andromeda galaxy, it's a really good approximation). Thus, the idea that $\psi(x,t)$ represents a the density of some material substance stops working as soon as you replace the approximation with the exact solution in which the wavefunction cannot be written in that form.

Another way of understanding stevendaryl's point about the wavefunction not existing in physical space is to rewrite it in the momentum basis (which you'll have to do for any field theory problem anyway). If $\psi(x,t)$ represented the density of some physical substance, what am I to make of $\phi(p,t)$? It describes a density-like distribution of momentum, but distributed through what?