# Type of wave functions

1. Nov 29, 2015

### Negatratoron

What's the type of wave functions? Is it just: function from a point in spacetime to Z; takes a location and returns an amplitude in discrete units?

(bonus question: according to your favorite theory, what is the type of points in spacetime (that is, topology of spacetime)? is it like r^n for some number of dimensions n, or is it a more discrete construction?)

2. Nov 30, 2015

### Staff: Mentor

Wave-functions are expansions in terms of position eigenfunctions.

Let O be an observable with orthronormal eigenfunctions |bi> so |bi><bi| = I. If |u> is any state then:
|u> = Σ |bi><bi|u> = ci |bi>.

The ci go over to a continuum for things like position and by definition is called the wave-function.

According to GR space-time is a pseudo-Riemannian manifold.

Thanks
Bill

Last edited: Nov 30, 2015
3. Nov 30, 2015

### Staff: Mentor

The wave function of a particle is typically written as $\psi(\vec{r},t)$; it's a complex-valued function of time and position in three-dimensional space. You can easily transform it to the momentum basis and write it as a function of momentum and time instead. I'm not sure what you mean by "discrete units" - probability amplitudes are continuous.

In non-relativistic QM (which is where we are if we're talking about "the wave function of a particle") we're working in three-dimensional Euclidean space. In relativistic theories, we work with the four-dimensional Minkowski spacetime. There's no quantum foam or other "more discrete construction".

4. Nov 30, 2015

### Staff: Mentor

Z is the set of integers, so I suspect he thinks that $\psi$ takes on discrete values.

5. Nov 30, 2015

### Negatratoron

Interesting, I did not know you could express a wave function using a momentum basis. And, probability amplitudes are continuous - got it.

Okay, makes sense. I know there are theories that take advantage of some additional dimensions (10, 11, 26 or however many). This may not be related to QM, but what do physicists introduce some of those extra dimensions for? And how are some of these extra dimensions shaped: closed intervals, open intervals, circles, something else?

Yeah, that's exactly what I was thinking.

6. Nov 30, 2015

7. Dec 1, 2015

### Staff: Mentor

There is an infinite amount of bases you can use to express the wave function. Position and momentum are just two that are often useful.
Especially for bound states, energy eigenstates are another useful basis.

The shape depends on the model. They lead to different physics, ideally to explain something that other models cannot. Sometimes they are just necessary for a model, e.g. string theory with just 3 spatial dimensions wouldn't work.
ψ doesn't have discrete values.

8. Dec 1, 2015

### Negatratoron

Infinite bases, I believe. You can even choose X, Y, and Z axes infinitely many ways.

Not sure I understand the energy eigenstates basis. Let me attempt to explain it: Given a context (should I call it a box? universe?), certain standing waves can exist in that context. These correspond to energy eigenstates. The wave function of a particle is given as a linear combination of those eigenstates.

Infinitely many standing waves can exist in a perfect rubber band - just harmonics of each other. But in an oddly-shaped box, are there still infinitely many or can there be only finitely many eigenstates?

Also: do we typically model particle interactions using the particles' wave functions, or using some other description of them?

9. Dec 1, 2015

### Staff: Mentor

In a box with finite volume and infinite potential walls, an infinite number of energy eigenstates exist*.
With finite potential walls, you always have an infinite number of unbound states. The number of bound states can be finite or infinite, depending on the potential.

*The eigenstates form a basis, therefore every possible wave function can be expressed as linear combination of those wave functions. Imagine a finite set of eigenstates: you would have a largest one which defines the maximal energy. It is possible to find a state that has a higher energy => contradiction.
In the case of finite walls you simply have unbound states beyond some energy threshold.
In particle physics, one usually considers particles in the momentum basis, evaluates their scattering amplitude there, and then uses this result for something that looks like colliding particles. An eigenstate of the momentum basis is not a proper particle (it is spread out over the whole universe, that is not a realistic model for colliding particles).