Position and momentum (basic stuff)

[broegger]

I was just reading about the free particle and how it can be represented as a superposition of momentum eigenstates and so on. I came to think about this:

Let's say we have free particle whose initial state is a Dirac delta function type of thing (localized position). The momentum distribution will then be a uniform distribution over all values. As time goes on the wave function will spread out according to the Schrödinger equation. But what happens to the momentum distribution; does it become more localized? It's just the Fourier Transform of the wave function, right?

This leads me to another question. The wave function, \Psi, seems to have some sort of preference for the "position" observable (I mean, when you take the norm-square, you get the position-distribution, you have to Fourier Transform it to get the momentum distribution and so on). Couldn't you have an equivalent "momentum-based" wave function, so that when you Fourier Transform this wave function you get the position distribution. Indeed, couldn't you have such equivalent wave functions for all observables (maybe there is, I've never met them)?

 

[dextercioby]

This leads me to another question. The wave function, \Psi, seems to have some sort of preference for the "position" observable (I mean, when you take the norm-square, you get the position-distribution, you have to Fourier Transform it to get the momentum distribution and so on). Couldn't you have an equivalent "momentum-based" wave function, so that when you Fourier Transform this wave function you get the position distribution. Indeed, couldn't you have such equivalent wave functions for all observables (maybe there is, I've never met them)?

I'll let someone else handle the first part.
There is no preference whatsoever,QM is very abstract.However,potentials (which would enter the Hamiltonian) are not.They depend in general on \hat{q} ...So that's why people prefer coordinate representation,especially in atomic physics...

Daniel.

 

[reilly]

Here's a hint or two about how to answer your first question. First, past t=0, a free particle wave function will be exp( +i px - i Et) -- take your pick of energy. E = p*p/2m is the easist. For general t, the wavefunction that's a delta at t=0, involves a Gaussian integral, which can be done.

People use momentum space representations all the time -- from Quantum Field Theory to basic QM. Configuration space methods tend to be associated with bound state problems, while momentum space methods tend to be associated with scattering problems.

 

[kanato]

...So that's why people prefer coordinate representation,especially in atomic physics...

In contrast, in condensed matter physics, almost everything is done in the momentum (or really the k-vector) basis. Usually the very first thing people do is Fourier transform the potential so we can write down Bloch states and such.

Couldn't you have an equivalent "momentum-based" wave function, so that when you Fourier Transform this wave function you get the position distribution. Indeed, couldn't you have such equivalent wave functions for all observables (maybe there is, I've never met them)?

Some problems are easier in the momentum basis. For instance, consider a particle in a uniform field, such as within a capacitor, then you have
E \Psi = (\frac{P^2}{2m} + qFx) \Psi
(using F for field so as to not confuse it with the energy E.)

In position basis, we'd take P -> -i hbar d/dx, but this generates a 2nd order differential equation. However, if we go to the momentum basis, where P is just a coordinate, then x becomes +i hbar d/dp, and this is a first order differential equation.

In condensed matter, working with many-particle wave functions, it's typically common to think of things in terms of an occupation number basis. Thanks to the Pauli exclusion principle, when dealing with electrons only one electron can occupy each quantum state, so you can express the state vector (or wave function) in terms of how many particles occupy each state (0 or 1).