In plain old QM, we can in principle take any state and expand it either in terms of momentum eigenstates or position eigentstates (i.e., the wave function). But in practice this usually means solving the Schrodinger equation, i.e., working in the postion representation from the start and then going over to the momentum representation later, if needed.(adsbygoogle = window.adsbygoogle || []).push({});

For instance for a free particle we start with

[tex]\frac{\hat{p}^2}{2m}|\psi(t)\rangle=i\frac{\partial}{\partial t}|\psi(t)\rangle[/tex]

and go over to

[tex]-\frac{1}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)=i\frac{\partial}{\partial t}\psi(x,t)[/tex]

This has the set of solutions [tex]\psi(x,t)\propto e^{i(kx-\frac{k^2}{2m}t)}[/tex] where k is an integration constant. Then we apply specific boundary conditions depending whether this is particle in a box, a free particle, the 2-slit experiment, etc.

But why not start in the momentum representation? Well the trouble is that in the momentum representation the momentum operator is represented by multiply-by-p. So we have simply

[tex]\frac{p^2}{2m}\phi(p,t)=i\frac{\partial}{\partial t}\phi(p,t)[/tex]

the solution to which is

[tex]\phi(p,t)\propto g(p)e^{-i\frac{p^2}{2m}t}[/tex]

where g(p) is any old function of p. How do we get that g must take the form [tex]A e^{i\beta p}+Be^{-i\beta p}[/tex], where beta is a constant (later to be identified with position)?

I was thinking that, hey, maybe some problems are easier to solve in the momentum representation from the start instead of the position rep, but I can't even figure out how to do the free particle.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# How to work in the momentum representation?

**Physics Forums | Science Articles, Homework Help, Discussion**