# Expectation values for Hydrogen

1. Apr 8, 2013

### TheRascalKing

Ok, so I'm a little confused about why <p> = 0 for Hydrogen in the ground state. If someone explain the reasoning behind this, I'd greatly appreciate it.

Also, and more importantly, does that mean that <p> = 0 for Hydrogen in other states as well? If not, how would you go about finding <p> for these excited states. I've searched google to no avail.

Sorry if it's a stupid question, I'm new to the QM game.

2. Apr 8, 2013

### The_Duck

In general, <p> = 0 for any bound state of any potential (as long as the potential is constant in time). One way to convince yourself of this is to recall that <p> = d<x>/dt (Ehrenfest's theorem). But an energy eigenstate is a stationary state: all expectation values should be constant in time. Therefore 0 = d<x>/dt = <p>.

Determining where the above argument goes wrong for unbound states, such as the energy eigenstates of a free particle, is left as an exercise to the reader.

3. Apr 9, 2013

### tom.stoer

think about what a non-vanishing expectation value <p> would mean; p is a vector-valued operator, so any non-vanishing momentum must necessarily break rotational invariance, it would mean that <p> points into some direction; but this is unreasonable for the ground state of a system with rotational invariance

of course for p² which is a scalar operator the expectation value <p²> is non-zero