# Time development of Expectation values

1. Oct 21, 2004

### Ed Quanta

Here is a question that I have found in my quantum text which I have been thinking about for a few days and am unable to make sense of.

If there is an operator A whose commutator with the Hamiltonian H is the constant c.

[H,A]=c

Find <A> at t>0, given that the system is in a normalized eigenstate of A at t=0 corresponding to the eigenvalue of a.

-So here is what I am thinking. Please tell me what I am doing wrong.

I know that where p is momentum operator [H,p]=ihdV/dx, and V is some potential in the Hamiltonian

and d<p>/dt=-<dV/dx>

So where [H,A]=c, wouldn't d<A>/dt=0? And then there would be no change in the eigenstate of A with time, and <A>=a which is its initial value at 0, which would be a(Psi) where Psi is my wave function or eigenvector or whatever I need it to be.

Am I understanding the question wrong?

2. Oct 21, 2004

### Tom Mattson

Staff Emeritus
Nope.

Check out Heisenberg's equation of motion:

(-i*hbar)(dA/dt)=[H,A]

So if [H,A]=c, then the operator A is evolving in time at a constant rate. The only way you will have dA/dt=0 is if c=0.

More http://farside.ph.utexas.edu/teaching/qm/fundamental/node31.html [Broken].

Last edited by a moderator: May 1, 2017
3. Oct 21, 2004

### Ed Quanta

Ok, I follow your argument as to why dA/dt=0, but what I am unclear about is what this does to <A>. And what is the significance of eigenvalue a of the eigenstate A in describing <A> where t>0? We know at t=0,A=a I believe.

4. Oct 22, 2004

### lalbatros

Solve the equation -i hbar dA/dt = c for A.
From this calculate the time evolution for <A> with the given initial value.

I did not try it. The fact that the initial average value is an eigenvalue will play a role in the second step: you know the initial state.

Last edited: Oct 22, 2004