# Hamiltonian and symmetries

1. Jun 5, 2010

### eoghan

[QM] Hamiltonian and symmetries

1. The problem statement, all variables and given/known data
Let there be the hamiltonian:
$$H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2(x^2+y^2+z^2)+kxyz+\frac{k^2}{\hbar \omega}x^2y^2z^2$$
Find the expectation value of the three components of $$\vec r$$ in the ground state using ONLY the symmetry properties of the hamiltonian.

2. Relevant equations

3. The attempt at a solution
I define this parity:
$$\Pi_{xy}: x\rightarrow-x\ \ \ \ y\rightarrow-y$$
Then the hamiltonian commutes with this parity: $$[H, \Pi_{xy}]=0$$
The ground state is not degenerate, so it has a definite parity with respect to $$\Pi_{xy}$$:
$$<gs|x|gs>=<gs|\Pi_{xy}\Pi_{xy}x\Pi_{xy}\Pi_{xy}|gs>=-<gs|\Pi_{xy}x\Pi_{xy}|gs>=-<gs|x|gs>$$
So <gs|x|gs>=0;
Is it right?

Last edited: Jun 5, 2010
2. Jun 5, 2010

### kuruman

Re: [QM] Hamiltonian and symmetries

That is not the definition of the parity operator I am familiar with. What happened to
$$z\rightarrow-z?$$
Only because you defined "this parity" so that it commutes.

Last edited: Jun 5, 2010
3. Jun 6, 2010

### eoghan

Yes, I know this is not the usual parity... but I don't really know any other way to solve this problem

4. Jun 6, 2010

### kuruman

As I mentioned earlier, your Hamiltonian is invariant under cyclic permutations. In other words, the system cannot distinguish x from y from z (it doesn't know the alphabet). What do you think that implies about the expectation values <x>, <y> and <z>?

5. Jun 7, 2010

### eoghan

That they are all the same?

6. Jun 7, 2010

### kuruman

Correct. Since they are all the same, calculating one of them will give you the others. So what do you think that value can be and why?