How Do Symmetries Determine Expectation Values in Quantum Mechanics?

[eoghan]

[QM] Hamiltonian and symmetries

Homework Statement

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.

Homework Equations

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?

 

[kuruman]

I define this parity:
$$\Pi_{xy}: x\rightarrow-x\ \ \ \ y\rightarrow-y$$

That is not the definition of the parity operator I am familiar with. What happened to
$$z\rightarrow-z?$$

Then the hamiltonian commutes with this parity: $$[H, \Pi_{xy}]=0$$

Only because you defined "this parity" so that it commutes.

 

[eoghan]

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

 

[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>?

 

[eoghan]

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>?

That they are all the same?

 

[kuruman]

That they are all the same?

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?