# Angular momentum of an electron

1. May 31, 2005

### broegger

Hi,

Say you have an electron in the hydrogen atom. Can this electron be in a state that is a superposition of the usual energyeigenstates (which are also eigenstates of L_z) AND have different expectation values for L_x and L_y? Are x and y symmetric in the sense that their expectation values are always equal in this situation?

2. May 31, 2005

### Kruger

You can't measure the expectation value of L_z and L_y simultaneously cause of the uncertainty. You can only measure total angular momentum and L_y or total angu. mom. and L_z precisely.

3. May 31, 2005

### dextercioby

You can measure them all (5) with arbitrary precision,if the quantum system is in a state with $l=0$.

Daniel.

4. Jun 1, 2005

### broegger

No no, I think you misunderstand. The EXPECTATION value is not the result of a single measurement. This is what I mean: Can you have a particle in a state, which is a superposition of eigenstates of L_z, but which - at some point in time (say t=0) - has different EXPECTATION values for L_x and L_y.

The reason I'm asking is that I found just that in my exam the other day. The assignment was this:

In this problem we consider a hydrogen atom whose normalized stationary states is denoted $$|nlm\rangle$$, where n is the main quantum number and l and m is the quantum numbers belonging to L^2 and L_z, respectively. Consider a hydrogen atom described by the state:
$$|\psi\rangle=\frac1{\sqrt2}(|210\rangle + |211\rangle)$$​

1. Blah blah
2. Blah blah
3. Blah blah
4. Calculate the expectation values of L_x and L_y in the state |psi>.

I got different answers for <L_x> and <L_y> (h/sqrt(2) and 0, respectively). Is this possible? When you choose z as reference axis the theory seems to treat x and y equally.

5. Jun 1, 2005

### dextercioby

There's no problem whatsoever.The two operators are different (see their expressions as a function of the ladder operators),so it's obvious to get different expectation values.

Daniel.

6. Jun 2, 2005

Ok, thanks.

7. Jun 2, 2005

### BlackBaron

That's right, but your specific state doesn't treat x and y equally. So, as dextercioby already pointed, there's no contradiction if you find different expectation values.

8. Jun 4, 2005

### BoTemp

Choosing z as the reference axis doesn't necessarily treat x and y equally, it's
just a convention. Only m=0 states have symmetry about the z-axis. So <L_x> = <L_y> (= 0 I think) whenever m= 0, but since $$|211\rangle$$ is part of
the wavefunction, it doesn't have the symmetry.