# Rotating eigenstates of J operator into each other?

1. Aug 2, 2014

### QualTime

1. The problem statement, all variables and given/known data

Consider the following set of eigenstates of a spin-J particle:

$$|j,j > , ... , |j,m > , ... | j , -j >$$

where
$$\hbar^2 j(j+1) , \hbar m$$
are the eigenvalues of J^2 and Jz, respectively. Is it always possible to rotate these states into each other? i.e. given |j,m> and |j,m'>, is it always possible to find a unitary rotation operator U^j such that
$$|j,m' > = U^{(j)} |j, m >$$

***

Not too sure how to approach this problem, although given that
$$U^\dagger |j,m' > = U^\dagger U |j,m >$$
and
$$< j,m' | j,m > = \delta_{mm'}$$
I would think that
$$< j,m' | U^\dagger | j,m' > = 0$$

which doesn't seem right hence the answer would be no.

Also the fact that the rotation matrix times a given eigenstate is in general a linear combination of 2j+1 independent states of the form |j,m'> makes me doubtful as well.

Any help would be appreciated (this isn't an actual hw question but taken from a practice exam so feel free to go into as much detail as possible as that would be really helpful).

2. Aug 3, 2014

### MisterX

What do you mean by rotation? I think it is always possible to construct such a unitary operator. One could just do it explicitly.

$U = \mathbf{I} - \mid j, m \rangle \langle j, m \mid - \mid j, m' \rangle \langle j, m' \mid + \mid j, m' \rangle \langle j, m \mid - \mid j, m \rangle \langle j, m' \mid$

I think the above would work. But do you consider that a rotation? I suppose one might consider all special unitary operators to be rotations in a sense (rotations in the Hilbert space). However I don't know if you mean rotations as in rotations in 3D space, that is $U = e^{-i\theta \hat{\mathbf{n}} \cdot \mathbf{J}}$. If this is what you mean, then I'm not sure of the answer. I know such rotations will often result in superpositions of m states, which is not what you want.

Last edited: Aug 3, 2014
3. Aug 4, 2014

### erogard

Rotation is a bit of a misnomer indeed. I think unitary transformation is what is meant here. I'll try your first suggestion (just need to check it's indeed unitary) but it looks good to me. Thanks.