Rotating eigenstates of J operator into each other?

[QualTime]

Homework Statement

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

<br /> |j,j &gt; , ... , |j,m &gt; , ... | j , -j &gt;<br />

where
<br /> \hbar^2 j(j+1) , \hbar m<br />
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
<br /> |j,m&#039; &gt; = U^{(j)} |j, m &gt;

***

Not too sure how to approach this problem, although given that
<br /> U^\dagger |j,m&#039; &gt; = U^\dagger U |j,m &gt;
and
<br /> &lt; j,m&#039; | j,m &gt; = \delta_{mm&#039;}<br />
I would think that
<br /> &lt; j,m&#039; | U^\dagger | j,m&#039; &gt; = 0<br />

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 homework question but taken from a practice exam so feel free to go into as much detail as possible as that would be really helpful).

 

[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&#039; \rangle \langle j, m&#039; \mid + \mid j, m&#039; \rangle \langle j, m \mid - \mid j, m \rangle \langle j, m&#039; \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.

 

[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.