# What determines whether or not $|jm\rangle$ can be rotated into $|jm'\rangle$?

This isn’t a full answer, but an easy way to see you can’t reach every state with a rotation is that the set of rotations has three real dimensions, while the set of normalized spin $s$ states has real dimensions $2(2s+1) - 2 = 4s$ real dimensions, where the subtraction is due to tossing out a global phase and demanding normalization. This means that for spin $s=1$ and higher, almost *all* states can’t be reached by a rotation, just on dimensional grounds. The classical intuition doesn’t work, because a “classical” spin always has a two-dimensional state space.

Are there any arguments on physical grounds we could have made to arrive at the above conclusion without going through the math?

**Yes.** This answer uses some mathematial *notation*, but it doesn't use any matrices. It uses only geometric intuition and the most basic general principles of quantum physics.

## Notation

Consider an irreducible representation of the spin group (the double cover of the rotation group), and use this notation:

$\hat u$ is a unit vector in 3d space.

$J(\hat u)$ is the generator of rotations about the $\hat u$-axis. For every axis $\hat u$, the generator $J(\hat u)$ is an observable represented by a self-adjoint operator on the Hilbert space.

$j$ is the largest eigenvalue of $J(\hat u)$ in the given irreducible representation.

$\big|\hat u\big\rangle$ is the eigenstate of $J(\hat u)$ with the largest eigenvalue $j$: $$ J(\hat u)\,\big|\hat u\big\rangle = j\,\big|\hat u\big\rangle. $$

We can think of $|\hat u\rangle$ as having an angular momentum with a specific orietnation in 3d space, namely along the $\hat u$-axis, but that interpretation isn't necessary for this answer to work.

## Intuition

Rotations affect observables in the obvious way: applying a rotation $\hat u\to R\hat u$ transforms $J(\hat u)\to J(R\hat u)$. This immediately implies two things:

All of the states $|\hat u\rangle$, for all directions $\hat u$, can be obtained from each other by rotations, up to a physically irrelevant overall phase factor. (A state can be represented by a vector in the Hilbert space, but the representation is not one-to-one: vectors related to each other by an overall nonzero complex constant represent the same state.)

For any given direction $\hat u$, the

*only*states that can be obtained by rotating $|\hat u\rangle$ are largest-eigenvalue eigenstates of the generators $J(R\hat u)$.

Rotating clockwise about $\hat u$ is the same as rotating counterclockwise about $-\hat u$, so the eigenstate of $J(\hat u)$ with eigenvalue $j$ corresponds to the eigenstate of $J(-\hat u)$ eigenvalue $-j$. This implies:

- The eigenstate of $J(\hat u)$ with eigenvalue $-j$ is a rotation of the eigenstate of $J(\hat u)$ with eigenvalue $j$.

The case $j=1/2$ is special, because then $J(\hat u)$ doesn't have any other eigenstates: the only two eigenstates it has are the ones with eigenvalues $\pm j$.

Now consider a representation with $j>1/2$, and consider an eigenstate $|m\rangle$ of $J(\hat u)$ with eigenvalue $m\neq \pm j$. The operator $J(\hat u)$ is the generator of rotations about the $\hat u$ axis, so $|m\rangle$ must be invariant under rotations about the $\hat u$-axis, up to a physically meaningless overall phase. On the other hand, if $|m\rangle$ could be obtained from $|\hat u\rangle$ by a rotation, then $|m\rangle$ would be an eigenstate of $J(R\hat u)$ for some other direction $R\hat u$ with the largest eigenvalue $j$. But $J(R\hat u)$ is not invariant under rotations about the $\hat u$-axis unless $R\hat u=\pm\hat u$, so that would contradict the fact that $|m\rangle$ must be invariant under rotations about the $\hat u$-axis. Altogether, this implies

- The eigenstate of $J(\hat u)$ with eigenvalue $-j$ is a rotation of the eigenstate of $J(\hat u)$ with eigenvalue $j$, and the other eigenstates of $J(\hat u)$ cannot be obtained by rotating the one with eigenvalue $j$.

## By the way

The answer above focused on eigenstates of $J(\hat u)$, but as knzhou's answer pointed out, we can also make a more general statement: in a representation with $j>1/2$, most states are not equal (or proportional) to $|\hat u\rangle$ for any direction $\hat u$. Every state can be expressed as a superposition of these, but in most cases the superposition needs to involve more than one direction in 3d space. Furthermore, the superposition is not unique: the same state may be expressed as different superpositions, involving different sets of directions in 3d space. This follows simply from the fact that the basis $\big\{|\hat u\rangle\big\}$ is overcomplete, which in turn follows from the fact that the parameter $\hat u$ is continuous: an irreducible representation has a basis with a finite number of vectors, so the basis parameterized continuously by $\hat u$ must be overcomplete.