Wavefunction of a space and spin rotated state

[NiRK20]

Homework Statement

Basically, I have the wavefunction of a particle with spin (not necessarly 1/2) given by $\psi(\textbf{r}, m) = \psi _{m} (\textbf{r}) = \langle\textbf{r}, m|\psi \rangle$. My task is to find the wavefunction of a rotated state $U(R)|\psi \rangle$, with $U$ being the product of a spacial rotation by a spin rotation.

Relevant Equations

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Since $U$ is a space and spin rotation, it would be

$$U(R) = e^{-i\textbf{L}\cdot \hat{\textbf{n}}\phi/\hbar}\cdot e^{-i\textbf{S}\cdot \hat{\textbf{n}}\phi/\hbar}$$

And, then

$$\psi'(\textbf{r}, m) = \langle\textbf{r}, m|e^{-i\phi(\textbf{L} + \textbf{S}) \cdot \hat{\textbf{n}}/\hbar}|\psi \rangle = \sum_{m'}\int \langle\textbf{r}, m|e^{-i\phi(\textbf{L} + \textbf{S}) \cdot \hat{\textbf{n}}/\hbar}|\textbf{r}', m'\rangle\psi _{m'}(\textbf{r}') d^{3}r'$$

The problem is where to go from here (if this is right until now). There is a way to compute these matrix elements? Does $|\textbf{r}\rangle$ equals to the eigenkets of angular momentum $|nlm\rangle$?

 

[DrClaude]

It is difficult to answer because it depends on what is expected. Can you post the full text of the question?

$$U(R) = e^{-i\textbf{L}\cdot \hat{\textbf{n}}\phi/\hbar}\cdot e^{-i\textbf{S}\cdot \hat{\textbf{n}}\phi/\hbar}$$

How does $U(R)$ become a function of $\phi$ only?

Does $|\textbf{r}\rangle$ equals to the eigenkets of angular momentum $|nlm\rangle$?

No, they are completely different. This is how you recover the wave function for a given $n,l,m$ from the ket:
$$
\psi_{n,l,m} (\mathbf{r}) = \braket{\mathbf{r} | nlm}
$$