How can I derive the fact that there are no "non-integral" raising and lowering operators for angular momentum?

Question

I understand the usual argument (as presented in introductory quantum mechanics books) on finding the eigenvalue spectrum for the angular momentum of a spin-1/2 particle. One crucial piece of this argument is the introduction of raising and lowering operators:

$$S_{\pm} = S_x\pm iS_y,$$ which raises/lowers the $S_z$ eigenvalue of a state by $\hbar$.

What I do not understand is: How can I prove that no other operators exists which raise or lower the $S_z$ eigenvalue of a state by a non-integral unit of $\hbar$?

Answer 1

To start with, if you are considering the operators which are linear combination of $S_x$, $S_y$, $S_z$, the answer is yes. In addition, with this condition, note that $S_{\pm}=S_x \pm iS_y$ are the only possible raising/lowering operators for particles of any spin, not just spin-${1 \over 2}$ ones.

The key idea to derive the exact form of $S_{\pm}$ lies in the properties of $S_x$, $S_y$, $S_z$. For simplicity, we let $\hbar=1$ and have the commutators $$[S_x,S_y]=iS_z, \ [S_y,S_z]=iS_x, \ [S_z,S_x]=iS_y \tag{1}$$ The raising/lowering operators $S_{\pm}=\alpha_{\pm}S_x+\beta_{\pm}S_y+\gamma_{\pm}S_z$ where $\alpha_{\pm}$, $\beta_{\pm}$, $\gamma_{\pm}$ are some constants are the operators which satisfy $$[S_z,S_{\pm}]=\rho_{\pm}S_{\pm} \tag{2}$$ where $\rho_{\pm}$ are some constants. Eq. (2) comes from the fact that for any $|\psi\rangle$ being a eigenstate of $S_z$, we require $S_{\pm}|\psi\rangle$ still be eigenstates of $S_z$ with the eigenvalues varying from that of $|\psi\rangle$ by some constants $\rho_{\pm}$. The constraint may sound quite harsh, but as we know, $S_{\pm}$ exist, so we do not have to worry about their existence.

Coming to the uniqueness of $S_{\pm}$, we define the set of all linear combinations of spin operators $H=\{\alpha S_x+\beta S_y+\gamma S_z|\alpha,\beta,\gamma \in \mathbb{C}\}$, and it is useful to consider the linear map $M:H \rightarrow H$ $$M(h)=[S_z,h] \tag{3}$$ And we can express the linear map $M$ in the basis of $S_x$, $S_y$, $S_z$, which is $$\begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \tag{4}$$ which means for any $h=\alpha S_x+\beta S_y+\gamma S_z$, $$M(h)= \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix}=-i\beta S_x+i\alpha S_y$$ If we compare Eq. (2) with Eq. (4), we can find $S_{\pm}=\alpha_{\pm}S_x+\beta_{\pm}S_y+\gamma_{\pm}S_z$ are nothing but the eigenvectors of the matrix in Eq. (4)! Solving them and comparing the solution to the raising/lowering operators, we have $S_{\pm}=S_x\pm iS_y$ as the unique solution to Eq. (2) with $\rho_{\pm}=\pm 1$. Another takeaway message here is that the similar procedure (although they may not be recognized in the same way as in the $\text{su}(2)$ system) occurs in classification of semisimple Lie algebra, which has important applications in particle physics.