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

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 $$\hbar$$?

• You could prove by induction that $[S_z,S^n_{+}]=n S_{+}^n$, and if your life dependent on it, and you were brave, you could analytically continue n to a non-integer real, like 1/2 or 1/3 ... people have. I suspect you don't mean to tread there... Nov 27, 2022 at 17:11
• @CosmasZachos What do you mean by analytically continue? As in it will lead to a nonsensical result or continue as in you could extend raising/lowering operators to cover non integer applications of raising/lowering operators? Nov 27, 2022 at 18:17
• As I said, judicious definitions of such operators are possible, provided you ensure the absence of nonsense… but this is a whole different question. You might be familiar with the freak operator whose eigenvalue is $\ell$… Nov 27, 2022 at 18:55
• I think what Silly Goose is alluding to is that there is a whole beautiful theory of what happens when you allow complex-valued angular momenta. In particle physics it is known as Regge Theory (en.wikipedia.org/wiki/Regge_theory) and it is useful when evaluating high-energy limits among other things. The approach is actually much older, popularized by Sommerfeld's Partial Differential Equations in Physics. It hails from 19th century studies of the performance of antennas in a conductive atmosphere. Which should give you an idea of the power of replacing integers with complex numbers. Nov 28, 2022 at 1:57

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.