Physical meaning of adding two spin operators

Question

When we add who spin operators, one in $x$ and the other in $z$ direction, dose it make any physical sense? Do the eigenvalues of this sum is also equal to a sum of their eigenvalues?

Answer 1

There are two ways of understanding the question.

Given the unit vector $\hat n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$, the general linear combination $\hat S_{\hat n}=\hat n\cdot \vec S= n_xS_x+n_yS_y+n_zS_z$ is a legitimate spin operator.

The eigenstates of $\hat S_{\hat n}$ are spin-states in the direction of $\hat n$. The eigenvalues of $\hat S_{\hat n}$ are... $\pm\hbar/2$. Mathematically, the operators $\hat S_i$ do not commute, so you cannot simply add the eigenvalues of each operator.

Physically, the direction of the quantization axis is arbitrary, so choosing to quantize along $\hat n$ will result in the same possible outcomes as quantizing along $\hat z$, i.e. $\pm \hbar/2$. (Remember that these possible outcomes are the eigenvalues.) By the same logic one concludes correctly that the eigenvalues of $\hat S_i$ are all the same, and in fact, in a $(2j+1)\times(2j+1)$ dimensional space of a particle with total angular momentum $j$, the eigenvalues of $J_i$ are all the same.

Note that specific combinations of the type above are extremely useful in quantum cryptography and at the core of the famous BB84 quantum crypto protocol.

The second way to understand your question is adding spin operators operating on different systems, i.e. $\hat S_i=\hat S^{(1)}_i+\hat S^{(2)}_i$ with $\hat S_i$ the total spin of the system along axis $i$. Since all operators acting in space 1 commute with all operators acting in space 2, the eigenvalues are then additive, in the same way that the total energy of a system with Hamiltonian $\hat H_1+\hat H_2$ is $E^{(1)}_{n_1}+E^{(2)}_{n_2}$. In your specific case, the eigenvalues of your total spin operators would be the $1,0,0,-1$.

Answer 2

If you consider the base $|j,m\rangle$ with $\hat{L}_z|j,m\rangle=\hbar m|j,m\rangle$ then you can consider a generic state as a combination of elements of this base, and impose it is an eigenstate of $\hat{L}_z+(\hat{L}_++\hat{L}_-)/2$ $$ \left( \hat{L}_z+\frac{\hat{L}_++\hat{L}_-}{2} \right) \sum\limits_{j,m} c_{jm}|j,m\rangle = \hbar \alpha \sum\limits_{j,m} c_{jm}|j,m\rangle $$ and this way you'll obtain the equation $$ \sum\limits_{j,m} c_{jm} \hbar m|j,m\rangle + \frac{1}{2} \sum\limits_{j,m} c_{jm} \hbar \sqrt{ (j-m)(j+m+1) } |j,m+1\rangle + \frac{1}{2} \sum\limits_{j,m} c_{jm} \hbar \sqrt{ (j+m)(j-m+1) } |j,m-1\rangle = \hbar \alpha \sum\limits_{j,m} c_{jm}|j,m\rangle $$ that you can rearrange as $$ \sum\limits_{j,m} \left( m c_{jm} + \frac{1}{2} \sqrt{(j-m+1)(j+m)} c_{j(m-1)} + \frac{1}{2} \sqrt{(j+m+1)(j-m)} c_{j(m+1)} - \alpha c_{jm} \right) |j,m\rangle = 0 $$ Since $|j,m\rangle$ are orthonormal you'll find $$ m c_{jm} + \frac{1}{2} \sqrt{(j-m+1)(j+m)} c_{j(m-1)} + \frac{1}{2} \sqrt{(j+m+1)(j-m)} c_{j(m+1)} - \alpha c_{jm} = 0 $$ that you can ideally solve as a recurrence relation. I stop here since I don't know if a solution exists and what is its form, but that's what I would do in absence of other ideas.