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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?

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    $\begingroup$ Related. $\endgroup$ May 28 at 11:27
  • $\begingroup$ Interesting question. Consider that if you consider the base $|m\rangle$ of $\hat{L}_z$ eigenstates, then $\hat{L}_x=\hat{L}_++\hat{L}_-$. So you have now $\hat{L}_z+(\hat{L}_++\hat{L}_-)/2$ where $\hat{L}_z$ has $|m\rangle$ eigenstates, $\hat{L}_-$ has coherent states as eigenstates and $\hat{L}_+$ has no eigenstates. I don't know so if there is an eigenstate of the sum of this operators, but it's interesting: good luck finding it! $\endgroup$
    – Rob Tan
    May 28 at 12:17
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    $\begingroup$ You already give a calculable example in your question. $\sigma_x+\sigma_z$ has eigenvalues $\sqrt 2 and -\sqrt 2$ which is not $1+1=2$ or $-1-1=-2$. Another information: density matrix, e.g. qubit case, can be written into form of $\frac{I+\vec{r}\cdot \vec{\sigma}}{2}$. $\endgroup$
    – narip
    May 28 at 12:28
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    $\begingroup$ @RobTan (angular momentum) coherent states are not eigenstates of $\hat L_-$. The lowering operators do not have eigenstates in a finite dimensional space except for the null vector. $\endgroup$ May 28 at 13:50
  • $\begingroup$ @ZeroTheHero finite dimensional because $-j<m<j$? Ok, I didn't know that $\endgroup$
    – Rob Tan
    May 28 at 14:42

2 Answers 2

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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 $1,0,0,-1$.

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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.

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    $\begingroup$ Note that if this equation implies $c_{jm}=0\,\forall\,j,m$ and forall $\alpha\in\text{R}$ there are not eigenstates and so no eigenvalues: in fact what I'm asking here is if eigenstates exist. Once responded, if someone is able to, we could determine eigenvalues and respond to your question. $\endgroup$
    – Rob Tan
    May 28 at 13:03

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