Let's consider two electrons. The spin operator of the system is (omitting vector symbols for simplicity):

$S=S_1\otimes\mathbb I_2+\mathbb I_1 \otimes S_2$

I want to show that an eigenbasis of {$S^2,S_z$} is given by {$\chi_u, \chi_+, \chi_-,\chi_d$} where

$$\chi_u=|uu\rangle$$ $$\chi_d=|dd\rangle$$ $$\chi_+=\frac{1}{\sqrt2}|ud+du\rangle$$ $$\chi_-=\frac{1}{\sqrt2}|ud-du\rangle$$ here $|u\rangle$ is an eigenvector of $S_i^2$ and of $S_{z,i}$ of eigenvalues $\frac{1}{2}$ and $\frac{1}{2}$ and $|d\rangle$ is an eigenvector of $S_i^2$ and of $S_{z,i}$ of eigenvalues $\frac{1}{2}$ and $-\frac{1}{2}$ respectively and for $i=1,2$.

I can figure out how $S_z$ works on this system of two particles, but I'm not quite sure what $S^2$ is here. My euristhic guess (I'm just doing a dot product) is that we could have:

$S^2=S_1^2\otimes\mathbb I_2+\mathbb I_1 \otimes S_2^2+2S_1\otimes S_2$


  1. I don't know if this is true

  2. If 1. is true, then the $2S_1\otimes S_2$ is very annoying and I don't really know how to deal with this and thus how to find the eigenbasis of {$S^2,S_z$}

  • $\begingroup$ Minor comment: If using $\hbar=1$, then $|u\rangle$ and $|d\rangle$ are eigenvectors of $S_i^2$, but with eigenvalues $s(s+1)=3/4$, not $1/2$. $\endgroup$
    – Urb
    Commented Jun 16, 2021 at 17:14
  • $\begingroup$ Related : Total spin of two spin-1/2 particles. $\endgroup$
    – Frobenius
    Commented Jun 16, 2021 at 17:25
  • $\begingroup$ @Urb Yes, sorry, I should have said quantum number instead $\endgroup$ Commented Jun 16, 2021 at 18:26

1 Answer 1

  1. This assertion is essentially true.
  2. To deal with the term $S_1\otimes S_2$, remember that this is a symplified notation for $$S_1\otimes S_2=S_{1x}\otimes S_{2x}+S_{1y}\otimes S_{2y}+S_{1z}\otimes S_{2z},$$ and following OP's notation this acts on vectors $|s_1s_2\rangle$ where $s_i\in\{u,d\}$ as $$(S_1\otimes S_2)|s_1s_2\rangle=S_{1x}|s_1\rangle\otimes S_{2x}|s_2\rangle+S_{1y}|s_1\rangle\otimes S_{2y}|s_2\rangle+S_{1z}|s_1\rangle\otimes S_{2z}|s_2\rangle.$$ OP knows how $S_z$ acts on $|u\rangle$ and $|d\rangle$, namely \begin{aligned}S_z|u\rangle=\frac{\hbar}{2}|u\rangle\\S_z|d\rangle=-\frac{\hbar}{2}|d\rangle\end{aligned} The action of $S_x$ and $S_y$ on the eigenvectors of $S_z$ ($|u\rangle$ and $|d\rangle$) can be found for example using their expression in terms of the Pauli matrices: \begin{aligned}S_x|u\rangle=\frac{\hbar}{2}|d\rangle\\S_x|d\rangle=\frac{\hbar}{2}|u\rangle\end{aligned} and \begin{aligned}S_y|u\rangle=i\frac{\hbar}{2}|d\rangle\\S_y|d\rangle=-i\frac{\hbar}{2}|u\rangle\end{aligned} Do you know how to prove that $\{\chi_u, \chi_+, \chi_-,\chi_d\}$ are eigenvectors of $S^2$ from here?
  • $\begingroup$ Thank you, now I understand all the above relations, but, for example, we have $[S_x, S_y]=S_z\neq 0$ so is it legit to calculate all those values simultaneously? $\endgroup$ Commented Jun 16, 2021 at 18:35
  • $\begingroup$ Yes, to calculate the action of $S^2$ on the $\chi$s we need to calculate all those values, there's nothing wrong with that. A different thing, which is where I think your confusion is coming, is that since $S_x$ and $S_y$ don't commute with each other and don't commute with $S_z$, the $\chi$s are not eigenstates of $S_x$ and $S_y$, you can check that by applying $S_x=S_{x,1} \otimes\mathbb I_2+\mathbb I_1\otimes S_{x,2}$ to one of the $\chi$s. You will see they are not eigenvectors. $\endgroup$
    – Urb
    Commented Jun 16, 2021 at 19:02

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