If $\boldsymbol{\sigma_{1}}$ is the Pauli vector for a particle and $\boldsymbol{\sigma_{2}}$ for the other particle, why is $\boldsymbol{\sigma_{1}}=-\boldsymbol{\sigma_{2}}$ in the singlet state? I read this in a book and I think there's something fundamental that I'm missing.

What would happen in the triplet states?

Edit: I'm trying to prove that the operator $S_{12}=3(\boldsymbol{\sigma_{1}}\cdot\boldsymbol{\hat{r}})(\boldsymbol{\sigma_{2}}\cdot\boldsymbol{\hat{r}})-\boldsymbol{\sigma_{1}}\cdot\boldsymbol{\sigma_{2}}$ is $0$ acting on a singlet state (and trying to calculate its eigenvalues in the triplet state). If $\boldsymbol{\sigma_{1}}=-\boldsymbol{\sigma_{2}}$ then $S_{12}=0$.

Page 77 of this book (Bertulani, Carlos A. Nuclear Physics in a Nutshell. Princeton University Press, 2007.): https://books.google.es/books?id=n51yJr4b_oQC&printsec=frontcover#v=onepage&q&f=false

  • $\begingroup$ The Pauli vector is the vector whose components are the three Pauli matrices. As such, it does not depend on the particle it is acting on. Is your book talking about the expectation values of the Pauli vector, which would depend on the two-particle state? Please specify. $\endgroup$ – flaudemus Feb 17 at 11:01
  • $\begingroup$ Edited the post. $\endgroup$ – Asier R. Feb 17 at 11:19
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    $\begingroup$ There's some context missing here: 1. When you say "a particle" and "the other particle", does this mean that you are considering two spin-1/2 particles and are decomposing their joint space of states into the triplet spin-1 part and the singlet spin-0 part? 2. What is the $\hat{r}$ in the $S_{12}$ expression you have? $\endgroup$ – ACuriousMind Feb 17 at 12:59
  • $\begingroup$ Yes, it's a proton and a neutron. $\boldsymbol{\hat{r}}$ is the unitary vector in the direction $\boldsymbol{r}$. Edited the post to include the link to the book. $\endgroup$ – Asier R. Feb 17 at 13:18
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    $\begingroup$ I don't mean to be annoying, but: To make questions more accessible and guard against link rot, please include all relevant information, such as the explanation of notation or specific terminology used, in your question. Also include the title and authors of any works cited, so that they can be reconstructed if the link is gone $\endgroup$ – ACuriousMind Feb 17 at 13:27

The eigenvalues of unit Pauli vectors are $\pm 1$ .

The easiest way for you to appreciate the structure is by inspection of the singlet and triplet states in the z-direction, whose highest - $\sigma_z$ states are $$ s = (\uparrow \downarrow - \downarrow\uparrow )/\sqrt{2}, \qquad t= \uparrow \uparrow . $$ (I somewhat perversely chose different $\sigma_z$ eigenstates to remind you the second piece of operator S is rotationally invariant, and for ease of computation below!)

By rotational invariance, you may choose $$\boldsymbol{\hat{r}}= \boldsymbol{\hat{z}},\Longrightarrow (\boldsymbol{\sigma_{1}}\cdot\boldsymbol{\hat{r}})(\boldsymbol{\sigma_{2}}\cdot\boldsymbol{\hat{r}})=\sigma_1^z \sigma_2^z,$$ with eigenvalues -1 for the singlet and 1 for the triplet.

The other term (Dirac exchange operator) resolves to the usual ladder operator (work them out!) and z-components of quadratic Casimir invariants, $$ \boldsymbol{\sigma_{1}}\cdot\boldsymbol{\sigma_{2}}= \sigma_1^z \sigma_2^z + (\sigma_1^+ \sigma_2^- + \sigma_1^- \sigma_2^+)/2 ,$$ with evident eigenvalues -3 for the singlet and 1 for the triplet.

Your operator S then has null eigenvalue for the singlet and eigenvalue 2 for the triplet.

  • $\begingroup$ Thanks, this was very helpful. To add to this, in the $s_{z}=0$ state of the triplet, the eigenvalue is $-4$ if my calculations are correct. $\endgroup$ – Asier R. Feb 17 at 16:24
  • $\begingroup$ Yes, you want me to use that state, instead? The eigenvalue of $\sigma_1\cdot \sigma_2$ is the same for each state of the triplet. It's only the first term that notices its alignment, and hence labeling of its states. $\endgroup$ – Cosmas Zachos Feb 17 at 19:30

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