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In the second quantization formalism of quantum chemistry, according to the book 1 on page 43, the creation operators $a_{p,\beta}^{\dagger},a_{p,\beta}^{\dagger}$ satisfy the relations

  • $[\hat S_\pm,a_{pm_s}^{\dagger}]=\sqrt{\frac{3}{4}-m_s(m_s\pm1)}a_{p,m_s\pm1}^{\dagger}$ (1)

  • $[\hat S_z,a_{pm_s}^{\dagger}]=m_s a_{pm_s}^{\dagger}$ (2).

The following form of the spin functions was used according to the author:

  • $\hat S_+ = \sum_p a^\dagger_{p\alpha}a_{p\beta}$ (3)

  • $\hat S_- = \sum_p a^\dagger_{p\beta}a_{p\alpha}$ (4)

  • $\hat S_z = \frac{1}{2}\sum_p (a^\dagger_{p\alpha}a_{p\alpha}-a^\dagger_{p\beta}a_{p\beta})$ (5)

By comparing with the equations which define spin tensor operators:

  • $[\hat S_\pm,\hat T^{S,M}]=\sqrt{S(S+1)-M(M\pm1)}\hat T^{S,M\pm1}$ (6)

  • $[\hat S_z,\hat T^{S,M}]=M\hat T^{S,M}$ (7)

it follows that the creation operators above are spin tensor operators.

How do equations (1) and (2) follow from the form of the spin operators given in equation (3)-(5)?

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    $\begingroup$ Have you tried to just compute those commutators directly using the definitions by plugging in for the S operators? $\endgroup$
    – march
    Dec 15, 2021 at 18:30
  • $\begingroup$ I have tried it, but was stuck. So thanks for posting the solution :-) $\endgroup$
    – Jadzia
    Dec 16, 2021 at 11:21

1 Answer 1

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You can just compute the commutators directly. For instance, \begin{align} [\hat S_z,\hat{a}_{qm}^{\dagger}] &= \left[ \frac{1}{2}\sum_p (\hat{a}^\dagger_{p\alpha}\hat{a}_{p\alpha}-\hat{a}^\dagger_{p\beta}\hat{a}_{p\beta}) ,\hat{a}_{qm}^{\dagger} \right] \\&= \frac{1}{2}\sum_p \left( \left[\hat{a}^\dagger_{p\alpha}\hat{a}_{p\alpha},\hat{a}_{qm}^{\dagger}\right] - \left[\hat{a}^\dagger_{p\beta}\hat{a}_{p\beta},\hat{a}_{qm}^{\dagger}\right] \right) \\&= \frac{1}{2}\sum_p \left( \hat{a}^\dagger_{p\alpha}\left[\hat{a}_{p\alpha},\hat{a}_{qm}^{\dagger}\right] + \left[\hat{a}^\dagger_{p\alpha},\hat{a}_{qm}^{\dagger}\right]\hat{a}_{p\alpha} - \left( \hat{a}^\dagger_{p\beta}\left[\hat{a}_{p\beta},\hat{a}_{qm}^{\dagger}\right] +\left[\hat{a}^\dagger_{p\beta},\hat{a}_{qm}^{\dagger}\right]\hat{a}_{p\beta} \right) \right) \\&= \frac{1}{2}\sum_p \left( \hat{a}^\dagger_{p\alpha}\delta_{qp}\delta_{\alpha m} + 0 - \left( \hat{a}^\dagger_{p\beta}\delta_{qp}\delta_{\beta m} +0 \right) \right) \\&= \frac{1}{2}\sum_p \left( \hat{a}^\dagger_{p\alpha}\delta_{qp}\delta_{\alpha m} - \hat{a}^\dagger_{p\beta}\delta_{qp}\delta_{\beta m} \right) \end{align} Then, performing the sum over $p$, this becomes \begin{align} [\hat S_z,\hat{a}_{qm}^{\dagger}] &= \frac{1}{2} \left( \hat{a}^\dagger_{q\alpha}\delta_{\alpha m} - \hat{a}^\dagger_{q\beta}\delta_{\beta m} \right)\,. \end{align} Note that if $m=\alpha$, then we get \begin{align} [\hat S_z,\hat{a}_{qm}^{\dagger}] &= \frac{1}{2} \hat{a}^\dagger_{q,\alpha}\,, \end{align} and if $m=\beta$, then we get \begin{align} [\hat S_z,\hat{a}_{qm}^{\dagger}] &= -\frac{1}{2} \hat{a}^\dagger_{q,\beta}\,, \end{align} indicating that $\alpha$ is acting as an $m=1/2$ quantum number and that $\beta$ is acting as a $m=-1/2$ quantum number, so we may as well relabel things so that $\alpha=1/2$ and $\beta = -1/2$. The other commutator is similar, although the calculation is more involved.


Note that, in the calculation above, we have used some standard commutator identities, which can be proved straight-forwardly by expanding out the commutators. These are \begin{align} [\hat{a},\hat{a}]&=0\,, \\ [\hat{a}+\hat{b},\hat{c}]&=[\hat{a},\hat{c}]+[\hat{b},\hat{c}]\,, \\ [\hat{a}\hat{b},\hat{c}]&=\hat{a}[\hat{b},\hat{c}]+[\hat{a},\hat{c}]\hat{b}\,. \end{align} In addition, we have also used the facts about the raising and lowering operators, where \begin{align*} \left[\hat{a}_{p\alpha},\hat{a}_{qm}^{\dagger}\right] = \delta_{pq}\delta_{\alpha m}\,. \end{align*}

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  • $\begingroup$ Thank you, I can see now that the second equation of my initial post holds. I still cannot see the first equation (1) from my post. Somehow I get a different result. $\endgroup$
    – Jadzia
    Dec 16, 2021 at 13:33
  • $\begingroup$ Would you be able to show the first equation (1) as well? $\endgroup$
    – Jadzia
    Dec 17, 2021 at 11:36

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