# Showing that the creation operator is a spin tensor operator

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)?

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

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*}