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)?
 A: 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*}
