Squared spin operators in second quantization

Question

Spin operator in second quantization can be written as: \begin{equation} \hat{\vec{S}}_{i} = \frac{1}{2} \sum_{\sigma \sigma'} \hat{c}^{\dagger}_{i\sigma} \hat{\vec{\sigma}}_{\sigma \sigma'} \hat{c}_{i\sigma'} \end{equation} where i - lattice multiindex, $\vec{\sigma}$ - Pauli matrices, $\hat{c}$ and $\hat{c}^{\dagger}$ - annihilation and creation fermion operators

I want to write $\left(\hat{S}^{x}_{i}\right)^{2}$, $\left(\hat{S}^{y}_{i}\right)^{2}$, $\left(\hat{S}^{z}_{i}\right)^{2}$, $\left(\hat{\vec{S}}_{i}\right)^{2}$.

For example, $\hat{S}^{x}_{i} = \frac{1}{2} \sum\limits_{\sigma\neq\sigma'} \hat{c}^{\dagger}_{i\sigma}\hat{c}_{i\sigma'}$

Then it should be correct that

\begin{equation} \left(\hat{S}^{x}_{i}\right)^{2} = \frac{1}{2} \sum\limits_{\sigma\neq\sigma', \delta\neq\delta'} \hat{c}^{\dagger}_{i\sigma}\hat{c}_{i\sigma'}\hat{c}^{\dagger}_{i\delta}\hat{c}_{i\delta'} \end{equation} The problem is, I struggle to simplify this expression. All I need to use is anticommutation relations, but every attempt leads to similar or even more complicated expressions. I failed to google answers. Is it even possible to combine somehow $\hat{c}$ and $\hat{c}^{\dagger}$ operators to get rid of some indexes or maybe transform this into function with operators $\hat{S}^{x}_{i}$, $\hat{S}^{y}_{i}$ , $\hat{S}^{z}_{i}$ ? If someone can give me a hint, I would appreciate it a lot.

Answer 1

I am not sure what you are really seeking, so I'd point out something specific for a given i, which I hitherto suppress, retaining just the two Pauli indices, $$\hat{S}^{x} = \frac{1}{2} (\hat{c}^{\dagger}_1 \hat{c}_2 + \hat{c}^{\dagger}_2 \hat{c}_1)\qquad \implies \\ \left(\hat{S}^{x}\right)^{2} = \frac{1}{4} (\hat{c}^{\dagger}_1 \hat{c}_2 + \hat{c}^{\dagger}_2 \hat{c}_1)^2= \frac{1}{4} ( \hat{c}^{\dagger}_1 \hat{c}_1+\hat{c}^{\dagger}_2 \hat{c}_2-2\hat{c}^{\dagger}_2 \hat{c}_2\hat{c}^{\dagger}_1 \hat{c}_1). $$

Your Fock space is 4-dimensional, with two bosons intercalating two fermions, $$ |0\rangle, \qquad \hat{c}^{\dagger}_1|0\rangle, \qquad \hat{c}^{\dagger}_2|0\rangle, \qquad \hat{c}^{\dagger}_1\hat{c}^{\dagger}_2|0\rangle.$$

It is then clear the eigenvalue of the two boson states, the first and the fourth, under $(S^x)^2$ are 0, so these are projected out; while the eigenvalues on the two fermions, the middle two states, are both 1/4, so four times your operator projects out the boson states. Can you see it is idempotent?

Answer 2

\begin{equation} \left(\hat{S}^{x}_{i}\right)^{2} = \frac{1}{2} \sum\limits_{\sigma\neq\sigma', \delta\neq\delta'} \hat{c}^{\dagger}_{i\sigma}\hat{c}_{i\sigma'}\hat{c}^{\dagger}_{i\delta}\hat{c}_{i\delta'} \end{equation} The problem is, I struggle to simplify this expression.

Write it out explicitly like: $$ \left(\hat{S_i}^{x}\right)^{2} = \frac{1}{4} \left(\hat{c}^{\dagger}_{i\uparrow} \hat{c}_{i\downarrow} + \hat{c}^{\dagger}_{i\downarrow} \hat{c}_{i\uparrow}\right)^2 = \frac{1}{4} \left( \hat{c}^{\dagger}_{i\uparrow} \hat{c}_{i\uparrow} +\hat{c}^{\dagger}_{i\downarrow} \hat{c}_{i\downarrow}-2\hat{c}^{\dagger}_{i\downarrow} \hat{c}_{i\downarrow}\hat{c}^{\dagger}_{i\uparrow} \hat{c}_{i\uparrow}\right) =\frac{1}{4}\left(n_{i\uparrow} - n_{i\downarrow}\right)^2\;, $$ where $$ n_{i\uparrow}=n_{i\uparrow}^2=c^\dagger_{i\uparrow}c_{i\uparrow} $$ and $$ n_{i\downarrow}=n_{i\downarrow}^2=c^\dagger_{i\downarrow}c_{i\downarrow}\;. $$

And the same expression results for $z$, much more easily: $$ \left(\hat{S_i}^{z}\right)^{2} = \frac{1}{4} \left(\hat{c}^{\dagger}_{i\uparrow} \hat{c}_{i\uparrow} - \hat{c}^{\dagger}_{i\downarrow} \hat{c}_{i\downarrow}\right)^2 =\frac{1}{4}\left(n_{i\uparrow} - n_{i\downarrow}\right)^2\;. $$

And, I'll leave it to you to check that the $y$ expression gives the same result...

Thus: $$ \left(\vec{S_i}\right)^{2}=\frac{3}{4}\left(n_{i\uparrow} - n_{i\downarrow}\right)^2 $$