# Squared spin operators in second quantization

Spin operator in second quantization can be written as: $$$$\hat{\vec{S}}_{i} = \frac{1}{2} \sum_{\sigma \sigma'} \hat{c}^{\dagger}_{i\sigma} \hat{\vec{\sigma}}_{\sigma \sigma'} \hat{c}_{i\sigma'}$$$$ where i - lattice multiindex, $$\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

$$$$\left(\hat{S}^{x}_{i}\right)^{2} = \frac{1}{2} \sum\limits_{\sigma\neq\sigma', \delta\neq\delta'} \hat{c}^{\dagger}_{i\sigma}\hat{c}_{i\sigma'}\hat{c}^{\dagger}_{i\delta}\hat{c}_{i\delta'}$$$$ The problem is, I struggle to simplify this expression. All I need to use is anticommutation relations, but every attempt leads to similar or even more complicated expressions. I failed to google answers. Is it even possible to combine somehow $$\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.

• Dec 12, 2023 at 21:58
• You might suppress the location index i, as different locations commute. Your Fock space then is 4 dimensional. Write the 4x4 matrix representing your operator. Dec 13, 2023 at 14:49
• "The problem is, I struggle to simplify this expression." Not sure if you consider this more simplified or less simplified, but you can also demand that $\sigma\neq\delta$ and that $\sigma'\neq\delta'$ in the last expression you wrote, since you can't create two electrons in the same state or destroy two in the same state...
– hft
Dec 14, 2023 at 0:44

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?

$$$$\left(\hat{S}^{x}_{i}\right)^{2} = \frac{1}{2} \sum\limits_{\sigma\neq\sigma', \delta\neq\delta'} \hat{c}^{\dagger}_{i\sigma}\hat{c}_{i\sigma'}\hat{c}^{\dagger}_{i\delta}\hat{c}_{i\delta'}$$$$ 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: $$\left(\vec{S_i}\right)^{2}=\frac{3}{4}\left(n_{i\uparrow} - n_{i\downarrow}\right)^2$$