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.

  • $\begingroup$ Jordan-Schwinger map. $\endgroup$ Dec 12, 2023 at 21:58
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 13, 2023 at 14:49
  • $\begingroup$ "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... $\endgroup$
    – hft
    Dec 14, 2023 at 0:44

2 Answers 2


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?


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.