Spin squared in terms of density and double occupancy

Question

I am having trouble reproducing Fradkin's Equation (2.23) in Field Theories of Condensed Matter.

$$ \sum_\vec{r} \left(\vec{S}(\vec{r})\right)^2 = \sum_\vec{r} \left(\frac{3}{4} n (\vec{r}) - \frac{3}{2} n_\uparrow(\vec{r}) n_\downarrow(\vec{r}) \right) $$

Where he invokes the completeness relation

$${\displaystyle {\vec {\sigma }}_{\alpha \beta }\cdot {\vec {\sigma }}_{\gamma \delta }\equiv \sum _{k=1}^{3}\sigma _{\alpha \beta }^{k}\,\sigma _{\gamma \delta }^{k}=2\,\delta _{\alpha \delta }\,\delta _{\beta \gamma }-\delta _{\alpha \beta }\,\delta _{\gamma \delta }.}$$

And from the Jordan map $\vec{S}(\vec{r}) = \frac{\hbar}{2} c^\dagger_\alpha(\vec{r}) \vec{\sigma}^{\alpha \beta} c_\beta(\vec{r})$ and $n_\alpha^2 = n_\alpha = c^\dagger_\alpha c_\alpha$ for fermions with the normal anti-commutation relations.

I first tried it using Fradkin's method: $$ \begin{equation} \begin{aligned} \left(\vec{S}(\vec{r})\right)^2 &= S^a(\vec{r}) S_a(\vec{r}) \\ &= \frac{\hbar^2}{4} \sum_a [c^\dagger_\alpha \sigma_a^{\alpha \beta} c_\beta c^\dagger_\gamma \sigma_a^{\gamma \delta} c_\delta] (\vec{r}) \\ &= \frac{\hbar^2}{4} (2 \delta^{\alpha \delta} \delta^{\beta \gamma} - \delta^{\alpha \beta} \delta^{\gamma \delta}) [c^\dagger_\alpha c_\beta c^\dagger_\gamma c_\delta] (\vec{r}) \\ &= \frac{\hbar^2}{4} \sum_{\alpha, \beta} [2 c^\dagger_\alpha c_\beta c^\dagger_\beta c_\alpha - c^\dagger_\alpha c_\alpha c^\dagger_\beta c_\beta] (\vec{r}) \\ &= \frac{\hbar^2}{4} \sum_{\alpha, \beta} [-2 c^\dagger_\alpha c_\alpha c^\dagger_\beta c_\beta + 4 c^\dagger_\alpha c_\alpha - c^\dagger_\alpha c_\alpha c^\dagger_\beta c_\beta] (\vec{r}) \\ &= - \frac{\hbar^2}{4} [3 (n_\uparrow + n_\downarrow)^2 - 4 n] (\vec{r}) \\ &= -\frac{\hbar^2}{4} [6 n_\uparrow n_\downarrow + 3 (n_\uparrow^2 + n_\downarrow^2) - 4 (n_\uparrow + n_\downarrow)] (\vec{r}) \\ &= \hbar^2 \left[\frac{n}{4} - \frac{3}{2} n_\uparrow n_\downarrow\right] (\vec{r}) \end{aligned} \end{equation} $$ I at first thought this was a mistake in Fradkin where he accidentally used $\{c, c\}$ as $1$ instead of $0$ when doing the other two $\{c^\dagger, c\} = 1$, but when I brute force it, it seems to work (I haven't checked this for errors): $$ \begin{aligned} \left(\vec{S}(\vec{r})\right)^2 &= S^a(\vec{r}) S_a(\vec{r}) \\ &= \frac{\hbar^2}{4} [( c_\downarrow^\dagger c_\uparrow + c_\uparrow^\dagger c_\downarrow, \mathrm{i} c_\downarrow^\dagger c_\uparrow - \mathrm{i} c_\uparrow^\dagger c_\downarrow, c_\uparrow^\dagger c_\uparrow - c_\downarrow^\dagger c_\downarrow )^2 ] (\vec{r}) \\ &= \frac{\hbar^2}{4} \left[ \begin{aligned} &\phantom{+}\,\, (c_\downarrow^\dagger c_\uparrow c_\downarrow^\dagger c_\uparrow + c_\downarrow^\dagger c_\uparrow c_\uparrow^\dagger c_\downarrow + c_\uparrow^\dagger c_\downarrow c_\downarrow^\dagger c_\uparrow + c_\uparrow^\dagger c_\downarrow c_\uparrow^\dagger c_\downarrow) \\ &- (c_\downarrow^\dagger c_\uparrow c_\downarrow^\dagger c_\uparrow - c_\downarrow^\dagger c_\uparrow c_\uparrow^\dagger c_\downarrow - c_\uparrow^\dagger c_\downarrow c_\downarrow^\dagger c_\uparrow + c_\uparrow^\dagger c_\downarrow c_\uparrow^\dagger c_\downarrow) \\ &+ (c_\uparrow^\dagger c_\uparrow c_\uparrow^\dagger c_\uparrow - c_\uparrow^\dagger c_\uparrow c_\downarrow^\dagger c_\downarrow - c_\downarrow^\dagger c_\downarrow c_\uparrow^\dagger c_\uparrow + c_\downarrow^\dagger c_\downarrow c_\downarrow^\dagger c_\downarrow) \end{aligned} \right] (\vec{r}) \\ &= \frac{\hbar^2}{4} \left[ 2c_\downarrow^\dagger c_\uparrow c_\uparrow^\dagger c_\downarrow + 2c_\uparrow^\dagger c_\downarrow c_\downarrow^\dagger c_\uparrow + c_\uparrow^\dagger c_\uparrow c_\uparrow^\dagger c_\uparrow - c_\uparrow^\dagger c_\uparrow c_\downarrow^\dagger c_\downarrow - c_\downarrow^\dagger c_\downarrow c_\uparrow^\dagger c_\uparrow + c_\downarrow^\dagger c_\downarrow c_\downarrow^\dagger c_\downarrow \right] (\vec{r}) \\ &= \frac{\hbar^2}{4} \left[ -2c_\downarrow^\dagger c_\downarrow c_\uparrow^\dagger c_\uparrow + 2c_\downarrow^\dagger c_\downarrow - 2c_\uparrow^\dagger c_\uparrow c_\downarrow^\dagger c_\downarrow + 2c_\uparrow^\dagger c_\uparrow + c_\uparrow^\dagger c_\uparrow c_\uparrow^\dagger c_\uparrow - c_\uparrow^\dagger c_\uparrow c_\downarrow^\dagger c_\downarrow - c_\downarrow^\dagger c_\downarrow c_\uparrow^\dagger c_\uparrow + c_\downarrow^\dagger c_\downarrow c_\downarrow^\dagger c_\downarrow \right] (\vec{r}) \\ &= \frac{\hbar^2}{4} \left[ -2n_\downarrow n_\uparrow + 2n_\downarrow - 2n_\uparrow n_\downarrow + 2n_\uparrow + n_\uparrow n_\uparrow - n_\uparrow n_\downarrow - n_\downarrow n_\uparrow + n_\downarrow n_\downarrow \right] (\vec{r}) \\ &= \frac{3\hbar^2}{4} \left[n_\uparrow + n_\downarrow - 2n_\downarrow n_\uparrow \right] (\vec{r}) \\ \end{aligned} $$

Not to mention, this latter attempt and Fradkin's version make more physical sense if one takes expectation values for some simple examples.

I either have a very silly mistake, or there is some property I am neglecting. Or am I doing something illegal with my contractions?

Answer 1

Okay, this is one of those silly index mistakes. One has to remember that $\sum_{\alpha, \beta} \delta^{\beta \beta} n_\alpha = 2 \sum_\alpha n_\alpha$ where $\alpha, \beta \in \{\uparrow,\downarrow\}$. Of course, the $2$ could be exchanged for whatever number of indices $\beta$ runs over.

Let me go through each anticommutation.

$$ 2 c^\dagger_\alpha c^\beta c^\dagger_\beta c^\alpha = -2c^\dagger_\alpha c^\dagger_\beta c^\beta c^\alpha + 2\delta^{\beta}_{\beta} c^\dagger_\alpha c_\alpha = -2c^\dagger_\alpha c^\alpha c^\dagger_\beta c^\beta + 2\delta^{\beta}_{\beta} c^\dagger_\alpha c^\alpha + 2{\delta}{^{\alpha}_{\phantom{\alpha}\beta}}c^\dagger_\alpha c^\beta = -2 n^2 + 6n $$

Propagating this change resolves the issue. Don't be sloppy, write out your Kronecker Deltas.