Hirsch's discrete Hubbard-Stratonovich transformation

Question

Suppose $$\hat{H}_1 = U\left(\hat{n}_{d,\uparrow} - \frac{1}{2}\right)\left(\hat{n}_{d,\downarrow} - \frac{1}{2}\right)\tag{1}$$ To decouple the many-body operator, many articles suggest using the discrete Hubbard-Stratonovich transformation $$e^{-\Delta \tau \hat{H}_1}=\frac{1}{2} \sum_{\sigma= \pm 1} e^{\lambda \sigma\left(\hat{n}_{d, \uparrow}-\hat{n}_{d, \downarrow}\right)}\tag{2}$$ where $\cosh\lambda=e^{\Delta\tau U/2}$.

All the articles make references to J. E. Hirsch's paper on the transformation, but the formula presented in the paper is $$\exp(-\Delta\tau U n_\uparrow n_\downarrow) = \frac{1}{2} \sum_{\sigma=\pm1} \exp\left[2a\sigma(n_\uparrow - n_\downarrow) - \frac{1}{2}U\Delta\tau(n_\uparrow + n_\downarrow)\right]\tag{3}$$ where $\tanh^2 a = \tanh(\Delta\tau U/4)$. The paper said it's easily to prove this, but I don't see where it comes from.

How does one derive equation (3), and use it to derive equation (2)?

Answer 1

The formula (3) does not work in general, it only works for fermions. You need to use the fact that $n_\uparrow,n_\downarrow\in\{0,1\}^2$. You therefore just need to check whether you can tune $a$ so that the RHS and LHS coincide for the four possible values. In practice, any function of $f(n_\uparrow,n_\downarrow)$ can be uniquely parametrised in terms of four coefficients $J,h_\uparrow,h_\downarrow,f_0$: $$ f(n_\uparrow,n_\downarrow) = Jn_\uparrow n_\downarrow+h_\uparrow n_\uparrow+h_\downarrow n_\downarrow+f_0 \\ f_0= f(0,0) \quad h_\uparrow = f(1,0)-f(0,0) \quad h_\downarrow = f(0,1)-f(0,0) \quad J = f(1,1)-f(0,1)-f(1,0)+f(0,0) $$ You can apply this to: $$ \begin{align} \ln\left(\frac12\sum_{\sigma=\pm}e^{\lambda\sigma(n_\uparrow-n_\downarrow)}\right) &= \ln[\cosh(\lambda(n_\uparrow-n_\downarrow))] \\ &= \begin{cases} 0 & n_\uparrow=n_\downarrow \\ \ln(\cosh\lambda) & n_\uparrow+n_\downarrow=1 \end{cases} \\ &= -2\ln(\cosh\lambda)n_\uparrow n_\downarrow+\ln(\cosh\lambda) n_\uparrow+\ln(\cosh\lambda) n_\downarrow \\ \end{align} $$ So you identify: $$ -2\ln\cosh\lambda = \Delta\tau U \\ \cosh(\lambda) = e^{\Delta\tau U/2} $$ and recover (1) without the constant term. Note that (2) is not quite correct, due to this unimportant forgotten constant term. However (3) is exact. You can apply the same method. Alternatively, you can replace $\lambda=2a$. The formulas match using the hyperbolic formulas: $$ \begin{align} \cosh(2a) &= e^{\Delta\tau U/2} \\ \cosh(2a) &= \frac{1+\tanh^2a}{1-\tanh^2a} \\ \tanh^2a &= \frac{1-e^{\Delta\tau U/2}}{1+e^{\Delta\tau U/2}}\\ &=\tanh\left(\frac{\Delta\tau U}4\right) \end{align} $$