Anderson Hamiltonian (product of number operators) in 1st quantization?

[AA1983]

In the Anderson model, it cost an energy $$Un_{\Uparrow}n_{\Downarrow}$$ for a quantum dot level to be occupied by two electrons. Here $$n_{\Uparrow}$$ is the second quantized number operator, counting the number of particles with spin $$\Uparrow$$. I need the term $$Un_{\Uparrow}n_{\Downarrow}$$ in first quantization. Here is what I know:

$$Un_{\Uparrow}n_{\Downarrow} =<br /> Ud_{\Uparrow}^{\dagger}d_{\Uparrow}d_{\Downarrow}^{\dagger}d_{\Downarrow}<br /> =<br /> -Ud_{\Uparrow}^{\dagger}d_{\Downarrow}^{\dagger}d_{\Uparrow}d_{\Downarrow}<br /> =<br /> \frac{1}{2}\sum_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}d_{\eta_{1}}^{\dagger}d_{\eta_{2}}^{\dagger}d_{\eta_{3}}d_{\eta_{4}}$$
where

$$V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}=\Big\{<br /> \begin{array}{c}<br /> -2U \qquad \text{for} \qquad \eta_{1}=\eta_{2}=\Uparrow,\: \eta_{2}=\eta_{4}=\Downarrow\\<br /> 0 \qquad \text{elsewhere}<br /> \end{array}$$.

V is also given by

$$V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}=\int dx_{j} dx_{k} \psi_{\eta_{1}}^{\ast}(x_{j})\psi_{\eta_{2}}^{\ast}(x_{k})V(x_{j}-x_{k})<br /> \psi_{\eta_{3}}(x_{j})\psi_{\eta_{4}}(x_{k})$$

Now, what is $$V(x_{j}-x_{k})$$ ?

 

[AA1983]

Is it $$V=-2U\delta(x_{j}-x_{k})\delta_{\eta_{1}\Uparrow}\delta_{\eta_{2}\Downarrow}\delta_{\eta_{3}\Uparrow}\delta_{\eta_{4}\Downarrow}$$ ?

 

[Lonewolf]

I agree in principle, but shouldn't $V_{\eta_{1}\eta_{2}\eta_{3}\eta_{4}}$ be non-zero for other combinations of indices? For example $\eta_{1}=\eta_{4}=\uparrow, \eta_{2}=\eta_{3}=\downarrow$ should probably be allowed, since you're not creating or annihilating two of the same type of spin. Also, since flipping spins means swapping two pairs of fermionic operators in $-Ud_{\Uparrow}^{\dagger}d_{\Downarrow}^{\dagger}d_{ \Uparrow}d_{\Downarrow}$, you won't pick up a minus sign, so it should probably be

$$V=-2U\delta(x_{j}-x_{k})\left(\delta_{\eta_{1}\eta_{3}}\delta_{\eta_{2}\eta_{4}} - \delta_{\eta_{1}\eta4}\delta_{\eta_{2}\eta_{3}}\right)$$

so $\eta_{1}$ is either the same as $\eta_{3}$ or $\eta_{4}$, and it picks up a minus sign in the latter. Similarly for $\eta_{2}$.