Equation (4) in Hohenberg-Kohn Paper

Question

In their landmark paper of 1964, Inhomogeneous Electron Gas, Hohenberg and Kohn wrote down:

We shall be considering a collection of an arbitrary number of electrons, enclosed in a large box and moving under the influence of an external potential $v(\mathbf{r})$ and the mutual Coulomb repulsion. The Hamiltonian has the form

$$H=T+V+U \tag{1}$$

where[...atomic units are used.]

\begin{eqnarray*} T &\equiv& \frac{1}{2} \int \nabla \psi^*(\mathbf{r}) \ \nabla\psi(\mathbf{r}) \ d \mathbf{r} \tag{2} \\ V &\equiv& \int v(\mathbf{r}) \ \psi^*(\mathbf{r}) \ \psi(\mathbf{r}) \ d\mathbf{r} \tag{3} \\ U &=& \frac{1}{2} \int \frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|} \ \psi^\ast(\mathbf{r}) \ \psi^*(\mathbf{r}^{\prime}) \ \psi(\mathbf{r}^{\prime}) \ \psi(\mathbf{r}) \ d \mathbf{r} \ d \mathbf{r}^{\prime} \tag{4} \end{eqnarray*}

After applying the Born-Oppenheimer approximation, one obtains the following $N$-electron Hamiltonian,

$$ \hat{H} = - \frac{1}{2} \sum_{i=1}^N \nabla_i^2 + \sum_{i=1}^N v(\hat{\mathbf{r}_i}) + \frac{1}{2} \sum_{i \neq j} \frac{1}{|\hat{\mathbf{r}}_i-\hat{\mathbf{r}}_j|} $$

and the time-independent Schrodinger equation,

$$ \hat{H} \ \psi(\mathbf{r}_1, \mathbf{r}_2, ..., \mathbf{r}_N) = E \ \psi(\mathbf{r}_1, \mathbf{r}_2, ..., \mathbf{r}_N) $$

My question is how to obtain Eq. (4) given the above Hamiltonian and Schrodinger equation?

I know Eq. (4) represents the classical Hartree energy and it can be deduced from Electrodynamics arguments (of course not in terms of the wavefunction form depicted above but in terms of the charge density of a charge distribution), BUT how to obtain it from the expectation value of the above Hamiltonian, i.e.,

$$ U = \int_{\mathbb{R}^3} \ldots \int_{\mathbb{R}^3} \ d \mathbf{r}_1 \ldots d \mathbf{r}_N \ \psi^*(\mathbf{r}_1, ..., \mathbf{r}_N) \ \left(\frac{1}{2} \sum_{i \neq j} \frac{1}{|\hat{\mathbf{r}}_i-\hat{\mathbf{r}}_j|}\right) \ \psi(\mathbf{r}_1, ..., \mathbf{r}_N) $$

Answer 1

You're confusing things: The operators are written in second quantization, so $\psi$ and $\psi^*$ (or $\psi^\dagger$) denote (fermionic) the field operators. There are no expectation values involved and in particular $\psi$ does not denote any wave function (and thus $\psi^*$ not some complex conjugated wave function). So your Hamiltonian is the very same as the Hamiltonian in the paper, term by term (if interpreted correctly).

For example, a (fermionic) two-body operator of the form

$$ W:=\frac{1}{2}\sum\limits_{i\neq j}^Nw(x_i,x_j) \tag{1} $$ can be written in second quantization as (modulo typos)

$$ W=\frac{1}{2} \int\mathrm dx_1 \int \mathrm dx_2\, \psi^\dagger(x_1)\, \psi^\dagger(x_2)\,w(x_1,x_2)\, \psi(x_2)\, \psi(x_1)\ \tag{2} \quad ,$$

where we've employed a notation which is more common in (many-body) physics (the $^*$ is used in more mathematical oriented texts). The derivation is given in pretty much any book concerning second quantization. Similar reasoning applies to the $T+V$ term.

Answer 2

The expectation value of two-electron operators such as the interelectronic Coulomb operator can be computed with the Slater-Condon rules, for example. Another alternative is to use second quantization and Wick's theorem.

For a one-determinant wave function $|\Psi\rangle$, the expectation value $\langle \Psi |r_{12}^{-1}|\Psi\rangle$ gives you the classical Coulomb term

$$ U = \frac 1 2 \int \frac {n({\bf r}) n({\bf r})} {|{\bf r}-{\bf r}'|} {\rm d}^3 r {\rm d}^3 r' $$

and the quantum mechanical exchange interaction.