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.,

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

  • $\begingroup$ +1. The V term is similar as it's the "expectation value" of v(r), so I suppose that you're not actually asking how to obtain the U term of Eq.4, but you're asking how to generalize it to N electrons? Is your last equation your hypothesis for this generalization? If so, I would not use the Latin term "i.e." or "id est" there. $\endgroup$ Oct 15, 2022 at 13:11
  • $\begingroup$ Dear @NikeDattani, $T$ is the kinetic energy (expectation value of the 1st term in the Hamiltonian), $V$ is the external potential energy (expectation value of the 2nd term, as you have pointed out), and $U$ is the expectation value of the last term of the Hamiltonian operator. This is how I always understood the many-electron Hamiltonian operator and its expectation value (please correct me if I am wrong). $\endgroup$
    – Sha
    Oct 15, 2022 at 13:46
  • $\begingroup$ @NikeDattani It is not an expectation value but an operator. Hohenberg and Kohn use second quantization in a notation not widely used in physics, tho. $\endgroup$
    – Jakob
    Oct 16, 2022 at 15:48
  • $\begingroup$ @Jakob I put the term "expectation value" inside quotation marks in my comment. The OP used that term, so I used the same term. $\endgroup$ Oct 17, 2022 at 4:16
  • $\begingroup$ @NikeDattani I see. Still someone had to point out that the whole confusion of the OP is simply due to notation. Nothing has to be calculated or generalized. $\endgroup$
    – Jakob
    Oct 17, 2022 at 6:44

2 Answers 2


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.

  • 1
    $\begingroup$ To keep things from getting too long here, the discussion has been moved to chat $\endgroup$
    – Tyberius
    Oct 17, 2022 at 14:11

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.

  • $\begingroup$ then obtaining Eq. (4) is not a simple matter of the expectation value of the last term in the above Hamiltonian? Could you please show explicitly the steps and calculations of how to obtian Eq. (4) $\endgroup$
    – Sha
    Oct 15, 2022 at 15:34
  • $\begingroup$ What? Eq. (4) is an ansatz. $\endgroup$ Oct 16, 2022 at 12:03
  • $\begingroup$ you wrote in your answer: "The expectation value of two-electron operators such as the interelectronic Coulomb operator can be computed with the Slater-Condon rules, for example." But now you say it is an ansatz! Eq. (4) must be come from somewhere, and my question is exactly this, how to get Eq. (4). $\endgroup$
    – Sha
    Oct 16, 2022 at 13:51

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