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In Hohenberg and Kohn's paper on the inhomogeneous electron gas they express the Hamiltonian for "a collection of an arbitrary number of electrons moving under the influence of an external potential v(r)" as:

$$H=T+V+U$$

Where,

$$T = \frac{1}{2}\int \nabla \psi^*(\boldsymbol{r}) \nabla \psi(\boldsymbol{r})d\bf{\boldsymbol{r}}$$

$$V = \int v(\boldsymbol{r})\psi^*(\boldsymbol{r}) \psi(\boldsymbol{r})d\bf{\boldsymbol{r}}$$

$$U = \frac{1}{2}\int \frac{1}{\left|\boldsymbol{r-r'} \right|}\psi^*(\boldsymbol{r})\psi^*(\boldsymbol{r'}) \psi(\boldsymbol{r'})\psi(\boldsymbol{r})d\bf{\boldsymbol{r}}$$

I am having difficultly reconciling these expressions with what I would regard as the more standard expressions:

$$T=\left<\Psi|-\frac{1}{2}\sum_{i=1}^N\nabla_i^2|\Psi\right>$$

$$V= \left<\Psi|v(\boldsymbol{r})|\Psi\right>$$

$$U= \left<\Psi|\sum_{i<j}\frac{1}{\left|\boldsymbol{r_i}-\boldsymbol{r_j}\right|}|\Psi\right>$$

In particular, since the multi-body wavefunction is,

$$\Psi=\Psi(\boldsymbol{r_1,r_2,r_3,r_4,...})$$

I don't understand why they appear to be using a single particle wave function $\psi(\boldsymbol{r})$ in their Hamiltonian..

Also, they define the density as:

$$ n(\boldsymbol{r})=\left(\Psi,\psi^*(\boldsymbol{r})\psi(\boldsymbol{r})\Psi\right)$$

Instead of:

$$ n(\boldsymbol{r})= \left<\Psi|\sum_{i=1}^N\delta(\boldsymbol{r-r_i})|\Psi\right> $$

Where $\Psi$ is presumably their many body wavefunction, $\psi^*(r)\psi(r)$ appears to be replacing the density operator $\delta(r-r_i)$, and the enclosing parentheses presumably represent the inner product instead of <>.

I am not sure how to interpret these expressions.

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Hohenberg and Kohn describe the system of electrons using the second-quantization form. Therefore $\psi({\bf r})$ is not a wavefunction but an operator ($\psi({\bf r})$ and $\psi^*({\bf r})$ respectively destroy an create a fermion at point ${\bf r}$). This is clear from the formula $(5)$ of their paper, where the one-electron density in the ground state $\Psi$ is written as $$ n({\bf r})=\langle\Psi,\psi^*({\bf r}) \psi({\bf r}) \Psi \rangle. $$ Details about the second-quantization formalism can be found in every advanced quantum mechanics textbook treating many-particle states, like Fetter and Walecka.

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    $\begingroup$ To add: In more "physicist-friendly" (and modern) texts in that field $\psi^*$ is usually denoted by $\psi^\dagger$ instead. $\endgroup$ Feb 11 at 8:04
  • $\begingroup$ @TobiasFünke Thanks, I think that was one of the reasons I did not guess that they were using second quantization. I've never seen * used instead of †. $\endgroup$
    – gigo318
    Feb 11 at 17:24

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