Commutator in the derivation of the Runge-Gross Theorem

Question

In the derivation of the Runge-Gross theorem, a theorem that states a one-to-one correspondence between potential and density in an evolving system, a commutator appears that I seem to have trouble with.

The equation of motion for the difference of the two current densities is set up as

$$ \frac{\partial }{\partial t} [{j (\vec r, t) - j'(\vec r, t)}_{t=0}] = -i \langle \Psi | [\hat j (r,t ), [\hat H - \hat H']] \Psi \rangle \tag{1}\label{cds}$$

The Hamiltonian in this is an n-electron, non relativistic Hamiltonian with a one body time dependent poetntial $V_{ext.}(\vec r,t)$, i.e.:

$$H(t) = - \frac{1}{2} \sum_{i=1}^N \nabla_i^2 + \frac{1}{2} \sum_{i \neq j}^{N} \frac{1}{|\vec r_i - \vec r_j|} + \sum_{i=1}^N v_{ext.}(\vec r_i, t ) \tag{2}\label{\Hamiltonian}$$

The second Hamiltonian $H'$ mentioned in Eq. \eqref{cds}, for the sake of this proof, differs only in its one-body potential from the first. Expression \eqref{cds} can be rewritten as:

$$ \frac{\partial }{\partial t} [{j (\vec r, t) - j'(\vec r, t)}_{t=0}] = -i \langle \Psi | [\hat j (r,t ), [\hat V(\vec r, 0) - \hat V'(\vec r, 0)]] \Psi \rangle \tag{3}\label{cds2}$$

The probability-current operator (afaik) can be written as:

$$\frac{i}{2} \sum_i \delta(\vec r_{i}- \vec r) \nabla_{\vec r_i} -\nabla_{\vec r} \delta(\vec r_{i}- \vec r) \tag{4}$$

This leads me to expressions of the following form:

$$ \begin{array} = \frac{\partial }{\partial t} [{j (\vec r, t) - j'(\vec r, t)}_{t=0}] = \langle \Psi | \sum_i \delta(\vec r_{i}-r) \nabla_{\vec r_i} v(\vec r_i, t) | \Psi \rangle- \langle \Psi | \sum_i \nabla_{\vec r_i} \delta(\vec r_{i} - \vec r) v(\vec r_i, t) | \Psi \rangle \\ -\langle \Psi | \sum_i v(\vec r_i, t) \delta(\vec r_{i}-r) \nabla_{\vec r_i} | \Psi \rangle + \langle \Psi | \sum_i v(\vec r_i, t) \nabla_{\vec r_i} \delta(\vec r_{i}- \vec r) | \Psi\rangle \cdots \end{array} \tag{5} $$

I am unsure how to resolve

$$ \langle \Psi | \sum_i \delta(\vec r_{i}-r) \nabla_{\vec r_i} v(\vec r_i, t) | \Psi \rangle \tag{6}$$

without the $\nabla$ I would assume: $$\begin{array} \; v_{ext.}(\vec r') \rho(\vec r') = \int dr \; \delta(\vec r - \vec r' ) v_{ext.} (\vec r ) \rho(\vec r)\nonumber \\ = \sum_i \langle \Psi | \delta(\vec r_i - \vec r') v_{ext.} (\vec r_i) | \Psi \rangle \end{array} \tag{7}$$

although here I already rely on some definitions.

Answer 1

You are on the right track with using the definition of the density operator. The commutator can essentially be handled an analogous way but the result is pretty general, $$ \langle \Psi \vert \sum_{i}^{N}\delta(\textbf{r} -\textbf{r}_{i}) f(\textbf{r}_{i}) \vert \Psi \rangle, $$ where the integral over a single $\textbf{r}_{i}$ can be computed, $$ \sum_{i}\int d^{3}r_{1}...d^{3}r_{i-1}d^{3}r_{i+1}..d^{3}r_{N} \vert \Psi(\textbf{r}_{1},...,\textbf{r}_{i} = \textbf{r},,...\textbf{r}_{N}) \vert^{2} f(\textbf{r}_{i} =\textbf{r} ). $$ Here I wrote out in some painful detail, but the key point now is that you can freely permute the coordinates of the wave-function to show that the sum over $i$ simply contributes a factor of $N$, $$ N\int d^{3}r_{2},...,d^{3}r_{n} \vert \Psi(\textbf{r},...,\textbf{r}_{N})\vert^{2} f(\textbf{r}), $$ which simplifies further by the definition of the reduced one-body density, $$ n(\textbf{r})f(\textbf{r}), $$ where this $f(\textbf{r})$ is essentially any one-body function that depends on a single $\textbf{r}_{i}$, i.e. any of the commutators that come out of the proof (which get sort of gnarly really fast).

$\textbf{Just to clarify:}$ for the case of the first derivative of the current density at $t=0$, $$\frac{\partial (j(\textbf{r},t) - j'(\textbf{r},t))}{\partial t}\bigg\rvert_{t=0} = -i \langle \Psi \vert [\hat{j}(\textbf{r}),\Delta \hat{V}] \vert \Psi \rangle,$$ where the $\Delta V$ is the difference between the two external potentials at the initial time. For the first derivative the commutator itself can be evaluated very easily, $$ [\hat{j}(\textbf{r}),\Delta V] = -i\sum_{i}^{N} \delta(\textbf{r}-\textbf{r}_{i})\nabla_{i} (v(\textbf{r}_{i})-v'(\textbf{r}_{i})). $$ This has the same form as my first expression and you get the result that you need for the first step in the proof, i.e, $$ \frac{\partial (j(\textbf{r},t) - j'(\textbf{r},t))}{\partial t}\bigg\rvert_{t=0} = - n(\textbf{r})\nabla (v(\textbf{r})-v'(\textbf{r})). $$ The remainder of the proof requires higher order derivatives, but the rest of the commutators will sort of do the same thing.