# Commutator in the derivation of the Runge-Gross Theorem

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.

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.
• do I understand you correctly that the "f" in your answer is $\nabla v$ ? May 14 at 23:39
• The $f$ is any function that depends on a single coordinate, so yes you can take it to be the gradient.
• Does this mean expression (6) results in $n(\textbf{r}) \nabla v(\textbf{r})$? May 15 at 9:26