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In Ullrich's TD-DFT book, the paramagnetic current-density operator is defined as $$\hat{\mathbf{j}}(\mathbf{r})=\frac{1}{2i}\sum_{a=1}^{N}\left[\nabla_a\delta(\mathbf{r}-\mathbf{r}_a)+\delta(\mathbf{r}-\mathbf{r}_a)\nabla_a\right]$$ The expectation value of this operator is $\langle\hat{\mathbf{j}}(\mathbf{r})\rangle=\mathbf{j}(\mathbf{r},t)$. I know that the current density is $$\mathbf{j}(\mathbf{r},t)=\frac{1}{2i}\left(\Psi^*\nabla\Psi-\Psi\nabla\Psi^*\right),$$ and this is the result I think I have to try to get with $\langle\hat{\mathbf{j}}(\mathbf{r})\rangle$.

At first sight, I have to find some property of the delta function that makes negative one of the terms in the integral of $\hat{\mathbf{j}}(\mathbf{r})$. So, I have $$ \langle\hat{\mathbf{j}}(\mathbf{r})\rangle=\frac{1}{2i}\sum_{a=1}^{N}\int\Psi^*(\mathbf{x}_1,\dots,\mathbf{x}_N,t)\left[\nabla_a\delta(\mathbf{r}-\mathbf{r}_a)+\delta(\mathbf{r}-\mathbf{r}_a)\nabla_a\right]\Psi(\mathbf{x}_1,\dots,\mathbf{x}_N,t)d\mathbf{x}_1\dots\mathbf{x}_N,$$ where $\mathbf{x}_i$ represents spacial and spin coordinates. I thought that this property could be useful: $$\int f(x)\delta'(x-a)dx=-f'(a),$$

Thus, for simplicity $N=1$, $$\begin{align} \langle\hat{\mathbf{j}}(\mathbf{r})\rangle&=\frac{1}{2i}\left(\int\Psi^*(\mathbf{x}_1,t)\nabla\delta(\mathbf{r}-\mathbf{r}_a)\Psi(\mathbf{x}_1,t)d\mathbf{x}_1+\int\Psi^*(\mathbf{x}_1,t)\delta(\mathbf{r}-\mathbf{r}_a)\nabla\Psi(\mathbf{x}_1,t)d\mathbf{x}_1\right)\\ &=\frac{1}{2i}\left(-\Psi^*(\mathbf{x},t)\nabla\Psi(\mathbf{x},t)+\Psi^*(\mathbf{x},t)\nabla\Psi(\mathbf{x},t)\right) \end{align} $$ and this is equal to zero. Maybe I used the property in a questionable way or there is something I am missing.

Also, my second issue is about obtaining a term with the gradient of $\Psi^*(\mathbf{x},t)$ and I have no clue of what property use (hermiticity?). Can you give me some hints to solve this? Further, this is useful for obtaining the continuity equation, so I need to resolve this problem.

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    $\begingroup$ You seem to be making a few mistakes. Notice that the gradient would be over the delta function as well in one of the terms and in order to get rid of it, you will have to use integration by parts, which will give you a gradient over the conjugate of $\Psi$. $\endgroup$ – Dvij D.C. May 18 '19 at 4:20
  • $\begingroup$ Thanks for the advice, solved it. $\endgroup$ – Verktaj May 18 '19 at 4:38
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This answer is posted as a request from some users.

The expected value of the paramagnetic current density of an $N$ electron system is

$$ \begin{align} \langle\hat{\mathbf{j}}(\mathbf{r})\rangle&=\frac{1}{2i}\sum_{a=1}^{N}\langle\Psi|\nabla_a\delta(\mathbf{r}-\mathbf{r}_a)+\delta(\mathbf{r}-\mathbf{r}_a)\nabla_a|\Psi\rangle \\ &=\frac{1}{2i}\sum_{a=1}^{N}\left[\int\Psi^{*}(\{\mathbf{x}_a\},t)\nabla_a\delta(\mathbf{r}-\mathbf{r}_a)\Psi(\{\mathbf{x}_a\},t)d\mathbf{x}_1\dots d\mathbf{x}_N\right.\\ &\qquad\qquad\quad\left.+\int\Psi^{*}(\{\mathbf{x}_a\},t)\delta(\mathbf{r}-\mathbf{r}_a)\nabla_a\Psi(\{\mathbf{x}_a\},t)d\mathbf{x}_1\dots d\mathbf{x}_N\right]\\ &=\frac{1}{2i}\sum_{a=1}^{N}\left[\int\Psi^{*}(\{\mathbf{x}_a\},t)\Psi(\{\mathbf{x}_a\},t)\nabla_a\delta(\mathbf{r}-\mathbf{r}_a)d\mathbf{x}_1\dots d\mathbf{x}_N \right.\\ &\qquad\qquad\quad\left.+2\int\Psi^{*}(\{\mathbf{x}_a\},t)\delta(\mathbf{r}-\mathbf{r}_a)\nabla_a\Psi(\{\mathbf{x}_a\},t)d\mathbf{x}_1\dots d\mathbf{x}_N \right] \end{align} $$ Using a property of the derivative of the Dirac delta (found in Zettili's Quantum Mechanics) on the first integral of the square brackets, $$\int_{-\infty}^\infty f(x)\delta'(x-a)dx=-f'(a),$$ we get (using a $b\neq a$ that runs from $1$ to $N$) $$ \begin{align} \langle\hat{\mathbf{j}}(\mathbf{r})\rangle&=\frac{1}{2i}\sum_{a=1}^{N}\left[\int(-1)\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_a d\mathbf{x}_b\right.\\ &\qquad\qquad\quad\left.-\int\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_ad\mathbf{x}_b\right.\\ &\qquad\qquad\quad+2\left.\int\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_ad\mathbf{x}_b\right]\\ &=\frac{1}{2i}\sum_{a=1}^{N}\left[\int\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_ad\mathbf{x}_b\right.\\ &\qquad\qquad\quad\left.-\int\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_ad\mathbf{x}_b\right]\\ &=\frac{N}{2i}\left[\int\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_ad\mathbf{x}_b\right.\\ &\qquad\quad\left.-\int\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)d\sigma_ad\mathbf{x}_b\right]\\ &=N\int\text{Im}\left[\Psi^{*}(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\nabla_\mathbf{r}\Psi(\mathbf{r},\sigma_a,\{\mathbf{x}_{b\neq a}\},t)\right]d\sigma_1d\mathbf{x}_b \end{align} $$ Obtaining the required value (see this, page 5 and 6): $$\langle\hat{\mathbf{j}}(\mathbf{r})\rangle=\mathbf{j}(\mathbf{r},t)$$

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