0
$\begingroup$

so I just read

Consider a particle (without spin) and spatial coordinate vector $\vec{q}$ and charge number Z. Classically, the >charge density of the particle is:

$$ \rho(\vec{r}) = Ze\delta(\vec{r}-\vec{q}) \tag{1}$$

whereas $\delta$ is the Dirac-Delta-Function.

From the corespondence principle it follows, that:

$$\hat{\rho(\vec{r})}=Ze\delta(\vec{r}-\hat{\vec{q}}) \tag{2}$$

The expectation value of the charge density $\hat{\rho(\vec{r})}$ is

$$\langle \hat{\rho(\vec{r})} \rangle = Ze\int \Psi^*(\vec{q})\delta(\vec{r}-\hat{\vec{q})}\Psi(\vec{q})dq^3 = Ze|\Psi(\vec{r})|^2 \tag{3}$$

whereas $dq^3=dxdydz$

Now I can't really follow the very last step in (3). How did they solve the integral?

The Dirac-Delta is rather new to me and my current interpretation of that integral is: The integrand is zero except at $\vec{q}$. So we basically "integrate" over a point. Whereas this "integration" is basically just evaluating the integrand at that specific point. Then we just get $\Psi^*\cdot \Psi=|\Psi|^2$ Which kind of collides with my current understanding of "integrating" but that's another topic and something I'll learn about in the next few weeks. But is that the argumentation used here?

$\endgroup$
2
$\begingroup$

The electron probability density $\rho(\mathbf{r})=|\Psi(\mathbf{r})|^2$ is the expectation value of the density operator $\hat{\rho}$, which can be written as

$$ \rho(\mathbf{r}) = \langle\hat{\rho}\rangle =\langle\Psi|\delta(\mathbf{r}-\mathbf{r}')|\Psi\rangle = \int\Psi^*(\mathbf{r}')\delta(\mathbf{r}-\mathbf{r}')\Psi(\mathbf{r}')d\mathbf{r}'=\Psi^*(\mathbf{r})\Psi(\mathbf{r})=|\Psi(\mathbf{r})|^2$$

If you look at your classical expression, $\rho_e(\mathbf{r})=Ze\delta(\mathbf{r}-\mathbf{r}')$, you can notice that its quantum analog is

$$ \langle \hat{\rho}_e\rangle=\langle Ze\hat{\rho}\rangle =Ze\langle\Psi|\delta(\mathbf{r}-\mathbf{r}')|\Psi\rangle = Ze|\Psi(\mathbf{r})|^2$$

In addition, the Dirac delta function has the property of translation. This property was used to evaluate the integral and get $|\Psi(\mathbf{r})|^2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot - my book didn't mentioned that property. (I should have looked it up - my bad - but I assumed my book stated all properties). $\endgroup$ – xotix Mar 12 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.