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When considering the Stark Effect, we consider the effect of an external uniform weak electric field which is directed along the positive $z$-axis, $\vec{\varepsilon} = \varepsilon \vec{k}$, on the ground state of a hydrogen atom. Then using nondegenerate perturbation theory it follows that we can approximate the energy of the ground state by $$E_{100} = E_{100}^{(1)} + \epsilon \varepsilon \langle 100| \hat{Z}| 100 \rangle + e^2 \varepsilon^2 \sum_{nlm \neq 100}\frac{|\langle nlm| \hat{Z}| 100 \rangle|^2}{E_{100}^{(0)}-E_{nlm}^{(0)}}.$$ We can show that the second term is zero i.e. $\langle 100| \hat{Z}| 100 \rangle = 0$.

How does it follow from this that the following conclusion can be made "The underlying physics behind this is that when the hydrogen atom is in the ground state, it has no permanent electric dipole moment"?

Thanks for any assistance.

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First of all, it should be $z$, not $\hat{z}$. You're taking the mean value of the $z$-coordinate in the ground state.

Now, the ground state has spherical symmetry. This means that whatever $\langle z \rangle$ is, it better be equal to $\langle x \rangle$ and $\langle y \rangle$, since in the ground state $|100\rangle$ there is nothing that makes the $z$ direction special.

The dipole moment operator for the electron is $q \mathbf{r}$, with $q$ its charge and $\mathbf{r}$ the position operator. So the mean value of the dipole moment in the ground state is $q \langle 100 | \mathbf{r} | 100 \rangle$. But $\mathbf{r} = x \mathbf{\hat{x}} + y \mathbf{\hat{y}} + z \mathbf{\hat{z}}$, and since we know that the mean value of the coordinates is zero, we get that the mean value of the dipole moment is zero too.

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  • $\begingroup$ Thanks for your answer, why wouldn't we, in analogy with classical electrodynamics, and since we are considering the state of the Hydrogen atom and noy just the electron, consider the dipole moment operator as $q \vec{r'}_{+} - q \vec{r'}_{-}$ since we consider both the positive charged nucleus and the negative charged electron? $\endgroup$ – user100411 Apr 1 '17 at 17:01
  • $\begingroup$ Yes, but since we consider the nucleus as being fixed, its position is zero so it doesn't contribute to the dipole moment. $\endgroup$ – Javier Apr 1 '17 at 18:09
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    $\begingroup$ Oh okay so you mean that the expectation value of the $q \vec{r}_{+}$ (which corresponds to the nucleus) would be zero since the position is assumed fixed, hence we get what you wrote? $\endgroup$ – user100411 Apr 1 '17 at 18:30
  • $\begingroup$ That's correct. $\endgroup$ – Javier Apr 1 '17 at 18:31
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The underlying physics behind this is...

...that only a (permanent) electric dipole can orient itself relative to the direction of the external field and therefore experience a shift in its energy: Turning towards the external field costs energy while giving way to the external force and orienting itself along the field lines minimizes its energy.

The ground state wavefunction of (the electron of) the hydrogen atom is spherically symmetric: The electron is just as likely to be found 'above' as well as 'beneath' the core. Therefore, there is no permanent dipole caused by the electric charge distribution.

Let's have again a look at your perturbation expansion for the energy shift,

$$E_{100} = E_{100}^{(0)} + \varepsilon E_{100}^{(1)} + \varepsilon^2 E_{100}^{(2)} + o(\varepsilon^3).$$

the first order term $E_{100}^{(1)}= \langle 100| e\hat{z}| 100 \rangle$ drops out because of the spherical symmetry. But the second order term $E_{100}^{(2)}$ does not! This means, an external electric field $\varepsilon$ can very well induce a dipole moment by deforming the wavefunction and thus the charge distribution.Then it acts on this induced dipole. Therefore, second order: $\varepsilon^2=\varepsilon\cdot\varepsilon$.

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