Estimating polarizability of Hydrogen Atom

Question

I am studying the example 9.2 from Zetilli's Quantum Mechanics book which is about the Stark Effect. There is a week electric field $\mathcal{E}$ directed along the positive z-axis and we want to study the effect of such a field on the ground state of an Hydrogen atom (ignoring the spin effect) and also find and approximation for the polarizability of it.

First we find an expression for the energy shift, which is given by

$$ \Delta E = e^{2} \mathcal{E}^{2} \sum_{nlm \neq 100} \frac{\vert{\langle nlm \vert \hat{Z}\vert 100\rangle} \vert^{2}}{E_{100}^{(0)}-E_{nlm}^{(0)}} $$

Then in order to find the polarizability $\alpha$ the book writes this equation below

$$\alpha = \frac{-2\Delta E}{\mathcal{E}^{2}}$$

Which I don't understand where it comes from. I know that $\hat{P}= \alpha E $ but I don't know how to proceed from there.

After this we use an inequality based on the energies of the states $E_{100}^{(0)}$ and $E_{200}^{(0)}$ and to find an upper limit for $\alpha$ and get to this result (where $a_{0}$ is the Bohr radius)

$$ \alpha \leq \frac{16}{3} a_{0} \sum_{nlm \neq 100} \vert{\langle nlm \vert \hat{Z}\vert 100\rangle} \vert^{2}$$

This sum is equal to $\langle 100 \vert \hat{Z}^{2}\vert 100 \rangle$, but when I calculate the value from this expression using the radial wave function $R_{10}$ and I substitute this result in the inequality I get

$$\alpha \leq \frac{a_{0}^{1/2}}{3}\sqrt{\frac{\pi}{2}}$$

While the author get to this result

$$\alpha \leq a_{0}^{3} \frac{16}{3}$$

I don't know where is my mistake.

Answer 1

Since the main issue seems to be where the expression for the polarizability comes from, I'll address that first.

The usual explanation (e.g. in PW Atkins Physical Chemistry) is as follows. The perturbation term in the Hamiltonian when a dipole moment $P$ couples with an electric field $\mathcal{E}$ is $-P\mathcal{E}$ (where I implicitly mean vector components in the direction of the field). But here the dipole is induced by the field. To work out the energy change we need to imagine turning on the field from zero, and integrating the resulting change in energy, using $P=\alpha \mathcal{E}$. So $$ \Delta E = -\int_0^\mathcal{E} P\, d\mathcal{E} = -\int_0^\mathcal{E} \alpha \mathcal{E}\, d\mathcal{E} = -\frac{1}{2} \alpha \mathcal{E}^2 . $$ This gives you your formula for $\alpha$. Naturally we are anticipating matching this up with a second-order perturbation theory expression for $\Delta E$, which will be proportional to $\mathcal{E}^2$, as in your first equation.

As for the integral, I can't say where you are going wrong exactly, but clearly the result should be proportional to $a_0^3$, just from the form of the matrix elements being calculated. Moreover, because all the wavefunctions will have a spacial extent of order $a_0$, you can expect the overall result (including the prefactor of $16 a_0/3$) to be $a_0^3$ multiplied by some numerical term of order $1$.

For a radially symmetric wavefunction $R_{10}(r)$, I would expect $$ \langle 100|\hat{Z}^2|100\rangle =\langle 100|\hat{Y}^2|100\rangle =\langle 100|\hat{X}^2|100\rangle =\frac{1}{3}\langle 100|r^2|100\rangle . $$ where $r$ is just the radial coordinate. I believe that this simplifies the integral and gives the right answer.

Answer 2

From 'Quantum Mechanics' by Shankar, 2nd ed., pg. 463-464;

Work done:

dW = -Fdx = -(qε)dx = -ε(qdx) = -ε dμ

Where ε is the electric field, μ is the dipole moment and dμ is an infinitesimal change in the dipole moment. If it is assumed that μ is linearly proportional to the electric field (this does not always hold, but disposing of this assumption would be to digress into more advanced theories than is necessary) we have

μ = αε

Where α is the polarization constant. An infinitesimal shift in μ is dμ = αdε. With this dW becomes,

dW = -εdμ = -αεdε

Changing W to E (energy) and integrating both sides give the desired relation.