The Stark eﬀect: hydrogen in an external electric ﬁeld

3.5 The Stark eﬀect: hydrogen in an external electric ﬁeld

With an external electric ﬁeld along the $z$ -axis, the perturbing Hamiltonian is ${\hat{H}}^{(1)} = - e ℰ z$ (we use $ℰ$ for the electric ﬁeld strength to distinguish it from the energy.) Now it is immediately obvious that $⟨ n l m | z | n l m ⟩ = 0$ for any state: The probability density is symmetric on reﬂection in the $x y$ -plane, but $z$ is antisymmetric. So for the ground state, the ﬁrst order energy shift vanishes. (We will return to excited states, but think now about why we can’t conclude the same for them.) This is not surprising, because an atom of hydrogen in its ground state has no electric dipole moment: there is no $p \cdot E$ term to match the $μ \cdot B$ one.

To calculate the second-order energy shift we need $⟨ n^{'} l^{'} m^{'} | z | 100 ⟩$ . We can write $z$ as $r cos θ$ or $\sqrt{4 π ∕ 3} r Y_{10} (θ, ϕ)$ . The lack of dependence on $ϕ$ means that $m$ can’t change, and in addition $l$ can only change by one unit, so $⟨ n^{'} l^{'} m^{'} | z | 100 ⟩ = δ_{l^{'} 1} δ m^{'} 0 ⟨ n^{'} 10 | z | 100 ⟩$ . However this isn’t the whole story: there are also states in the continuum, which we will denote $| k ⟩$ (though these are not plane waves, since they see the Coulomb potential). So we have

E_{100}^{(2)} = {(e ℰ)}^{2} \sum_{n > 1} \frac{{|⟨ n 10 | z | 100 ⟩|}^{2}}{E_{1}^{(0)} - E_{n}^{(0)}} + {(e ℰ)}^{2} \int d^{3} k \frac{{|⟨ k | z | 100 ⟩|}^{2}}{E_{1}^{(0)} - E_{k}^{(0)}}

(We use $E_{1}$ for $E_{100}$ ). This is a compact expression, but it would be very hard to evaluate directly. We can get a crude estimate of the size of the eﬀect by simply replacing all the denominators by $E_{1}^{(0)} - E_{2}^{(0)}$ ; this overestimates the magnitude of every term but the ﬁrst, for which it is exact, so it will give an upper bound on the magnitude of the shift. Then

\begin{array}{rcl} E_{1}^{(2)} & > & \frac{{(e ℰ)}^{2}}{E_{1}^{(0)} - E_{2}^{(0)}} (\sum_{n \geq 1} \sum_{l m} ⟨ 100 | z | n l m ⟩ ⟨ n l m | z | 100 ⟩ + \int d^{3} k ⟨ 100 | z | k ⟩ ⟨ k | z | 100 ⟩) \\ = & \frac{{(e ℰ)}^{2}}{E_{1}^{(0)} - E_{2}^{(0)}} ⟨ 100 | z^{2} | 100 ⟩ = - \frac{8 {(e ℰ)}^{2} a_{0}^{3}}{3 ℏ c α} \end{array}

where we have included $n = 1$ and other values of $l$ and $m$ in the sum because the matrix elements vanish anyway, and then used the completeness relation involving all the states, bound and unbound, of the hydrogen atom.

There is a trick for evaluating the exact result, which gives $9 ∕ 4$ rather than $8 ∕ 3$ as the constant (See Shankar.) So our estimate of the magnitude is fairly good. (For comparison with other ways of writing the shift, note that ${(e ℰ)}^{2} ∕ ℏ c α = 4 π ε_{0} ℰ^{2}$ —or in Gaussian units, just $ℰ^{2}$ .)

Having argued above that the hydrogen atom has no electric dipole, how come we are getting a ﬁnite eﬀect at all? The answer of course is that the ﬁeld polarises the atom, and the induced dipole can then interact with the ﬁeld.

Now for the ﬁrst excited state. We can’t conclude that the ﬁrst-order shift vanishes here, of course, because of degeneracy: there are four states and ${\hat{H}}^{(1)}$ is not diagonal in the usual basis $| 2 l m ⟩$ . In fact as we argued above it only connects $| 200 ⟩$ and $| 210 ⟩$ , so the states $| 21 \pm 1 ⟩$ decouple and their ﬁrst order shifts do vanish. Using $⟨ 210 | z | 200 ⟩ = - 3 a_{0}$ , we have in this subspace (with $⟨ 200 | = (1, 0)$ and $⟨ 210 | = (0, 1)$ )

{\hat{H}}^{(1)} = 3 a_{0} e ℰ (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}),

and the eigenstates are $\sqrt{1 ∕ 2} (| 200 ⟩ \pm | 210 ⟩)$ with eigenvalues $\mp 3 a_{0} e ℰ$ . So the degenerate quartet is split into a triplet of levels (with the unshifted one doubly degenerate).

In reality the degeneracy of the $n = 2$ states is lifted by the ﬁne-structure splitting; are these results then actually relevant? They will be approximately true if the ﬁeld is large; at an intermediate strength both ﬁne-structure and Stark eﬀects should be treated together as a perturbation on the pure Coulomb states. For very weak ﬁelds degenerate perturbation theory holds in the space of $j = \frac{1}{2}$ states, which are split by $\mp \sqrt{3} a_{0} e ℰ$ .

Gasiorowicz ch 11.3
Shankar ch 17.2,3
Townsend ch 11.4

[prev] [up] [top]