The quadratic Stark effect

Suppose that a one-electron atom [i.e., either a hydrogen atom, or an alkali metal atom (which possesses one valance electron orbiting outside a closed, spherically symmetric shell)] is subjected to a uniform electric field in the positive $z$-direction. The Hamiltonian of the system can be split into two parts. The unperturbed Hamiltonian,

$$H_0 = \frac{{\bf p}^2}{2 m_e} + V(r),$$ (621)

and the perturbing Hamiltonian,

$$H_1= e \vert{\bf E}\vert z.$$ (622)

It is assumed that the unperturbed energy eigenvalues and eigenstates are completely known. The electron spin is irrelevant in this problem (since the spin operators all commute with $H_1$), so we can ignore the spin degrees of freedom of the system. This implies that the system possesses no degenerate energy eigenvalues. This is not true for the $n\neq 1$ energy levels of the hydrogen atom, due to the special properties of a pure Coulomb potential. It is necessary to deal with this case separately, because the perturbation theory presented in Sect. 6.3 breaks down for degenerate unperturbed energy levels.

An energy eigenket of the unperturbed Hamiltonian is characterized by three quantum numbers--the radial quantum number $n$, and the two angular quantum numbers $l$ and $m$ (see Sect. 5.6). Let us denote such a ket $\vert n,l,m\rangle$, and let its energy level be $E_{nlm}$. According to Eq. (618), the change in this energy level induced by a small electric field is given by

$$\Delta E_{nlm} = e \vert{\bf E}\vert \langle n,l,m \vert z\vert n,l,m\rangle$$

$$+ e^2 \vert{\bf E}\vert^2 \sum_{n',l',m'\neq n,l,m} \frac{\vert\langle n,l,m\vert z\vert n,'l',m'\rangle\vert^2}{ E_{nlm}-E_{n'l'm'} }.$$ (623)

Now, since

$$L_z = x p_y - y p_x,$$ (624)

it follows that

$$[L_z, z]= 0.$$ (625)

Thus,

$$\langle n,l, m\vert [L_z, z] \vert n',l',m'\rangle = 0,$$ (626)

giving

$$(m - m') \langle n,l, m\vert z\vert n',l',m'\rangle = 0,$$ (627)

since $\vert n,l,m\rangle$ is, by definition, an eigenstate of $L_z$ with eigenvalue $m \hbar$. It is clear, from the above relation, that the matrix element $\langle n,l, m\vert z\vert n',l',m'\rangle$ is zero unless $m'=m$. This is termed the selection rule for the quantum number $m$.

Let us now determine the selection rule for $l$. We have

$$[L^2, z] = [L_x^{ 2}, z] + [L_y^{ 2}, z]$$

$$= L_x[L_x, z] + [L_x, z] L_x + L_y[L_y, z] + [L_y, z] L_y$$

$$= {\rm i} \hbar\left( -L_x y - y L_x + L_y x + x L_y\right)$$

$$= 2 {\rm i} \hbar ( L_y x - L_x y + {\rm i} \hbar z)$$

$$= 2 {\rm i} \hbar ( L_y x - y L_x ) = 2 {\rm i} \hbar ( x L_y - L_x y),$$ (628)

where use has been made of Eqs. (297)-(302). Similarly,

$$[L^2, y] = 2 {\rm i} \hbar ( L_x z - x L_z ),$$ (629)

$$[L^2, x] = 2 {\rm i} \hbar ( y L_z - L_y z).$$ (630)

Thus,

$$[L^2, [L^2, z]] = 2 {\rm i} \hbar \left( L^2, L_y x - L_x y + {\rm i} \hbar z \right)$$

$$= 2 {\rm i} \hbar \left( L_y [L^2, x] - L_x [ L^2, y] + {\rm i} \hbar [L^2, z]\right),$$

$$= - 4 \hbar^2 L_y(y L_z - L_y z) + 4 \hbar^2 L_x(L_x z - x L_z)$$

$$- 2 \hbar^2(L^2 z - z L^2).$$ (631)

This reduces to

$$[L^2, [L^2, z]] = - \hbar^2 \left[4 (L_x x + L_y y + L_z z) L_z - 4 (L_x^{ 2} + L_y^{ 2} + L_z^{ 2}) z\right.$$

$$\left. + 2 (L^2 z - z L^2)\right].$$ (632)

However, it is clear from Eqs. (297)-(299) that

$$L_x x + L_y y + L_z z = 0.$$ (633)

Hence, we obtain

$$[L^2, [L^2, z]] = 2 \hbar^2 (L^2 z + z L^2).$$ (634)

Finally, the above expression expands to give

$$L^4 z - 2 L^2 z L^2 + z L^4 - 2 \hbar^2 (L^2 z +z L^2) = 0.$$ (635)

Equation (635) implies that

$$\langle n,l,m\vert L^4 z - 2 L^2 z L^2 + z L^4 - 2 \hbar^2 (L^2 z +z L^2) \vert n',l',m' \rangle = 0.$$ (636)

This expression yields

$$\left[l^2 (l+1)^2 - 2 l (l+1) l' (l'+1) + l'^2 (l'+1)^2 \right.$$

$$\left.- 2 l (l+1) - 2 l' (l'+1)\right] \langle n,l,m\vert z\vert n',l',m' \rangle = 0,$$ (637)

which reduces to

$$(l+l'+2) (l+l') (l-l'+1) (l-l'-1)\langle n,l,m\vert z\vert n',l',m' \rangle = 0.$$ (638)

According to the above formula, the matrix element $\langle n,l, m\vert z\vert n',l',m'\rangle$ vanishes unless $l=l'=0$ or $l' = l\pm 1$. This matrix element can be written

$$\langle n,l,m\vert z\vert n',l',m' \rangle = \int\!\int\!\in... ...') r'\cos\theta' \psi_{n'm'l'}(r',\theta',\varphi') dV',$$ (639)

where $\psi_{nlm}({\bf r}') = \langle {\bf r}'\vert n,l,m\rangle$. Recall, however, that the wave-function of an $l=0$ state is spherically symmetric (see Sect. 5.3): i.e., $\psi_{n00}({\bf r}') = \psi_{n00}(r')$. It follows from Eq. (639) that the matrix element vanishes by symmetry when $l=l'=0$. In conclusion, the matrix element $\langle n,l, m\vert z\vert n',l',m'\rangle$ is zero unless $l' = l\pm 1$. This is the selection rule for the quantum number $l$.

Application of the selection rules to Eq. (623) yields

$$\Delta E_{nlm} = e^2 \vert{\bf E}\vert^2 \sum_{n'}\sum_{l'... ...,l,m\vert z\vert n',l',m\rangle\vert^2}{E_{nlm} - E_{n'l' m}}.$$ (640)

Note that all of the terms in Eq. (623) which vary linearly with the electric field-strength vanish by symmetry, according to the selection rules. Only those terms which vary quadratically with the field-strength survive. The polarizability of an atom is defined in terms of the energy-shift of the atomic state as follows:

$$\Delta E = - \frac{1}{2} \alpha \vert{\bf E}\vert^2.$$ (641)

Consider the ground state of a hydrogen atom. (Recall, that we cannot address the $n>1$ excited states because they are degenerate, and our theory cannot handle this at present). The polarizability of this state is given by

$$\alpha = 2 e^2 \sum_{n>1} \frac{\vert\langle 1,0,0\vert z\vert n,1,0\rangle\vert^2}{E_{n00}-E_{100}}.$$ (642)

Here, we have made use of the fact that $E_{n10} = E_{n00}$ for a hydrogen atom.

The sum in the above expression can be evaluated approximately by noting that [see Eq. (416)]

$$E_{n00} = - \frac{e^2}{8\pi\epsilon_0 a_0 n^2}$$ (643)

for a hydrogen atom, where

$$a_0 = \frac{4\pi \epsilon_0 \hbar^2}{\mu e^2} = 5.3\times 10^{-11} {\rm meters}$$ (644)

is the Bohr radius. We can write

$$E_{n00}-E_{100} \geq E_{200} - E_{100} = \frac{3}{4} \frac{e^2}{8\pi\epsilon_0 a_0}.$$ (645)

Thus,

$$\alpha < \frac{16}{3} 4\pi \epsilon_0 a_0 \sum_{n>1} \vert\langle 1,0,0\vert z\vert n,1,0\rangle\vert^2.$$ (646)

However,

$$\sum_{n>1} \vert\langle 1,0,0\vert z\vert n,1,0\rangle\vert^2 = \sum_{n',l',m'} \langle 1,0,0\vert z\vert n',l',m'\rangle\langle n',m',l'\vert z\vert 1,0,0\rangle$$

$$= \langle 1,0,0\vert z^2\vert 1,0,0\rangle,$$ (647)

where we have made use of the fact that the wave-functions of a hydrogen atom form a complete set. It is easily demonstrated from the actual form of the ground state wave-function that

$$\langle 1,0,0\vert z^2\vert 1,0,0\rangle = a_0^{ 2}.$$ (648)

Thus, we conclude that

$$\alpha < \frac{16}{3} 4\pi \epsilon_0 a_0^{ 3} \simeq 5.3 4\pi \epsilon_0 a_0^{ 3}.$$ (649)

The true result is

$$\alpha = \frac{9}{2} 4\pi \epsilon_0 a_0^{ 3} = 4.5 4\pi \epsilon_0 a_0^{ 3}.$$ (650)

It is actually possible to obtain this answer, without recourse to perturbation theory, by solving Schrödinger's equation exactly in parabolic coordinates.