The quadratic Stark effect

Suppose that a hydrogen atom is subject to a uniform external electric field, of magnitude $\vert{\bf E}\vert$, directed along the $z$-axis. The Hamiltonian of the system can be split into two parts. Namely, the unperturbed Hamiltonian,

$$H_0 =\frac{p^2}{2\,m_e} -\frac{e^2}{4\pi\,\epsilon_0\,r},$$ (890)

and the perturbing Hamiltonian

$$H_1 = e\,\vert{\bf E}\vert\,z.$$ (891)

Note that the electron spin is irrelevant to this problem (since the spin operators all commute with $H_1$), so we can ignore the spin degrees of freedom of the system. Hence, the energy eigenstates of the unperturbed Hamiltonian are characterized by three quantum numbers--the radial quantum number $n$, and the two angular quantum numbers $l$ and $m$ (see Sect. 9). Let us denote these states as the $\psi_{nlm}$, and let their corresponding energy eigenvalues be the $E_{nlm}$. According to the analysis in the previous subsection, the change in energy of the eigenstate characterized by the quantum numbers $n,l,m$ in the presence of a small electric field is given by

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

$$+ e^2\,\vert{\bf E}\vert^2\sum_{n',l',m'\neq n,l,m}\frac{\vert\langle n,l,m\vert z\vert n',l',m'\rangle\vert^2}{E_{nlm}-E_{n'l'm'}}.$$ (892)

This energy-shift is known as the Stark effect.

The sum on the right-hand side of the above equation seems very complicated. However, it turns out that most of the terms in this sum are zero. This follows because the matrix elements $\langle n,l,m\vert z\vert n',l',m'\rangle$ are zero for virtually all choices of the two sets of quantum number, $n,l,m$ and $n',l',m'$. Let us try to find a set of rules which determine when these matrix elements are non-zero. These rules are usually referred to as the selection rules for the problem in hand.

Now, since [see Eq. (511)]

$$L_z = x\,p_y - y\,p_x,$$ (893)

it follows that [see Eqs. (463)-(465)]

$$[L_z,z]= 0.$$ (894)

Thus,

$$\langle n,l,m\vert[L_z,z]\vert n',l',m'\rangle = \langle n,l,m\vert L_z\,z-z\,L_z\vert n',l',m'\rangle$$

$$= \hbar\,(m-m')\,\langle n,l,m\vert z\vert n',l',m'\rangle = 0,$$ (895)

since $\psi_{nlm}$ is, by definition, an eigenstate of $L_z$ corresponding to the eigenvalue $m\,\hbar$. Hence, it is clear, from the above equation, that one of the selection rules is that the matrix element $\langle n,l,m\vert z\vert n',l',m'\rangle$ is zero unless

$$m' = m.$$ (896)

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

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

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

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

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

$$= 2\,{\rm i}\,\hbar\,(L_y\,x - y\,L_x) = 2\,{\rm i}\,\hbar\,(x\,L_y-L_x\,y),$$ (897)

where use has been made of Eqs. (463)-(465), (509)-(511), and (517). Thus,

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

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

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

$$-2\,\hbar^2\,(L^2\,z-z\,L^2),$$ (898)

which reduces to

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

$$\left.+2\,(L^2\,z-z\,L^2)\right\}$$

$$= -\hbar^2\,\left\{4\,(L_x\,x+ L_y\,y+L_z\,z)\,L_z-2\,(L^2\,z+z\,L^2)\right\}.$$ (899)

However, it is clear from Eqs. (509)-(511) that

$$L_x\,x+L_y\,y+L_z\,z = 0.$$ (900)

Hence, we obtain

$$[L^2,[L^2,z]] = 2\,\hbar^2\,(L^2\,z+z\,L^2).$$ (901)

Finally, the above expression expands to give

$$L^4\,z-2\,L^2\,z\,L^2 + z\,L^4 - 2\,\hbar^2\,(L^2\,z+z\,L^2) = 0.$$ (902)

Equation (902) implies that

$$\langle n,l,m\vert L^4\,z-2\,L^2\,z\,L^2 + z\,L^4 - 2\,\hbar^2\,(L^2\,z+z\,L^2) \vert n',l',m\rangle = 0.$$ (903)

Since, by definition, $\psi_{nlm}$ is an eigenstate of $L^2$ corresponding to the eigenvalue $l\,(l+1)\,\hbar^2$, this expression yields

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

$$\left.-2\,l\,(l+1) - 2\,l'\,(l'+1)\right\}\langle n,l,m\vert z\vert n',l',m\rangle = 0,$$ (904)

which reduces to

$$(l+l'+2)\,(l+l')\,(l-l'+1)\,(l-l'-1)\,\langle n,l,m\vert z\vert n',l',m\rangle = 0.$$ (905)

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 $l'=l\pm 1$. [Of course, the factor $l+l'+2$, in the above equation, can never be zero, since $l$ and $l'$ can never be negative.] Recall, however, from Sect. 9, that an $l=0$ wave-function is spherically symmetric. It, therefore, follows, from symmetry, that the matrix element $\langle n,l,m\vert z\vert n',l',m\rangle$ is zero when $l=l'=0$. In conclusion, the selection rule for $l$ is that the matrix element $\langle n,l,m\vert z\vert n',l',m\rangle$ is zero unless

$$l' = l\pm 1.$$ (906)

Application of the selection rules (896) and (906) to Eq. (892) yields

$$\Delta E_{nlm} = e^2\,\vert{\bf E}\vert^2\sum_{n',l'=l\pm 1}... ...e n,l,m\vert z\vert n',l',m\rangle\vert^2}{E_{nlm}-E_{n'l'm}}.$$ (907)

Note that, according to the selection rules, all of the terms in Eq. (892) which vary linearly with the electric field-strength vanish. Only those terms which vary quadratically with the field-strength survive. Hence, this type of energy-shift of an atomic state in the presence of a small electric field is known as the quadratic Stark effect. Now, the electric polarizability of an atom is defined in terms of the energy-shift of the atomic state as follows:

$$\Delta E = -\frac{1}{2}\,\alpha\,\vert{\bf E}\vert^2.$$ (908)

Hence, we can write

$$\alpha_{nlm} = 2\,e^2\sum_{n',l'=l\pm 1} \frac{\vert\langle n,l,m\vert z\vert n',l',m\rangle\vert^2}{E_{n'l'm}-E_{nlm}}.$$ (909)

Unfortunately, there is one fairly obvious problem with Eq. (907). Namely, it predicts an infinite energy-shift if there exists some non-zero matrix element $\langle n,l,m\vert z\vert n',l',m\rangle$ which couples two degenerate unperturbed energy eigenstates: i.e., if $\langle n,l,m\vert z\vert n',l',m\rangle\neq 0$ and $E_{nlm}=E_{n'l'm}$. Clearly, our perturbation method breaks down completely in this situation. Hence, we conclude that Eqs. (907) and (909) are only applicable to cases where the coupled eigenstates are non-degenerate. For this reason, the type of perturbation theory employed here is known as non-degenerate perturbation theory. Now, the unperturbed eigenstates of a hydrogen atom have energies which only depend on the radial quantum number $n$ (see Sect. 9). It follows that we can only apply the above results to the $n=1$ eigenstate (since for $n>1$ there will be coupling to degenerate eigenstates with the same value of $n$ but different values of $l$).

Thus, according to non-degenerate perturbation theory, the polarizability of the ground-state (i.e., $n=1$) of a hydrogen atom is given by

$$\alpha = 2\,e^2\sum_{n>1}\frac{\vert\langle 1,0,0\vert z\vert n,1,0\rangle\vert^2}{E_{n00}-E_{100}}.$$ (910)

Here, we have made use of the fact that $E_{n10}=E_{n00}$. The sum in the above expression can be evaluated approximately by noting that (see Sect. 9.4)

$$E_{n00} = -\frac{e^2}{8\pi\,\epsilon_0\,a_0\,n^2},$$ (911)

where

$$a_0 = \frac{4\pi\,\epsilon_0\,\hbar^2}{m_e\,e^2}$$ (912)

is the Bohr radius. Hence, we can write

$$E_{n00}-E_{100} \geq E_{200}-E_{100} = \frac{3}{4}\,\frac{e^2}{8\pi\,\epsilon_0\,a_0},$$ (913)

which implies that

$$\alpha < \frac{16}{3}\,4\pi\,\epsilon_0\,a_0\,\sum_{n>1} \vert\langle 1,0,0\vert z\vert n,1,0\rangle\vert^2.$$ (914)

However, [see Eq. (853)]

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

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

$$= \langle 1,0,0\vert z^2\vert 1,0,0\rangle = \frac{1}{3}\,\langle 1,0,0\vert r^2\vert 1,0,0\rangle,$$ (915)

where we have made use of the selection rules, the fact that the $\psi_{n',l',m'}$ form a complete set, and the fact the the ground-state of hydrogen is spherically symmetric. Finally, it follows from Eq. (672) that

$$\langle 1,0,0\vert r^2\vert 1,0,0\rangle = 3\,a_0^{\,2}.$$ (916)

Hence, we conclude that

$$\alpha < \frac{16}{3}\,4\pi\,\epsilon_0\,a_0^{\,3}\simeq 5.3\,\,4\pi\,\epsilon_0\,a_0^{\,3}.$$ (917)

The exact result (which can be obtained by solving Schrödinger's equation in parabolic coordinates) is

$$\alpha = \frac{9}{2}\,4\pi\,\epsilon_0\,a_0^{\,3} = 4.5\,\,4\pi\,\epsilon_0\,a_0^{\,3}.$$ (918)