11.5: Quadratic Stark Effect

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

$$\label{e12.58} H_0 =\frac{p^{\,2}}{2\,m_e} -\frac{e^{\,2}}{4\pi\,\epsilon_0\,r},$$ and the perturbing Hamiltonian $$H_1 = e\,|{\bf E}|\,z.$$

Note that the electron spin is irrelevant to this problem (because 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 Chapter [scent].) 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 section, 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

$$\begin{aligned} {\mit\Delta} E_{nlm}&= e\,|{\bf E}|\,\langle n,l,m|z|n,l,m\rangle\nonumber\\[0.5ex] &\phantom{=}+ e^{\,2}\,|{\bf E}|^{\,2}\sum_{n',l',m'\neq n,l,m}\frac{|\langle n,l,m|z|n',l',m'\rangle|^{\,2}}{E_{nlm}-E_{n'l'm'}}.\label{e12.59}\end{aligned}$$ This energy-shift is known as the Stark effect .

The sum on the right-hand side of the previous equation seems very complicated. However, it turns out that most of the terms in this sum are zero. This follows because the matrix elements $\langle n,l,m|z|n',l',m'\rangle$ 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 that determine when these matrix elements are non-zero. These rules are usually referred to as the selection rules for the problem in hand.

Now, because [see Equation ([e8.3])]

$$L_z = x\,p_y - y\,p_x,$$ it follows that [see Equations ([commxx])–([commxp])] $$[L_z,z] = 0.$$ Thus, $$\begin{aligned} \langle n,l,m|[L_z,z]|n',l',m'\rangle& = \langle n,l,m|L_z\,z-z\,L_z|n',l',m'\rangle\nonumber\\[0.5ex] &= \hbar\,(m-m')\,\langle n,l,m|z|n',l',m'\rangle = 0,\end{aligned}$$ because $\psi_{nlm}$ is, by definition, an eigenstate of $L_z$ corresponding to the eigenvalue $m\,\hbar$. Hence, it is clear, from the previous equation, that one of the selection rules is that the matrix element $\langle n,l,m|z|n',l',m'\rangle$ is zero unless $$\label{e12.63} m' = m.$$

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

$$\begin{aligned} [L^2,z]&= [L_x^{\,2},z] + [L_y^{\,2},z]\nonumber\\[0.5ex] & = L_x\,[L_x,z] + [L_x,z]\,L_x + L_y\,[L_y,z]+[L_y,z]\,L_y\nonumber\\[0.5ex] &= {\rm i}\,\hbar\,(-L_x\,y-y\,L_x+L_y\,x+x\,L_y)\nonumber\\[0.5ex] &= 2\,{\rm i}\,\hbar\,(L_y\,x -L_x\,y + {\rm i}\,\hbar\,z)\nonumber\\[0.5ex] &= 2\,{\rm i}\,\hbar\,(L_y\,x - y\,L_x) = 2\,{\rm i}\,\hbar\,(x\,L_y-L_x\,y),\end{aligned}$$ where use has been made of Equations ([commxx])–([commxp]), ([e8.1])–([e8.3]), and ([e8.10]). Thus, $$\begin{aligned} [L^2,[L^2,z]]&= 2\,{\rm i}\,\hbar\,\left(L^2, L_y\,x-L_x\,y + {\rm i}\,\hbar\,z\right)\nonumber\\[0.5ex] &= 2\,{\rm i}\,\hbar\,\left(L_y\,[L^2,x] - L_x\,[L^2,y] + {\rm i}\,\hbar\,[L^2,z]\right)\nonumber\\[0.5ex] &=-4\,\hbar^{\,2}\,L_y\,(y\,L_z-L_y\,z) + 4\,\hbar^{\,2}\,L_x\,(L_x\,z-x\,L_z)\nonumber\\[0.5ex] &\phantom{=}-2\,\hbar^{\,2}\,(L^2\,z-z\,L^2),\end{aligned}$$ which reduces to $$\begin{aligned} [L^2,[L^2,z]]&= -\hbar^{\,2}\,\left\{4\,(L_x\,x+ L_y\,y+L_z\,z)\,L_z -4\,(L_x^{\,2}+L_y^{\,2}+L_z^{\,2})\,z\right.\nonumber\\[0.5ex] &\left.\phantom{=}+2\,(L^2\,z-z\,L^2)\right\}\nonumber\\[0.5ex] &= -\hbar^{\,2}\,\left\{4\,(L_x\,x+ L_y\,y+L_z\,z)\,L_z-2\,(L^2\,z+z\,L^2)\right\}.\end{aligned}$$ However, it is clear from Equations ([e8.1])–([e8.3]) that $$L_x\,x+L_y\,y+L_z\,z = 0.$$ Hence, we obtain $$[L^2,[L^2,z]] = 2\,\hbar^{\,2}\,(L^2\,z+z\,L^2).$$ Finally, the previous expression expands to give

$$\label{e12.69} L^4\,z-2\,L^2\,z\,L^2 + z\,L^4 - 2\,\hbar^{\,2}\,(L^2\,z+z\,L^2) = 0.$$

Equation ([e12.69]) implies that

$$\langle n,l,m|L^4\,z-2\,L^2\,z\,L^2 + z\,L^4 - 2\,\hbar^{\,2}\,(L^2\,z+z\,L^2) |n',l',m\rangle = 0.$$ Because, by definition, $\psi_{nlm}$ is an eigenstate of $L^2$ corresponding to the eigenvalue $l\,(l+1)\,\hbar^{\,2}$, this expression yields $$\begin{aligned} \left\{l^{\,2}\,(l+1)^2-2\,l\,(l+1)\,l'\,(l'+1) + l'^{\,2}\,(l'+1)^2\right.&\nonumber\\[0.5ex] \left.-2\,l\,(l+1) - 2\,l'\,(l'+1)\right\}\langle n,l,m|z|n',l',m\rangle& =0,\end{aligned}$$ which reduces to $$(l+l'+2)\,(l+l')\,(l-l'+1)\,(l-l'-1)\,\langle n,l,m|z|n',l',m\rangle = 0.$$ According to the previous formula, the matrix element $\langle n,l,m|z|n',l',m\rangle$ vanishes unless $l=l'=0$ or $l'=l\pm 1$. [Of course, the factor $l+l'+2$, in the previous equation, can never be zero, because $l$ and $l'$ can never be negative.] Recall, however, from Chapter [scent], that an $l=0$ wavefunction is spherically symmetric. It, therefore, follows, from symmetry, that the matrix element $\langle n,l,m|z|n',l',m\rangle$ is zero when $l=l'=0$. In conclusion, the selection rule for $l$ is that the matrix element $\langle n,l,m|z|n',l',m\rangle$ is zero unless $$\label{e12.73} l' = l\pm 1.$$

Application of the selection rules ([e12.63]) and ([e12.73]) to Equation ([e12.59]) yields

$$\label{e12.74} {\mit\Delta} E_{nlm} = e^{\,2}\,|{\bf E}|^{\,2}\sum_{n',l'=l\pm 1} \frac{|\langle n,l,m|z|n',l',m\rangle|^2}{E_{nlm}-E_{n'l'm}}.$$ Note that, according to the selection rules, all of the terms in Equation ([e12.59]) that 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 : $${\mit\Delta} E = -\frac{1}{2}\,\alpha\,|{\bf E}|^{\,2}.$$ Hence, we can write $$\label{e12.76} \alpha_{nlm} = 2\,e^{\,2}\sum_{n',l'=l\pm 1} \frac{|\langle n,l,m|z|n',l',m\rangle|^{\,2}}{E_{n'l'm}-E_{nlm}}.$$

Unfortunately, there is one fairly obvious problem with Equation ([e12.74]). Namely, it predicts an infinite energy-shift if there exists some non-zero matrix element $\langle n,l,m|z|n',l',m\rangle$ that couples two degenerate unperturbed energy eigenstates: that is, if $\langle n,l,m|z|n',l',m\rangle\neq 0$ and $E_{nlm}=E_{n'l'm}$. Clearly, our perturbation method breaks down completely in this situation. Hence, we conclude that Equations ([e12.74]) and ([e12.76]) 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. The unperturbed eigenstates of a hydrogen atom have energies that only depend on the radial quantum number $n$. (See Chapter [scent].) It follows that we can only apply the previous results to the $n=1$ eigenstate (because 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

$$\alpha = 2\,e^{\,2}\sum_{n>1}\frac{|\langle 1,0,0|z|n,1,0\rangle|^{\,2}}{E_{n00}-E_{100}}.$$ Here, we have made use of the fact that $E_{n10}=E_{n00}$. The sum in the previous expression can be evaluated approximately by noting that (see Section [s10.4]) $$E_{n00} = -\frac{e^{\,2}}{8\pi\,\epsilon_0\,a_0\,n^{\,2}},$$ where $$a_0 = \frac{4\pi\,\epsilon_0\,\hbar^{\,2}}{m_e\,e^{\,2}}$$ is the Bohr radius. Hence, we can write $$E_{n00}-E_{100} \geq E_{200}-E_{100} = \frac{3}{4}\,\frac{e^{\,2}}{8\pi\,\epsilon_0\,a_0},$$ which implies that $$\alpha < \frac{16}{3}\,4\pi\epsilon_0\,a_0\,\sum_{n>1} |\langle 1,0,0|z|n,1,0\rangle|^{\,2}.$$ However, [see Equation ([e12.20])] $$\begin{aligned} \sum_{n>1}|\langle 1,0,0|z|n,1,0\rangle|^{\,2} &= \sum_{n>1}\langle 1,0,0|z|n,1,0\rangle\,\langle n,1,0|z|1,0,0\rangle\nonumber\\[0.5ex] &=\sum_{n',l',m'}\langle 1,0,0|z|n',l',m'\rangle\,\langle n',l',m'|z|1,0,0\rangle\nonumber\\[0.5ex] &=\langle 1,0,0|z^{\,2}|1,0,0\rangle = \frac{1}{3}\,\langle 1,0,0|r^{\,2}|1,0,0\rangle,\end{aligned}$$ 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 Equation ([e9.73]) that $$\langle 1,0,0|r^{\,2}|1,0,0\rangle = 3\,a_0^{\,2}.$$ Hence, we conclude that $$\alpha < \frac{16}{3}\,4\pi\,\epsilon_0\,a_0^{\,3}\simeq 5.3\,\,4\pi\,\epsilon_0\,a_0^{\,3}.$$ The exact result (which can be obtained by solving Schrödinger’s equation in parabolic coordinates) is $$\alpha = \frac{9}{2}\,4\pi\,\epsilon_0\,a_0^{\,3} = 4.5\,\,4\pi\,\epsilon_0\,a_0^{\,3}.$$

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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