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Next: Degenerate Perturbation Theory Up: Time-Independent Perturbation Theory Previous: Non-Degenerate Perturbation Theory

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},
\end{displaymath} (911)

and the perturbing Hamiltonian
H_1 = e \vert{\bf E}\vert z.
\end{displaymath} (912)

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 Cha. 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 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

$\displaystyle \Delta E_{nlm}$ $\textstyle =$ $\displaystyle e \vert{\bf E}\vert \langle n,l,m\vert z\vert n,l,m\rangle$  
    $\displaystyle + e^2 \vert{\bf E}\vert^2\sum_{n',l',m'\neq n,l,m}\frac{\vert\langle n,l,m\vert z\vert n',l',m'\rangle\vert^2}{E_{nlm}-E_{n'l'm'}}.$ (913)

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. (529)]

L_z = x p_y - y p_x,
\end{displaymath} (914)

it follows that [see Eqs. (481)-(483)]
\begin{displaymath}[L_z,z]= 0.
\end{displaymath} (915)

$\displaystyle \langle n,l,m\vert[L_z,z]\vert n',l',m'\rangle$ $\textstyle =$ $\displaystyle \langle n,l,m\vert L_z z-z L_z\vert n',l',m'\rangle$  
  $\textstyle =$ $\displaystyle \hbar (m-m') \langle n,l,m\vert z\vert n',l',m'\rangle = 0,$ (916)

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.
\end{displaymath} (917)

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

$\displaystyle [L^2,z]$ $\textstyle =$ $\displaystyle [L_x^{ 2},z] + [L_y^{ 2},z]$  
  $\textstyle =$ $\displaystyle L_x [L_x,z] + [L_x,z] L_x + L_y [L_y,z]+[L_y,z] L_y$  
  $\textstyle =$ $\displaystyle {\rm i} \hbar (-L_x y-y L_x+L_y x+x L_y)$  
  $\textstyle =$ $\displaystyle 2 {\rm i} \hbar (L_y x -L_x y + {\rm i} \hbar z)$  
  $\textstyle =$ $\displaystyle 2 {\rm i} \hbar (L_y x - y L_x) = 2 {\rm i} \hbar (x L_y-L_x y),$ (918)

where use has been made of Eqs. (481)-(483), (527)-(529), and (535). Thus,
$\displaystyle [L^2,[L^2,z]]$ $\textstyle =$ $\displaystyle 2 {\rm i} \hbar \left(L^2, L_y x-L_x y + {\rm i} \hbar z\right)$  
  $\textstyle =$ $\displaystyle 2 {\rm i} \hbar \left(L_y [L^2,x] - L_x [L^2,y] + {\rm i} \hbar [L^2,z]\right)$  
  $\textstyle =$ $\displaystyle -4 \hbar^2 L_y (y L_z-L_y z) + 4 \hbar^2 L_x (L_x z-x L_z)$  
    $\displaystyle -2 \hbar^2 (L^2 z-z L^2),$ (919)

which reduces to
$\displaystyle [L^2,[L^2,z]]$ $\textstyle =$ $\displaystyle -\hbar^2 \left\{4 (L_x x+ L_y y+L_z z) L_z
-4 (L_x^{ 2}+L_y^{ 2}+L_z^{ 2}) z\right.$  
    $\displaystyle \left.+2 (L^2 z-z L^2)\right\}$  
  $\textstyle =$ $\displaystyle -\hbar^2 \left\{4 (L_x x+ L_y y+L_z z) L_z-2 (L^2 z+z L^2)\right\}.$ (920)

However, it is clear from Eqs. (527)-(529) that
L_x x+L_y y+L_z z = 0.
\end{displaymath} (921)

Hence, we obtain
\begin{displaymath}[L^2,[L^2,z]] = 2 \hbar^2 (L^2 z+z L^2).
\end{displaymath} (922)

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.
\end{displaymath} (923)

Equation (923) 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.
\end{displaymath} (924)

Since, by definition, $\psi_{nlm}$ is an eigenstate of $L^2$ corresponding to the eigenvalue $l (l+1) \hbar^2$, this expression yields
$\displaystyle \left\{l^2 (l+1)^2-2 l (l+1) l' (l'+1) + l'^2 (l'+1)^2\right.$      
$\displaystyle \left.-2 l (l+1) - 2 l' (l'+1)\right\}\langle n,l,m\vert z\vert n',l',m\rangle$ $\textstyle =$ $\displaystyle 0,$ (925)

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.
\end{displaymath} (926)

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 Cha. 9, that an $l=0$ wavefunction 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.
\end{displaymath} (927)

Application of the selection rules (917) and (927) to Eq. (913) 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}}.
\end{displaymath} (928)

Note that, according to the selection rules, all of the terms in Eq. (913) 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.
\end{displaymath} (929)

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}}.
\end{displaymath} (930)

Unfortunately, there is one fairly obvious problem with Eq. (928). 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. (928) and (930) 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 Cha. 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}}.
\end{displaymath} (931)

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},
\end{displaymath} (932)

a_0 = \frac{4\pi\epsilon_0 \hbar^2}{m_e e^2}
\end{displaymath} (933)

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},
\end{displaymath} (934)

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.
\end{displaymath} (935)

However, [see Eq. (874)]
$\displaystyle \sum_{n>1}\vert\langle 1,0,0\vert z\vert n,1,0\rangle\vert^2$ $\textstyle =$ $\displaystyle \sum_{n>1}\langle 1,0,0\vert z\vert n,1,0\rangle \langle n,1,0\vert z\vert 1,0,0\rangle$  
  $\textstyle =$ $\displaystyle \sum_{n',l',m'}\langle 1,0,0\vert z\vert n',l',m'\rangle \langle n',l',m'\vert z\vert 1,0,0\rangle$  
  $\textstyle =$ $\displaystyle \langle 1,0,0\vert z^2\vert 1,0,0\rangle = \frac{1}{3} \langle 1,0,0\vert r^2\vert 1,0,0\rangle,$ (936)

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. (693) that
\langle 1,0,0\vert r^2\vert 1,0,0\rangle = 3 a_0^{ 2}.
\end{displaymath} (937)

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}.
\end{displaymath} (938)

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}.
\end{displaymath} (939)

next up previous
Next: Degenerate Perturbation Theory Up: Time-Independent Perturbation Theory Previous: Non-Degenerate Perturbation Theory
Richard Fitzpatrick 2010-07-20