MARCO LEGAL

Substituting the above expression into Eq. (12.46), evaluated for m = n, and neglecting O(ǫ3_{) terms, we obtain}

(E_n − E + e_nn) −X

k_6=n

|enk|2

Ek− En ≃ 0.

(12.55)

Thus, the modified nth energy eigenstate possesses an eigenvalue

E_n′ = En + enn+ X k6=n |e_nk|2 En− Ek +_O(ǫ3), (12.56) and a wave-function ψ_n′ = ψn + X k_6=n ekn En − Ek ψk +O(ǫ2). (12.57)

Incidentally, it is easily demonstrated that the modified eigenstates remain or- thonormal to O(ǫ2_).

12.5 The Quadratic Stark Effect 12 TIME-INDEPENDENT PERTURBATION THEORY

12.5 The Quadratic Stark Effect

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

H0 = p2 2 me − e 2 4π ǫ0r , (12.58)

and the perturbing Hamiltonian

H1 = e |E| z. (12.59)

Note that the electron spin is irrelevant to this problem (since the spin oper- ators all commute with H1), 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 ψnlm, and let their corresponding energy eigenvalues be the Enlm.

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

∆Enlm = e |E|hn, l, m|z|n, l, mi +e2|E|2 X n′_,l′_,m′_6=n,l,m |_{hn, l, m|z|n}′, l′, m′_i|2 Enlm − En′_l′_m′ . (12.60)

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 hn, l, m|z|n′_{, l}′_{, m}′_{i 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.

12.5 The Quadratic Stark Effect 12 TIME-INDEPENDENT PERTURBATION THEORY

Now, since [see Eq. (8.4)]

Lz = x py− y px, (12.61)

it follows that [see Eqs. (7.15)–(7.17)]

[Lz, z] = 0. (12.62)

Thus,

hn, l, m|[Lz, z]|n′, l′, m′i = hn, l, m|Lzz − z Lz|n′, l′, m′i

= ¯h (m − m′)_{hn, l, m|z|n}′, l′, m′_{i = 0, (12.63)} since ψnlm is, by definition, an eigenstate of Lz corresponding to the eigenvalue

m¯h. Hence, it is clear, from the above equation, that one of the selection rules is that the matrix element hn, l, m|z|n′_{, l}′_{, m}′_{i is zero unless}

m′ = m. (12.64)

Let us now determine the selection rule for l. We have [L2, z] = [L_x2, z] + [L_y2, z]

= L_x[L_x, z] + [Lx, z] Lx+ Ly[Ly, z] + [Ly, z] Ly

= i ¯h (−Lxy − y Lx + Lyx + x Ly)

= 2i ¯h (L_yx − Lxy +i ¯h z)

= 2i ¯h (Lyx − y Lx) = 2i ¯h (x Ly− Lxy), (12.65)

where use has been made of Eqs. (7.15)–(7.17), (8.2)–(8.4), and (8.10). Thus, [L2, [L2, z]] = 2i ¯h L2, Lyx − Lxy +i ¯h z  = 2i ¯h  Ly[L2, x] − Lx[L2, y] +i ¯h [L2, z]  = −4¯h2Ly(y Lz − Lyz) + 4¯h2Lx(Lxz − x Lz) −2¯h2(L2z − z L2), (12.66)

12.5 The Quadratic Stark Effect 12 TIME-INDEPENDENT PERTURBATION THEORY which reduces to [L2, [L2, z]] = −¯h2 4 (Lxx + Lyy + Lzz) Lz− 4 (Lx2 + Ly2 + Lz2) z +2 (L2z − z L2) = −¯h2 4 (Lxx + Lyy + Lzz) Lz− 2 (L2z + z L2) . (12.67) However, it is clear from Eqs. (8.2)–(8.4) that

Lxx + Lyy + Lzz = 0. (12.68)

Hence, we obtain

[L2, [L2, z]] = 2¯h2(L2z + z L2). (12.69) Finally, the above expression expands to give

L4z − 2 L2z L2 + z L4 − 2¯h2(L2z + z L2) = 0. (12.70)

Equation (12.70) implies that

hn, l, m|L4z − 2 L2z L2 + z L4 − 2¯h2(L2z + z L2)|n′, l′, m_{i = 0.} (12.71) Since, by definition, ψnlm is an eigenstate of L2 corresponding to the eigenvalue

l (l + 1)¯h2, this expression yields 

l2(l + 1)2 − 2 l (l + 1) l′(l′ + 1) + l′2(l′ + 1)2

−2 l (l + 1) − 2 l′(l′ + 1)}_{hn, l, m|z|n}′, l′, m_{i = 0,} (12.72) which reduces to

(l + l′+ 2) (l + l′) (l − l′ + 1) (l − l′ − 1)_{hn, l, m|z|n}′, l′, m_{i = 0.} (12.73) According to the above formula, the matrix element hn, l, m|z|n′_{, l}′_{, mi vanishes}

unless l = l′ _{= 0 or l}′ _{= l± 1. [Of course, the factor l + l}′_{+ 2, in the above equa-}

tion, 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 hn, l, m|z|n′_{, l}′_{, mi is zero when}

l = l′ = 0. In conclusion, the selection rule for l is that the matrix element hn, l, m|z|n′, l′, m_{i is zero unless}

12.5 The Quadratic Stark Effect 12 TIME-INDEPENDENT PERTURBATION THEORY

Application of the selection rules (12.64) and (12.74) to Eq. (12.60) yields

∆Enlm = e2|E|2 X n′_,l′_=l±1 |_{hn, l, m|z|n}′, l′, m_i|2 Enlm− En′_l′_m . (12.75)

Note that, according to the selection rules, all of the terms in Eq. (12.60) 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:

∆E = −1 2α |E|

2_{. (12.76)}

Hence, we can write

αnlm = 2 e2 X n′_,l′_=l±1 |_{hn, l, m|z|n}′, l′, m_i|2 En′_l′_m − E_nlm . (12.77)

Unfortunately, there is one fairly obvious problem with Eq. (12.75). Namely, it predicts an infinite energy-shift if there exists some non-zero matrix element hn, l, m|z|n′, l′, m_{i which couples two degenerate unperturbed energy eigenstates:}

i.e., if_{hn, l, m|z|n}′, l′, m_{i 6= 0 and E}nlm = En′_l′_m. Clearly, our perturbation method

breaks down completely in this situation. Hence, we conclude that Eqs. (12.75) and (12.77) 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 = 1eigenstate (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

α = 2 e2X

n>1

|_{h1, 0, 0|z|n, 1, 0i|}2 En00 − E100

12.5 The Quadratic Stark Effect 12 TIME-INDEPENDENT PERTURBATION THEORY

Here, we have made use of the fact that En10 = En00. The sum in the above

expression can be evaluated approximately by noting that (see Sect. 9.4) En00 = − e2 8π ǫ0a0n2 , (12.79) where a0 = 4π ǫ0¯h2 mee2 (12.80) is the Bohr radius. Hence, we can write

En00− E100 ≥ E200 − E100 = 3 4 e2 8π ǫ0a0 , (12.81)

which implies that

α < 16

3 4π ǫ0a0 X

n>1

|_{h1, 0, 0|z|n, 1, 0i|}2. (12.82)

However, [see Eq. (12.21)] X n>1 |_{h1, 0, 0|z|n, 1, 0i|}2 = X n>1 h1, 0, 0|z|n, 1, 0i hn, 1, 0|z|1, 0, 0i = X n′_,l′_,m′ h1, 0, 0|z|n′, l′, m′_{i hn}′, l′, m′|z|1, 0, 0_i = _{h1, 0, 0|z}2|1, 0, 0_{i =} 1 3h1, 0, 0|r 2_{|1, 0, 0} i, (12.83) where we have made use of the selection rules, the fact that the ψn′_,l′_,m′ form a

complete set, and the fact the the ground-state of hydrogen is spherically sym- metric. Finally, it follows from Eq. (9.72) that

h1, 0, 0|r2|1, 0, 0_{i = 3 a0}2. (12.84) Hence, we conclude that

α < 16

3 4π ǫ0a

3

0 ≃ 5.3 4π ǫ0a03. (12.85)

The exact result (which can be obtained by solving Schr¨odinger’s equation in parabolic coordinates) is

α = 9

24π ǫ0a

3