11.5: Quadratic Stark Effect
( \newcommand{\kernel}{\mathrm{null}\,}\)
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=p22me−e24πϵ0r, and the perturbing Hamiltonian H1=e|E|z.
Note that the electron spin is irrelevant to this problem (because the spin operators 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 Chapter [scent].) Let us denote these states as the ψnlm, and let their corresponding energy eigenvalues be the Enlm. 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 ΔEnlm=e|E|⟨n,l,m|z|n,l,m⟩=+e2|E|2∑n′,l′,m′≠n,l,m|⟨n,l,m|z|n′,l′,m′⟩|2Enlm−En′l′m′. 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 ⟨n,l,m|z|n′,l′,m′⟩ 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])] Lz=xpy−ypx, it follows that [see Equations ([commxx])–([commxp])] [Lz,z]=0. Thus, ⟨n,l,m|[Lz,z]|n′,l′,m′⟩=⟨n,l,m|Lzz−zLz|n′,l′,m′⟩=ℏ(m−m′)⟨n,l,m|z|n′,l′,m′⟩=0, because ψnlm is, by definition, an eigenstate of Lz corresponding to the eigenvalue mℏ. Hence, it is clear, from the previous equation, that one of the selection rules is that the matrix element ⟨n,l,m|z|n′,l′,m′⟩ is zero unless m′=m.
Let us now determine the selection rule for l. We have =[L2x,z]+[L2y,z]=Lx[Lx,z]+[Lx,z]Lx+Ly[Ly,z]+[Ly,z]Ly=iℏ(−Lxy−yLx+Lyx+xLy)=2iℏ(Lyx−Lxy+iℏz)=2iℏ(Lyx−yLx)=2iℏ(xLy−Lxy), where use has been made of Equations ([commxx])–([commxp]), ([e8.1])–([e8.3]), and ([e8.10]). Thus, ]=2iℏ(L2,Lyx−Lxy+iℏz)=2iℏ(Ly[L2,x]−Lx[L2,y]+iℏ[L2,z])=−4ℏ2Ly(yLz−Lyz)+4ℏ2Lx(Lxz−xLz)=−2ℏ2(L2z−zL2), which reduces to ]=−ℏ2{4(Lxx+Lyy+Lzz)Lz−4(L2x+L2y+L2z)z=+2(L2z−zL2)}=−ℏ2{4(Lxx+Lyy+Lzz)Lz−2(L2z+zL2)}. However, it is clear from Equations ([e8.1])–([e8.3]) that Lxx+Lyy+Lzz=0. Hence, we obtain [L2,[L2,z]]=2ℏ2(L2z+zL2). Finally, the previous expression expands to give
L4z−2L2zL2+zL4−2ℏ2(L2z+zL2)=0.
Equation ([e12.69]) implies that ⟨n,l,m|L4z−2L2zL2+zL4−2ℏ2(L2z+zL2)|n′,l′,m⟩=0. Because, by definition, ψnlm is an eigenstate of L2 corresponding to the eigenvalue l(l+1)ℏ2, this expression yields {l2(l+1)2−2l(l+1)l′(l′+1)+l′2(l′+1)2−2l(l+1)−2l′(l′+1)}⟨n,l,m|z|n′,l′,m⟩=0, which reduces to (l+l′+2)(l+l′)(l−l′+1)(l−l′−1)⟨n,l,m|z|n′,l′,m⟩=0. According to the previous formula, the matrix element ⟨n,l,m|z|n′,l′,m⟩ vanishes unless l=l′=0 or l′=l±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 ⟨n,l,m|z|n′,l′,m⟩ is zero when l=l′=0. In conclusion, the selection rule for l is that the matrix element ⟨n,l,m|z|n′,l′,m⟩ is zero unless l′=l±1.
Application of the selection rules ([e12.63]) and ([e12.73]) to Equation ([e12.59]) yields ΔEnlm=e2|E|2∑n′,l′=l±1|⟨n,l,m|z|n′,l′,m⟩|2Enlm−En′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 : ΔE=−12α|E|2. Hence, we can write αnlm=2e2∑n′,l′=l±1|⟨n,l,m|z|n′,l′,m⟩|2En′l′m−Enlm.
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 ⟨n,l,m|z|n′,l′,m⟩ that couples two degenerate unperturbed energy eigenstates: that is, if ⟨n,l,m|z|n′,l′,m⟩≠0 and Enlm=En′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 α=2e2∑n>1|⟨1,0,0|z|n,1,0⟩|2En00−E100. Here, we have made use of the fact that En10=En00. The sum in the previous expression can be evaluated approximately by noting that (see Section [s10.4]) En00=−e28πϵ0a0n2, where a0=4πϵ0ℏ2mee2 is the Bohr radius. Hence, we can write En00−E100≥E200−E100=34e28πϵ0a0, which implies that α<1634πϵ0a0∑n>1|⟨1,0,0|z|n,1,0⟩|2. However, [see Equation ([e12.20])] ∑n>1|⟨1,0,0|z|n,1,0⟩|2=∑n>1⟨1,0,0|z|n,1,0⟩⟨n,1,0|z|1,0,0⟩=∑n′,l′,m′⟨1,0,0|z|n′,l′,m′⟩⟨n′,l′,m′|z|1,0,0⟩=⟨1,0,0|z2|1,0,0⟩=13⟨1,0,0|r2|1,0,0⟩, 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 symmetric. Finally, it follows from Equation ([e9.73]) that ⟨1,0,0|r2|1,0,0⟩=3a20. Hence, we conclude that α<1634πϵ0a30≃5.34πϵ0a30. The exact result (which can be obtained by solving Schrödinger’s equation in parabolic coordinates) is α=924πϵ0a30=4.54πϵ0a30.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)