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11.4: Quadratic Stark Effect

  • Page ID
    15795
  • 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

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

      \( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)