Question 11.5: (a) What is the effect of applying a uniform electric field ......
(a) What is the effect of applying a uniform electric field on the energy spectrum of
an atom?
(b) If spin effects are neglected, the four states of the hydrogen atom with quantum number n = 2 have the same energy, E^{0}. Show that when a uniform z-directed electric field E is applied to hydrogen atoms in these states, the resulting first-order energies are E^{0} \pm 3a_{B} e\left|E\right|, E^{0}, E^{0}.
Treat the z-directed electric field as a perturbation on the separable, orthonormal, unperturbed electron wave functions \psi _{nlm} (r,\theta ,\phi )=R_{n} (r)\Theta ^{m}_{l} (\theta )\Phi _{m} (\phi ), where r, \theta, and \phi are the standard spherical coordinates. You may use the unperturbed wave functions
\psi _{200} =\frac{2}{(2a_{B} )^{3/2} } \left(1-\frac{r}{2a_{B}} \right) e^{-r/2a_{B}} \left(\frac{1}{4\pi } \right)^{1/2}\psi _{210} =\frac{1}{\sqrt{3}(2a_{B} )^{3/2} }\frac{r}{a_{B}} e^{-r/2a_{B}} \frac{1}{2} \left(\frac{3}{\pi } \right)^{1/2} \cos (\theta )
(a) We are asked to predict what the effect of applying a uniform electric field is on the energy spectrum of an atom. It is reasonable to expect that the atom is in its lowest-energy ground state. Using perturbation theory, the perturbation due to an electric field applied in the z direction is \hat{W} =e\left|E\right|\hat{z}, and the matrix element between the ground state and the first excited state is W_{01} =e\left|E\right|\left\langle0\left|\hat{z}\right|1 \right\rangle=e\left|E\right|z_{01}.The matrix element is the expectation value of the position operator, and so it is at most of the order of the size of an atom \sim 10^{-8} cm. Maximum electric fields in a laboratory are \sim 10^{6} Vcm^{-1} on a macroscopic scale. Transitions can only take place between energy levels that differ at most by the potential difference induced by the field – that is, \sim 10^{-2} eV. This will not be enough to induce transitions between principal quantum numbers n, but may break the degeneracy between states with different l belonging to the same n. The simple theory of the atom in Section 2.2.3.2 indicated that the degeneracy of an electronic state \psi _{nlm} in the hydrogen atom is determined by the principle quantum number n to be n² since
\sum\limits_{l=0}^{l=n-1}{(2l+1)} =n^{2}In our assessment of the influence that an electric field has on the energy levels of an atom, it was assumed that the electric field at the atom is the same as the macroscopic field. However, on a microscopic scale, electric fields can be dramatically enhanced locally depending on the exact geometry.
(b) If we ignore the effects of electron spin, the time-independent Schrödinger equation for an electron in the hydrogen atom is
\hat{H}^{(0)}\psi =E^{(0)} \psiwhich has the solutions
\psi _{nlm} (r,\theta ,\phi )=R_{nl} (r)\Theta ^{m}_{l} (\theta )\Phi _{m} (\phi )=\mid nlm〉The principle quantum number n specifies the energy of a state, the orbital quantum number is l = 0, 1, 2, … ,(n−1), and the quantum number m=±l, … ,±2, ±1, 0.
The electron states specified by n, l, and m are n² degenerate, and each state has definite parity. For n = 2, there are four states with the same energy. They are \mid 200〉, \mid 21-1〉, \mid 210〉, and \mid 211〉.
We now apply a uniform electric field E in the z direction and seek solutions to the time-independent Schrödinger equation
(\hat{H}^{(0)}+\hat{W} )\psi =E\psiwhere
\hat{W} =e\left|E\right|\hat{z}Because the degeneracy of the n = 2 state is 4, we wish to find solutions to the 4×4 matrix
\sum\limits_{j=1,k=1}^{N=4}{\left[H_{jk}-E\delta _{jk} \right]a_{k} } =0The solutions are given by the secular equation
The diagonal matrix elements are zero, because the odd-parity of z forces the integrand to odd parity:
e\left|E\right| \left\langle nlm\left|\hat{z} \right|nlm \right\rangle =e\left|E\right| \int{\psi ^{*}_{nlm}\hat{z}\psi _{nlm} d^{3} r } =0The perturbation is in the z direction which in spherical coordinates only involves r and θ via the relation z=r\cos (\theta ). Hence, because the eigenfunctions are separable into orthonormal functions of r, θ, and \phi in such a way that \psi _{nlm} (r,\theta ,\phi )=R_{nl} (r)\Theta ^{m}_{l} (\theta )\Phi _{m} (\phi ), it follows that the matrix elements between states with different m for a z-directed perturbation are zero. This is because states described by \Phi _{m} (\phi ) are orthogonal, and the perturbation in z has no \phi dependence. Hence, the only possible nonzero off-diagonal matrix elements are those that involve states with the same value of m:
\left\langle200\left|\hat{W} \right| 210\right\rangle =\left\langle210\left|\hat{W} \right| 200\right\rangle=e\left|E\right| \int{\psi ^{*}_{210} \hat{z}\psi _{200} d^{3}r }The wave functions are
\psi _{200} =\frac{2}{(2a_{B} )^{3/2} } \left(1-\frac{r}{2a_{B} } \right)e^{-r/2a_{B}} \left(\frac{1}{4\pi } \right) ^{1/2}which has even parity, and
\psi _{210} =\frac{1}{\sqrt{3}(2a_{B} )^{3/2} }\frac{r}{a_{B} }e^{-r/2a_{B} } \frac{1}{2} \left(\frac{3}{\pi } \right)^{1/2} \cos (\theta )which has angular dependence and odd parity. Using z=r\cos (\theta ), we now calculate the matrix element
e\left|E\right| \int{\psi ^{*}_{210}\hat{z}\psi _{200}d^{3}r} =-3e\left|E\right| a_{B}
To solve the integral, we used
\int\limits_{0}^{\infty }{r^{n}e^{-r/a_{B} }dr } =n!a^{n+1}_{B}The secular equation may now be written
\left | \begin{matrix} -E & 0 & -3e\left|E\right| a_{B} & 0 \\ 0 & -E & 0 & 0 \\ -3e\left|E\right| a_{B} & 0 & -E & 0 \\ 0 & 0 & 0 & -E \end{matrix} \right |=0This has four solutions \pm 3a_{B} e\left|E\right|, 0, 0, so that the new first-order corrected energy levels are E=E^{(0)} \pm 3a_{B} e\left|E\right|, E^{(0)}, E^{(0)}. This partial lifting of degeneracy in hydrogen due to application of an electric field is called the Stark effect.
The effect of applying an electric field is to break the symmetry of the potential and partially lift the degeneracy of the state. For the states for which degeneracy is lifted, it is as if the atom has an electric dipole moment of magnitude 3e\left|E\right| a_{B}.
To find the wave functions for the perturbed states, we need only consider the 2×2 matrix that relates \mid 200〉 and \mid 210〉:
\left [ \begin{matrix} 0 & -3e\left|E\right| a_{B} \\-3e\left|E\right| a_{B} & 0 \end{matrix} \right ]This has eigenfunctions \psi _{+} =\frac{1}{\sqrt{2} } (\psi _{210}-\psi _{200} ) with eigenvalue E_{+} =E^{(0)} +3e\left|E\right|a_{B} and \psi _{-} =\frac{1}{\sqrt{2} } (\psi _{210}+\psi _{200} ) with eigenvalue E_{-} =E^{(0)} -3e\left|E\right|a_{B}. Because the eigenfunctions \mid 200〉 and \mid 210〉 are of different parity, the eigenfunctions \psi _{+} and \psi _{-} are of mixed parity.