(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 quan tum number $n = 2$ have the same energy, $E^{0}$ . Show that when an electric field E is applied to hydrogen atoms in these states, the resulting first-order energies are $E^{0} \pm3a_{B} e|{\bf E}|,\,E^{0},\,E^{0}.$

Question

(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 quan tum number [latex]n = 2[/latex] have the same energy, [latex]E^{0}[/latex]. Show that when an electric field E is applied to hydrogen atoms in these states, the resulting first-order energies are [latex]E^{0} \pm3a_{B} e|{\bf E}|,\,E^{0},\,E^{0}.[/latex]

Treat the [latex]{z}[/latex]-directed electric field as a perturbation on the separable, orthonormal, unperturbed electron wave functions [latex]\psi_{n l m}(r, heta,\phi)=R_{n}(r){\Theta_{l}}(\stackrel{}{ heta}){\Phi_{m}}(\phi),[/latex] where [latex]r, heta,[/latex] and [latex]\boldsymbol{\phi}[/latex] are the standard spherical coordinates. You may use the unperturbed wave functions

[latex]\psi_{200}=\frac{2}{(2a_{\mathrm{B}})^{3/2}}\cdot\left(1-\frac{r}{2a_{\mathrm{B}}} ight)\cdot e^{-r/2a_{\mathrm{B}}} \cdot\left(\frac{1}{4\pi} ight)^{1/2}[/latex]

[latex]\psi_{210}={\frac{1}{\sqrt{3}(2a_{\mathrm{B}})^{3/2}}} \cdot {\frac{r}{a_{\mathrm{B}}}}\cdot e^{-r/2a_{\mathrm{B}}} \cdot {\frac{1}{2}}{\biggl(}{\frac{3}{\pi}}{\biggr)}^{1/2}\cos( heta)[/latex]

Accepted Answer

(a) We are asked to predict the effect that applying a uniform electric field has 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 [latex]\scriptstyle{Z}[/latex] direction is [latex]W=e|{\bf E}|z,[/latex] and the matrix element between the ground state and the first excited state is [latex]W_{01}=e|{\bf E}|\langle0|z|1 angle=e|{\bf E}|z_{01}.[/latex] 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 [latex]\sim10^{-8}\,\mathrm{cm}.[/latex] Maximum electric fields in a laboratory are [latex]\sim10^{6}{\mathrm {~V}}{\mathrm{~cm}}^{-1}[/latex] on a macroscopic scale. Transitions can only take place between energy levels that differ at most by the potential difference induced by the field [latex]-\mathbf{i}.\mathbf {e}., \sim10^{-2}\mathbf{e}\mathbf{V}.[/latex] 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 [latex]\psi_{n l m}[/latex] in the hydrogen atom is determined by the principal quantum number n to be [latex] extstyle n^{2}[/latex] since

[latex]\sum\limits_{l=0}^{l=n-1}(2l+1)=n^{2}[/latex]

In our assessment of the influence 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

[latex]H^{(0)}\psi=E^{(0)}\psi[/latex]

which has the solutions

[latex]\psi_{n l m}(r, heta,\phi)=R_{n}(r)\Theta_{l}( heta) \Phi_{ m}(\phi)=|n l m angle[/latex]

The principal quantum number n specifies the energy of a state, the orbital quantum number is [latex]l=0,1,2,\cdot\cdot\cdot,(n-1),[/latex] and the azimuthal quantum number is [latex]m =\pm l,\cdot\cdot \cdot\,,\pm2,\pm1,0.[/latex]

The electron states specified by [latex]n l m{\mathrm{~are~}}n^{2}[/latex] degenerate, and each state has definite parity. For n = 2, there are four states with the same energy. They are [latex]|200 angle,|21-1 angle,[/latex][latex]|210 angle,\mathrm{and}\;|211 angle.[/latex]

We now apply an electric field [latex]\mathbf{E}[/latex] and seek solutions to the time-independent Schrödinger equation

[latex]\left(H^{\left(0 ight)}+W ight)\psi=E\psi[/latex]

where, taking the electric field to be in the z direction, we have

[latex]W=e|{\mathrm{E}}|{\hat{\mathbf{z}}}[/latex]

Because the degeneracy of the n = 2 state is 4, we wish to find solutions to the 4 × 4 matrix

[latex]\sum\limits_{k=1}^{N=4}[H_{j k}-E\delta_{j k}]a_{k}=0 [/latex]

The solutions are given by the secular equation

[latex]\left|\begin{array}{c c c}{{\left\langle200\vert W\vert200 ight angle-E}}&{{\langle200\vert W\vert21-1 angle}}&{{\langle200\vert W\vert210 angle}}&{{\langle200\vert W\vert211 angle}}\ {{\left\langle21-1\vert W\vert200 ight angle}} &{{\left\langle21-1\vert W\vert21-1 ight angle-E}}&{{\left\langle21-1\vert W\vert210 ight angle}}&{{\left\langle21-1\vert W\vert211 ight angle}}\{{\left\langle210\vert W ight\vert200 angle}}&{{\langle210\vert W\vert21-1 angle}}&{{\langle210\vert W \vert210 angle-E}}&{{\langle210\vert W\vert211 angle}}\ {{\langle211\vert W\vert200 angle}}&{{\langle211\vert W\vert21-1 angle}}&{{ \langle211\vert W\vert210 angle}}&{{\langle211\vert W\vert211 angle-E}}\end{array} ight|=0[/latex]

The diagonal matrix elements are zero, because the odd parity of z forces the integrand to odd parity:

[latex]e|{\bf E}|\langle n l m|{\hat{\bf z}}|n l m angle=e|{\bf E}|\int\psi_{n l m}^{\ast}{\hat{\bf z}}\psi_{n l m}d^{3}r=0[/latex]

The perturbation is in the [latex]{{z}}[/latex] direction which in spherical coordinates only involves [latex]r \,{and}\, θ[/latex] via the relation [latex]z=r \cos( heta).[/latex] Hence, because the eigenfunctions are separable into orthonormal functions of [latex]r, heta,\,\mathrm{and} \,\phi[/latex] in such a way that

[latex]\psi_{n l m}(r, heta,\phi)=R_{n}(r)\Theta_{l}( heta) \Phi_{ m}(\phi)[/latex]

it follows that the matrix elements between states with different m for a z-directed perturbation are zero. This is because states described by [latex]\Phi_{m}(\phi)[/latex] are orthogonal, and the perturbation in z has no [latex]\boldsymbol{\phi}[/latex] dependence. Hence, the only possible nonzero off-diagonal matrix elements are those that involve states with the same value of m:

[latex]\langle200|W|210 angle=\langle210|W|200 angle=e|\mathbf{E}|\psi_{210}^{*}{\hat{z}}\psi_{200}d^{3}r[/latex]

The wave functions are

[latex]\psi_{200}=\frac{2}{(2a_{\mathrm{B}})^{3/2}}\cdot\left(1-\frac{r}{2a_{\mathrm{B}}} ight)\cdot e^{-r/2a_{\mathrm {B}}} \cdot\left(\frac{1}{4\pi} ight)^{1/2}[/latex]

which has even parity, and

[latex]\psi_{210}=\frac{1}{\sqrt{3}(2a_{\mathrm{B}})^{3/2}} \cdot \frac{r}{a_{\mathrm{B}}}\cdot e^{-r/2a_{\mathrm{B}}}\cdot\frac{1}{2}{\left(\frac{3}{\pi} ight)}^{1/2}\cos( heta)[/latex]

which has angular dependence and odd parity. Using [latex]z=r\cos ( heta),[/latex] we now calculate the matrix element

[latex]e|{\bf E}|\int\psi_{210}^{\ast}{\hat{\bf z}}\psi_{200} d^{3} r=\frac{e a_{\mathrm{B}}|{\bf E}|}{32\pi}\int\limits_{-\infty }^{\infty}r^{4}\biggl(2-\frac{r}{a_{\mathrm{B}}}\biggr)\cdot e^{-r/2a_{\mathrm{B}}}d r\int\limits_{\mathrm{0}}^{\pi}cos^{2}( heta)\sin( heta)d heta\int\limits_{0}^{2\pi}d\phi[/latex]

[latex]e|{\bf E}|\int\psi_{210}^{*}{\hat{\bf z}}\psi_{200} d^{3} r=-3e|{\bf E}|a_{\mathrm{B}}[/latex]

To solve the integral, we used

[latex]\int\limits_{0}^{\infty}r^{n}e^{-r/a_{\mathrm{B}}}d r= n!a_{ \mathrm{B}}^{n+1}[/latex]

The secular equation may now be written

[latex]\left|\begin{array}{c c c c}{{-E}}&{{0}}&{{-3e|{\bf E}|a_{\bf B}}}&{{0}}\ {{0}}&{{-E}}&{{0}}&{{0}}\ {{-3e|{\bf E}|a_{\bf B}}}&{{0}}&{{-E}}&{{0}}\ {{0}}&{{0}}&{{0}}&{{-E}}\end{array} ight| =0[/latex]

This has four solutions [latex]\pm3{\mathrm{a_B}}e|{\mathrm{E}}|,0, 0,[/latex] so that the newfirst-order corrected energy levels are [latex] E=E^{(0)}\pm3a_{\mathrm B}e\vert{\bf E}\vert,E^{(0) },E^{ (0)}.[/latex] 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 [latex]3e|\mathbf{E}|a_{\mathrm{B}}.[/latex]

To find the wave functions for the perturbed states, we need only consider the 2 × 2 matrix that relates [latex]|200 angle{\mathrm {~and~}}|210 angle:[/latex]

[latex]\left[{\begin{array}{c c c}{0}&{-3e|\mathbf{E}|a_{ \mathrm {B}}}\ {-3e|\mathbf{E}|a_{\mathrm{B}}}&{0}\end{array}} ight][/latex]

This has eigenfunctions

[latex]\psi_{+}=\frac{1}{\sqrt2}(\psi_{210}-\psi_{200})[/latex] with eigenvalue [latex]E_{+}=E^{(0)}+3e|\mathrm{E}|a_{\mathrm{B}}[/latex]

and

[latex]\psi_{-}=\frac{1}{\sqrt{2}}(\psi_{210}-\psi_{200})\;\mathrm {with\;eigenvalue}\;E_{-}=E^{(0)}-3e|\mathrm{E}|a_{\mathrm{B}}[/latex]

Because the eiegnfunctions [latex]|200 angle{\mathrm {~and~}}|210 angle[/latex] are of different parity, the eigenfunctions [latex] \psi_{ +}\operatorname{and}\psi_{-}[/latex] are of mixed parity.

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