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In perturbation theory, what is the first order correction to the energy of a hydrogen atom (Bohr radius $a_0$) in its ground state due to the presence of a static electric field E?

  1. 0
  2. $eEa_0$
  3. $3eEa_0$
  4. $\frac{8e^2Ea_0^3}{3}$
  5. $\frac{8e^2E^2a_0^3}{3}$

Quantum Mechanics}Perturbation Theory

The perturbed Hamiltonian is given by \Delta H = eEz = eEr\cos\theta, where the last substitution is made for spherical coordinates.

The first-order energy-shift is given by \langle \psi_0 | \Delta H | \psi_0 \rangle, where \psi_0\propto e^{-kr}.

dV=r^2drd\Omega=r^2\sin^2\theta d\phi d\theta dr.

Thus, E_0 = \int_0^\infty \int_0^{\pi} \int_0^{2\pi} e^{-kr}\cos\theta\sin^2\theta d\theta d\phi =0, since the integral of \cos\theta\sin^2\theta over 0 to \pi is 0.

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See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
2008-06-27 14:46:04
The wave function of 1s is even, the perturbative operator = - q * Efield * Z is odd.

Thus the first-order matrix element must be zero.
Alternate Solution - Unverified
2018-03-31 21:36:41
In the ground state , The wavefunction is rotationally invariant. An Electric field points at some definite direction. The energy will depend on the angle between the Electric field and the dipole moment of the hydrogen. Obviously this dipole can\'t point at any definite direction because our atom is in a ground state. So , there shouldn\'t be any corrrection to the energy. Corrections should occur to P states because they have Y(cos(\\theta)) dependence in their wave functions. Otherwise , you will get \\int_{-1}^{+1}dcos(\\theta) cos(\\theta)=0 if there is no other factors in the integrand to prevent it from being an odd function in cos(theta)NEC
2011-11-11 19:46:02
No math necessary:

Bohr's energy calculation in a hydrogen atom was calculated solely on the presence of E field. QED.
2013-07-16 02:41:54
yep. solvable system. no correction.
2014-10-17 21:28:23
I think that the question is referring to an external (and uniform) electric field. However, the question fails to make that clear.
2011-10-11 23:07:06
V ' = V0 * z

The first order correction to the energy is the expectation value of the perturbing Hamiltonian. In other words, it's the expectation value of V0*z, or

The ground state is symmetric in z. = O

Answer is A.
2011-10-11 23:09:44
The site does not allow me to use brackets. Sorry.

The answer is equal to the expectation value of V0*z. The expectation value of z is zero. Answer is A
2008-06-27 14:46:04
The wave function of 1s is even, the perturbative operator = - q * Efield * Z is odd.

Thus the first-order matrix element must be zero.
2009-10-30 23:05:54
I want to post the same thing as you did. But you are one year ahead of me :)
Alternate Solution - Unverified
2007-11-01 14:26:58
None of the other answers have the correct units.
2007-11-02 00:48:27
No, electric field times charge is force, and force times distance is work, a form of energy. You have to do it the proper way.
2006-09-14 23:08:00
Minor typo: In spherical coordinates dV = r^2 sin{\theta} dr d \theta d \phi (not sin^2{\theta}) and the limits for theta are 0 to 2*pi. And of course the integral of sin{\theta} cos{\theta} from 0 to 2*pi is 0.
2006-11-30 19:34:06
The limit is still 0 to pi ... theta is azimuthal angle.... but integral of Sin[theta]Cos[theta] from 0 to pi is still zero.....
2013-05-25 08:45:43
theta is the polar angle
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