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GR9277 #99
Problem
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\prob{99}
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$\Rightarrow$}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.

Search this site for keyword perturbation for more on this.

Alternate Solutions
 marching_band_to_nowhere2019-09-02 22:42:59 This is a question about the Stark Effect, which is known to not affect the ground state electrons in any orbital. Thus, we know the first order correction to the energy of the ground state is 0 by definition.Reply to this comment evanb2008-06-27 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.Reply to this comment
marching_band_to_nowhere
2019-09-02 22:42:59
This is a question about the Stark Effect, which is known to not affect the ground state electrons in any orbital. Thus, we know the first order correction to the energy of the ground state is 0 by definition.
marching_band_to_nowhere
2019-09-02 22:42:44
This is a question about the Stark Effect, which is known to not affect the ground state electrons in any orbital. Thus, we know the first order correction to the energy of the ground state is 0 by definition.
enterprise
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)
bagel
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.
 liliapunto2013-07-16 02:41:54 yep. solvable system. no correction.
 James6M2014-10-17 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
V0*.

The ground state is symmetric in z. = O

 keradeek2011-10-11 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
evanb
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.
 nyuko2009-10-30 23:05:54 I want to post the same thing as you did. But you are one year ahead of me :)
zzzzort
2007-11-01 14:26:58
None of the other answers have the correct units.
 jburkart2007-11-02 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.
herrphysik
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.
 E^2Pi-1=02006-11-30 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.....
 p_e_duffy2013-05-25 08:45:43 theta is the polar angle

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$