GR9277 #99



Alternate Solutions 
marching_band_to_nowhere 20190902 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.   evanb 20080627 14:46:04  The wave function of 1s is even, the perturbative operator =  q * Efield * Z is odd.
Thus the firstorder matrix element must be zero.  

Comments 
marching_band_to_nowhere 20190902 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 20190902 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 20180331 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 if there is no other factors in the integrand to prevent it from being an odd function in cos(theta)   bagel 20111111 19:46:02  No math necessary:
Bohr's energy calculation in a hydrogen atom was calculated solely on the presence of E field. QED.
liliapunto 20130716 02:41:54 
yep. solvable system. no correction.

James6M 20141017 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.

  keradeek 20111011 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
Answer is A.
keradeek 20111011 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 20080627 14:46:04  The wave function of 1s is even, the perturbative operator =  q * Efield * Z is odd.
Thus the firstorder matrix element must be zero.
nyuko 20091030 23:05:54 
I want to post the same thing as you did. But you are one year ahead of me :)

  zzzzort 20071101 14:26:58  None of the other answers have the correct units.
jburkart 20071102 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 20060914 23:08:00  Minor typo: In spherical coordinates (not ) and the limits for theta are 0 to 2*pi. And of course the integral of from 0 to 2*pi is 0.
E^2Pi1=0 20061130 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_duffy 20130525 08:45:43 
theta is the polar angle

 

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