9277#99

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9277#99
http://grephysics.net/ans/9277/99
how is the change in hamiltonian eErcos(theta) the static electric field will also distort the orbital, ie to say the KE and PE both will change..
how is the change in hamiltonian eErcos(theta) the static electric field will also distort the orbital, ie to say the KE and PE both will change..
Re: 9277#99
Correct, adding an external electric field modifies all sorts of things. But this is the first order correction in perturbation theory, which only rarely is an exact correction to the energy.

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Re: 9277#99
WakkaDojo is probably right, very often perturbations do not effect the energy to first order, but do effect the energy to second order.
That page also has a typo, dV = r^2sin(theta) dr dtheta dphi, not r^2sin^2(theta) dr dtheta dphi.
That page also has a typo, dV = r^2sin(theta) dr dtheta dphi, not r^2sin^2(theta) dr dtheta dphi.

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: 9277#99
nathan12343 wrote:WakkaDojo is probably right, very often perturbations do not effect the energy to first order, but do effect the energy to second order.
That page also has a typo, dV = r^2sin(theta) dr dtheta dphi, not r^2sin^2(theta) dr dtheta dphi.
ok right i understand, but how do you get this eqn eErcos(theta)Correct, adding an external electric field modifies all sorts of things. But this is the first order correction in perturbation theory, which only rarely is an exact correction to the energy.
Re: 9277#99
The force on the electron due to the static external field E is e E. The work done is e E z where z is displacement distance chosen along the arbitrary vertical axis. In spherical coordinates, z = r cos(t) where t is the azimuthal angle. You then integrate over all space in spherical coordinates.blackcat007 wrote:ok right i understand, but how do you get this eqn eErcos(theta)
Feel free to leave it in the form \Delta H = e E z, but you won't have any luck integrating over all space, since you'll need to transform the wave function into cartesian or cylindrical coordinates, which will be a mess.

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Re: 9277#99
WakkaDojo:
Why are we assuming E along Zaxis only ?, It could be in any of ( x, y ) in cartesian or ( r ) in Cylindrical cordianate system.The question did not specify any direction right ?
Why are we assuming E along Zaxis only ?, It could be in any of ( x, y ) in cartesian or ( r ) in Cylindrical cordianate system.The question did not specify any direction right ?
Re: 9277#99
Yeah true, but physics is the same in all inertial coordinates (and this problem has spherical symmetry) so to make the math easier you just change your coordinates to make z be in the direction of E. Since the field is uniform there is no problem in doing this. In other words, if you chose to put the E field pointing in some crazy direction and then worked out the horrible amount of math that ensued, you would end up with the same answer as if you just made it easy on yourself and picked E in the Z direction.sravanskarri wrote:WakkaDojo:
Why are we assuming E along Zaxis only ?, It could be in any of ( x, y ) in cartesian or ( r ) in Cylindrical cordianate system.The question did not specify any direction right ?
This sort of thing was always kinda downplayed at my university so don't feel bad about not being comfortable with it. Just ask yourself how changing the coordinates would affect your answer. In this case you are looking for energy which is not a vector, it is just a scalar. And so there won't be a direction anyway at the end. Think about it a little bit...

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 Joined: Sat Jun 14, 2008 10:19 pm
Re: 9277#99
Thanks, I got what you are saying.To be a little formal if we choose any complicated direction we end up having " Integral[ E *( a(n,m) * phi(n)* phi (m) )]" ; a(n,m) can be pulled out by a suitable tansformation and we end up with zero. ..