Eigenfunction of electron in E and B fields

AI Thread Summary
The discussion revolves around finding the energy eigenvalues and eigenfunctions of a free electron in a constant magnetic field and a perpendicular electric field. Participants explore the Hamiltonian using the Landau gauge and the scalar potential, leading to a complex expression that resembles a harmonic oscillator. The challenge lies in transforming the Hamiltonian to isolate terms that can be related to harmonic oscillator solutions, particularly addressing the impact of constant terms and momentum in the y-direction. Suggestions include coordinate transformations to adjust the equilibrium position and the notion that constant terms can be managed without affecting eigenfunctions. The conversation highlights the intricacies of quantum mechanics in the presence of electromagnetic fields.
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Homework Statement


Consider a free electron in a constant magnetic field \vec{B}=B\hat{z} and a perpendicular electric field \vec{E}=\varepsilon\hat{y}. Find the energy eigenvalues and eigenfunctions in terms of harmonic oscillator eigenfunctions
Hint: Use Landau gauge \vec{A}=-By\hat{x}

What I actually don't understand is at the end... read on

Homework Equations


The Hamiltonian of a charged particle in an external em field is
H=\frac{1}{2m}\left(\vec{p}-\frac{q}{c}\vec{A}\right)^2+q\phi
The hint says to use \vec{A}=-By\hat{x} and since \vec{E}=-\nabla\phi I can make \phi=-\varepsilon y

The Attempt at a Solution


Plug in expressions for A and \phi into H, which gives
H=\frac{1}{2m}\left(p_x^2+p_y^2+p_z^2+\left(\frac{qB}{c}\right)^2y^2+\frac{qB}{c}p_x y\right)-q\varepsilon y

I know I just have to play around with this expression to make it look like a harmonic oscillator, but I have no idea how... In another problem that I solved, I used an A potential with both an x and a y components, but I didn't have the scalar potential in that case. It's that term that ruins everything!

Any ideas??
 
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The scalar potential, which is just proportional to y, should only serve to change your equilibrium position for your harmonic oscillator. Kind of like gravity acting on a spring. I think you can make coordinate transformations to this new equilibrium position.

I'm not 100% sure on this, but see if you can do something with it.
 
Ok, let's see...

Let y=y'+\frac{\varepsilon m c^2}{q B^2}. Now the Hamiltonian reads (ignoring the unimportant parts):

H&=&\frac{1}{2m}\left(p_x+\frac{qB}{c}\left(y'+\frac{\varepsilon mc^2}{qB^2}\right)\right)^2-q\varepsilon\left(y'+\frac{\varepsilon mc^2}{qB^2}\right)

&=& \frac{1}{2m}\left(p_x+\frac{qB}{c}y'+\frac{\varepsilon mc}{B}\right)^2-q\varepsilon y'-\frac{\varepsilon^2mc^2}{B^2}

=\frac{1}{2m}\left(p_x+\frac{qB}{c}y'\right)^2+\frac{m}{2}\left(\frac{\varepsilon c}{B}\right)^2+\left(p_x+\frac{qB}{c}y'\right)\frac{\varepsilon c}{B}-q\varepsilon y'-\frac{\varepsilon^2 mc^2}{B^2}

=\frac{1}{2m}\left(p_x+\frac{qB}{c}y'\right)^2-\frac{mc^2\varepsilon^2}{2B^2}+p_x\frac{\varepsilon c}{B}

I think I know what to do with the stuff in parenthesis, but I don't know how to get rid of the constant term and the p_x term... Do they matter? I guess I don't really know how to continue after all...
 
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I'm not sure I follow your substitution (also, where'd all the py and pz's go?). Actually, the momentum in the y direction is the momentum you want to couple with the y's to get a harmonic oscillator.

Hmm, perhaps someone better at this can help you. It's been a while since I've actually worked out problems like this.
 
Matterwave said:
(also, where'd all the py and pz's go?)

I just didn't write it, since I was working only with those two terms...

Matterwave said:
Actually, the momentum in the y direction is the momentum you want to couple with the y's to get a harmonic oscillator.

I know, but I have the solution to this one problem where they then go on to define raising and lowering operators as x\pm i p_x and the same for y, and they get the Hamiltonian to look like the product of a raising and a lowering operator plus a half, so I thought if I could make it look like that I could just use that solution, but I don't know what to do with the constant term, and the p_x term, more importantly.
 
Well, the constant isn't going to mess with the eigenfunctions right, you can just bring it to the other side of the eigenfunction equation.

H*psi=E*psi

So, you can just bring the constant over to the right side of the equation, and it would only adjust your eigenvalue.

The px part also stumps me...
 
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