Eigenfunction of electron in E and B fields

[Yaelcita]

Homework Statement

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

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
[tex]H=\frac{1}{2m}\left(\vec{p}-\frac{q}{c}\vec{A}\right)^2+q\phi[/tex]
The hint says to use [tex]\vec{A}=-By\hat{x}[/tex] and since [tex]\vec{E}=-\nabla\phi[/tex] I can make [tex]\phi=-\varepsilon y[/tex]

The Attempt at a Solution

Plug in expressions for A and [tex]\phi[/tex] into H, which gives
[tex]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[/tex]

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??

 

[Matterwave]

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.

 

[Yaelcita]

Ok, let's see...

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

[tex] 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)[/tex]

[tex]&=& \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}[/tex]

[tex]=\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}[/tex]

[tex]=\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}[/tex]

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 [tex]p_x[/tex] term... Do they matter? I guess I don't really know how to continue after all...

 

[Matterwave]

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.

 

[Yaelcita]

(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...

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 [tex]x\pm i p_x[/tex] 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.

 

[Matterwave]

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...