# Eigenfunction of electron in E and B fields

1. Feb 10, 2010

### Yaelcita

1. The problem statement, all variables and given/known data
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

2. Relevant 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 $$\phi=-\varepsilon y$$

3. 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??

2. Feb 11, 2010

### 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.

3. Feb 11, 2010

### Yaelcita

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....

Last edited: Feb 11, 2010
4. Feb 11, 2010

### 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.

5. Feb 11, 2010

### Yaelcita

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

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.

6. Feb 11, 2010

### 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...