Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Eigenfunction of electron in E and B fields

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    The Hamiltonian of a charged particle in an external em field is
    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]

    3. 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??
  2. jcsd
  3. Feb 11, 2010 #2


    User Avatar
    Science Advisor
    Gold Member

    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.
  4. Feb 11, 2010 #3
    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....
    Last edited: Feb 11, 2010
  5. Feb 11, 2010 #4


    User Avatar
    Science Advisor
    Gold Member

    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.
  6. Feb 11, 2010 #5
    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 [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.
  7. Feb 11, 2010 #6


    User Avatar
    Science Advisor
    Gold Member

    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.


    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...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook