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Path integral for a particle coupled to a magnetic field

  1. Oct 25, 2011 #1
    Hi all,

    I am currently having trouble with an exercise: writing the propagator of a particle coupled to a magnetic field.

    So the lagrangian is [itex]L_A (\vec{x},\dot{\vec{x}}^2) = \frac{m}{2}\dot{\vec{x}} + e\vec{A}.\dot{\vec{x}}[/itex]
    And it says that I should solve it in two different ways:
    -by writing [itex]e\vec{A}.\dot{\vec{x}}[/itex] has a derivative relative to t
    -by completing the square in the Lagrangian, ie something like [itex]L_A (\vec{x},\dot{\vec{x}}) = \frac{m}{2}(\dot{\vec{x}}+\frac{e}{m}\vec{A})^2 - \frac{e^2}{2m}\vec{A}^2[/itex] I guess.

    But still, I develop a few lines more of calculus but I am stuck. I would be really happy if someone could explain how to proceed.
    Last edited: Oct 25, 2011
  2. jcsd
  3. Oct 25, 2011 #2
    I think I'm going to try and solve this with you :)

    I can't do it tonight cause I have a test tomorrow that I have to study for, but this looks cool and full of learning.

    let me get this straight we have
    L=\frac{m}{2}\dot{x}^{2}_{i} +eA_j \dot{x}_j
    and we should put this in two different forms:
    L=\frac{1}{2m}(m\dot{x}_j+e A_i )^2-e^2 A_{i}^2
    L=\frac{m}{2}\dot{x}_{i}^2-e\dot{A}_j x_j +e\frac{\partial}{\partial t}(A_i x_i)
    Do those look ok to you?
  4. Oct 25, 2011 #3
    Yes, it does look good. But in the second lagrangian, because we don't want to have an electrical field, we have [itex]\dot{\vec{A}}=0[/itex] so the second terms vanishes.
    Last edited: Oct 25, 2011
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