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Homework Help: Canonical perturbation theory

  1. Oct 27, 2007 #1
    1. The problem statement, all variables and given/known data

    An electron is inside a magnetic field oriented in the z-direction. No measurement of the electron has been made. A magnetic field in the x-direction is now switched on. Calculate the first-order change in the energy levels as a result of this perturbation.

    3. The attempt at a solution

    I got this wrong, claiming that the first order perturbation is zero. I said this because if you evaluate the expectation of the [itex]S_x[/itex] matrix in the states up/down individually, you obviously get zero since [itex]S_x[/itex] is anti-diagonal.

    Apparently you are supposed to evaluate the expectation of [itex]S_x[/itex] in the superposition of states [itex]\psi = 1/\sqrt{2}(1,0)^T + 1/\sqrt{2}(0,1)^T[/itex]. I have trouble understanding this because [itex]\psi[/itex] is not an eigenstate of the orginial Hamiltonian, which it should be for canonical perturbation theory.

    Any help is greatly appreciated.
  2. jcsd
  3. Oct 27, 2007 #2

    Physics Monkey

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    Hi noospace,

    Something is very strange here. Assuming the z-direction magnetic field is large and the x-direction magnetic field is small, your calculation is correct. There is no first order change in the energy levels in such a situation. (One could just solve the complete problem trivially and see this fact.) Using eigenstates of the perturbation to calculate expectation values is certainly wrong. I suspect there is some kind of confusion here. Any ideas?
  4. Nov 2, 2007 #3
    Hi Physics Monkey,

    Thanks for your response. My lecturer's approach to this question is to evaluate the expectation of the operator [itex]S_x[/itex] in the superposition of states [itex]\psi = \frac{1}{\sqrt{2}} | 1/2 1/2 \rangle + \frac{1}{\sqrt{2}}}|1/2 \pm 1/2\rangle[/itex]. The answer given is

    [itex] \pm\frac{B_x\hbar}{2m_e}[/itex]

    My lecturer claims that since [itex]| 1/2 1/2 \rangle, | 1/2 -1/2 \rangle[/itex] are separately eigenstates of the Hamiltonian then so should their superposition. I claim this is false since the sum of two eigenvectors need not be an eigenvector.

    Who is correct?
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