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Hydrogen in Magnetic Field, Interaction Representation

  1. Dec 8, 2016 #1
    The hydrogen is placed in the external magnetic field:
    $$ \textbf{B}=\hat{i}B_1 cos(\omega t) + \hat{j} B_2 sin(\omega t) + \hat{k} B_z ,$$

    Using the relation ## H = - \frac{e\hbar}{2mc} \mathbf \sigma \cdot \mathbf B ##, then I got the form

    $$ H = H_0 + H' , $$

    where

    $$ H'= - \frac{e \hbar}{2 m c} (B_1 cos(\omega t) \sigma_1 + B_2 sin(\omega t) \sigma_2 ) .$$

    To find the spin state in the interaction representation, I substitute the ## H' ## in

    $$ i\hbar\frac{\partial}{\partial t} | \psi \rangle = H' | \psi \rangle , $$

    with ## | \psi \rangle = \begin{pmatrix} a \\ b \end{pmatrix} .##

    Next, I tried to solving it, e.g. for ## a ##

    $$\frac{\partial^2 a}{{\partial t}^2} + \lambda^2 a = 0 ,$$

    with

    $$ \lambda^2 = \big(\frac{e}{2mc}\big)^2[B_1^2 cos^2(\omega t)+B_2^2(sin^2(\omega t))] .$$

    Then I'm heading for the boundary condition.

    But then I realized that the ## \lambda^2 ## is depend on ## t ##.

    So, my question, is it "ok" to continue to the boundary condition, or there is something wrong with my attempt?
     
  2. jcsd
  3. Dec 8, 2016 #2

    kuruman

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    How did you get the differential equation for a? It seems to me you should get two coupled differential equations involving a and b. Is what you show the decoupled result? Also, should you not include the term ##B_z \sigma_3## in your expression for ##H'##?
     
  4. Dec 8, 2016 #3
    I don't include the ## B_z \sigma_3 ## because it time independent.

    I realized that I don't differentiate with time the ##{B_1 cos (\omega t) \mp i B_2 sin (\omega t)}## term in the ##\partial a/\partial t## and ##\partial b/\partial t##.

    Here's my result after differentiating it (for ## \partial a/\partial t ##),

    $$ \frac{\partial^2 a}{{\partial t}^2} = - \frac{e\hbar}{2mc}\Big[ \frac{ie}{2mc} [B_1^2 cos^2{\omega t}+B_2^2 sin^2{\omega t}] a + \frac{2mc}{e\hbar} \omega i \frac{B_1 cos \omega t + i B_2 sin \omega t}{B_1 cos \omega t - i B_2 sin \omega t} \frac{\partial a}{\partial t} \Big] .$$
     
  5. Dec 8, 2016 #4

    kuruman

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    So what? Is it not a perturbation to ##H_0##? As long as ##B_z \neq 0##, you cannot ignore it.
     
  6. Dec 9, 2016 #5
    Thanks for your clue.

    I am using Schwinger-Tomonaga equation, where the Hamiltonian that evolve with time is only the time dependent. CMIIW
     
  7. Dec 27, 2016 #6
    update, I have finish this work.
     
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