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Hamiltonian (electron in an electro-m field)

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Given [tex]H=\frac{1}{2m}\left[ \vec{P}-q\vec{A}\right] ^{2}+qU+\frac{q\hbar }{2m}\vec{\sigma}.\vec{B} ..(1) [/tex]
    show that it can be written in this form;
    [tex]H=\frac{1}{2m}\left\{ \vec{\sigma}.\left[ \vec{P}-q\vec{A}\right] \right\}^{2}+qU ....(2) [/tex]
    2. Relevant equations

    In my case, I am going to use only;

    [tex]\left( \vec{\sigma}.\vec{R}\right) ^{2}=\vec{R^2}+i\vec{\sigma}.\left( \vec{R}\times \vec{R}\right) ...(3) [/tex]

    3. The attempt at a solution
    Writing equation (1) in this form;
    [tex]H=\frac{1}{2m}\left[\left[ \vec{P}-q\vec{A}\right]^{2}-i\vec{\sigma}(iq\hbar\vec{B})\right]+qU[/tex]

    using the equation (3) and identifying terms;

    [tex]\vec{R}=\vec{P}-q\vec{A}[/tex]

    then, to replace [tex]\left[\left[ \vec{P}-q\vec{A}\right]^{2}-i\vec{\sigma}(iq\hbar\vec{B})\right][/tex] by [tex]\left\{ \vec{\sigma}.\left[ \vec{P}-q\vec{A}\right] \right\}^{2}[/tex]

    the following must satisfy;

    [tex]\left( \vec{R}\times\vec{R}\right)=iq\hbar\vec{B}[/tex]

    to do that, I write RxR like;
    [tex]\left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] \times \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] =\vec{P}\times \vec{P}+q^{2}\vec{A} \times \vec{A}-\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}[/tex]

    the first two term (PxP and AxA) on the right hand are equal to 0. But I dont know what to do with the rest;
    [tex]\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}[/tex]
     
  2. jcsd
  3. Nov 22, 2015 #2

    blue_leaf77

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    Remember that ##\mathbf{p} = -i\hbar\nabla## and that in the actual Schroedinger equation, the Hamiltonian will be multiplied by the wavefunction. Thus, instead of only [tex]\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P}[/tex], you should consider [tex](\vec{P}\times q\vec{A} -q\vec{A}\times \vec{P})\psi[/tex] where ##\psi## is an arbitrary wavefunction.
     
  4. Nov 22, 2015 #3
    Thank you for replying.

    Actually, I knew that. And that's where I'm stucked. Could you please develop a bit more?
     
  5. Nov 22, 2015 #4

    blue_leaf77

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    Consider the first term which is proportional to ##\mathbf{p}\times (\mathbf{A}\psi) = i\hbar\nabla \times (\mathbf{A}\psi)##, At this point, you can use the identity involving the curl of a product between a scalar function and a vector like the one in https://en.wikipedia.org/wiki/Curl_(mathematics).
     
  6. Nov 22, 2015 #5
    Ok, thank you !
     
    Last edited: Nov 22, 2015
  7. Nov 22, 2015 #6

    blue_leaf77

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    Yes, that's indeed how you should proceed.
     
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