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Why isn't the Pauli equation equivalent to the Schrodinger equation?

  1. Jun 2, 2010 #1
    The Pauli equation (seen here) contains its spin dependence in the term which reads

    [tex]\frac{1}{2m}\left[ \sigma\cdot\left(p-\frac{e}{c}A\right)\right]^2[/tex]

    So let B be any vector. Then

    [tex]\left( \sigma\cdot B\right)^2[/tex]
    [tex]=\left(\sigma_1 B_1 +\sigma_2 B_2 + \sigma_3 B_3\right)\left(\sigma_1 B_1 +\sigma_2 B_2 + \sigma_3 B_3\right)[/tex]
    [tex]=\sigma_1^2 B_1^2 +\sigma_2^2 B_2^2+\sigma_3^2 B_3^2 + (\sigma_1\sigma_2+\sigma_2\sigma_1)B_1 B_2 + (\sigma_1\sigma_3+\sigma_3\sigma_1)B_1 B_3 + (\sigma_3\sigma_2+\sigma_2\sigma_3)B_3 B_2[/tex]
    [tex]=1\cdot B_1^2 + 1\cdot B_2^2 + 1\cdot B_3^2 + 0 + 0 +0[/tex]
    [tex]=B^2[/tex]

    So isn't the sigma-dependent term in the Pauli equation identically equal to

    [tex]\frac{1}{2m}\left(p-\frac{e}{c}A\right)^2[/tex]

    ?

    if yes, then in what sense is it spin-dependent? If no, then where did I go wrong?
     
  2. jcsd
  3. Jun 2, 2010 #2
    I have only briefly looked at your calculation and it looks correct, but I don't think you are calculating what you really want. (By the way, I assume that the identity matrix is implied in your notation.)

    Try to do the same, but now assume that B is a vector operator. So in general [tex]\sigma_1\sigma_2B_1B_2 + \sigma_2\sigma_1B_2B_1 \neq (\sigma_1\sigma_2 + \sigma_2\sigma_1)B_1B_2[/tex], since you don't always have [tex][B_1,B_2] = 0[/tex].
     
  4. Jun 2, 2010 #3
    Thanks, element4. I get it now. The momentum operator and the EM potential do not commute since the latter is a function of position, so the B^2 terms in my derivation are not so simple. Haven't worked it out yet, but I'm pretty sure that's it.
     
  5. Jun 3, 2010 #4
    Ok. So I worked through it and have one more nagging question. The final form I get is

    [tex]\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\vec{B}+e\Phi-i\hbar\frac{\partial}{\partial t}\right\}\Psi-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0[/tex]

    The part in the braces is familiar and can also be seen on the wikipedia page: http://en.wikipedia.org/wiki/Pauli_equation#Special_Cases What about the last term? Is it something typically neglected? Why? Or did I screw up the derivation?
     
  6. Jun 3, 2010 #5

    Meir Achuz

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    The correct result is not AXdel. It should be curl A, which equals B because the curl does not act on psi. This just gives -mu.B, the interaction of the electron magnetic moment with B. This result also derives the g=2 for electron magnetic moment. Be more careful with the combination -[del.A + A.del]psi.
     
  7. Jun 3, 2010 #6
    Thanks, Meir Achuz. But the term containing

    [tex]\frac{1}{2m}\left[ \sigma\cdot\left(p-\frac{e}{c}A\right)\right]^2[/tex]

    when expanded results in both [tex]A\times\nabla[/tex] and [tex]\nabla\times A[/tex] terms (I think). I have the [tex]\nabla\times A[/tex] covered above in the term containing B. But I don't know what to do with the [tex]A\times\nabla[/tex] term
     
  8. Jun 7, 2010 #7
    Bump.

    Just one little bump.
     
  9. Sep 15, 2010 #8
    I did not work this through, but I think the del X A term does not equal B. It's the same way with [P, X] = (d/dx)*x - x*(d/dx) != 1 - x*(d/dx).
     
    Last edited: Sep 15, 2010
  10. Sep 17, 2010 #9
    thank you for your reply, yangjong. And welcome to the board!

    I think what you are suggesting is that when I see [tex]\nabla\times A[/tex] I should not read this as "multiply by B" but rather consider it an operator: when acting on a function [tex]\Psi[/tex] it yields [tex]\nabla\times (A\Psi)[/tex].

    That is, where I wrote

    [tex]\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\vec{B}+e\Phi \Psi -i\hbar\frac{\partial}{\partial t}\right\}\Psi-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0[/tex]

    above, it should be instead

    [tex]\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2\Psi-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\left(\nabla\times [\vec{A}\Psi]\right)+e\Phi\Psi -i\hbar\frac{\partial\Psi}{\partial t}\right\}-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0[/tex]

    [tex]\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2\Psi-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\left(\nabla\Psi\times \vec{A} +\Psi\nabla\times \vec{A}\right)+e\Phi\Psi -i\hbar\frac{\partial\Psi}{\partial t}\right\}-\frac{e\hbar}{mc}\vec{\sigma}\cdot\left(\vec{A}\times\nabla\right)\Psi=0[/tex]

    [tex]\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2\Psi-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\left(\Psi\nabla\times \vec{A}\right)+e\Phi\Psi -i\hbar\frac{\partial\Psi}{\partial t}\right\}=0[/tex]

    [tex]\left\{\frac{1}{2m}\left(\vec{p}-\frac{e}{c}\vec{A}\right)^2-\frac{e\hbar}{2mc}\vec{\sigma}\cdot\vec{B}+e\Phi-i\hbar\frac{\partial}{\partial t}\right\}\Psi =0[/tex]

    That's it. Thank you!
     
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