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Pauli matrix with del operator

  1. Oct 14, 2014 #1
    1. The problem statement, all variables and given/known data
    In the Pauli theory of the electron, one encounters the expresion:

    (p - eA)X(p - eA

    where ψ is a scalar function, and A is the magnetic vector potential related to the magnetic induction B by B = ∇XA. Given that p = -i∇, show that this expression reduces to ieBψ.

    2. Relevant equations


    3. The attempt at a solution

    I'm first working this matrix determinant out without substitution, although I know the answer should include the B vector, and look something like this: ieBψ

    I split (p - eA) into its vector components and used the determinant, but it's just getting messy and I don't know what to do.

    My first question is, do I need to work this matrix out or is there an identity I should use to avoid all this? I've done this problem almost 10 times by brute force and can't seem to get it.

    This problem has been asked about previously but it wouldn't let me re-open the thread.
     
  2. jcsd
  3. Oct 14, 2014 #2
    Ok, I got it into this form -e(pXA + AXp)ψ, trying to work it out from here.
     
  4. Oct 14, 2014 #3
    Ok, having some issues. I used the identity on the first cross product PxA and got

    [itex]\frac{1}{2}B ( \nabla \bullet r) + \frac{1}{2} ( r \bullet \nabla )B [/itex]

    But I don't know what to do with the second cross product, it's not simplifying well.

    I am wondering, can I treat these three as scalars because the del operator is not operating on anything? Can I use the scalar triple product identity?
     
  5. Oct 14, 2014 #4

    vela

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    How is ##\vec{r}## appearing?

    You start with
    $$(-i\nabla - e\vec{A})\times (-i\nabla - e\vec{A})\psi = (i\nabla + e\vec{A})\times (i\nabla \psi + e\vec{A}\psi).$$ What terms do you get when you multiply that out?
     
  6. Oct 14, 2014 #5
    Ohh, I wrote something down wrong in the question,

    I was given [itex]A = \frac{1}{2}(B \times r)[/itex]

    I managed to get the whole thing down to [itex]ei(3/2)B\psi[/itex] but that was using the scalar triple product identity for

    [itex]Axp = (\frac{1}{2}B \times r) \times p[/itex]
     
  7. Oct 14, 2014 #6
    Multiply it out? You mean matrix cross-products component-wise?
     
  8. Oct 14, 2014 #7

    vela

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    The simplification holds in general. You don't need to know ##\vec{A}## to show it, so I'd ignore that info for now.
     
  9. Oct 14, 2014 #8

    vela

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    No, nothing that painful. Just distribute the cross product across the sums.
     
  10. Oct 14, 2014 #9
    I have gone through painful cross product component wise stuff for a whole day and still can't get it,

    If I just foil I get

    [itex]-\nabla^2 \psi + i\nabla eA\psi + eAi\nabla\psi + e^2 AA\psi [/itex] ?
     
  11. Oct 14, 2014 #10
    This is the last problem in some excruciating HW that's due today and I feel like I'm losing some marbles...
     
  12. Oct 14, 2014 #11

    vela

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    You need to be more careful. The first term isn't ##\nabla^2\psi = \nabla \cdot \nabla\psi##. Likewise, the other terms should still be cross products.
     
  13. Oct 14, 2014 #12
    So I see that [itex]\nabla \times \nabla = 0 , and, A \times A = 0[/itex] ?
     
  14. Oct 14, 2014 #13

    vela

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    Yup. For the second term, remember that ##\nabla## acts on the product ##\vec{A}\psi##.
     
  15. Oct 14, 2014 #14
    I am getting [itex]ie((\nabla \times A) + (A \times \nabla))\psi[/itex] and it seems I should find an identity now but I can't seem to find it.
     
  16. Oct 14, 2014 #15
    Ohhh, I see, product rule via the determinant. I am hoping some of this cancels out.
     
  17. Oct 14, 2014 #16
    And my [itex] i \nabla \times eA \psi [/itex] should relinquish two terms, one of them being a [itex] -(eA \times i \nabla \psi) [/itex]?
     
  18. Oct 14, 2014 #17

    vela

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    You need to get over your obsession with determinants. ;)

    http://en.wikipedia.org/wiki/Vector_calculus_identities

    Right.
     
  19. Oct 14, 2014 #18
    lol, Ok, I'm getting a result as [itex]i \nabla \times eA \psi[/itex]

    I think my extra linear algebra classes made me WANT to do all this algebra :/

    Now I will plug in for A and use an identity again!?

    Which gives me [itex]ie \frac{1}{2}(B( \nabla r)+(r \nabla)B) \psi[/itex]
     
    Last edited: Oct 14, 2014
  20. Oct 14, 2014 #19

    vela

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    Do you mean ##i(\nabla\times (e\vec{A}))\psi## as opposed to ##i(\nabla\times (e\vec{A}\psi))##? Recall that ##\vec{B} = \nabla \times \vec{A}##.
     
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