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Stern Gerlach term in the Pauli Equation

  1. Mar 7, 2015 #1
    Where does the Stern Gerlach term in the Pauli equation come from? Taken from http://en.wikipedia.org/wiki/Pauli_equation. Following wikipedia's steps the Stern Gerlach term pops out when you apply the Pauli vector identity. I don't understand this step. It seems as if there should be no Stern Gerlach term.

    Here are my steps starting with the Pauli vector identity,
    [itex](\boldsymbol{\sigma} \cdot \boldsymbol{a})(\boldsymbol{\sigma} \cdot \boldsymbol{b}) = \boldsymbol{a} \cdot \boldsymbol{b} + i\boldsymbol{\sigma} \cdot (\boldsymbol{a} \times \boldsymbol{b})[/itex]

    [itex](\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A}))^2 = (\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A}))(\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A})) = (\boldsymbol{p} - e\boldsymbol{A})^2 + i\boldsymbol{\sigma} \cdot ((\boldsymbol{p} - e\boldsymbol{A}) \times (\boldsymbol{p} - e\boldsymbol{A}))[/itex]


    [itex]\boldsymbol{v} \times \boldsymbol{v} = \boldsymbol{0}[/itex]


    [itex](\boldsymbol{\sigma} \cdot (\boldsymbol{p} - e\boldsymbol{A}))^2 = (\boldsymbol{p} - e\boldsymbol{A})^2[/itex]

    I did it out in individual components as well, and came to the same conclusion. What am I missing?
  2. jcsd
  3. Mar 7, 2015 #2

    king vitamin

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    Gold Member

    You almost get zero, but you end up with one term because momentum does not commute with the vector potential in general.

    i\boldsymbol{\sigma} \cdot ((\boldsymbol{p} - e\boldsymbol{A}) \times (\boldsymbol{p} - e\boldsymbol{A})) = -\hbar e \epsilon_{jkl}\sigma_j \partial_k A_l = - e \hbar \mathbf{\sigma} \cdot \mathbf{B}

    I used p = -i hbar ∇. If you look back over your work when you did the calculation by components, you'll probably find the step where you moved the momentum operator past the vector potential.
  4. Mar 7, 2015 #3
    Note that unlike in conventional vector algebra, in this algebra of vector operators
    ##\mathbf{p}\times \mathbf{A} + \mathbf{A} \times \mathbf{p} \neq 0 ##

    $$[\mathbf{A} \times \mathbf{p}]_i = \sum_{jk} \epsilon_{ijk}A_jp_k = \sum_{jk} \epsilon_{ijk}p_kA_j - [p_k, A_j] = \sum_{jk} \epsilon_{ijk}p_kA_j +i\hbar \frac{\partial A_j}{\partial x_k} $$ $$\mathbf{p}\times \mathbf{A} + \mathbf{A} \times \mathbf{p}= i\hbar \nabla \times \mathbf{A} $$

    An important takeaway from this is that identities from vector calculus will not always be true for vector operators.

    Edit: Have I made a sign error?
    Last edited: Mar 7, 2015
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