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A Help to rewrite Dirac equation

  1. Dec 29, 2016 #1
    $$i\frac{\partial \phi}{\partial t} = \frac{1}{2m} (\sigma .P)(\sigma .P)\phi + eφ\phi$$

    Rewriting the equation by using B = ∇ × A and e = −|e| (electron charge) leads to a Schr¨odinger like equation:
    $$i\frac{\partial \phi}{\partial t} =[ \frac{1}{2m} (-i∇ + |e|A)^2 + \frac{|e|}{2m} σ.B - |e|φ ]\phi$$

    How did the B suddenly appear in the second equation? Alll help will be greatly appreciated.
    Source: http://physics.sharif.edu/~qmech/puppel.pdf ,Page 21.
     
  2. jcsd
  3. Dec 29, 2016 #2

    Dr Transport

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    use the identity.....[itex] (\vec{\sigma}\cdot\vec{a})(\vec{\sigma}\cdot\vec{b}) = \vec{a}\cdot\vec{b}+i(\vec{a}\times\vec{b})\cdot\vec{\sigma}[/itex]
     
  4. Dec 29, 2016 #3

    vanhees71

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    Also note that this is not the Dirac but the Pauli equation, which can be derived as the 0th order non-relativistic expansion of the Dirac equation (throwing away the antiparticles in the process).
     
  5. Dec 29, 2016 #4
    Thanks! One last question is why the author lose the mass in this equation, $$i\frac{\partial \phi}{\partial t} = \frac{1}{2m} (\sigma .P)(\sigma .P)\phi + eφ\phi$$

    Shouldnt it be like this $$i\frac{\partial \phi}{\partial t} = \frac{1}{2m} (\sigma .P)(\sigma .P)\phi + m\phi + eφ\phi$$?
     
  6. Dec 30, 2016 #5

    vanhees71

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    The mass term is just a constant potential, which you can absorb in an overall phase factor ##\exp(-\mathrm{i} m t)##, which cancels the term on the left- and right-hand side of your equation.
     
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