Help to rewrite Dirac equation

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

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.
 

Answers and Replies

  • #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]
 
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  • #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).
 
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  • #4
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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]
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 $$i\frac{\partial \phi}{\partial t} = \frac{1}{2m} (\sigma .P)(\sigma .P)\phi + m\phi + eφ\phi$$?
 
  • #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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