# A Help to rewrite Dirac equation

Tags:
1. Dec 29, 2016

### TimeRip496

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

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. Dec 29, 2016

### Dr Transport

use the identity.....$(\vec{\sigma}\cdot\vec{a})(\vec{\sigma}\cdot\vec{b}) = \vec{a}\cdot\vec{b}+i(\vec{a}\times\vec{b})\cdot\vec{\sigma}$

3. Dec 29, 2016

### vanhees71

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).

4. Dec 29, 2016

### TimeRip496

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. Dec 30, 2016

### vanhees71

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.