Non-relativistic limit of the coupled Dirac equation

Question

In Greiner's relativistic quantum mechanics textbook he has a derivation of the Pauli equation as a nonrelativistic limit of the coupled Dirac equation. Just below Eq. (2.81) he makes the following substitution: \begin{align} (\hat{\boldsymbol{\sigma}} \cdot \hat{\boldsymbol{\Pi}})(\hat{\boldsymbol{\sigma}} \cdot \hat{\boldsymbol{\Pi}})&=\hat{\boldsymbol{\Pi}}^{2}+\mathbf{i} \hat{\boldsymbol{\sigma}} \cdot(\hat{\boldsymbol{\Pi}} \times \hat{\boldsymbol{\Pi}}) \tag{1}\\ &=\left(\hat{{\bf p}}-\frac{e}{c} {\bf A}\right)^{2}+\mathbf{i} \hat{\boldsymbol{\sigma}} \cdot\left[\left(-\mathrm{i} \hbar \boldsymbol{\nabla}-\frac{e}{c} {\bf A}\right) \times\left(-\mathrm{i} \hbar \boldsymbol{\nabla}-\frac{e}{c} {\bf A}\right)\right] \tag{2} \\ &=\left(\hat{{\bf p}}-\frac{e}{c} {\bf A}\right)^{2}-\frac{e}{c} \hbar \hat{\boldsymbol{\sigma}} \cdot(\boldsymbol{\nabla} \times {\bf A}) \tag{3} \end{align} where $\hat{\boldsymbol{\Pi}}=\left(\hat{{\bf p}}-\frac{e}{c} {\bf A}\right)$. I don't understand why he has been able to drop the other terms in Eq. (3). I assume it is something to do with the non-relatavistic limit. On page 124 he gives the relations $$ | i\hbar \, \partial \chi / \partial t|\ll| m_{0} c^{2} \chi | \text { and }\left|e A_{0} \chi\right| \ll\left|m_{0} c^{2} \chi\right| $$ where the four component spinor $\psi$ is given by $$ \psi=\left(\begin{array}{c}{\tilde{\varphi}} \\ {\tilde{\chi}}\end{array}\right)=\left(\begin{array}{c}{\varphi} \\ {\chi}\end{array}\right) \exp \left[-\mathrm{i}\left(m_{0} c^{2} / \hbar\right) t\right]. $$ But I don't see how this can be used to derive Eq. (1). Any assistance would be greatly appreciated.

Answer 1

There are no terms missing. The term with the Pauli matrix involves the cross product of $\hat{\boldsymbol{\Pi}}$ with itself. For an ordinary vector ${\bf V}$, this would vanish ${\bf V}\times{\bf V}=0$. However, the different components of the mechanical momentum $\hat{\boldsymbol{\Pi}}$ do not commute with each other. Specifically, the ${\boldsymbol{\nabla}}$ does not commute with the spacetime-dependent vector potential ${\bf A}({\bf x},t)$. The noncommutativity of these terms yields the magnetic moment term that appears in your final equation; there is no approximation in going from (1) to (3).

The noncommutation leads to there being a term proportional to $$\hat{\boldsymbol{\sigma}}\cdot{\boldsymbol{\nabla}}\times{\bf A}+\hat{\boldsymbol{\sigma}}\cdot{\bf A}\times{\boldsymbol{\nabla}}.$$ As an operator (acting on a wave fucntion $\psi$) ${\boldsymbol{\nabla}}\times{\bf A}$ is just $${\boldsymbol{\nabla}}\times{\bf A}\,\psi=({\boldsymbol{\nabla}}\times{\bf A})\psi+{\boldsymbol{\nabla}}\psi\times{\bf A}=({\boldsymbol{\nabla}}\times{\bf A})\psi-{\bf A}\times{\boldsymbol{\nabla}}\psi.$$ (Getting this involves commuting the derivatives past ${\bf A}$, as well as using the antisymmetry of the cross product: ${\bf V}\times{\bf W}=-{\bf W}\times{\bf V}$.) From this result it follows that $$\left(\hat{\boldsymbol{\sigma}}\cdot{\boldsymbol{\nabla}}\times{\bf A}+\hat{\boldsymbol{\sigma}}\cdot{\bf A}\times{\boldsymbol{\nabla}}\right)\psi=\hat{\boldsymbol{\sigma}}\cdot\left\{\left[({\boldsymbol{\nabla}}\times{\bf A})\psi-{\bf A}\times{\boldsymbol{\nabla}}\psi\right]+{\bf A}\times{\boldsymbol{\nabla}}\psi \right\}=\hat{\boldsymbol{\sigma}}\cdot({\boldsymbol{\nabla}}\times{\bf A})$$

There are, of course, approximations required to get to equation (1). That equation represents a nonrelativistic limit, which can be obtained by eliminating terms with more than one power of $c$ (or $m$) in a denominator.