0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you also quote the full equation, i.e., the equation before the "other terms" are dropped? $\endgroup$ – Dvij Mankad Jul 5 at 20:56
  • $\begingroup$ @FeynmansOutforGrumpyCat I have tried to edit it as you asked. Please, let me know if more needs to be added. $\endgroup$ – Virgo Jul 5 at 23:17
  • $\begingroup$ See en.wikipedia.org/wiki/Dirac_equation. $\endgroup$ – Cinaed Simson Jul 5 at 23:29
  • $\begingroup$ @CinaedSimson thanks but I don't think they give the details I need to understand the derivation of Eq. (1) in my post. $\endgroup$ – Virgo Jul 5 at 23:41
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.