Non-relativistic limit of the Dirac equation

Question

How to recover non-relativistic limit of the Dirac equation $$\left( i\gamma^{\mu}\mathcal{D}_{\mu} - m \right)\Psi(x) = 0$$ where $\mathcal{D}_{\mu} = \partial_{\mu} + iqA_{\mu}$. I do not assume that the vector potential $A_{\mu}$ is zero, but I have a particle in central potential and magnetic field. During derivation please highlight where does the non-relativistic assumptions is used and how. I have found some notes here, but I don't understand why can we Taylor expand this expression $$\frac{1}{E + m -qA^0}\vec{\sigma}\cdot(\vec{p} - q\vec{A}) \approx \frac{1}{E+m}\vec{\sigma}\cdot(\vec{p} - q\vec{A}) + \frac{1}{(E+m)^2}qA^0\vec{\sigma}\cdot\vec{p}$$ around $qA^0 = 0$ and neglect $\vec{A}$ in the second term.

Answer 1

The essential reason is that the "$E$" in your equation has no non-relativistic limit. This can be seen more clearly by separating out the rest-mass-energy $mc^2$ and writing $E = mc^2 + H$, with $H$ comprising the total of the kinetic and internal energy. The "$mc^2$" part blows up to infinity as $c$ goes to infinity.

On the other hand $H$ has a non-relativistic limit. At the same time, the mass in non-relativistic theory can be treated as the non-relativistic limit of the "moving mass" $M = E/c^2$. So, to get a non-relativistic version of Dirac requires actually extending the Poincaré group from 10 dimensions to 11, by replacing $E$ by $M$ and $H$, treating them as independent.

So, the confusion arises from confusing $E$ and $H$. The shifting from $E$ to $H$ is precisely what is done with the Foldy-Wouthuysen transformation.

In place of the mass shell identity $$\frac{E^2}{c^2} - |𝐩|^2 = (mc)^2,$$ you then have two identities: $$|𝐩|^2 - 2MH + \frac{1}{c^2}H^2 = 0, \quad M - \frac{1}{c^2}H = m.$$ If you solve for $H$, you will see more clearly the connection to non-relativistic theory ... and you will also see more clearly what that extra 11th dimension in the symmetry group reflects: $$H = \frac{|𝐩|^2}{m + M}.$$ Compare to the non-relativistic case: $$H = \frac{|𝐩|^2}{2m} + U$$ where $U$ is the internal energy - i.e. the "kinetic energy" arising from the kinetics of the internal constituency of a body ... except now, it's being generalized and treated as a thing in its own right that potentially applies even to elementary bodies. So, it's the rest frame value of $H$, just as $m$ (the rest-mass) is the rest frame value of $M$. Substituting into the equations for the two invariants leads to the following modified form: $$|𝐩|^2 - 2MH + \frac{1}{c^2}H^2 = -2mU + \frac{1}{c^2}U^2, \quad μ ≡ M - \frac{1}{c^2}H = m - \frac{1}{c^2}U.$$ The drop-down from extended Poincaré-11 to ordinary Poincaré-10 then corresponds to setting $U = 0$.

On the other hand, the non-relativistic limit corresponds to setting $1/c^2$ to $0$. The non-relativistic limit of $M$, $m$ and $μ$ are all $m$, the mass, with all 11 dimensions of the symmetry group kept intact.

In Relativity, the following is invariant: $$E dt - 𝐩·d𝐫,$$ where $d𝐫 = (dx,dy,dz)$. If you make the above replacements, you get $$H dt - 𝐩·d𝐫 + μc^2 dt.$$ To make a meaningful non-relativistic limit out of this, you have to throw in the proper time, which we can denote $s$ and rewrite it as: $$H dt - 𝐩·d𝐫 + μc^2 dt - μc^2 ds.$$ Then, you can define a new coordinate $u = c^2(s - t)$ - i.e. the time dilation, itself, ramped up by $c^2$ and write: $$H dt - 𝐩·d𝐫 - μdu.$$

Now, you have a new invariant that you can take the non-relativistic limit for. In place of the Minkowski invariant $$ds^2 = dt^2 - \frac{|d𝐫|^2}{c^2}$$ are now two invariants: $$|d𝐫|^2 + 2 dt du + \frac{1}{c^2}du^2, \quad ds = dt + \frac{1}{c^2}du.$$ This, too, has a non-relativistic limit: $$|d𝐫|^2 + 2 dt du, \quad ds = dt.$$ The 11th generator in the extended Poincaré group is actually the translation on the $u$ coordinate. Since everything else is $u$-invariant, then ... by the Noether Theorem ... there is a conserved quantity. That's $μ$.

In the non-relativistic limit, $μ$ is the "central charge" that arises when you centrally extend the Galilei group to the Bargmann group - which is actually what the correct symmetry group for non-relativistic theory is, not Galilei!

Extend the operator correspondences appropriately: $$H ⇔ iħ \frac{∂}{∂t}, \quad 𝐩 ⇔ -iħ∇, \quad μ ⇔ -iħ\frac{∂}{∂u},$$ replace the $𝝰$ and $β$ matrices to rewrite $$δH - 𝝲·𝐩 - εμ ⇔ iħ\left(δ\frac{∂}{∂t} + 𝝲·∇ + ε\frac{∂}{∂u}\right),$$ for units $δ$, $𝝲 = \left(γ^1, γ^2, γ^3\right)$, $ε$, endowing them with a suitably-defined algebra (which will - in fact - be isomorphic to the complexified Dirac algebra), and add in the eigenvalue equation $$μψ = -iħ\frac{∂ψ}{∂u},$$ for the central charge. Restricting to the ordinary Poincaré group, for Relativity, by setting $U = 0$ and $μ = m$ produces: $$iħ\left(δ\frac{∂ψ}{∂t} + 𝝲·∇ψ + ε\frac{∂ψ}{∂u}\right) = 0, \quad mψ = -iħ\frac{∂ψ}{∂u}$$ which will be equivalent to the Dirac equation. Taking the limit $1/c^2 = 0$, on the other hand, will produce a result equivalent to the Schrödinger equation.

You can also add in a coupling with the electromagnetic potential. In that case, the non-relativistic version will be the Pauli-Schrödinger equation.

Note that shifting from $E$ to $H$ by subtracting off $mc^2$ corresponds adding a phase factor $\exp(±imc^2/ħ)$ to $ψ$, depending on which direction you're going in.

I demonstrate this, in detail - here Morphing The Dirac-Maxwell Equation To The Pauli-Schrödinger Equation and in the opposite direction, here Morphing The Schrödinger Equation To The Klein-Gordon Equation.

Needless to say, none of this was available in during the time Schrödinger devised his equation. He didn't know that Bargmann was the correct symmetry group for non-relativistic theory, rather than Galileo, nor that its natural geometric interpretation is 5D, not 4D; and even up to the present day there is still not a proper understanding that the correct symmetry group for Relativity is not the Poincaré group, but its (trivial) central extension to the extended Poincaré group.