I'm trying to derive the full and correct Hamiltonian for spin$\frac{1}{2}$ particles from Dirac equation up to second order in $v/c$. For a potential and magnetic field constant in time.

In particular, there should be Zeeman term, relativistic kinetic term, spin-orbit term and Darwin term:

$$\hat{H}=\frac{\mathbf{P}^2}{2m}+U-e \hbar \mathbf{ \sigma} \mathbf{B}-\frac{\mathbf{P}^4}{8m^3 c^2}- \frac{e \hbar}{4 m^2 c^2} \mathbf{ \sigma} (\mathbf{F} \times \mathbf{P})- \frac{e \hbar^2}{8 m^2 c^2} (\nabla \mathbf{F}) \tag{1}$$


$$\mathbf{P}=- i \hbar \nabla-e \mathbf{A} \\ \mathbf{B}= \nabla \times \mathbf{A} \\ \mathbf{F}=- \frac{1}{e} \nabla U$$

This is taken from various sources, for example, Berestetsky, Lifshitz, Pitaevsky "Quantum electrodynamics", (33.12) though I have added the Zeeman term and changed some notation.

Now, assuming the above is correct, I want to derive this expression from Dirac equation directly.

I use the following form of the equation as a starting point:

$$\begin{cases} \displaystyle \Psi_+ = - \frac{\sigma \mathbf{P}}{mc} \Psi_-+\frac{E}{mc^2} \Psi_+ \\ \displaystyle \Psi_- = \frac{\sigma \mathbf{P}}{mc} \Psi_+ -\frac{E}{mc^2} \Psi_- \end{cases} \tag{2}$$

Where $\Psi_{\pm}$ are two component spinors and $\sigma$ is the vector of Pauli matrices.

Introducing a new energy:

$$E=mc^2+\mathcal{E} \\ \mathcal{E}=i \hbar \frac{\partial}{\partial t} -U $$

We obtain:

$$\begin{cases} \displaystyle \frac{\mathcal{E}}{mc^2} \Psi_+= \frac{\sigma \mathbf{P}}{mc} \Psi_- \\ \displaystyle \left(2+ \frac{\mathcal{E}}{mc^2} \right)\Psi_- = \frac{\sigma \mathbf{P}}{mc} \Psi_+ \end{cases} \tag{3}$$

Assuming $$\Vert \mathcal{E} \Vert \ll mc^2$$

We obtain, up to first order in $\Vert \mathcal{E} \Vert / mc^2$:

$$\mathcal{E} \Psi_+ = \frac{\sigma \mathbf{P}}{2m} \left(1- \frac{\mathcal{E}}{2mc^2} \right) \sigma \mathbf{P} \Psi_+ \tag{4}$$

I assume that this should directly give us (1).

However, after some simplifications, while I get most of the terms above right (including the most important SO term), I get an incorrect Darwin term (twice as large) and also a bunch of extra terms.

My full derivation is lengthy, so I wanted to ask first if I missed something? Is (4) correct (up to second order in $v/c$) and should it lead to (1) without any extra terms?

I have extensively used the identity:

$$(\sigma \mathbf{a}) (\sigma \mathbf{b})= \mathbf{a} \mathbf{b}+i \sigma (\mathbf{a} \times \mathbf{b})$$

though it's not always clear how to apply it to operators. In any case, it gives a lot of extra terms not present in (1) when applied to (4).

There's a related question, but only about Pauli equation, which is simple enough to derive.

  • $\begingroup$ Can you explain explicitly the meaning of every term in Eq.(1)? $\endgroup$
    – Jack
    May 3, 2019 at 3:28
  • $\begingroup$ @Jack, I can, but in this question, the meaning of the terms doesn't really matter, as they appear as a pure mathematical result of taking the approximation of Dirac equation. However, they do have (commonly known) physical interpretation. The first two terms are kinetic and potential energy, the second is spin magnetic moment interacting with external magnetic field, the third is the relativistic connection to kinetic energy, the fourth is spin-orbit coupling (interaction between spin and orbital angular momenta), the last one (Darwin) is proportional to external charge density $\endgroup$
    – Yuriy S
    May 3, 2019 at 9:39

1 Answer 1


Let me provide some steps of the derivation here to see where the problem had been.

$$\mathcal{E} \Psi_+ = \frac{\sigma \mathbf{P}}{2m} \left(1- \frac{\mathcal{E}}{2mc^2} \right) \sigma \mathbf{P} \Psi_+$$

Let's change the notation a little:


$$E_c \Psi_+ = \frac{(\sigma \mathbf{P})^2}{2m} \Psi_+ +U \Psi_+ -\frac{E_c}{4m^2 c^2} (\sigma \mathbf{P})^2 \Psi_+ + \frac{1}{4m^2 c^2} ( \sigma \mathbf{P} U \sigma \mathbf{P} )\Psi_+ $$

$$E_c \Psi_+ = \left(1- \frac{(\sigma \mathbf{P})^2}{4m^2 c^2}\right) \left( \frac{(\sigma \mathbf{P})^2}{2m} +U + \frac{1}{4m^2 c^2} ( \sigma \mathbf{P} U \sigma \mathbf{P} ) \right)\Psi_+ $$

$$E_c \Psi_+ = \left( \frac{(\sigma \mathbf{P})^2}{2m} +U - \frac{(\sigma \mathbf{P})^4}{8m^3 c^2}- \frac{(\sigma \mathbf{P})^2 U}{4m^2 c^2} + \frac{1}{4m^2 c^2} ( \sigma \mathbf{P} U \sigma \mathbf{P} ) \right)\Psi_+ $$

It's easy to identify several correct terms from (1) immediately.

One important note: Berestetsky et al derive the expression (1) sans the Zeeman term for the case when no vector potential is present. Which is probably why I initially got all the extra terms.

In this simple case we can write (defining $\mathbf{P}=\mathbf{p}-e \mathbf{A}$):

$$(\sigma \mathbf{P})^2=\mathbf{p}^2$$

Another important note: the probability density up to $v^2/c^2$ has the following form:

$$\rho=|\Psi_+|^2+|\Psi_-|^2=|\Psi_+|^2+ \frac{1}{4m^2 c^2} |\sigma \mathbf{p} \Psi_+|^2$$

To preserve the usual definition, according to Berestetsky et al, we have to definte another wavefunction:

$$\Psi_+= \left(1- \frac{\mathbf{p}^2}{8m^2 c^2}\right) \Psi$$

Which leaves us with:

$$E_c \Psi =\left(1+ \frac{\mathbf{p}^2}{8m^2 c^2}\right) \left( \frac{\mathbf{p}^2}{2m} +U - \frac{\mathbf{p}^4}{8m^3 c^2}- \frac{\mathbf{p}^2 U}{4m^2 c^2} + \frac{1}{4m^2 c^2} ( \sigma \mathbf{p} U \sigma \mathbf{p} ) \right) \left(1- \frac{\mathbf{p}^2}{8m^2 c^2}\right)\Psi $$

Up to second order we have:

$$E_c \Psi = \left( \frac{\mathbf{p}^2}{2m} +U - \frac{\mathbf{p}^4}{8m^3 c^2}- \frac{\mathbf{p}^2 U}{8m^2 c^2}- \frac{ U \mathbf{p}^2}{8m^2 c^2} + \frac{1}{4m^2 c^2} ( \sigma \mathbf{p} U \sigma \mathbf{p} ) \right) \Psi $$

Which coincides with the expression provided in Berestetsky et al before they derive (1) (without the Zeeman term I added).

So I guess my question is answered.

If someone adds a more in depth and/or intuitive explanation for all this stuff, I will gladly award them the bounty.


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