In natural units, the Dirac Equation is $$i \frac{\text{d}}{\text{d}t} \psi = \left[\vec \alpha \cdot \vec P +\beta m + e \Phi\right]\psi.$$ I use Pauli-Dirac basis for matrices, \begin{align*} \vec \alpha = \begin{pmatrix} 0 & \vec \sigma\\ \vec \sigma & 0 \end{pmatrix} && \text{and} && \vec \beta = \begin{pmatrix} I & 0\\ 0 & -I \end{pmatrix}. \end{align*}

When the fields and momentum are $0$, the states and energies are

(1) $\tilde \phi=a (e^{-imt},0,0,0)$ and $b (0,e^{-imt},0,0)$ corresponding to a positive energy $m$

(2) $\tilde\chi=c (0,0,e^{imt},0)$ and $d (0,0,0,e^{imt})$ corresponding to a negative energy $-m$

My goal is to derive the non-relativistic limit (Pauli-Schroedinger Equation) using the above. I consider all the fields/momentum to be nonzero now but small now. I am stuck on one single point:

It is standard to use the ansatz $\psi = (\tilde\phi_1,\tilde\phi_2,\tilde\chi_3,\tilde\chi_4)=(\tilde\psi,\tilde\chi)\equiv e^{-imt}(\phi,\chi)$. In the last statement, we are effectively factoring out the positive energy phase in our solution. From this, why is $\frac \partial {\partial t} \chi\sim 0$, in the limit that $P \ll m$ and $\Phi\ll m$?

My confusion is that clearly from (1) and (2), $\tilde\phi\sim e^{-imt}$ while $\tilde\chi\sim e^{imt}$. So doesn't this mean $\phi\sim 1$ while $\chi\sim e^{2imt}$? In this case, it seems that we do not get $\frac \partial {\partial t} \chi\sim 0$, but instead we get $\frac \partial {\partial t} \phi\sim 0$, which is the opposite of what I wanted, and is the opposite of what is in the literature (see "From Dirac Equation")

  • $\begingroup$ The ansatz seems to be incorrect. $\endgroup$ Feb 19 at 18:00
  • 1
    $\begingroup$ I really dont think so, but could you elaborate if you have a good argument? $\endgroup$
    – Mondo Duke
    Feb 20 at 2:32

1 Answer 1


To get the non-relativistic limit, that is the Pauli equation, one can use the ansatz $\psi = e^{-imt}\begin{pmatrix}\phi\\ \chi\end{pmatrix}$, as you mentioned. In the ansatz, however, $\phi$ and $\chi$ are generic functions that you will find after plugging the ansatz into Dirac's equation and taking the non-relativistic limit, which is $|\partial\chi/\partial t|\ll|m\chi|$ and $|e\Phi\chi|\ll|m\chi|$ (i.e, kinetic and potential energy respectively much smaller then the rest energy).

It's not clear to me how you wish to use (1) and (2).


Following your comment, it seems to me that you're plugging into the ansatz the form of the free-particle solutions (1) and (2). What I mean with

In the ansatz, however, $\phi$ and $\chi$ are generic functions[...]

Is that their form is not known when using the ansatz, so assuming they behave as in (1) and (2) would be incorrect. Indeed the aim is to retrieve the Pauli equation, whose solutions are not in the form of (1) and (2).

  • $\begingroup$ I use (1) and (2) in the first sentences of the last paragraph: "My confusion is that clearly from (1) and (2), ϕ~$e^{−imt}$ while χ$∼e^{imt}$. So doesn't this mean ϕ∼1 while χ∼$e^{2imt}$?". Is this a correct statement? I might not have been as clear as possible following this statement, so I will be clearer: if this quoted argument is correct, then the energy of $\chi$ goes as $i \partial_t \chi=i \frac{\partial \chi }{\partial t}=i \cdot 2im = -2m\neq 0$. I wanted $i \partial_t \chi$ to be 0, but I get that is goes as $2m$. This is my confusion. $\endgroup$
    – Mondo Duke
    Feb 22 at 8:58
  • $\begingroup$ @MondoDuke I have edited the answer. I hope it is clearer what I meant now. $\endgroup$ Feb 22 at 10:14
  • $\begingroup$ thanks. A followup, with that logic why cant we also assume $\partial_t \phi << m \phi$ in the nonrelativistic limit? Clearly this would be invalid $\endgroup$
    – Mondo Duke
    Feb 22 at 20:01
  • $\begingroup$ @MondoDuke Yes, you can assume that, but luckily the terms $m\phi$ cancel out in the equation for $\phi$, as expected. You can see this being done in the Wiki link in your post or in Greiner's "Relativistic Quantum Mechanics". $\endgroup$ Feb 23 at 9:14
  • $\begingroup$ We know that each component of the Dirac bi-spinor verifies the Klein-Gordon equation*, you replace your 'ansatz' (1) in the K-G equation, possible that you find the equation that you seek to demonstrate and the same thing for the relation (2) with the consideration that one can restore (1) with exp(mt)=exp(-m(-t)), a particle (e) of negative energy which goes back in time is a particle (-e) of positive energy moving forward in time (Feynman-Stuckelberg interpretation). *en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation $\endgroup$
    – The Tiler
    Feb 23 at 16:46

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