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")
Best Answer
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).
EDIT
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
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).