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I just have a question regarding assessing the non-relativistic limit of the Klein-Gordon equation. In the book I'm following (Quantum Mechanics by Bransden & Joachain) they use the limits (Chpt. 15.1 pg 681);

\begin{align} |q\phi|&\ll mc^2\\ \left|\frac{\hbar}{2mc^2}\frac{d\phi}{dt}\right|&\ll|\phi|\\ \left|\frac{\hbar^2}{2mc^2}\frac{d^2\Psi}{dt^2}\right|&\ll\left|\hbar\frac{d\Psi}{dt}\right| \end{align}

in order to investigate the non-relativistic limit of the KG equation for a spinless particle, charge $q$ in an EM field given by $A$ and $\phi$. My question is, how do we know we can use these limits? I'm okay with all other steps of the derivation and the Schrodinger equation, but can't understand where these limits come from as they are just brought in without explanation in the book.

I am unfamiliar with QFT and also tensor notation, but am okay with taking the energy approx. equal to the rest mass and $v\ll c$. I also don't know very much about electromagnetism and consequently am struggling to see how these limits arise. Any help is very much appreciated, thank you!

Qmechanic
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1 Answers1

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  • $|q\phi|\ll mc^2$ means that the rest mass of the particle is way larger than the electrostatic potential energy. This is the usual assumption of non-relativistic mechanics, which is equivalent to $v\ll c$. You can see this with the virial theorem: $$ \frac{1}{2}mv^2\sim q\phi $$ which means that $q\phi\ll mc^2$ iff $v\ll c$.

  • $\frac{\hbar}{2mc^2}\frac{d\phi}{dt}\ll\phi$ is a restriction on the potential energy: it cannot vary very fast in time because otherwise the electromagnetic radiation would dominate compared to the electrostatic field. The limit $\frac{\hbar}{2mc^2}\frac{d\phi}{dt}\ll\phi$ says nothing about the speed of the particles, but about the field they feel. If you write $$ \phi\sim V_0\mathrm e^{i\omega t} $$ youll find that $\frac{\hbar}{2mc^2}\frac{d\phi}{dt}\ll\phi$ is equivalent to $\hbar \omega\ll 2mc^2$. This again means that the radiation energy ("the energy of photons") is small compared to the rest mass of the particle.

  • $\frac{\hbar^2}{2mc^2}\frac{d^2\Psi}{dt^2}\ll\hbar\frac{d\Psi}{dt}$ means again that kinetic energy of the electrons is small compared to their rest mass. If you write $$ \Psi\sim \Psi_0 \mathrm e^{-i(E t-px)/\hbar} $$ you'll see that $\frac{\hbar^2}{2mc^2}\frac{d^2\Psi}{dt^2}\ll\hbar\frac{d\Psi}{dt}$ is equivalent to $E\ll 2mc^2$.

AccidentalFourierTransform
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