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!

  • $\begingroup$ General remark: For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$
    – Qmechanic
    Commented Apr 26, 2016 at 18:08
  • $\begingroup$ @Qmechanic Sorry, I should have mentioned I'm approaching this from a pre-QFT pov as I actually checked out that post before asking and can't understand too much! I'm okay with the relation between the two equations, it's just why we may take these limits that I don't understand. $\endgroup$
    – user115561
    Commented Apr 26, 2016 at 18:26

1 Answer 1

  • $|q\phi|\ll mc^2$ 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 $E\ll 2mc^2$.

  • $\begingroup$ Thank you, this is exactly what I was looking for and pitched precisely at the right level for me to understand! $\endgroup$
    – user115561
    Commented Apr 26, 2016 at 21:08
  • $\begingroup$ @ellaf I'm glad I could help :-) if you have any more questions feel free to ask. $\endgroup$ Commented Apr 26, 2016 at 21:18
  • $\begingroup$ The second point is not about radiation, but particle-antiparticle pair production. If $\omega>2mc^2$, the EM field has enough energy to create pairs (the factor of 2 is important here) $\endgroup$
    – Adam
    Commented Apr 27, 2016 at 12:04
  • $\begingroup$ @Adam what you say is essentially the same thing I said in my answer: if the energy of the photons is high enough they can induce pair production. But, strictly speaking, in this context one should not speak of pair production (OP is asking about "pre-QFT": see the comments above) but of non-linear electromagnetic effects instead. $\endgroup$ Commented Apr 27, 2016 at 12:26
  • $\begingroup$ You are not saying that in your answer: "otherwise the electromagnetic radiation would dominate compared to the electrostatic field" and " says nothing about the particles", whereas pair production has everything to do with the particles. And it is not because the OP ask about pre-QFT that the answer must be too... when the effect is clearly coming from QM + SR, which is QFT... $\endgroup$
    – Adam
    Commented Apr 27, 2016 at 12:38

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