Question about non-relativistic limit of QFT

[guillefix]

In pages 41-42 of these notes: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf , it is said that |\vec{p}|\ll m implies |\ddot{\tilde{\phi}}|\ll m|\dot{\tilde{\phi}}|

Why is this so?

 

[Avodyne]

Take a plane-wave solution of the KG equation,
\phi=\exp(i\vec p\cdot\vec x - i E t)
where
E =\sqrt{\vec p^2+m^2}
Now assume $|\vec p|\ll m$. Then we have
E \simeq m + {\vec p^2\over 2m}
and the solution can be written as
\phi=\exp(-imt)\tilde\phi
where
\tilde\phi=\exp\bigl[i\vec p\cdot\vec x - i(\vec p^2\!/2m)t\bigr]
Now we can check that
\left|\ddot{\tilde\phi}\right|\ll m\left|\dot{\tilde\phi}\right|
as claimed. This will also apply to superpositions of different plane waves, provided that only plane waves with $|\vec p|\ll m$ are included in the superposition.

 

[guillefix]

Thanks, and another question

Thanks for that! I've got another question though. In the same document, a bit later, he says that one can also derive the Schrödinger Lagrangian by taking the non-relativistic limit of the (complex?) scalar field Lagrangian. And for that he uses the condition \partial_{t} \Psi \ll m \Psi, which in fact I suppose he means |\partial_{t} \tilde{\Psi}| \ll |m \tilde{\Psi}|, otherwise I don't get it. In any case, starting with the Lagrangian:

\mathcal{L}=\partial^{\mu}\tilde{\psi} \partial_{\mu} \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}

Using the inequationI think it's correct, I can only get to:

\mathcal{L}=-\nabla\tilde{\psi} \nabla \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}

And from that I've tried relating \tilde{\psi} or \psi (as we can write the above Lagrangian with both, as it's invariant under multiplying by a pure phase), to \dot{\psi}

 

[Avodyne]

Yes, start with the lagrangian for a complex field,
{\cal L}=\partial^\mu\psi^*\partial_\mu\psi-m^2\psi^*\psi Let
\psi=e^{-imt}\tilde\psi Then we have
\partial_t\psi=e^{-imt}(-im\tilde\psi+\partial_t\tilde\psi) \quad\hbox{and}\quad \partial_t\psi^*=e^{+imt}(+im\tilde\psi^*+\partial_t\tilde\psi^*) Multiply these together, and drop the \partial_t\tilde\psi^*\partial_t\tilde\psi term as "small", but do not drop the cross terms. If you like, then integrate by parts to move the time derivative off \tilde\psi^* and onto \tilde\psi.