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 Schrodinger 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 $\tilde\psi$.