# Non-relativistic limit of complex scalar field

In page 42 of David Tong's lectures on Quantum Field Theory, 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 inequation I 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 $\dot{\psi}$

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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. –  Qmechanic Sep 14 '13 at 10:23
Note that Klein-Gordon equation is Poincare-covariant, and Schrodinger equation is not. This means, that the second can't be the direct consequence of the first. –  user8817 Sep 14 '13 at 21:47

You cannot derive it "directly" from a Klein-Gordon equation, or from a Klein-Gordon Lagrangian.

Starting from a Klein-Gordon equation for $\psi$, and defining $\psi(\vec x,t) = e^{-imt} \tilde \psi(\vec x,t)$ ($2.103$), you get a new equation for $\tilde \psi$, which is not a Klein-Gordon equation :

$$\ddot {\tilde \psi} - 2 im \dot {\tilde \psi} - \nabla^2\tilde \psi = 0 \tag{2.104}$$

By Fourier transform , this is equivalent to the condition :

$$(E'^2 + 2mE'-\vec p^2) = 0 \tag{1}$$

What does that mean?

We begin with a Klein-Gordon equation for $\psi$, which, by Fourier Transform, is equivalent to the condition $$(E^2 -\vec p^2 - m^2) = 0 \tag{2}$$

Now, the transformation $\psi(\vec x,t) = e^{-imt} \tilde \psi(\vec x,t)$ $(2.103)$, gives the link between $E'$ and $E$, this is $E' = E - m$, this is a shift in the definition of the energy.

So, from $(2)$, we have simply : $((E'+m)^2-\vec p^2 - m^2)=0$, which is just the condition $(1)$

Now, if we suppose $|\vec p| \ll m$, this means $|E-m| \ll m$ (with $E \sim m)$, that is $E'\ll m$, so $E'^2 \ll mE'$.

Turning back to the equation $(2.104)$, which is not a Klein-Gordon equation, we see, by Fourier Transform, that we can neglect the first term relatively to the second term, and finally, you get :

$$i \dot {\tilde \psi}= - \frac{1}{2m}\nabla^2\tilde \psi = 0 \tag{2.105}$$

About lagrangians, you will have, I think, a problem, if you want to define a lagrangian giving $(2.104)$, with only a real scalar field $\tilde \psi$, because of the $\dot {\tilde \psi}$ term.

So, you have to consider a complex scalar field, and in the Lagrangian, you will have terms like $\dot{ \bar {\tilde \psi}} \dot{ {\tilde \psi}}$ and $im \bar {\tilde \psi}\dot{ {\tilde \psi}}, im \dot{\bar {\tilde \psi}}{ {\tilde \psi}}$, and the first term, in the approximation we discussed, is negligible relatively to the other terms.

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But how do you get those last two terms $m\bar{\tilde{\psi}}\dot{\tilde{\psi}}$, $m\tilde{\psi}\dot{\bar{\tilde{\psi}}}$ ? And also, that's not exactly what you want, is it? You want a $i\bar{\tilde{\psi}}\dot{\tilde{\psi}}$ –  guillefix Sep 16 '13 at 10:16
Yes, these are terms $im \bar {\tilde \psi}\dot{ {\tilde \psi}}, im \dot{\bar {\tilde \psi}}{ {\tilde \psi}}$ (I skipped the $i$) –  Trimok Sep 16 '13 at 10:32
It is just a modification of Lagrangian $1.15$, where we multiply by $m$ the first 2 terms, we skip the last term, and we add the quadratic time derivative term. –  Trimok Sep 16 '13 at 10:34
I think I've done it now. My problem was saying that "as we can write the above Lagrangian with both, as it's invariant under multiplying by a pure phase", that it's true but the pure phase can't depend on x, and the $e^{-mt}$ does. So when I rightly substituted that on the KG complex scalar field Lagrangian, I get indeed the terms you say, and after applying the non-rel. condition I do get the Lagrangian 1.15 with the modifications you say (and with a $\frac{1}{2}$ in front of the gradients); and that gives the right sort of Schrodinger's equation. Thanks! –  guillefix Sep 16 '13 at 16:50
@user29621 : OK, nice. –  Trimok Sep 16 '13 at 16:52