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 $\dot{\psi}$

  • $\begingroup$ 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 Sep 14, 2013 at 10:23
  • $\begingroup$ 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. $\endgroup$
    – user8817
    Commented Sep 14, 2013 at 21:47
  • $\begingroup$ It can be after ignoring terms that are small in nonrelativistic situations $\endgroup$
    – Riemann
    Commented Apr 7 at 15:23

1 Answer 1


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.

  • $\begingroup$ 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}}$ $\endgroup$
    – guillefix
    Commented Sep 16, 2013 at 10:16
  • $\begingroup$ Yes, these are terms $im \bar {\tilde \psi}\dot{ {\tilde \psi}}, im \dot{\bar {\tilde \psi}}{ {\tilde \psi}}$ (I skipped the $i$) $\endgroup$
    – Trimok
    Commented Sep 16, 2013 at 10:32
  • $\begingroup$ 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. $\endgroup$
    – Trimok
    Commented Sep 16, 2013 at 10:34
  • $\begingroup$ 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! $\endgroup$
    – guillefix
    Commented Sep 16, 2013 at 16:50
  • $\begingroup$ @user29621 : OK, nice. $\endgroup$
    – Trimok
    Commented Sep 16, 2013 at 16:52

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