I am trying to derive the non-relativistic Lagrangian for a complex scalar field from taking the non-relativistic limit of the complex scalar field Lagrangian. I am following the steps in "QFT for the Gifted Amateur," section 12.3, which uses the mostly-minus metric convention.

They start out with the complex scalar field Lagrangian (with interaction term),

$$\mathcal{L} = \partial^{\mu}\psi^{\dagger}\partial_{\mu}\psi-m^2\psi^{\dagger}\psi -\lambda(\psi^{\dagger}\psi)^2\tag{1} $$

and take the non-relativistic limit,

$$\psi \rightarrow \frac{1}{\sqrt{2m}}e^{-imt}\Psi(\textbf{x},t),\tag{2}$$

and the corresponding adjoint expression and plug them into $\mathcal{L}$. The book first studied the time derivative part, and argues:

As in the last example, the guts of taking the [non-relativistic] limit may be found in the time derivatives. We find

$$ \partial_{0}\psi^{\dagger}\partial_{0}\psi = \frac{1}{2m} \left[ \partial_0 \Psi^{\dagger}\partial_0\Psi + im \left( \Psi^{\dagger}\partial_0\Psi - (\partial_0\Psi^{\dagger})\Psi \right) +m^2\Psi^{\dagger}\Psi \right]. \tag{3}$$ The first term going as $1/m$ is negligible in comparison with the others.

I agree with the equation they derived, however, I don't know why the first term that goes as $(1/m)$ is negligible, because if you expand out the rest of the Lagrangian (this is now my work, not the textbook's,) you get

$$ \mathcal{L} = \frac{m}{2}\Psi^{\dagger}\Psi + \frac{i}{2}\left( \Psi^{\dagger}\partial_0\Psi-(\partial^0\Psi^{\dagger})\Psi \right) + \frac{1}{2m}\partial^0\Psi^{\dagger}\partial_0\Psi$$$$ - \frac{1}{2m}\nabla\Psi^{\dagger} \cdot \nabla\Psi - \frac{m}{2}\Psi^{\dagger}\Psi - \frac{\lambda}{4m^2}(\Psi^{\dagger}\Psi)^2. \tag{4} $$

Then it is claimed that $\Psi$ and its adjoint may be expanded in plane waves, like

$$ \Psi = \sum a_{\textbf{p}}e^{-ip\cdot x},\tag{5} $$

so $\left( \Psi^{\dagger}\partial_0\Psi - \Psi \partial_0\Psi^{\dagger} \right) $ can be replaced with $2\Psi^{\dagger}\partial_0\Psi$, which leaves

$$ \mathcal{L} = i\Psi^{\dagger}\partial_0\Psi + \frac{1}{2m}\partial^0\Psi^{\dagger}\partial_0\Psi - \frac{1}{2m}\nabla\Psi^{\dagger} \cdot \nabla \Psi - \frac{\lambda}{4m^2} (\Psi^{\dagger}{\Psi})^2.\tag{6} $$

The algebra I am fine with, but I say that if you look at the full Lagrangian, not just the time derivative part, both the time derivative and spatial derivative parts go as $1/m$, and even worse, the interaction term goes as $1/m^2$. The book neglected the time derivative part and uses as its final answer

$$ \mathcal{L} = i\Psi^{\dagger}\partial_0\Psi - \frac{1}{2m}\nabla\Psi^{\dagger} \cdot \nabla\Psi - \frac{\lambda}{4m^2} \left( \Psi^{\dagger}\Psi \right)^2. \tag{7}$$

Is there a reason why the $\frac{1}{2m}\partial_0\Psi^{\dagger}\partial_0\Psi$ time-derivative term is neglected because "it is small compared to the other two terms in the time derivative $(\partial_0\Psi^{\dagger}\partial_0\Psi)$ expression", even though it is $\mathbf{not}$ small compared to the other terms in the Lagrangian (at least, I don't think it is).


1 Answer 1


This is perhaps best seen via the dispersion relation after doing a Fourier transformation to the energy-momentum representation. The time derivative $\partial_0$ corresponds (up to a factor of $i$) to multiplication with energy $E$ (which is the total energy minus the rest energy that was subtracted in eq. (2).).

In the non-relativistic (NR) limit, we are only interested in energy excitations $|E|\ll mc^2$, where $m$ is the rest mass. Therefore the quadratic time derivative term $\frac{1}{2m}\partial^0\Psi^{\dagger}\partial_0\Psi$ is small compared to the single time derivative term $i\Psi^{\dagger}\partial_0\Psi$, and can be neglected. (If we keep it, we will also find solutions $|E|\gg mc^2$ associated with antiparticles.)

Therefore we arrive at the Schroedinger Lagrangian (7). For this reduction from the complex Klein-Gordon eq. to the Schr. eq., see also e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.

In order to quantize the Schroedinger Lagrangian (7), one needs to consider the Hamiltonian formulation. See e.g. this Phys.SE post. The long story short is that we recognize the last two terms in eq. (7) as minus the Hamiltonian density $-{\cal H}$. The energy of a field configuration therefore comes from these two terms.

  • $\begingroup$ Ok, so if I plug the plane wave expansion into the Lagrangian and keep the terms in the same order as they are in my second equation from the bottom, then I get $\mathcal{L} = E\Psi^{\dagger}\Psi + \frac{E^2}{2m}\Psi^{\dagger}\Psi - \frac{p^2}{2m}\Psi^{\dagger}\Psi -\frac{\lambda}{4m^2}(\Psi^{\dagger}\Psi)^2. $ I can clearly see how the second term is negligible compared to the first. Is there a reason why the second term is negligible compared to the third term, with the square of the momentum? $\endgroup$ Jun 27, 2015 at 17:48
  • $\begingroup$ Only indirectly. $\endgroup$
    – Qmechanic
    Jun 27, 2015 at 18:10
  • $\begingroup$ Oh, I might see it now. If the energy E that is remaining is not including the rest energy, then $E = \frac{p^2}{2m}$. Since we are investigating the case where $E\ll m$, then given E is just kinetic energy, then we get $E \ll \frac{p^2}{2E}$, or $E \ll p$. $\endgroup$ Jun 27, 2015 at 18:45

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