Reading about fermion gas in a paper they used the following Lagrangian, which describes an effective field theory for nonrelativistic fermions (I neglect the four point interaction term).

$$ L = \psi^{\dagger}(\partial_{\tau} - \frac{\nabla^2}{2m})\psi. $$

They work in two-dimensional Euclidean space-time, where $\psi_{\sigma}(\tau,x)$ is a four component Grassmann-valued spinor with spin components labelled by the index $\sigma$.

Somehow it looks very familiar, but I am not able to deduce this from something I know. Since this is an Effective Field Theory, means this that it can be deduced with an expansion of the relativistic Lagrangian?


1 Answer 1


To see that this is the simplest possible non-relativistic quantum field theory for fermions, it's useful to derive the dynamics. The canonical momentum for $\psi(x,y,z)$ is the Lagrangian's derivative with respect to $\partial_\tau \psi(x,y,z)$ – and it is $\psi(x,y,z)^\dagger$ (up to signs and $\pm i$ which depend on conventions).

At any rate, the Hamiltonian is $H = p\dot q - L$ and the first term just cancels against the derivative term in the Lagrangian, so what we are left is $$ {\mathcal H} = \psi^\dagger \frac{\nabla^2}{2m} \psi $$ But up to a sign which may be wrong, this is just the factor of the kinetic energy operator $p^2/2m$ inserted into the operator of the total number of particles, so in the momentum basis and in a box, the total Hamiltonian is nothing else than $$ H = \sum_{\vec p} \frac{p^2}{2m} c^\dagger_p c_p $$ where $c^\dagger_p c_p=N_p$ is simply the number of particles with the momentum vector $\vec p$. We multiply each of them by the kinetic energy and sum over allowed $\vec p$ to get the total energy of the multi-fermion system.

We may also sum over the spin polarizations but the operator $\nabla^2$ commutes with the spin degrees of freedom, so adding the spin is just like having several independent spin states for each $\vec p$ that are simply summed over.

Let me emphasize again that while the Hamiltonian of the "simplest fermionic field theory" only has one term, with the spatial derivatives, the Lagrangian also has to have the term that depends on the time derivative of $\psi$ – all Lagrangians for dynamical degrees of freedom have to depend on time derivatives for the canonical momenta to exist and for discontinuous histories to be disfavored when the least action is looked for.

Yes, the simple Lagrangian above may also be deduced from the relativistic Lagrangians, i.e. the Dirac Lagrangian, as long as we exploit the fact that the two 2-component parts of the Dirac spinor are nearly equal for low velocities. This allows one to express one of them as a function of the derivatives of the other, and in this way, first spatial derivatives in the Dirac Lagrangian become second derivatives. The time derivative term in your Lagrangian is pretty much exactly the same term as the first time derivative in the Dirac Lagrangian. Otherwise the limiting reduction of the spatial derivative terms is the usual derivation of the "Pauli" non-relativistic equation from the Dirac theory, applied to the (second-quantized) fields producing multi-particle states. This derivation may be done in terms of the Lagrangian – which you are interested in – or in other ways.

  • $\begingroup$ Thank you for your explanation! In the last paragraph you mentioned what I want to achieve. So I will start with $L=\bar{\psi}\gamma_{\mu}\partial^{\mu}\psi + m\bar{\psi}\psi$, but I am uncertain about the spinors, holds the relation $\bar{\psi} = \psi^{\dagger}\gamma_0$? It should since we are dealing with Dirac spinors. Additionally the Grassmann notation confueses me, if the spinor is a Grassmann valued spinor, then the whole spinor is grassmannian, am I right? - I just dont know how to start solving the problem right know. $\endgroup$
    – nerdizzle
    Aug 11, 2014 at 18:12
  • $\begingroup$ Fermionic fields are always Grassmannian, yes, $\bar\psi$ always means $\psi^\dagger \gamma_0$ for Dirac spinors and so on, and so on. There are clearly way too many things you would be asking now. Isn't it a good idea to learn these things systematically from a textbook? $\endgroup$ Aug 11, 2014 at 19:43
  • $\begingroup$ true that, but I do already know some things, so I would prefer to do this exercise and read in books about, otherwise by only reading it is not possible to reach my goal. I just struggle with starting points, but time will give me the answer :) $\endgroup$
    – nerdizzle
    Aug 11, 2014 at 21:14

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