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It is frequently stated that the continuum limit of a lattice model is equivalent to the low-energy limit, e.g. here, here and section IIB of this. I do not know how to show this for myself. Take for example the tight-binding Hamiltonian

$$ H = -t \sum_n c_{n+1}^\dagger c_n + \text{h.c.},\quad t \in \mathbb{R}$$

with periodic boundary conditions, where $\{ c_n \}$ are fermionic modes that obey $\{ c_n,c_m^\dagger \} = \delta_{nm}$, indices label lattice sites and h.c. denotes hermitian conjugate.

Continuum limit

We can express the lattice fermions as a continuous field sampled at discrete lattice points as $c_n \equiv \sqrt{a}c(x_n)$, where $x_n = na$ with lattice spacing $a$. We approximate the Hamiltonian to first order in $a$ as

$$ H = -t \sum_n a\left[ c^\dagger(x_n) + a\partial_x c^\dagger(x_n) + O(a^2)\right]c(x_n) + \text{h.c.} $$

Now take the continuum limit by setting $a \rightarrow 0$ to map the sums to integrals. We get

$$ H \approx -t \int dx c^\dagger(x) c(x) - v_F\int dx \left[ (\partial_x c^\dagger(x)) c(x)+c^\dagger(x) \partial_x c(x) \right] = -t \int dx c^\dagger(x) c(x) $$

where $v_F = 2at$ and going to the last equality I noted the second term can be written as a total derivative so integrates to zero.

Low energy limit

We Fourier transform the fields with the definition $ c_n = \frac{1}{\sqrt{N}} \sum_p e^{ipan} c_p$, in which case the Hamiltonian is diagonalised:

$$ H = \sum_p E(p) c^\dagger_p c_p, \quad E(p) = -2t\cos(ap)$$

The low-energy theory is described by expanding about the points where $E(p) = 0$. There are two unique points located at $p_\pm = \pm \frac{\pi}{2a}$ in the Brillouin zone. The low energy limit is therefore obtained as:

$$ H \approx \sum_k E(p_+ + k) c^\dagger_+(k) c_+(k) + E(p_- + k) c^\dagger_-(k)c_-(k) $$

where $c_\pm(k) = c_{p_\pm + k}$. Using $E(p_\pm + k) = \pm v_F k + O(a^2)$, we can inverse Fourier transform to find the Hamiltonian back in position space as

$$ H \approx H_+ + H_- ,\quad H_\pm = \pm v_F (-i) \int dx c^\dagger_\pm(x) \partial_x c_\pm(x)$$

This method is the method taken for deriving the continuum limit of graphene for example.

My question

It is clear that the continuum limit obtained by setting $a \rightarrow 0$ in position space and the low-energy limit by expanding about Fermi points $p_\pm$ in momentum space yield different results. Why is it commonly stated that the continuum limit is the same as the low-energy limit and why didn't it work for me?

Hermitian_hermit
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  • One clear way to see that you’ve taken the continuum limit incorrectly in the first approach: your final continuum Hamiltonian has no dispersion! Every site is decoupled from one another, which is clearly not capturing the desired physics. – Zack Mar 01 '22 at 00:45
  • @Zack yes I see; but I could remedy this by expanding to second order in $a$ to obtain something. My question remains still even after this – Hermitian_hermit Mar 01 '22 at 00:48
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    You’re right. The more general lesson is that you need to be careful about taking the continuum limit of a fermion model, due to the presence of the Fermi sea. Your first continuum limit (when properly accounting for the dispersion minimum) will be identical to expanding near $k=0$, which is the correct low energy limit for an empty band. On the other hand, your second limit amounts to expanding about the Fermi points, whose states will actually have the relatively fast oscillation $e^{\pm i k_F x}$. – Zack Mar 01 '22 at 01:00
  • Often one writes $c(x)=e^{i k_F x} \psi_R(x) + e^{-i k_F x} \psi_R(x)$, so that the low energy limit is given in terms of slowly varying fields $\psi_{R/L}$. – Zack Mar 01 '22 at 01:01
  • @Zack ah I see, so if my model was bosonic, I would actually expand about $k=0$ as there is no Fermi sea to account for, in which case this procedure would give me a quadratic term, in agreement with the $a \rightarrow 0$ continuum limit? Alternatively, a single particle theory of Fermions where the Dirac sea is not full would also be expanded about $k=0$ too, again giving us agreement? – Hermitian_hermit Mar 01 '22 at 08:14
  • @Zack I recommend considering to write an answer. :) – AlmostClueless Mar 01 '22 at 09:56

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