Suppose I had a simple $1D$ tight-binding Hamiltonian

$$ H = -t \sum_i a^\dagger_n a_{n +1} + \text{h.c.}$$

with $N$ sites and lattice spacing $a$. This Hamiltonian can be diagonalised with a discrete Fourier transform

$$ a_n = \frac{1}{N} \sum_{k \in \text{BZ}} e^{i k x_n} a_k \tag{1}$$

where the Brillouin zone (BZ) is given by $\text{BZ} = [-\pi/a,\pi/a]$ and the momenta are quantised as $k = 2n\pi/L$, where $L = Na$ is the length of the lattice. Substituting this into the Hamiltonian gives us

$$ H = \frac{1}{N} \sum_{k \in \text{BZ}} E(k) a^\dagger_k a_k, \quad E(k) = -2t \cos(ka) \tag{2}$$

If I was to take the thermodynamic limit where $N \rightarrow \infty$, I can use the fact that $\Delta k = 2\pi/L$, so we can use this to construct an integral via a Riemann sum:

$$ \frac{1}{L} \sum_k = \frac{1}{2\pi} \sum_k \Delta k \rightarrow \frac{1}{2 \pi} \int dk \tag{3}$$

$L = Na$ for us, so $1/N = a/L$. When I substitute this into my Hamiltonian eq. (2) I am able to set up a Riemann sum and apply the relationship of Eq. (3), however I am going to get a stray factor of $a$ in front:

$$ H = a \int_{- \pi/a}^{\pi/a} \frac{dk}{2\pi} E(k) a^\dagger(k) a(k) \tag{4} $$

Continuum limit $a \rightarrow 0$

The ground state of this Hamiltonian is where all the negative energy states are filled. Now I project this Hamiltonian onto the low-energy states by restricting to $k$ with small energy $E(k)$. I achieve this by Taylor expanding about $E(k) = 0$, where $ k = \pm \pi/2a \equiv k_\pm$ are the two Fermi points. I get $E(k_\pm + p) = \pm 2atp + O(a^2)$, so substituting this into Eq. (4) gives me

$$ H_\pm \approx \pm a v_f \int_{|p|<\Lambda} \frac{dp}{2\pi} p a_\pm^\dagger(p) a_\pm(p) +O(a^3)$$

where now the integration has a cutoff $\Lambda$ which I suspect is $O(1/a)$ (I use the cutoff $\Lambda$ to restrict to low-energy subspace), $v_f = 2at$ is the Fermi velocity and $a_\pm(p) = a(P_\pm + p)$ etc.

When I take the continuum limit $a \rightarrow 0$ of this I would expect the Fermi velocity $v_f$ to remain constant, however I still have that extra factor of $a$ in front that I do not know what to do with. The additional factor of $a$ can be traced back to my original definition of the Fourier transform in Eq. (1) with $1/N$ in front. An alternative definition is to take $1/\sqrt{N}$ instead in Eq. (1), which means we do not get a factor of $1/N$ in front of the Hamiltonian Eq. (2). However, without this factor of $1/N$, I am unable to set up a Riemann sum like in Eq. (3) as I do not have a factor I can use to construct the integration measure $\Delta k$.

The method I have taken of choosing $1/N$ normalisation for Fourier transforms is not uncommon and is seen on page 18 of Field Theories of Condensed Matter Physics (2nd Edition) by E. Fradkin for example. However, the author appears to have taken $a=1$. This appears to me that this restricts us from taking the continuum limit then as $a$ is now fixed, or if I reinstate $a$ I come across the problems of this question I am writing.

The second answer of this question is very similar to what I am trying to do but the Fourier transforms have been set up differently. I am unsure how one would set up a Riemann sum to turn sums over momentum into integrals as I am trying to do here. This is the essence of my problem.

Any comments would be greatly appreciated. Thanks


1 Answer 1


You don't have any choice about the normalization of the Fourier transform. It should read:

$$ a_n = \frac{1}{\sqrt{N}} \sum_{k \in BZ} e^{ik x_n} a_k$$

in order to have the correct anti-commutation relations $\{ a_i, a_j \} = \delta_{ij}$.

I would suggest simply working with finite $L$ ("box quantization'') to avoid issues of how to turn the sum into an integral. This is still compatible with the continuum limit $a \to 0$. However, if you really want to send $L \to \infty$ in order to make $k$ a continuous variable, note that the correct normalization for the fermion operators $a(k)$ when $k$ is a continuous variable needs to be $$ \{ a(k), a(k') \} = \delta(k - k')$$ The $a(k)$'s as you have defined them don't satisfy this property. I will leave it as an exercise to work out how to change the normalization of the $a(k)$'s to satisfy this condition.

  • $\begingroup$ I have been thinking about your exercise and I believe that if we take the discrete F.T. definition with $1/\sqrt{N}$ then for continuum limit $a \rightarrow 0$ I would have to redefine my fields as $\psi(x) = \lim_{a \rightarrow 0} a_n/\sqrt{a} $, while for thermodynamic limit $N \rightarrow 0$ (giving us a continuous $k$) I would define $\psi(k) = \lim_{N \rightarrow \infty} a_k/\Delta k$, where $\Delta k = 2 \pi/Na$. With these definitions the discrete F.T.s become continuous F.T.s with $1/\sqrt{2 \pi}$ in front which indeed obeys the canonical commutation relations for both $x$ and $k$. $\endgroup$ Apr 30, 2021 at 9:16

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