# Continuum limit of a free bosonic theory

Suppose you have a bosonic theory in a lattice

\begin{equation} H=\sum_{<i,j>}c_1a^+_ia^-_j+c_2(a^+_ia^+_j+a^-_ia^-_j\big) \end{equation}

Where $\{a^{\pm}_i\}$ are bosonic creation and anihilation operators, $c_1$ and $c_2$ are arbitrary constants and $<i,j>$ means sum over neighbouring sites in a lattice. This is just an example, it could have interaction terms too.

I was wondering if there was a way to take this Hamiltonian to the Path Integral Formalism. Something similar to the Hubbard-Stratonovich transformation for the Ising model, where one replaces the ising variables in the lattice $\sigma_i$ for a continuous field $\phi (\vec{r})$ and the partition function becomes a path integral for a $\lambda \phi^4$ theory.

Is there a way to take the continuum limit to a quantum field theory from a bosonic theory in a lattice?

• There seem to be three distinct things you’re asking about: the derivation of the path integral for bosons, the continuum limit, and the $c_2$ term which creates and annihilates pairs. Could you clarify which you’re mainly interested in? – Stephen Powell Apr 7 '18 at 18:58
• Hi! I know how to derive the path integral formalism for bosons when one starts with a lagrangian for a field theory. My interest is how to do that when one has a hamiltonian for a system in a lattice. I assumed it had to do with the continuum limit... – P. C. Spaniel Apr 7 '18 at 19:59
• The usual way of getting to a path integral from this kind of Hamiltonian is to use the coherent state path integral; see these lecture notes by Eduardo Fradkin or these by Nicolas Dupuis. This can be done on the lattice and gives an action that looks similar but with the operators replaced by c-numbers (and with some time derivatives). – Stephen Powell Apr 9 '18 at 10:24
• Re the continuum limit: Unlike in the Ising model, you don't need any tricks because the degrees of freedom in the path integral are continuous. If you are interested only in long-wavelength properties (which is what taking the continuum limit means) then you can get rid of the short-wavelength ("fast") modes of the bosonic field. For your model, they can just be dropped. With interactions, in principle you need to use the renormalization group to integrate out the fast modes. (See this question.) – Stephen Powell Apr 9 '18 at 10:34
• By “continuum limit”, you mean to remove the underlying microscopic lattice (denoted by indices i and j)? – AlQuemist Apr 9 '18 at 14:26

Yes there is a way. since you ask how to do it for a lattice I won't explain it indetail I only tell that part.

you have to use this following completeness relation;

$$\prod_{i}d\bar{\psi_i}d\psi_i/\pi\exp[-\sum_i\bar{\psi_i}\psi_i]|\psi\rangle\langle\psi|=1_{fock}$$ this is how to write a $1$ in fock space in terms of coherent states. $i$'s are the lattice sites, $|\psi\rangle$ is the coherent states and we also have , $a_i|\psi\rangle=\psi_i|\psi\rangle$. The rest is straight forward you just need to write your partition function and then you need to make a Suzuki-Trotter expansion using these $1$'s I provided, then you will have you partition function in path integral formalism in a lattice. The key point is to use bosonic coherent states, and using the completeness relation.

• Thanks for the answer! I search and understood everything. There is something I still don't get, though. Can this method be applied to Hamiltonians that have first-neighbour interactions? I only saw it applied to Hamiltonians that only had products of a^{\pm}_i without mixing sites. – P. C. Spaniel Apr 10 '18 at 22:27
• The crucial part of the question is how to obtain the long-wavelength continuum limit, not merely a path-integral representation. – AlQuemist Apr 11 '18 at 10:22

To simplify the matters and reduce notational clutter, let's assume we have a 1d chain with nearest-neighbour hopping/interaction, described by the Hamiltonian, $$H = t \sum_{\langle i,j \rangle} a_i^\dagger a_j + \Big( V \sum_{\langle i,j \rangle} a_i a_j + \text{c.c.} \Big) ~.$$

To systematically obtain the long-wavelength (low-energy) continuum limit of this Hamiltonian, we should first perform a Fourier transformation to momentum representation, using

\begin{align*} a_k &= \frac{1}{\sqrt{N}} \sum_{l} e^{i k x_l} a_l \\ a_l &= \frac{1}{\sqrt{N}} \sum_{k} e^{i k x_l} a_k \\ \delta(k) &= \frac{1}{N} \sum_{l} e^{i k x_l} \\ \delta(x) &= \frac{1}{N} \sum_{k} e^{i k x} ~, \end{align*}

where $l$ is the lattice index, $k$ is the momentum, $N$ is the number of lattice sites, and we have used the ‘natural’ units where $\hbar = 1$.

The first term of the Hamiltonian can be rewritten as $$\sum_{\langle i,j \rangle} a_i^\dagger a_j = \sum_{i, n = \pm 1} a_i^\dagger a_{i + n} ~,$$

because of our assumption of nearest-neighbour hopping on a 1d chain. This can be indeed generalized with little effort to any kind of lattice in any dimension ($n$ becomes a vector then). This paves the way for the Fourier transform:

\begin{align*} \sum_{\langle i,j \rangle} a_i^\dagger a_j &= \sum_{i, n = \pm 1} a_i^\dagger a_{i + n} \\ &= \frac{1}{N} \sum_{i,n} \sum_{k, q} e^{i (q - k) x_i} e^{i q n a} \, a_k^\dagger a_q \\ &= \sum_{k, q, n = \pm 1} \delta(q - k) \, e^{i q n a} \, a_k^\dagger a_q \\ &= \sum_{k} ( e^{i k a} + e^{-i k a} ) \, a_k^\dagger a_k = \sum_{k} 2 \cos(k a)\, a_k^\dagger a_k ~, \end{align*}

where we have used the Fourier transform relations above, and $x_{i + n} \equiv x_i + n a$ with $a$ denoting the lattice spacing.

Similarly, one can show that

$$\sum_{\langle i,j \rangle} a_i a_j = \sum_{k} 2 \cos(k a)\, a_k a_{-k} ~.$$

Therefore, the Hamiltonian in momentum representation reads

$$H = 2t \sum_{k} \cos(k a)\, a_k^\dagger a_k + \Big( 2V \sum_{k} \cos(k a)\, a_k a_{-k} + \text{c.c.} \Big) ~.$$

Now we can take the long-wavelength limit by assuming that the characteristic wavelength of the excitations, $\lambda$, is much larger than the lattice spacing, $a$; ie., $\lambda \gg a$ or $k a \ll 1$ which allows an expansion of the dispersions in terms of $k a$:

$$\cos(ka) = 1 - \frac{1}{2} (k a)^2 + \mathcal{O}\left( (k a)^4 \right) ~.$$

Neglecting higher-order terms, we end up with an effective long-wavelength Hamiltonian,

\begin{align*} H_{\lambda \gg a} &\approx 2t \sum_{k} a_k^\dagger a_k - t \sum_{k} (k a)^2 \, a_k^\dagger a_k \\ &+ \Big( 2V \sum_{k} a_k a_{-k} - V \sum_{k} (k a)^2 \, a_k a_{-k} + \text{c.c.} \Big) ~. \end{align*}

Such neglects of higher-orders can be substantiated via RG arguments; consult eg., Kopietz et al. “Introduction to the Functional Renormalization Group” (2013) [wcat].

It is straightforward to show, via a Fourier transform back to position representation, that

\begin{align*} \sum_{k} a_k^\dagger a_k &= \sum_{l} a_l^\dagger a_l \\ \sum_{k} (k a)^2 a_k^\dagger a_k &= a^2 \sum_{l} \frac{\partial}{\partial x_l} a_l^\dagger \, \frac{\partial}{\partial x_l} a_l \\ \sum_{k} a_k a_{-k} &= \sum_{l} a_l a_l \\ \sum_{k} (k a)^2 a_k a_{-k} &= a^2 \sum_{l} \frac{\partial}{\partial x_l} a_l \, \frac{\partial}{\partial x_l} a_l ~. \end{align*}

Hence,

$$H_{\lambda \gg a} \approx 2 t \sum_l a_l^\dagger a_l - t a^2 \sum_l \partial_{x_l} a_l^\dagger \partial_{x_l} a_l + \Big( 2 V \sum_l a_l a_l - V a^2 \sum_l \partial_{x_l} a_l \partial_{x_l} a_l + \text{c.c.} \Big) ~.$$

or better, using a suggestive notation,

\begin{align*} H_{\lambda \gg a} &\approx 2 t \sum_l \frac{\Delta x}{a} \, \varphi^\dagger(x_l) \varphi(x_l) - t \sum_l \frac{\Delta x}{a} \, \partial_{x_l} \varphi^\dagger(x_l) \partial_{x_l} \varphi(x_l) \\ &+ \Big( 2 V \sum_l \frac{\Delta x}{a} \, \varphi(x_l) \varphi(x_l) - V \sum_l \frac{\Delta x}{a} \, \partial_{x_l} \varphi(x_l) \partial_{x_l} \varphi(x_l) + \text{c.c.} \Big) ~, \end{align*}

where $\Delta x \equiv a$, $\varphi(x_l) := a_l$ and we have introduced the dimensionless coordinates, $x \mapsto \frac{x}{a}$. Now we can take the continuum limit ($a \rightarrow 0$), so that

\begin{align*} H_{\lambda \gg a} &\approx \int {\mathrm{d} x} \Big\{ 2 t \, \varphi^\dagger(x) \varphi(x) - t \, \partial_x \varphi^\dagger(x) \partial_x \varphi(x) + \Big( 2 V \, \varphi(x) \varphi(x) - V \, \partial_x \varphi(x) \partial_x \varphi(x) + \text{c.c.} \Big) \Big\} ~. \end{align*}

An analogous procedure can be performed in the functional-integral/action formalism, but the steps are the same.

This is essentially a “gradient expansion”. Notice that since the Hamiltonian is quadratic, there is no need for a Hubbard-Stratonovich transformation.

• Hi! Grest answer. Question: how would you interpret the terms that dont conserve particle number such as phi * phi or phi dagger * phi dagger? Wouldnt it be better to first do a bogoliubov rotation to rewrite the hamiltonian in terms of new operators that do have that symmetry? – P. C. Spaniel Apr 11 '18 at 17:47
• @P.C.Spaniel : The particle-nonconserving terms like $\varphi \varphi$ are the consequence of analogous terms in the original Hamiltonian, like $a_i a_j$. Their precise physical meaning depends on the context and how one has derived the original Hamiltonian, but usually they represent some sort of “bath” into which particles can disappear, or from which particles can be created. A good example is the theory of lattice phonons. – AlQuemist Apr 11 '18 at 19:54
• @P.C.Spaniel : If you could use a Bogoliubov transformation to diagonalize the (quadratic) Hamiltonian, then it is better to perform such a transformation first. However, if you have interaction (quartic or higher terms) in the Hamiltonian, such a transformation is out of question. – AlQuemist Apr 11 '18 at 19:58
• @P.C.Spaniel : do you have any other questions regarding the answer? – AlQuemist Apr 13 '18 at 17:46