Quantum Field Theory from Continuum Limit
This article is one of Quantum Field Theory.
2.1 Quantum Fields
2.1.1 The Free Boson
For real scalar field $\varphi(x,t)$, let action became: $$\begin{align}S[\varphi]=\int dx\, dt\, \mathcal{L}(\varphi,\dot{\varphi},\nabla\varphi)\quad \dot{\varphi}\equiv \frac{\partial\varphi}{\partial t}\\ \mathcal{L}=\frac{1}{2}\left\{\frac{1}{c^2}\dot{\varphi}^2-(\nabla\varphi)^2-m^2\varphi^2\right\}\nonumber\end{align}$$ $\mathcal{L}$ is the Lagrangian density and $m$ is the mass of the field.
We replacing space with a discrete lattice of points at positions $x_n=an$, where $a$ is the lattice spacing and $n$ is an integer. We shall assume that this one-dimensional lattice is finite in extent (with $N$ sites) and that the variables defined on it obey periodic boundary conditions ($\varphi_N=\varphi_0$). The above Lagrangian $L=\int dx\, \mathcal{L}$ is then replaced by the following expression: $$\begin{align}L=\sum^{N-1}_{n=0}\frac{1}{2}a\left\{\dot{\varphi}^2_n-\frac{1}{a^2}(\varphi_{n+1}-\varphi_n)^2-am^2\varphi^2_n\right\}\end{align}$$
The canonical quantization of such a system is done by replacing the classical variables $\varphi_n$ and their conjugate momenta $\pi_n$ by operators, and by imposing the following commutation relations at equal times: $$\begin{align}[\varphi_n,\pi_m]=i\delta_{nm},\quad [\pi_n,\pi_m]=[\varphi_n,\varphi_m]=0\quad (t_n=t_m)\end{align}$$ It is customary in quantum field theory to work in the Heisenberg picture, that is, to give operators a dependence upon time, while keeping the quantum states timeindependent.
The Hamiltonian (4) does not explicitly depend upon position: it is invariant under translations. This motivates the use of discrete Fourier transforms: $$\begin{align}\tilde{\varphi}_k=\frac{1}{\sqrt{N}}\sum^{N-1}_{n=0}e^{-2\pi ikn/N}\varphi_n,\quad \tilde{\pi}_k=\frac{1}{\sqrt{N}}\sum^{N-1}_{n=0}e^{-2\pi ikn/N}\pi_n\end{align}$$ where the index $k$ takes integer values from $0$ to $N - 1$, since $\tilde{\varphi}_{k+N}=\tilde{\varphi}_k$. However, this range is arbitrary, the important point being to restrict summations over $k$ to any range of $N$ consecutive integers. Since $\varphi_n$ and $\pi_n$ are real, the Hermitian conjugates are $$\begin{align}\tilde{\varphi}^\dagger_k=\tilde{\varphi}_{-k}\quad \tilde{\pi}^\dagger_k=\tilde{\pi}_{-k}\end{align}$$
The Fourier modes $\tilde{\varphi}_k$ and $\tilde{\pi}_k$ obey the following commutation rules: $$\begin{align}[\tilde{\varphi}_k,\tilde{\pi}^\dagger_q]=\frac{1}{N}\sum^{N-1}_{m,n=0}e^{-2\pi i(km-qn)/N}[\varphi_m,\pi_n]=\frac{i}{N}\sum^{N-1}_{n=0}e^{-2\pi in(k-q)/N}=i\delta_{kq}\end{align}$$ In terms of these modes, the Hamiltonian (4) becomes $$\begin{align}H=\frac{1}{2}\sum^{N-1}_{k=0}\left\{\frac{1}{a}\tilde{\pi}_k\tilde{\pi}^\dagger_k+a\tilde{\varphi}_k\tilde{\varphi}^\dagger_k\left[m^2+(2/a^2)\left(1-\cos\frac{2\pi K}{N}\right)\right]\right\}\end{align}$$ Since $\tilde{\varphi}_k$ and $\tilde{\pi}_k$ obey canonical commutation relations, this is exactly the Hamiltonian for a system of uncoupled harmonic oscillators, with frequencies $\omega_k$ defined by $$\begin{align}\omega^2_k=m^2+\frac{2}{a^2}\left(1-\cos\frac{2\pi K}{N}\right)\end{align}$$ The inverse lattice spacing here plays the role of the harmonic oscillator's mass. Following the usual methods, we define raising and lowering operators $$\begin{align}a_k=\frac{1}{\sqrt{2a\omega_k}}\left(a\omega_k\tilde{\varphi}_k+i\tilde{\pi}_k\right) ,\quad a^\dagger_k=\frac{1}{\sqrt{2a\omega_k}}\left(a\omega_k\tilde{\varphi}^\dagger_k+i\tilde{\pi}^\dagger_k\right) \end{align}$$ obeying the commutation rules $$\begin{align}[a_k,a^\dagger_q]=\delta_{kq}\end{align}$$ When expressed in terms of these operators, the Hamiltonian takes the form $$\begin{align}H=\frac{1}{2}\sum^{N-1}_{k=0}(a^\dagger_k a_k+a_k a^\dagger_k)\omega_k=\sum^{N-1}_{k=0}(a^\dagger_k a_k+\frac{1}{2})\omega_k\end{align}$$ The ground state $|0\rangle$ of the system is defined by the condition $$\begin{align}a_k|0\rangle=0\quad \forall k\end{align}$$ and the complete set of energy eigenstates is obtained by applying on $|0\rangle$ all possible combinations of raising operators: $$\begin{align}|k_1,k_2,\cdots,k_n\rangle=a^\dagger_{k_1}a^\dagger_{k_2}\cdots a^\dagger_{k_n}|0\rangle\end{align}$$ where the $k_i$ are not necessarily different (as written, these states are not necessarily normalized). The energy of such a state is $$\begin{align}E[k]=E_0+\sum_i \omega_{k_i}\end{align}$$ where $E_0$ is the ground state energy: $$\begin{align}E_0=\frac{1}{2}\sum^{N-1}_{k=0}\omega_k\end{align}$$ When $N$ is large and $ma\ll 1$, $E_0$ behaves like $N/a$.
The time evolution of the operators ak is determined by the Heisenberg relation: $$\begin{align}\dot{a}_k=i[H,a_k]=-i\omega_ka_k\end{align}$$ whose solution is $$\begin{align}a_k(t)=a_k(0)e^{-i\omega_kt}\end{align}$$ From this, (6) and (11) follows the time dependence of the field itself: $$\begin{align}\varphi_n(t)=\sum^{N-1}_{k=0}\sqrt{\frac{2}{Na\omega_k}}\left[e^{-(2\pi kn/N-\omega_kt)}a_k(0)+e^{-i(2\pi kn/N-\omega_kt)}a_k^\dagger(0)\right]\end{align}$$
The continuum limit is obtained by sending the lattice spacing a to zero, and the number $N$ of sites to $\infty$, while keeping the volume $V = Na$ constant. The infrared limit is taken in sending $V$ to $\infty$, while keeping a constant. We now translate the relations found above in terms of continuous field operators. The continuum limits of the field and conjugate momentum are $$\begin{align}\varphi_n\rightarrow \varphi(x)\quad \frac{1}{a}\pi_n\rightarrow \pi(x)=\dot{\varphi}(x)\quad (x=na)\end{align}$$ Sums over sites and Kronecker deltas become $$\begin{align}a\sum^{N-1}_{n=0}\rightarrow \int dx\quad \delta_{nn'}\rightarrow a\delta(x-x')\end{align}$$ Therefore, the canonical commutation relations of the field with its conjugate momentum become $$\begin{align}[\varphi(x),\pi(x')]=i\delta(x-x')\end{align}$$ The discrete Fourier index $k$ is replaced by the physical momentum $p=2\pi k/V$. Sums over Fourier modes and Kronecker deltas in mode indices become $$\begin{align}\frac{1}{V}\sum^{N-1}_{k=0}\rightarrow \int \frac{dp}{2\pi}\quad \delta_{kk'}\rightarrow \frac{2\pi}{V}\delta(p-p')\end{align}$$ We define the continuum annihilation operator and the associated frequency as $$\begin{align}a(p)=a_k\sqrt{V}\quad \omega(p)=\sqrt{m^2+p^2}\end{align}$$ whose commutation relations are therefore $$\begin{align}[a(p),a^\dagger(p')]=(2\pi)\delta(p-p')\end{align}$$ The field $\varphi(x)$ admits the following expansion in terms of the continuum creation and annihilation operators: $$\begin{align}\varphi(x)=\int \frac{dp}{2\pi}\left\{a(p)e^{i(px-\omega(p)t)}+a^\dagger(p)e^{-i(px-\omega(p)t)}\right\}\end{align}$$
The simplest excited states, the so-called elementary excitations, are of the form $a^\dagger(p)|0\rangle$ with energy $$\begin{align}\omega(p)=\sqrt{m^2+p^2}\end{align}$$ This dispersion relation (i.e., the functional relation between energy and momentum) is characteristic of relativistic particles. We thus interpret these elementary excitations as particles of mass m and momentump. The states (15) physically represent a collection of independent particles. The momenta of these particles are conserved separately (they are "good quantum numbers").
$$\begin{align}:a(p)a^\dagger(p):= a^\dagger(p)a(p)\end{align}$$
2.1.2 The Free Fermion
In the context of a free-field theory, and in terms of mode operators $a(p)$ and $a^\dagger(p)$, this property follows from anticommutation relations: $$\begin{align} \{a(p),a^\dagger(q)\}=(2\pi)2\omega_p\delta(p-q)\\ \{a(p),a(q)\}=\{a^\dagger(p),a^\dagger(q)\}=0\nonumber\end{align}$$ where $\{a,b\}=ab+ba$ is the anticommutator. However, the canonical quantization of a field taking its values in the set of real or complex numbers can lead only to commutation relations, as opposed to anticommutation relations.
We apply to Grassmann variables the same canonical formalism as for real or complex variables, except that their anticommuting properties forbid the existence in the Lagrangian of terms quadratic in derivatives. Specifically, let us consider a discrete set $\{\psi_i\}$ of real Grassmann variables with the Lagrangian
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2.2 Path Integrals
2.2.1 System with One Degree of Freedom
$$\begin{align}\end{align}$$
2.2.2 Path Integration for Quantum Fields
$$\begin{align}\end{align}$$
2.3 Correlation Functions
2.3.1 System with One Degree of Freedom
$$\begin{align}\end{align}$$
2.3.2 The Euclidian Formalism
$$\begin{align}\end{align}$$
2.3.3 The Generating Functional
Correlation functions may be formally generated through the so-called generating functional: $$\begin{align}Z[j]=\int [dx(t)]\exp -\left\{S[x(t)]-\int dt\, j(t)x(t)\right\}\end{align}$$ where $j(t)$ is an auxiliary "current" coupled linearly to the dynamical variables $x$. Formula (85) may recast into $$\begin{align}Z[j]=Z[0]\langle \exp \int dt\, j(t)x(t)\rangle=Z[0]\sum^\infty_{n=0}\int dt_1\cdots dt_n\frac{1}{n!}j(t_1)\cdots j(t_n)\langle x(t_1)\cdots x(t_n)\rangle\end{align}$$ or, equivalently, $$\begin{align}\langle x(t_1)\cdots x(t_n)\rangle=Z[0]^{-1}\left. \frac{\delta}{\delta j(t_1)}\cdots \frac{\delta}{\delta j(t_n)}Z[j]\right|_{j=0}\end{align}$$
This definition is easily extended to a quantum field $\phi(x)$. The current is then a function $j(x)$ of Euclidian space-time: $$\begin{align}Z[j]=Z[0]\langle \exp \int d^dx\, j(x)\phi(x)\rangle\end{align}$$
2.3.4 Example: The Free Boson
In two dimensions, the free boson has the following Euclidian action: $$\begin{align}S=\frac{1}{2}g\int d^2x\left\{\partial_\mu\varphi\partial^\mu\varphi+m^2\varphi^2\right\}\end{align}$$ where $g$ is some normalization parameter that we leave unspecified at the moment. We first calculate the two-point function, or propagator: $$\begin{align}K(x,y)=\langle \varphi(x)\varphi(y)\rangle \end{align}$$ If we write the action as $$\begin{align}S=\frac{1}{2}\int d^2x\, d^2y\, \varphi(x)A(x,y)\varphi(y)\end{align}$$ where $A(x,y)=g\delta(x-y)(-\partial^2+m^2)$, the propagator is then $K(x,y)=A^{-1}(x,y)$, or $$\begin{align}g(-\partial^2_x+m^2)K(x,y)=\delta (x-y)\end{align}$$ This follows from a continuous generalization of the results of App. 2.A on Gaussian integrals. This differential equation may also be derived from the quantum equivalent of the equations of motion, as done in Ex. (2.2). Because of rotation and translation invariance, the propagator $K(x, y)$ should depend only on the distance $r=|x-y|$ separating the two points, and we set $K(x,y) = K(r)$. Integrating (99) over $x$ within a disk $D$ of radius $r$ centered around $y$, we find $$\begin{align}1=2\pi g\int^r_0d\rho\, \rho\left(-\frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho K'(\rho))+m^2K(\rho)\right)=2\pi g\left\{-rK'(r)+m^2\int^r_0d\rho\, \rho K(\rho)\right\}\end{align}$$ where $K'(r)=dK/dr$. The massless case ($m=0$) can be solved immediately, the solution being, up to an additive constant, $$\begin{align}K(r)=-\frac{1}{2\pi g}\ln r\end{align}$$ or, in other words, $$\begin{align}\langle \varphi(x)\varphi(y)\rangle=-\frac{1}{4\pi g}\ln (x-y)^2\end{align}$$
The massive case is solved by taking one more derivative with respect to $r$, which leads to the modified Bessel equation of order $0$: $$\begin{align}K''+\frac{1}{r}K'-m^2K=0\end{align}$$ On physical grounds we are interested in solutions that decay at infinity, and therefore $$\begin{align}K(r)=\frac{1}{2\pi g}K_0(mr)\end{align}$$ where $K_0$ is the modified Bessel function of order $0$: $$\begin{align}K_0(x)=\int^\infty_0dt\frac{\cos(xt)}{\sqrt{t^2+1}}\quad (x>0)\end{align}$$ The constant factor $1/2\pi g$ may be checked by taking the limit $r\rightarrow 0$. At large distances (i.e., when $mr\gg 1$) the modified Bessel function decays exponentially and $$\begin{align}K(r)\sim e^{-mr}\end{align}$$ This is also obvious from (103) when the second term is neglected. It is a generic feature of massive fields that correlation functions decay exponentially, with a characteristic length (the correlation length) equal to the inverse mass.
From the elementary Gaussian integral (2.209), it is a simple matter to argue that the generating functional (2.95) for the free boson is equal to $$\begin{align}Z[j]=Z[0]\exp \left\{\frac{1}{2}\int d^dx\, d^dy\, j(x)K(x,y)j(y)\right\}\end{align}$$
2.3.5 Wick's Theorem
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