# Week 4, Monday

## Wick Rotation

We mentioned already the imaginary time path integral, where we make the time formally imaginary. Now that we have seen the Feynman propagator $$\Delta(x-y) = \lim_{\epsilon \to 0^+} \int \frac{d^4k}{(2\pi)^4} \frac{1}{k^2 + m^2 – i \epsilon} e^{ik(x-y)},$$ we can give it a better justification. If we want to deform a time-integral into the complex plane, we must never cross a pole of the integrand. But the propagator does have two poles for complex $t$. Rotating the time contour to go from $-i \infty$ to $+i\infty$ does not cross either pole, whereas rotating the integration contour the other way would have crossed both.

Definition: A Wick Rotation is the analytic continuation of the time-integration contour in the path integral to run from $-i \infty$ to $+i\infty$ along the imaginary axis. It is equivalent to replacing $t\mapsto -i t$.

In practice, nearly every QFT calculation is done after Wick rotation. It changes the space-time signature $(-+++) \mapsto (++++)$, and $e^{iS} \mapsto e^{-S_E}$ with the Euclidean action $$S_E = \int d^4x \left( \frac{1}{2} \partial_\mu\phi \partial^\mu \phi + \frac{1}{2} m^2 \phi^2 + \frac{1}{4!} \lambda \phi^4 \right),$$ which is bounded below as long as the parameters $m^2$, $\lambda$ are such that the vacuum exists. By interpreting it as the energy of some statistical mechanical system, we see that the Wick rotation maps relates the quantum mechanical path integral to the partition function $$\int \mathcal{D}\phi e^{\frac{i}{\hbar}S} \mapsto \int \mathcal{D}\phi e^{-\frac{1}{k_B T}S_E}$$ with $k_B T=\hbar$. This is a deep and often-used connection: Quantum field theory in $(1,d)$-dimensions is statistical mechanics in $(d+1)$-dimensional Euclidean space.

## Lattice Action

In order to simulate QFT on a computer we need truncate the infinite-dimensional space of fields to something finite dimensional. The easiest way to do so is to allow only discrete positions $t, x \in a \mathbb{Z}$ with the lattice spacing $a$. For simplicity and to speed up the computations we will only consider $(1,1)$-dimensions, that is, one time and one space direction.

The Lagrangian contains derivatives, which we need to discretize somehow. There is more than one way of doing so, for example $$\frac{\partial \phi}{\partial \phi} \approx \frac{\phi(t,x+a) – \phi(t,x-a)}{2a} \approx \frac{\phi(t,x+a) – \phi(t,x)}{a} \approx \frac{\phi(t,x) – \phi(t,x-a)}{a}$$ are all approximations that yield the derivative in the limit $a\to 0$. We can multiply the two asymmetric versions to get a symmetric expression for the square, $$\left(\frac{\partial \phi}{\partial \phi}\right)^2 = \frac{1}{a^2} \left( \phi(t, x+a)^2 – 2 \phi(t,x-a) \phi(t,x+a) + \phi(t, x-a)^2 \right).$$ If we furthermore sum over all lattice sites then we can combine the two $\phi^2$ terms. That way, we arrive at the lattice action $$S_E = \sum_{n\in \mathbb{Z}^2} \left[ \frac{1}{2} \sum_{i=1}^d \big(\phi(n+e_i) – \phi(n)\big)^2 + \frac{1}{2} m_L^2 \phi(n)^2 + \frac{1}{4!} \lambda_L \phi(n)^4 \right]$$ where $e_1=(1,0)$ and $e_2=(0,1)$ are the two lattice basis vectors, $m_L^2=a^2 m^2$ is the lattice mass-squared, and $\lambda_L=a^2\lambda$ is the lattice coupling constant. This also matches dimensional analysis, in $d=2$ the field is dimensionless and $m^2$, $\lambda$ have mass dimension $2$. A bit more compact is the alternate expression $$S_E = -\sum_{\text{adjacent }i,j} \phi(n_i) \phi(n_j) + \sum_{n\in\mathbb{Z}^2}\Big[ \underbrace{\left(2+\tfrac{1}{2}m_L^2\right)}_{=\tilde\mu^2} \phi(n)^2 + \underbrace{\frac{1}{4!} \lambda_L}_{=\tilde\lambda} \phi(n)^4 \Big]$$ The continuum limit is $a\to 0$ with $\frac{m_L^2}{\lambda_L}$ fixed.

The computer can of course not keep an infinite number of lattice sites in memory, so we have to approximate the lattice with a finite number $N$ of sites in each direction. It is convenient for the implementation to use periodic boundary conditions, then we do not have to implement separate differentials on the edges and vertices.