# Week 3, Thursday

## Wick’s Theorem For Operators

There is a version of Wick’s theorem directly for operators, which roughly states that the difference between time- and normal-ordered products is a bunch of terms with pairs of operators replaced by the propagator. This replacement process is called a Wick contraction:

Definition: A Wick contraction, usually written as over/under braces or line $\overline{\phi(x) \phi(y)}$ is the process of replacing the product $\phi(x)\phi(y)$ with the Feynman propagator $\tfrac{1}{i}\Delta(x-y)$.

and

Wick’s Theorem The time-ordered product of operators is the normal ordered sum of all contractions, $$T\big(\phi(x_1) \cdots \phi(x_n)\big) = \sum_\text{all contractions} :\left(\text{contracted }\prod \phi(x_i)\right):$$

For example, consider the field at $4$ positions. For brevity, I’ll write $\phi_i = \phi(x_i)$. Then the time ordered product equals the sum of the normal ordering of terms with no contraction (1 term), one contraction (6 terms), and two contractions (3 terms): $$\begin{split} T(\phi_1 \phi_2 \phi_3 \phi_4) =& \; :\phi_1 \phi_2 \phi_3 \phi_4: +\; \overline{\phi_1 \phi_2} :\phi_3 \phi_4: +\; \overline{\phi_1 \phi_3} :\phi_2 \phi_4: +\; \overline{\phi_1 \phi_4} :\phi_2 \phi_3: \\ &\quad +\; \overline{\phi_2 \phi_3} :\phi_1 \phi_4: +\; \overline{\phi_2 \phi_4} :\phi_1 \phi_3: +\; \overline{\phi_3 \phi_4} :\phi_1 \phi_2: \\ &\quad +\; \overline{\phi_1 \phi_2} ~ \overline{\phi_3 \phi_4} + \overline{\phi_1 \phi_3} ~ \overline{\phi_2 \phi_4} + \overline{\phi_1 \phi_4} ~ \overline{\phi_2 \phi_3}. \end{split}$$ We note that the propagator evaluated at zero is infinite, this is where the delta function source term is located after all. So, at least naively, $$\overline{\phi(x) \phi(x)} = \tfrac{1}{i} \Delta(x-x) = \tfrac{1}{i} \Delta(0) = \infty$$ does not make sense. On the other hand, if the coincident fields are normal-ordered as in $$T\big(\phi(x_1) \cdots \phi(x_n) :\phi(y)^m:\big)$$ then it turns out that there is no singularity. That is, the time ordered product can be rewritten, using only Wick’s theorem, into terms involving normal orders and propagators but never the propagator evaluated at zero. An example can be found in the homework. Finally, note that Wick’s theorem only holds for the free field. However, it will turn out to be a useful tool for perturbation theory around a free field theory.

## Interacting QFT

Of course free fields are boring at the end of the day. Any kind of truly interesting quantum field theory has interactions. We have already seen that the coefficient $\lambda_n$ of a $\phi^n$ term in the potential has dimension $4-n$, so only $n=3$ and $n=4$ have non-negative mass dimension. As we will see later, negative dimensional terms will lead to problems with renormalization, so we will avoid them for now.

First, consider a $\tfrac{1}{3!}\lambda_3 \phi^3$ term in the action. This means that the potential $V(\phi) = \tfrac{1}{2} m^2 \phi^2 + \tfrac{1}{3!}\lambda_3 \phi^3$ is unbounded below, regardless of the sign of $\lambda_3$. So even though you might arrange for a metastable vacuum, that is, a local minimum of the potential, sooner or later your field will tunnel through any potential barrier and release an infinite amount of energy as it runs off $\phi(x)\to -\mathop{\mathrm{sign}}(\lambda_3)\infty$. Hence this theory cannot be defined for all times. Nevertheless, one should be able to describe it for small enough time intervals. Furthermore, it is a useful example for any calculation that does not describe the tunneling process, so you will find it in many books as an example. Hence the situation is perhaps not completely hopeless. Still, we will not consider the $\phi^3$ interaction in the following.

This leaves us with a single possible interaction:

Definition: The theory of a single real-valued scalar field $\phi(x)$ and Lagrange density $$\mathcal{L} = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi -\frac{1}{2} m^2 \phi^2 -\frac{1}{4!} \lambda \phi^4$$ is called $\phi^4$ theory. It depends on two real parameters $m^2$ and $\lambda$.

This is also the theory that you get if you take the Standard Model and set all fields to zero except for the Higgs field, so its study is a kind of toy model for the Higgs particle. We note that $m^2$ is just a name for the coefficient of $\phi^2$, and not necessarily the square of a real number. At least classically, we hence distinguish $3$ regimes:

• $\lambda < 0$: Unstable vacuum.
• $\lambda \geq 0$, $m^2 > 0$: Deformed harmonic oscillator.
• $\lambda \geq 0$, $m^2 < 0$: Two distinct minima of the potential. In particular, the minimum is not at $\phi=0$.

It turns out that the quantum theory has the same basic features, although the transition point between the different phases receives quantum corrections.

There are two particularly useful limits of the parameters:

• $\lambda\to 0$ is the free field limit
• $\lambda\to \infty$, $m^2\to -\infty$ creates two infinitely-deep potential wells. The field can no longer take any value, but is constrained to $\phi(x) = \pm \phi_0$ for some some constant $\phi_0\in\mathbb{R}$. This might remind you of a spin-$\tfrac{1}{2}$ particle. In fact, in this limit the $\phi^4$ theory becomes the Ising model.

The bad news is that nobody has found an analytic solution to the $\phi^4$ theory so far, so we will have to do a certain amount of hand-waving to say anything about it. One thing that we can say much about is special limits, for example the Ising model can be solved exactly (at least in $1$ and $2$ dimensions). Similarly, we will have a lot to say about the perturbation expansion around the free field theory. However, there are limits to what perturbation theory can do for you. Essentially, perturbation theory is a Taylor series expansion $$\langle 0|T\prod \phi(x_i)|0\rangle = \sum a_n \lambda^n.$$ For example, we computed $a_0$ in the previous lecture as the sum of all Wick contractions of pairings. Although not physical, we can think of $\lambda \in \mathbb{C}$ and think of the correlators as complex functions. The radius of convergence of the power series should be as far as the first singularity in the complex $\lambda$-plane. But for any negative real $\lambda$, the vacuum is unstable and surely the correlation functions diverge. Hence the radius of convergence ought to be zero. Fortunately, perturbation theory turns out to be much more useful that what this argument suggests. What happens is that the first couple of terms in the series in fact do provide excellent approximations, even though it eventually diverges. But, since we anyways can only compute a limited number of terms in perturbation theory, this is in practice not much of a concern. This phenomenon is known as asymptotic series expansion

# Week 3, Wednesday

The Lagrange density and path integral in field theory is $$\begin{gathered} \mathcal{L} = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi – \frac{1}{2} m^2 \phi^2 + J(x) \phi, \\ Z = \int \mathcal{D}\phi \; e^{i\int d^4x \; \mathcal{L}} = \langle 0 | 0 \rangle, \end{gathered}$$ which equals one if the path integral measure is properly normalized. To get something non-trivial we need, for example, boundary conditions on fields. It turns out that a more fruitful way is to include insert fields into the integrand. By going back to the definition of the path integral as product over time slices, we get $$\int \mathcal{D}\phi \; \phi(x) \phi(y) e^{i\int d^4x \mathcal{L}} = \langle 0 | T \phi(x) \phi(y) |0\rangle$$ and similar. A useful trick is to think of the extra fields as coming from a source term $$Z(J) = \int \mathcal{D}\phi \; e^{i\int d^4x \; [\mathcal{L} + J \phi]}$$ Acting with the functional derivative on the path integral and then setting $J=0$ results in $$\tfrac{1}{i} \frac{\delta}{\delta J(x)} \; \tfrac{1}{i} \frac{\delta}{\delta J(y)} \; Z(J)|_{J=0} = \int \mathcal{D}\phi \; \phi(x) \phi(y) e^{i\int d^4x \mathcal{L}},$$ which is precisely what we want. The analogous formula holds for arbitrary numbers of fields in the integrand.

## Solving the Path Integral

The free Lagrangian is quadratic in $\phi$, so we can again use the Gaussian integral to solve it. In momentum space $$\tilde\phi(k) = \int d^4x e^{-ikx} \phi(x) ,\quad \phi(x) = \int \frac{d^4k}{(2\pi)^4} e^{ikx} \tilde\phi(k),$$ the action including the source term is $$\begin{split} S(J) =&\; \int d^4x \; (\mathcal{L} + J\phi) = \\ =&\; \frac{1}{2} \int \frac{d^4k}{(2\pi)^4} \left[ \frac{\tilde{J}(k) \tilde{J}(-k)}{k^2+m^2-i\epsilon} – \tilde\chi(k) (k^2+m^2) \tilde{\chi}(-k) \right] \end{split}$$ where $$\tilde\chi(k) = \tilde\phi(k) – \frac{\tilde{J}(k)}{k^2 + m^2}$$ is the “midpoint” variable of the quadratic exponent. The path integral measure is invariant under the shift, $\mathcal{D}\phi = \mathcal{D}\chi$, so we obtain $$\begin{split} Z(J) =&\; \exp\left[\frac{i}{2} \int \frac{d^4k}{(2\pi)^4} \frac{\tilde{J}(k)\tilde{J}(-k)}{k^2 + m^2} \right] \\ &\qquad \times \int \mathcal{D}\chi \exp\left[ i\int d^4x \left(-\tfrac{1}{2} \partial_\mu \chi \partial^\mu \chi – \tfrac{1}{2} m^2\chi^2\right) \right] \\ =&\; \exp\left[ \frac{i}{2} \int d^4x \; d^4y \; J(x) \Delta(x-y) J(y) \right] \end{split}$$ where $\Delta$ is again the Feynman propagator.

Hence, the path integral derivation of the propagator is $$\begin{split} \langle 0|T \phi(x) \phi(y) |0\rangle =&\; \tfrac{1}{i} \frac{\delta}{\delta J(x)} \; \tfrac{1}{i} \frac{\delta}{\delta J(y)} \; Z(J)|_{J=0} \\ =&\; \left( \tfrac{1}{i} \Delta(x-y) + O(J) \right) Z(J) |_{J=0} = \tfrac{1}{i} \Delta(x-y) \end{split}$$ which of course matches what we found in the previous lecture using the canonical quantization. The generalization to an arbitrary number of fields is known as Wick’s Theorem: $$\langle 0 | T \phi(x_1) \phi(x_2) \cdots \phi(x_{2n}) |0\rangle = \frac{1}{i^n} \sum_\text{pairings} \Delta(x_{i_1} – x_{i_2}) \cdots \Delta(x_{i_{2n-1}} – x_{i_{2n}})$$

# Week 3, Monday

## Green’s Functions

The Klein-Gordon equation describes just propagating plane waves. A more interesting question is how does the field react to external influences. We can model this with a source term $$(-\partial_\mu \partial^\mu + m^2) \phi = J(x) = J(t, \vec{x})$$ This is an inhomogeneous linear differential equation, so any solution is the sum of a solution of the homogeneous differential equation plus one particular solution of the inhomogeneous equation.

Definition: A solution of $$(-\partial_x^2 + m^2) G(x,y) = \delta^4(x-y)$$ is called a Green’s function.

In momentum space $$\tilde\phi(k) = \int d^4x e^{-ikx} \phi(x) ,\quad \phi(x) = \int \frac{d^4k}{(2\pi)^4} e^{ikx} \tilde\phi(k)$$ we can formally solve for the Green’s function. However, the Fourier transformation integral of the formal solution back to position space diverges because of poles at $k_0 = \pm \sqrt{\vec{k}+ m^2} = \pm \omega$. The correct way of dealing with the Fourier transform of distributions requires a choice of integration contour in the complex $k_0\in\mathbb{C}$ plane that evades the two poles. The two natural contours are

• Just above the two poles at $k_0=\pm \omega$, this is the retarded Green’s function. It vanishes if $x$ is in the past of $y$.
• Just below the two poles, this is the advanced Green’s function. It vanishes if $x$ is in the future of $y$.

In addition, any Green’s function vanishes if $x$ and $y$ are space-like separated. Instead of defining the integration in prose, we can also use the limit of a small positive $\epsilon$ to describe the integration contour as $$G_{\text{Adv}/\text{Ret}}(x, y) = \lim_{\epsilon\to 0^+} \int \frac{d^4k}{(2\pi)^4} \frac{1}{-(k_0 \pm i \epsilon)^2 + \vec{k}^2 + m^2} e^{ikx}$$ The picture below is the propagator in the $x_2=x_3=0$ plane. At the light cone front $x_0 = \pm x_1$ the propagator is a delta function, which is not drawn.

## Propagator of the Free Field

The propagator of a free scalar field is also a Green’s function: $$\begin{split} \langle 0| \phi(x) \phi(y) |0\rangle =&\; \int \widetilde{dk} \int \widetilde{dk}’ \langle 0| a(\vec{k}) a^\dagger(\vec{k}’) |0\rangle \; e^{i(kx-k’y)} \\ =&\; \int \frac{d^3\vec{k}}{(2\pi)^3} \frac{1}{2\omega} e^{ik(x-y)} \\ =&\; \frac{1}{2} \int \frac{d^3\vec{k}}{(2\pi)^3} \oint \frac{dk_0}{2\pi i} \frac{1}{k^2+m^2} e^{ik(x-y)} \\ =&\; \frac{1}{2i} \Big( G_\text{Ret}(x-y) – G_\text{Adv}(x-y) \Big) \end{split}$$ where the $\oint$-contour consists of two small circles around the two poles, which is equals to the retarded Green’s function contour minus the advanced Green’s function contour.

## Feynman Propagator

There is a third choice of integration contour which turns out to be the most useful for quantum field theory. of course we need the vacuum expectation value of time-ordered products in quantum field theory, so let us define

Definition: The Green’s function $\Delta$ defined as $$\tfrac{1}{i} \Delta(x-y) = \langle 0 | T \phi(x) \phi(y) |0\rangle$$ is called the Feynman propagator.

In terms of integration contour in the $k_0\in \mathbb{C}$ plane, this is the path that goes below at $k_0=-\omega$ and above at $k_0=\omega$

Depending on whether $x$ is in the future or the past of $y$, this contour selects the advanced or retarded Green’s function. Finally, in terms of an algebraic $i\epsilon$ prescription, the Feynman propagator is $$\Delta(x-y) = \lim_{\epsilon \to 0^+} \int \frac{d^4k}{(2\pi)^4} \frac{1}{k^2 + m^2 – i \epsilon} e^{ik(x-y)}$$