3 Interacting scalar field

Now that we understand free quantum fields, we can turn our attention to interacting fields. Unfortunately, there are only a few QFTs that permit an exact analytic solution such as the Schwinger model, a two-dimensional description of a photon and fermion. Most of these are only useful in very limited circumstances or as toy models. Another approach is to solve the theory numerically, usually by placing it on a finite lattice. While this allows the study of realistic models like quantum chromodynamics (QCD), it is extremely complicated to do from scratch.

Here we will instead focus on perturbation theory, i.e. we will consider the interaction to be a small perturbation on top of the free field which we already understand. This means we split the full Lagrangian

\displaystyle\mathcal{L}=\mathcal{L}_{0}+\mathcal{L}_{I}

(124)

into a free theory $\mathcal{L}_{0}$ and an interaction term $\mathcal{L}_{I}$ which is hopefully small. Perturbation theory allows us to study theories like quantum electrodynamics (QED) or QCD (at least in the high-energy limit). The calculation of the anomalous magnetic moment $g-2$ of the electron I have mentioned at the beginning of the course is (almost completely) done this way. Similarly, calculations that are used at Large Hadron Collider (LHC) and that, for example, helped discover the Higgs boson in 2012, are also mostly done using perturbation theory. This is valid because the coupling strength between different particles, i.e. $\mathcal{L}_{I}$ , is small in these regimes. In the example for $g-2$ , we expand in $\alpha_{em}=1/137$ and the theory value includes effects up to $\alpha^{5}$ . At the LHC we usually expand in the strong coupling $\alpha_{s}\approx 0.1$ and a few cutting-edge calculations have reached $\alpha_{s}^{3}$ accuracy.

In this chapter we will only consider the self-interaction of a scalar particle $\phi$ . Nature has a fundamental scalar particle, namely the Higgs boson, and the theory we develop here can (almost) be used to describe the Higgs boson. The difference is that the real Higgs field is not a single real-valued field (like the one we discussed) but a doublet of two complex-valued fields. The relevant part of the Lagrangian is usually written as (like on that mug)

\displaystyle\mathcal{L}=|\partial_{\mu}\phi|^{2}-V(\phi)=|\partial_{\mu}\phi|% ^{2}+\mu^{2}\phi^{\dagger}\phi-\frac{\lambda}{4!}(\phi^{\dagger}\phi)^{2}\,.

(125)

We will modify this slightly for our single real-valued scalar field

\displaystyle\mathcal{L}=\underbrace{\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac% {1}{2}m^{2}\phi^{2}}_{\mathcal{L}_{0}}-\underbrace{\frac{\lambda}{4!}\phi^{4}}% _{\mathcal{L}_{i}}\,.

(126)

I want to stress that you could repeat everything we are about to do for the more general case of the Higgs boson and I would encourage you to try. This theory is often called the $\phi^{4}$ model due to its interaction term. Another related model would be the $\phi^{3}$ theory which would have $\mathcal{L}_{i}=\lambda/3!\ \phi^{3}$ which can serve as a simplified model of QED.

It is still possible to derive CCRs for an interacting Lagrangian but we will not be able to write $\phi$ in terms of ladder operators because this relied on the free field’s EoM. The important concept behind perturbative QFT is that the interaction is not just small but usually also short-ranged. This means that at the beginning and end of the experiment, we can consider the particles practically free. For example at the $27\,{\rm km}$ big LHC which has $25\,{\rm m}$ big detectors, the actual perturbative interaction takes place in a region that is smaller than $10^{-18}\,{\rm m}$ (based on a typical hard interaction scale of $500\,{\rm GeV}$ ). For $t\to\pm\infty$ the particles involved in the collision are just too far away from each other to feel each other’s influence, meaning that we can use the free-particle solution in this regime. We now need to try and find a way to formalise this.

3.1 Interaction picture

In Section 2.1 we have seen the Heisenberg and Schrödinger pictures. In the former, the time dependence fully resides in the operators while in the latter it is in the states. We will now develop a third picture, the interaction picture that splits the time dependence between the two: the operators will follow the free Hamiltonian $H_{0}$ ’s Heisenberg equation (70) while the states have a time dependence from the interaction Hamiltonian $\mathcal{H}_{I}$ . From (124), it follows immediately that

\displaystyle H=H_{0}+H_{I}=H_{0}-\int{\rm d}^{3}\mathcal{L}_{I}\,.

(127)

Recall how we wrote

	$\displaystyle\|{\Psi_{H}}\rangle$	$\displaystyle=U^{-1}(t)\|{\Psi_{S}(t)}\rangle=U^{\dagger}(t)\|{\Psi_{S}(t)}% \rangle=\|{\Psi_{S}(0)}\rangle\,.$		(67)
	$\displaystyle O_{H}$	$\displaystyle=U^{\dagger}(t)O_{S}U(t)\,,$		(68)

with

\displaystyle i\frac{\partial U}{\partial t}=H_{S}U\,.

(66)

Rather than taking the full $H$ in (66) we now only use the free $H_{0}$ to define

	$\displaystyle\|{\Psi_{I}(t)}\rangle$	$\displaystyle={\rm e}^{iH_{0}t}\|{\Psi_{S}(t)}\rangle$		(128)
	$\displaystyle O_{I}(t)$	$\displaystyle={\rm e}^{iH_{0}t}O_{S}{\rm e}^{iH_{0}t}$		(129)

The time evolution of $O_{I}$ is still governed by the free Heisenberg equation (70)

\displaystyle i\frac{{\rm d}O_{I}}{{\rm d}t}=[O_{I},H_{0}]\,,

(130)

while the states follow a modified Schrödinger equation

\displaystyle\begin{split}i\frac{{\rm d}}{{\rm d}t}|{\Psi_{I}(t)}\rangle&=i% \frac{{\rm d}}{{\rm d}t}\Big{(}{\rm e}^{iH_{0}t}|{\Psi_{S}(t)}\rangle\Big{)}=-% {\rm e}^{iH_{0}t}H_{0}|{\Psi_{S}(t)}\rangle+{\rm e}^{iH_{0}t}H|{\Psi_{S}(t)}% \rangle={\rm e}^{iH_{0}t}H_{I}{\rm e}^{-iH_{0}t}|{\Psi_{I}(t)}\rangle\\ &=\tilde{H}_{I}|{\Psi_{I}(t)}\rangle\,,\end{split}

(131)

where we have used the Schrödinger picture Schrödinger equation for $|{\Psi_{S}(t)}\rangle$ and defined the interaction picture interaction Hamiltonian $\tilde{H}_{I}$ .

These rules tell us how to transform from the Schrödinger picture to the interaction picture. We still need to translate between the Heisenberg and interaction pictures. At some instant $t_{0}$ , we define the pictures to be identical. At any other time $t$ , we translate via the Schrödinger picture

\displaystyle\phi_{H}(t)=U^{\dagger}(t,t_{0})\phi_{I}(t)U(t,t_{0})\qquad\text{% with}\qquad U(t,t_{0})={\rm e}^{iH_{0}(t-t_{0})}{\rm e}^{-iH(t-t_{0})}\,.

(132)

It is easy to see that the time evolution of $U(t,t_{0})$ is

\displaystyle i\frac{{\rm d}}{{\rm d}t}U(t,t_{0})=\tilde{H}_{I}U(t,t_{0})\,.

(133)

3.2 The $\mathcal{S}$ matrix

We are now ready to describe a scattering process. The time evolution of the field operator $\phi$ is given by $U(t,t_{0})$ . Let us take the limit $t_{0}\to-\infty$ , well before the scattering takes place. This is where we require the interaction picture and Heisenberg pictures to agree. The field operators in this limit are just the free field which we refer to as the “in” contribution and write in terms of ladder operators

\displaystyle\phi_{\rm in}=\lim_{t\to-\infty}\phi(t)=\int\frac{{\rm d}^{3}k}{(% 2\pi)^{3}}\frac{1}{\sqrt{2E_{\vec{k}}}}\Big{[}a_{\rm in}(\vec{k}){\rm e}^{-ik% \cdot x}+a_{\rm in}^{\dagger}(\vec{k}){\rm e}^{ik\cdot x}\Big{]}\,.

(134)

Note that $\phi_{\rm in}$ still has the time dependence from the free Hamiltonian. This is the case despite the limit since there is still the dynamics of the free field. At a later time, including all the way through the scattering, we have

\displaystyle\phi(t)=U^{\dagger}(t,-\infty)\phi_{\rm in}U(t,-\infty)\,.

(135)

Experimentally, we do not observe $\phi(t)$ but rather the outcome of the scattering in the far future $t\to\infty$

\displaystyle\phi_{\rm out}=\lim_{t\to+\infty}\phi(t)=U^{\dagger}(+\infty,-% \infty)\phi_{\rm in}U(+\infty,-\infty)\,.

(136)

This “out” field is once again free and we can write it again in terms of ladder operators

\displaystyle\phi_{\rm out}=\lim_{t\to+\infty}\phi(t)=\int\frac{{\rm d}^{3}k}{% (2\pi)^{3}}\frac{1}{\sqrt{2E_{\vec{k}}}}\Big{[}a_{\rm out}(\vec{k}){\rm e}^{-% ik\cdot x}+a_{\rm out}^{\dagger}(\vec{k}){\rm e}^{ik\cdot x}\Big{]}\,.

(137)

Both $\phi_{\rm in}$ and $\phi_{\rm out}$ are free field solutions but they are different free field solutions. This is because, due to the scattering, the ladder operators $a_{\rm in}$ and $a_{\rm out}$ are different. We refer to these states as the asymptotic states to indicate the limit $t\to\pm\infty$ .

The relation between the two sets of asymptotic states is governed by the time evolution operator $U$ . For simplicity, let us define the $\mathcal{S}$ matrix

\displaystyle\mathcal{S}=U(+\infty,-\infty)

(138)

Specifically,

\displaystyle a_{\rm out}(\vec{k})=\mathcal{S}^{\dagger}a_{\rm in}(\vec{k})% \mathcal{S}\qquad\text{and}\qquad a_{\rm out}(\vec{k})^{\dagger}=\mathcal{S}^{% \dagger}a_{\rm in}^{\dagger}(\vec{k})\mathcal{S}\,.

(139)

This means that $\mathcal{S}$ also transform between in and out states. We begin our experiment with a prepared in state $|{\rm in}\rangle_{i}$ by applying $a^{\dagger}_{\rm in}$ on the vacuum. During the scattering this gets transformed into an out state $|{\rm out}\rangle_{o}$ which is made up through $a^{\dagger}{\rm out}$ . We will use a subscripts $i$ and $o$ to indicate the ladder operators we have used. These two states are related through the $\mathcal{S}$ matrix

\displaystyle|{\rm out}\rangle_{o}=\mathcal{S}^{\dagger}|{\rm in}\rangle_{i}\,.

(140)

To understand the scattering process we first need write $|{\rm in}\rangle_{i}$ in terms of the basis of the out states $|{n}\rangle_{o}$

\displaystyle|{\rm in}\rangle_{i}=\sum_{n}c_{n}|{n}\rangle_{o}\,.

(141)

Experimentally we will measure the transition probability between our prepared $|{\rm in}\rangle_{i}$ and a given out state $|{n}\rangle_{o}$

\displaystyle P\sim\big{|}\,{}_{o}\langle{n}|{\rm in}\rangle_{i}\,\big{|}^{2}=% \big{|}\,{}_{i}\langle{n}|\mathcal{S}|{\rm in}\rangle_{i}\,\big{|}^{2}=\big{|}% \,{}_{o}\langle{n}|\mathcal{S}|{\rm in}\rangle_{o}\,\big{|}^{2}\,.

(142)

This means our goal will be to find an expression for the $\mathcal{S}$ matrix.

To do this, we would first need to find $U$ to use (138). The definition (132) is not very helpful because of how complicated it is. Instead, we will use the differential equation (133) which defines this solution in the first place. Integrating from $t_{0}=-\infty$ to $t$

\displaystyle U(t,-\infty)=1-i\int_{-\infty}^{t}{\rm d}t_{1}\ \tilde{H}_{I}(t_% {1})\cdot U(t_{1},-\infty)\,.

(143)

Note that, because we fixed $t_{0}=-\infty$ , the interaction Hamiltonian $\tilde{H}_{I}$ is defined in terms of in states. Unfortunately, this expression still has a $U$ on the right-hand side so let us iterate this

	$\displaystyle U(t,-\infty)$	$\displaystyle=1-i\int_{-\infty}^{t}{\rm d}t_{1}\ \tilde{H}_{I}(t_{1})+(-i)^{2}% \int_{-\infty}^{t}{\rm d}t_{1}\int_{-\infty}^{t_{1}}{\rm d}t_{2}\ \tilde{H}_{I% }(t_{1})\cdot\tilde{H}_{I}(t_{2})\cdot U(t_{2},-\infty)$		(144)
		$\displaystyle=1+\sum_{n=1}^{\infty}(-i)^{n}\int_{-\infty}^{t}{\rm d}t_{1}\int_% {-\infty}^{t_{1}}{\rm d}t_{2}\cdots\int_{-\infty}^{t_{n-1}}{\rm d}t_{n}\ % \tilde{H}_{I}(t_{1})\cdot\tilde{H}_{I}(t_{2})\cdots\tilde{H}_{I}(t_{n})\,.$		(145)

Per our construction of the interaction picture, $\tilde{H}_{I}$ only contains the interaction and not the dynamics of the free field. For example, in the theory we defined in (126), we had $\tilde{H}_{I}\sim\lambda$ . Since we further requested that the coupling $\lambda$ of interaction is small, we would be justified to assume that $\tilde{H}_{I}(t_{1})\cdot\tilde{H}(t_{2})\sim\lambda^{2}$ is smaller than $\tilde{H}_{I}(t_{1})\sim\lambda$ . We therefore often choose to truncate the summation at a finite $n$ (in practice is this rarely more than $n=2$ or $n=3$ because of the complexity of the calculation).

One important feature of the iterated integrals in (145) is that we go further into the past in the product of $\tilde{H}_{I}$ since

\displaystyle t_{n}\leq t_{n-1}\leq\cdots\leq t_{1}\leq t\,,

(146)

which makes for awkward integration boundaries. Instead, let us define the time-ordered product similar to the normal ordering we have used before. Specifically,

\displaystyle T\big{\{}O(t_{1})\cdot O(t_{2})\cdots O(t_{n})\big{\}}=O(t_{\rho% (1)})\cdot O(t_{\rho(2)})\cdots O(t_{\rho(n)})\,,

(147)

where $\rho$ is the permutation of $\{1,...,n\}$ such that time is ordered, i.e.

\displaystyle t_{\rho(i)}\geq t_{\rho(j)}\qquad\text{if}\qquad i<j\,.

(148)

We can now change the integration domain to $(-\infty,t]$ for each integral at the cost of a factor of $n!$ . Explicitly for $n=2$ , we split the integral into two equal pieces and swap $t_{1}\leftrightarrow t_{2}$

$\displaystyle\int_{-\infty}^{t}{\rm d}t_{1}\int_{-\infty}^{t_{1}}{\rm d}t_{2}% \ \tilde{H}_{I}(t_{1})\cdot\tilde{H}_{I}(t_{2})=$
$\displaystyle\frac{1}{2}\int_{-\infty}^{t}{\rm d}t_{1}\int_{-\infty}^{t_{1}}{% \rm d}t_{2}\ \tilde{H}_{I}(t_{1})\cdot$	$\displaystyle\tilde{H}_{I}(t_{2})+\frac{1}{2}\int_{-\infty}^{t}{\rm d}t_{2}% \int_{-\infty}^{t_{2}}{\rm d}t_{1}\ \tilde{H}_{I}(t_{2})\cdot\tilde{H}_{I}(t_{% 1})$
$\displaystyle=$	$\displaystyle\frac{1}{2}\int_{-\infty}^{t}{\rm d}t_{1}{\rm d}t_{2}\ T\big{\{}% \tilde{H}_{I}(t_{1})\cdot\tilde{H}_{I}(t_{2})\big{\}}\,.$	(149)

Doing the same for all orders, we can rewrite $U$ as

	$\displaystyle U(-\infty,t)$	$\displaystyle=1+\sum_{n=1}^{\infty}\frac{(-i)^{n}}{n!}\int_{-\infty}^{t}{\rm d% }t_{1}\cdots{\rm d}t_{n}\ T\big{\{}\tilde{H}_{I}(t_{1})\cdot\tilde{H}_{I}(t_{2% })\cdots\tilde{H}_{I}(t_{n})\big{\}}$		(150)
		$\displaystyle\equiv T\exp\Bigg{(}-i\int_{-\infty}^{t}{\rm d}t^{\prime}\ \tilde% {H}_{I}(t^{\prime})\Bigg{)}\,.$		(151)

Here we have introduce the time-ordered exponential as a short-hand. We can make one more simplification by taking the limit $t\to\infty$ and by realising that

\displaystyle\tilde{H}_{I}=\int{\rm d}^{3}x\ \mathcal{H}_{I}=-\int{\rm d}^{3}x% \ \mathcal{L}_{I}

(152)

to arrive at our most compact solution for the $\mathcal{S}$ matrix

\displaystyle\mathcal{S}=T\exp\Bigg{(}i\int{\rm d}x^{4}\mathcal{L}_{I}\Bigg{)}% =T{\rm e}^{iS_{I}}

(153)

with the interaction part of the action $S_{I}$ . Looking at this you might think there is some deep interpretation of this expression in terms of the action, similar to the principle of least action we had in the classical case. And you would be right to think this, it is possible to perform the entire quantisation procedure for free and interacting fields by defining the path integral

\displaystyle Z=\int\mathcal{D}\phi\ {\rm e}^{iS[\phi]/\hbar}\,,

(154)

which integrates over all possible field configurations $\phi$ and weights them according to ${\rm e}^{iS[\phi]/\hbar}$ . In the classical limit $\hbar\to 0$ only the field configuration of the least action contributes to the integral. While very elegant, the path-integral formalism is more complicated since we would have to define what $\mathcal{D}\phi$ means. Therefore, we will not use this method going forward instead relying on the canonical quantisation we have used so far.

3.3 Wick theorem

If we want to make a prediction about a scattering process we need to calculate $\mathcal{S}$ matrix elements to a given order. This means calculating correlators like

\displaystyle{}_{i}\langle{\rm out}|\int{\rm d}^{4}x_{1}\cdots{\rm d}^{4}x_{n}% \ T\big{\{}\mathcal{L}_{I}(x_{1})\cdots\mathcal{L}_{I}(x_{n})\big{\}}|{\rm in}% \rangle_{i}\,.

(155)

Since both $\mathcal{L}_{I}$ and the external states involves a number of $\phi$ fields, we want to be able to calculate general objects like

\displaystyle T\big{\{}\phi(x_{1})\cdots\phi(x_{m})\big{\}}\,.

(156)

We have already seen a simple case of this with the Feynman propagator $D_{F}=\langle{0}|T\{\phi(x)\phi(y)\}|{0}\rangle$ in (123). To make calculating this easier, let us define

\displaystyle\phi_{I}=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{% \vec{k}}}}\Big{[}\underbrace{a(\vec{k}){\rm e}^{-ik\cdot x}}_{\to\phi^{(+)}}+% \underbrace{a^{\dagger}(\vec{k}){\rm e}^{+ik\cdot x}}_{\to\phi^{(-)}}\Big{]}=% \phi^{(+)}(x)+\phi^{(-)}(x)\,.

(157)

This decomposition into positive ( $\phi^{(+)}$ ) and negative ( $\phi^{(-)}$ ) frequency modes is helpful because

\displaystyle\phi^{(+)}|{0}\rangle=\langle{0}|\phi^{(-)}=0\,.

(158)

It also makes it easier to define normal ordering which moves the $a$ , and therefore $\phi^{(+)}$ , to the right of the $a^{\dagger}$ , and therefore $\phi^{(-)}$ . It follows that the vev of a normal-ordered list of fields is zero

\displaystyle\langle{0}|\ :\phi(x_{1})\cdots\phi(x_{m}):\ |{0}\rangle=0\,.

(159)

To see why this is so useful, consider again the two-particle case $m=2$ that we considered when defining the Feynman propagator. For $x\neq y$ ,

$\displaystyle T\{\phi_{I}(x)\phi_{I}(y)\}$	$\displaystyle=\begin{cases}\phi^{(+)}(x)\phi^{(+)}(y)+\phi^{(+)}(x)\phi^{(-)}(% y)+\phi^{(-)}(x)\phi^{(+)}(y)+\phi^{(-)}(x)\phi^{(-)}(y)&x^{0}>y^{0}\\ \phi^{(+)}(y)\phi^{(+)}(x)+\phi^{(+)}(y)\phi^{(-)}(x)+\phi^{(-)}(y)\phi^{(+)}(% x)+\phi^{(-)}(y)\phi^{(-)}(x)&x^{0}<y^{0}\end{cases}$
	$\displaystyle=\begin{cases}\phi^{(+)}(x)\phi^{(+)}(y)+\phi^{(-)}(y)\phi^{(+)}(% x)+\phi^{(-)}(x)\phi^{(+)}(y)+\phi^{(-)}(x)\phi^{(-)}(y)&x^{0}>y^{0}\\ \phi^{(+)}(y)\phi^{(+)}(x)+\phi^{(-)}(x)\phi^{(+)}(y)+\phi^{(-)}(y)\phi^{(+)}(% x)+\phi^{(-)}(y)\phi^{(-)}(x)&x^{0}<y^{0}\end{cases}$
	$\displaystyle\qquad\qquad+\begin{cases}[\phi^{(+)}(x),\phi^{(-)}(y)]&x^{0}>y^{% 0}\\ [\phi^{(+)}(y),\phi^{(-)}(x)]&x^{0}<y^{0}\end{cases}\,.$	(160)

Every term expect for the commutator is now a normal-ordered product of interaction-picture operators. The commutator is the only term with a non-vanishing vev. Because the interaction-picture fields $\phi$ follow the time evolution of the free Hamiltonian, we can use what we discovered in the previous section. Especially, we can use that the commutator corresponds to the Feynman propagator $D_{F}(x-y)$ .

To simplify our notation, let us define a Wick contraction as

\displaystyle\includegraphics{bmlimages/notes-3.svg}\bml@image@depth{3}% \bmlDescription{A Wick contraction of two terms \phi(x) and \phi(y)}=\begin{% cases}[\phi^{(+)}(x),\phi^{(-)}(y)]&x^{0}>y^{0}\\ [\phi^{(+)}(y),\phi^{(-)}(x)]&x^{0}<y^{0}\end{cases}=D_{F}(x-y)\,,

={[ϕ(+)⁢(x),ϕ(−)⁢(y)]x0>y0[ϕ(+)⁢(y),ϕ(−)⁢(x)]x0<y0=DF⁢(x−y),\displaystyle\includegraphics{bmlimages/notes-3.svg}\bml@image@depth{3}% \bmlDescription{A Wick contraction of two terms \phi(x) and \phi(y)}=\begin{% cases}[\phi^{(+)}(x),\phi^{(-)}(y)]&x^{0}>y^{0}\\ [\phi^{(+)}(y),\phi^{(-)}(x)]&x^{0}<y^{0}\end{cases}=D_{F}(x-y)\,,

(161)

to indicate which two terms are part of the propagator. Here we drop the $I$ subscript and assume that Wick-contracted terms are always in the interaction picture

\displaystyle T\{\phi(x)\phi(y)\}=\ :\phi(x)\phi(y):\ +\includegraphics{% bmlimages/notes-4.svg}\bml@image@depth{4}\bmlDescription{A Wick contraction of% two terms \phi(x) and \phi(y)}\,.

.\displaystyle T\{\phi(x)\phi(y)\}=\ :\phi(x)\phi(y):\ +\includegraphics{% bmlimages/notes-4.svg}\bml@image@depth{4}\bmlDescription{A Wick contraction of% two terms \phi(x) and \phi(y)}\,.

(162)

This is the simplest case of the Wick theorem. The more general case is

\displaystyle T\{\phi(x_{1})\phi(x_{2})\cdots\phi(x_{m})\}=\ :\phi(x_{1})\phi(% x_{2})\cdots\phi(x_{m})+\text{all possible contractions}\ :\,.

(163)

To calculate time-ordered products like this we need to sum over all possible ways of grouping fields into pairs using the Wick contractions. As an example, let us consider the case of four fields $m=4$

\displaystyle\begin{split}T\{\phi_{1}\phi_{2}\phi_{3}\phi_{4}\}=\ :\ &% \includegraphics{bmlimages/notes-5.svg}\bml@image@depth{5}{\bmlDescription{A % Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. }}+\includegraphics{bmlimages/notes-6.svg}\bml@image@depth{6}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3, D: \phi_4. A is connected to B.}}+\includegraphics{bmlimages/notes% -7.svg}\bml@image@depth{7}{\bmlDescription{A Wick contraction involving three % terms A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is connected to C.}}+% \includegraphics{bmlimages/notes-8.svg}\bml@image@depth{8}{\bmlDescription{A % Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. A is connected to D.}}\\ &+\includegraphics{bmlimages/notes-9.svg}\bml@image@depth{9}{\bmlDescription{A% Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. B is connected to C.}}+\includegraphics{bmlimages/notes-10.svg}% \bml@image@depth{10}{\bmlDescription{A Wick contraction involving three terms % A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. B is connected to D.}}+% \includegraphics{bmlimages/notes-11.svg}\bml@image@depth{11}{\bmlDescription{A% Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. C is connected to D.}}\\ &+\includegraphics{bmlimages/notes-12.svg}\bml@image@depth{12}{\bmlDescription% {A Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. A is connected to B.C is connected to D.}}+\includegraphics{bmlimages/% notes-13.svg}\bml@image@depth{13}{\bmlDescription{A Wick contraction involving% three terms A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is connected to C.B% is connected to D.}}+\includegraphics{bmlimages/notes-14.svg}\bml@image@depth% {14}{\bmlDescription{A Wick contraction involving three terms A: \phi_1, B: % \phi_2, C: \phi_3, D: \phi_4. B is connected to C.A is connected to D.}}\ :\,,% \end{split}

:,\displaystyle\begin{split}T\{\phi_{1}\phi_{2}\phi_{3}\phi_{4}\}=\ :\ &% \includegraphics{bmlimages/notes-5.svg}\bml@image@depth{5}{\bmlDescription{A % Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. }}+\includegraphics{bmlimages/notes-6.svg}\bml@image@depth{6}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3, D: \phi_4. A is connected to B.}}+\includegraphics{bmlimages/notes% -7.svg}\bml@image@depth{7}{\bmlDescription{A Wick contraction involving three % terms A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is connected to C.}}+% \includegraphics{bmlimages/notes-8.svg}\bml@image@depth{8}{\bmlDescription{A % Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. A is connected to D.}}\\ &+\includegraphics{bmlimages/notes-9.svg}\bml@image@depth{9}{\bmlDescription{A% Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. B is connected to C.}}+\includegraphics{bmlimages/notes-10.svg}% \bml@image@depth{10}{\bmlDescription{A Wick contraction involving three terms % A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. B is connected to D.}}+% \includegraphics{bmlimages/notes-11.svg}\bml@image@depth{11}{\bmlDescription{A% Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. C is connected to D.}}\\ &+\includegraphics{bmlimages/notes-12.svg}\bml@image@depth{12}{\bmlDescription% {A Wick contraction involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: % \phi_4. A is connected to B.C is connected to D.}}+\includegraphics{bmlimages/% notes-13.svg}\bml@image@depth{13}{\bmlDescription{A Wick contraction involving% three terms A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is connected to C.B% is connected to D.}}+\includegraphics{bmlimages/notes-14.svg}\bml@image@depth% {14}{\bmlDescription{A Wick contraction involving three terms A: \phi_1, B: % \phi_2, C: \phi_3, D: \phi_4. B is connected to C.A is connected to D.}}\ :\,,% \end{split}

(164)

where we use $\phi_{i}\equiv\phi(x_{i})$ to save space. We already know how to evaluate Wick-contracted terms using $D_{F}$ , i.e.

\begin{split}:\,\includegraphics{bmlimages/notes-15.svg}\bml@image@depth{15}{% \bmlDescription{A Wick contraction involving five terms A: \phi_1 \cdots\phi_{i-1}\cdot, B: \phi_i, C: \cdot\phi_{i+1} \cdots\phi_{j-1}\cdot, D: \phi_j, D: \cdot\phi_{j+1} \cdots\phi_m . B is connected to D.}}:\,=\qquad\qquad\qquad\\ D_{F}(x_{i}-x_{j})\,:\phi_{1}\cdots\phi_{i-1}\cdot\phi_{i+1}\cdots\phi_{j-1}% \cdot\phi_{i+1}\cdots\phi_{m}:\,.\end{split}

:=DF⁢(xi−xj):ϕ1⁢⋯⁢ϕi−1⋅ϕi+1⁢⋯⁢ϕj−1⋅ϕi+1⁢⋯⁢ϕm:.\begin{split}:\,\includegraphics{bmlimages/notes-15.svg}\bml@image@depth{15}{% \bmlDescription{A Wick contraction involving five terms A: \phi_1 \cdots\phi_{i-1}\cdot, B: \phi_i, C: \cdot\phi_{i+1} \cdots\phi_{j-1}\cdot, D: \phi_j, D: \cdot\phi_{j+1} \cdots\phi_m . B is connected to D.}}:\,=\qquad\qquad\qquad\\ D_{F}(x_{i}-x_{j})\,:\phi_{1}\cdots\phi_{i-1}\cdot\phi_{i+1}\cdots\phi_{j-1}% \cdot\phi_{i+1}\cdots\phi_{m}:\,.\end{split} (165)

If we are considering vevs the uncontracted terms drop out and we only need to consider all $m$ terms contracted, i.e. for example the last line of (164).

Proof of the Wick theorem

We will use a proof by induction. Our base case is $m=2$ which we have already shown. For the induction step we assume the theorem holds for $m-1$ fields and assume that w.l.o.g. the fields are time-ordered, i.e. $x_{1}^{0}\leq x_{2}^{0}\leq...\leq x_{m}^{0}$ . We have

\displaystyle T\{\phi_{1}\cdot\phi_{2}\cdots\phi_{m}\}=\phi_{1}\cdot\phi_{2}% \cdots\phi_{m}=\big{(}\phi^{(+)}_{1}+\phi^{(-)}_{1}\big{)}\ \cdot\ :\phi_{2}% \cdots\phi_{m}+\Big{(}\text{contractions \textbackslash$\ \phi_{1}$}\Big{)}:\,,

(166)

where we have applied the Wick theorem for $\phi_{2}\cdots\phi_{m}$ . We now need to move the $\phi^{(\pm)}_{1}$ into the normal ordering. The $\phi^{(-)}_{1}$ is trivial because it is already where it is supposed to be. To get the $\phi^{(+)}_{1}$ in we need to commute it all the way through the product. For fields that are already involved in a contraction, this is trivial as $D_{F}$ is just a number and commutes with everything. This means it is sufficient to only consider uncontracted fields. To simplify the notation a bit, we will write down the case without contractions but the generalisation is trivial once a suitable notation is developed

	$\displaystyle\phi^{(+)}_{1}:\phi_{2}\cdots\phi_{m}:\$	$\displaystyle=\ :\phi_{2}\cdots\phi_{m}:\ \phi^{(+)}_{1}+[\phi^{(+)}_{1},\ :% \phi_{2}\cdots\phi_{m}:\ ]$
		$\displaystyle=\ :\phi^{(+)}_{1}\cdot\phi_{2}\cdots\phi_{m}+[\phi^{(+)}_{1},% \phi_{2}]\cdot\phi_{3}\cdots\phi_{m}+\phi_{2}\cdot[\phi^{(+)}_{1},\phi_{3}]% \cdot\phi_{4}\cdots\phi_{m}+\cdots\ :\,.$		(167)

Since $\phi^{(+)}_{1}$ commutes with the $+$ part of $\phi_{i}$ , we have

\displaystyle\phi^{(+)}_{1}:\phi_{2}\cdots\phi_{m}:\

\displaystyle=\ :\phi^{(+)}_{1}\cdot\phi_{2}\cdots\phi_{m}+\includegraphics{% bmlimages/notes-16.svg}\bml@image@depth{16}{\bmlDescription{A Wick contraction% involving five terms A: \phi_1, B: \phi_2, C: \cdot, D: \phi_3, D: \cdots\phi% _m. A is connected to B.}}+\includegraphics{bmlimages/notes-17.svg}% \bml@image@depth{17}{\bmlDescription{A Wick contraction involving five terms A% : \phi_1, B: \phi_2, C: \cdot, D: \phi_3, D: \cdot\phi_4\cdots\phi_m. A is % connected to D.}}+\cdots\ :\,,

+⋯:,\displaystyle=\ :\phi^{(+)}_{1}\cdot\phi_{2}\cdots\phi_{m}+\includegraphics{% bmlimages/notes-16.svg}\bml@image@depth{16}{\bmlDescription{A Wick contraction% involving five terms A: \phi_1, B: \phi_2, C: \cdot, D: \phi_3, D: \cdots\phi% _m. A is connected to B.}}+\includegraphics{bmlimages/notes-17.svg}% \bml@image@depth{17}{\bmlDescription{A Wick contraction involving five terms A% : \phi_1, B: \phi_2, C: \cdot, D: \phi_3, D: \cdot\phi_4\cdots\phi_m. A is % connected to D.}}+\cdots\ :\,,

(168)

which is exactly what we wanted to show.

Let us know use what we know to find a graphic representation of these contractions. Consider the vev of (164)

	$\displaystyle\langle{0}\|T\{\phi_{1}\phi_{2}\phi_{3}$	$\displaystyle\phi_{4}\}\|{0}\rangle=\langle{0}\|\Big{(}\ :\ \includegraphics{% bmlimages/notes-18.svg}\bml@image@depth{18}{\bmlDescription{A Wick contraction% involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is % connected to B.C is connected to D.}}+\includegraphics{bmlimages/notes-19.svg}% \bml@image@depth{19}{\bmlDescription{A Wick contraction involving three terms % A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is connected to C.B is connected% to D.}}+\includegraphics{bmlimages/notes-20.svg}\bml@image@depth{20}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3, D: \phi_4. B is connected to C.A is connected to D.}}\ :\ +\text{% uncontracted}\Big{)}\|{0}\rangle$ + + :+uncontracted)\|0⟩\displaystyle\phi_{4}\}\|{0}\rangle=\langle{0}\|\Big{(}\ :\ \includegraphics{% bmlimages/notes-18.svg}\bml@image@depth{18}{\bmlDescription{A Wick contraction% involving three terms A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is % connected to B.C is connected to D.}}+\includegraphics{bmlimages/notes-19.svg}% \bml@image@depth{19}{\bmlDescription{A Wick contraction involving three terms % A: \phi_1, B: \phi_2, C: \phi_3, D: \phi_4. A is connected to C.B is connected% to D.}}+\includegraphics{bmlimages/notes-20.svg}\bml@image@depth{20}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3, D: \phi_4. B is connected to C.A is connected to D.}}\ :\ +\text{% uncontracted}\Big{)}\|{0}\rangle		(169)
		$\displaystyle=D_{F}(x_{1}-x_{2})D_{F}(x_{3}-x_{4})+D_{F}(x_{1}-x_{3})D_{F}(x_{% 2}-x_{4})+D_{F}(x_{1}-x_{4})D_{F}(x_{2}-x_{3})\,.$		(170)

Remember that the $x_{i}$ are points in spacetime and $D_{F}(x_{i}-x_{j})$ contains the part of the amplitude that moves a particle from $x_{i}$ to $x_{j}$ (or vice versa). We can therefore draw diagrams to represent these terms

\displaystyle\langle{0}|T\{\phi_{1}\phi_{2}\phi_{3}

\displaystyle\phi_{4}\}|{0}\rangle=\begin{gathered}\scalebox{0.9}{ \includegraphics{bmlimages/notes-21.svg}\bml@image@depth{21} \bmlDescription{A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x2% are connected. x3 and x4 are connected}}\end{gathered}+\begin{gathered}% \scalebox{0.9}{ \includegraphics{bmlimages/notes-22.svg}\bml@image@depth{22} \bmlDescription{A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x3% are connected. x2 and x4 are connected}}\end{gathered}+\begin{gathered}% \scalebox{0.9}{ \includegraphics{bmlimages/notes-23.svg}\bml@image@depth{23} \bmlDescription{A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x4% are connected. x2 and x3 are connected}}\end{gathered}\,.

A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x2 are connected. x3 and x4 are connected

A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x3 are connected. x2 and x4 are connected

A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x4 are connected. x2 and x3 are connected

.\displaystyle\phi_{4}\}|{0}\rangle=\begin{gathered}\scalebox{0.9}{ \includegraphics{bmlimages/notes-21.svg}\bml@image@depth{21} \bmlDescription{A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x2% are connected. x3 and x4 are connected}}\end{gathered}+\begin{gathered}% \scalebox{0.9}{ \includegraphics{bmlimages/notes-22.svg}\bml@image@depth{22} \bmlDescription{A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x3% are connected. x2 and x4 are connected}}\end{gathered}+\begin{gathered}% \scalebox{0.9}{ \includegraphics{bmlimages/notes-23.svg}\bml@image@depth{23} \bmlDescription{A Feynman diagram with four vertices x1, x2, x3, x4. x1 and x4% are connected. x2 and x3 are connected}}\end{gathered}\,.

(174)

This type of diagram is known as a Feynman diagram and they will soon get more interesting. The three diagrams in (174) merely encode all three ways particles can move between the four positions.

Note that if we had three fields we would have found

\displaystyle\langle{0}|T\{\phi_{1}\phi_{2}\phi_{3}\}|{0}\rangle=\langle{0}|% \Big{(}\ :\ \includegraphics{bmlimages/notes-24.svg}\bml@image@depth{24}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3. }}+\includegraphics{bmlimages/notes-25.svg}\bml@image@depth{25}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3. A is connected to B.}}+\includegraphics{bmlimages/notes-26.svg}% \bml@image@depth{26}{\bmlDescription{A Wick contraction involving three terms % A: \phi_1, B: \phi_2, C: \phi_3. A is connected to C.}}+\includegraphics{% bmlimages/notes-27.svg}\bml@image@depth{27}{\bmlDescription{A Wick contraction% involving three terms A: \phi_1, B: \phi_2, C: \phi_3. B is connected to C.}}% \ :\ \Big{)}|{0}\rangle=0\,,

:)|0⟩=0,\displaystyle\langle{0}|T\{\phi_{1}\phi_{2}\phi_{3}\}|{0}\rangle=\langle{0}|% \Big{(}\ :\ \includegraphics{bmlimages/notes-24.svg}\bml@image@depth{24}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3. }}+\includegraphics{bmlimages/notes-25.svg}\bml@image@depth{25}{% \bmlDescription{A Wick contraction involving three terms A: \phi_1, B: \phi_2,% C: \phi_3. A is connected to B.}}+\includegraphics{bmlimages/notes-26.svg}% \bml@image@depth{26}{\bmlDescription{A Wick contraction involving three terms % A: \phi_1, B: \phi_2, C: \phi_3. A is connected to C.}}+\includegraphics{% bmlimages/notes-27.svg}\bml@image@depth{27}{\bmlDescription{A Wick contraction% involving three terms A: \phi_1, B: \phi_2, C: \phi_3. B is connected to C.}}% \ :\ \Big{)}|{0}\rangle=0\,,

(175)

because all of the terms have an uncontracted field, i.e. they are $\propto\langle{0}|\ :\ \phi_{i}\ :\ |{0}\rangle=0$ .

3.4 Asymptotic states and the interacting vacuum

Before we can develop Feynman diagrams further, we need to go to two diversions: one of them relevant, one less so.

The first point is related to the asymptotic states that we defined in (134)

\displaystyle\phi_{\rm in}(x)=\lim_{t\to-\infty}\phi(t)=\int\frac{{\rm d}^{3}k% }{(2\pi)^{3}}\frac{1}{\sqrt{2E_{\vec{k}}}}\Big{[}a_{\rm in}(\vec{k}){\rm e}^{-% ik\cdot x}+a_{\rm in}^{\dagger}(\vec{k}){\rm e}^{ik\cdot x}\Big{]}\,.

(134)

This operator creates a particle at position $x$ . However, it is often more useful to think in momentum space and instead create a particle of momentum $p$ as we did in Section 2.5. For this, we defined (102) for the free field which we translate to the $\phi_{\rm in}$ fields

\displaystyle|{p}\rangle=\sqrt{2E_{\vec{p}}}\ a^{\dagger}_{\rm in}(p)|{0}% \rangle\,.

(176)

Applying $\phi^{(+)}(x)$ on this state

	$\displaystyle\phi^{(+)}(x)\|{p}\rangle$	$\displaystyle=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}{\rm e}^{-ik\cdot x}\sqrt{% \frac{2E_{\vec{p}}}{2E_{\vec{k}}}}a_{\rm in}(\vec{k})a_{\rm in}^{\dagger}(p)\|{% 0}\rangle=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}{\rm e}^{-ik\cdot x}\sqrt{\frac{2% E_{\vec{p}}}{2E_{\vec{k}}}}[a_{\rm in}(\vec{k}),a_{\rm in}^{\dagger}(p)]\|{0}\rangle$		(177)
		$\displaystyle={\rm e}^{-ip\cdot x}\|{0}\rangle\,,$		(178)

where we have used that $[a_{\rm in},a_{\rm in}^{\dagger}]|{0}\rangle=a_{\rm in}a_{\rm in}^{\dagger}|{0% }\rangle-a_{\rm in}^{\dagger}a_{\rm in}|{0}\rangle$ and $a_{\rm in}|{0}\rangle=0$ . Of course we can repeat the same construction for the out states as well. This is the connection between $S$ matrix elements and the vevs of time-ordered products that we have been calculating with the Wick theorem.

The nature of the vacuum

This diversion is not particularly important for the applications of QFT but is quite a fundamental building block of the theory. So far we have been using $|{0}\rangle$ both for the ground state of the free theory and of the interacting theory. Unfortunately, since the theories are not the same, there is no reason that the two ground states should be the same (or even that there should be a relation between the two). Here, and only here, we will distinguish between the free theory’s vacuum $|{0}\rangle$ and that of the interacting theory $|{\Omega}\rangle$ .

We define the energy zero as $H_{0}|{0}\rangle=0$ which means that, in general, the energy of the interacting vacuum will be different $E_{0}=\langle{\Omega}|H|{\Omega}\rangle$ . To relate the two Heisenberg states, we apply the full theory’s time evolution operator ${\rm e}^{-iHt}$ to the free vacuum

\displaystyle{\rm e}^{-iHt}|{0}\rangle=\sum_{n}{\rm e}^{-iE_{n}t}|{n}\rangle% \langle{n}|{0}\rangle={\rm e}^{-iE_{0}t}|{\Omega}\rangle\langle{\Omega}|{0}% \rangle+\sum_{n>0}{\rm e}^{-iE_{n}t}|{n}\rangle\langle{n}|{0}\rangle

(179)

with eigenstates $|{n}\rangle$ and eigenenergies $E_{n}$ of the full theory which includes the ground state $|{\Omega}\rangle$ . Since per definition, the ground state has the lowest energy $E_{n}>E_{0}$ for all $n>0$ . In the limit $t\to\infty(1-i\delta)$ all terms vanish since $\delta>0$ but the ground state’s contribution will vanish the slowest

\displaystyle{\rm e}^{-iHt}|{0}\rangle\stackrel{{\scriptstyle t\to\infty(1-i% \delta)}}{{=}}{\rm e}^{-iE_{0}t}|{\Omega}\rangle\langle{\Omega}|{0}\rangle+% \text{terms that vanish faster}\,.

(180)

We can now solve this for $|{\Omega}\rangle$ and obtain

\displaystyle|{\Omega}\rangle=\lim_{t\to\infty(1-i\delta)}\Big{(}{\rm e}^{-iE_% {0}t}\langle{\Omega}|{0}\rangle\Big{)}^{-1}{\rm e}^{-iHt}|{0}\rangle\,.

(181)

To make this a bit easier to use, let us shift $t\to t+t^{\prime}$ with $t^{\prime}\ll t$ and add a factor ${\rm e}^{iH_{0}(t+t^{\prime})}$ which, if applied to $|{0}\rangle$ , will give one

	$\displaystyle\|{\Omega}\rangle$	$\displaystyle=\lim_{t\to\infty(1-i\delta)}\Big{(}{\rm e}^{-iE_{0}(t+t^{\prime}% )}\langle{\Omega}\|{0}\rangle\Big{)}^{-1}{\rm e}^{-iH(t+t^{\prime})}\|{0}\rangle$		(182)
		$\displaystyle=\lim_{t\to\infty(1-i\delta)}\Big{(}{\rm e}^{-iE_{0}(t^{\prime}-(% -t))}\langle{\Omega}\|{0}\rangle\Big{)}^{-1}\underbrace{{\rm e}^{-iH(t^{\prime}% -(-t))}{\rm e}^{iH_{0}(t+t^{\prime})}}_{U(t^{\prime},-t)}\|{0}\rangle\,.$		(183)

This means that we can obtain the interacting vacuum from the free vacuum by evolving it from the distant past ( $-t\to\infty$ ) to the present $t^{\prime}$ . A similar construction is possible for $\langle{\Omega}|$ where we need to choose $t$ to be the opposite sign

\displaystyle\langle{\Omega}|

\displaystyle=\lim_{t\to\infty(1-i\delta)}\Big{(}{\rm e}^{-iE_{0}(t-t^{\prime}% )}\langle{0}|{\Omega}\rangle\Big{)}^{-1}\langle{0}|U(t,t^{\prime})\,.

(184)

We can now write the correlator for $x^{0}>y^{0}>t^{\prime}$

\displaystyle\begin{split}\langle{\Omega}|\phi(x)\phi(y)|{\Omega}\rangle=\lim_% {t\to\infty(1-i\delta)}&\underbrace{\Big{(}{\rm e}^{-iE_{0}(t-t^{\prime})}% \langle{0}|{\Omega}\rangle\Big{)}^{-1}\langle{0}|U(t,t^{\prime})}_{\langle{% \Omega}|}\\ &\underbrace{U^{\dagger}(x^{0},t^{\prime})\phi_{I}(x)U(x^{0},t^{\prime})}_{% \phi(x)}\underbrace{U^{\dagger}(y^{0},t^{\prime})\phi_{I}(y)U(y^{0},t^{\prime}% )}_{\phi(y)}\\ &\underbrace{U(t^{\prime},-t)|{0}\rangle\Big{(}{\rm e}^{-iE_{0}(t^{\prime}-(-t% ))}\langle{\Omega}|{0}\rangle\Big{)}^{-1}}_{|{\Omega}\rangle}\,.\end{split}

(185)

We can simplify things using $U^{\dagger}(t_{1},t_{2})=U(t_{2},t_{1})$ and $U(t_{1},t_{2})U(t_{2},t_{3})=U(t_{1},t_{3})$ assuming the times are properly ordered

\displaystyle\langle{\Omega}|\phi(x)\phi(y)|{\Omega}\rangle=\lim_{t\to\infty(1% -i\delta)}

\displaystyle\Big{(}{\rm e}^{-iE_{0}\,2t}|\langle{\Omega}|{0}\rangle|^{2}\Big{% )}^{-1}\langle{0}|U(t,x^{0})\phi_{I}(x)U(x^{0},y^{0})\phi_{I}(y)U(y^{0},-t)|{0% }\rangle\,.

(186)

Assuming $\langle{\Omega}|{\Omega}\rangle=1$ , we can write

\displaystyle\langle{\Omega}|{\Omega}\rangle=\Big{(}{\rm e}^{-iE_{0}\,2t}|% \langle{\Omega}|{0}\rangle|^{2}\Big{)}^{-1}\langle{0}|U(t,t^{\prime})U(t^{% \prime},-t)|{0}\rangle

(187)

to cancel the prefactor phase and $E_{0}$

\displaystyle\langle{\Omega}|\phi(x)\phi(y)|{\Omega}\rangle=\lim_{t\to\infty(1% -i\delta)}

\displaystyle\frac{\langle{0}|U(t,x^{0})\phi_{I}(x)U(x^{0},y^{0})\phi_{I}(y)U(% y^{0},-t)|{0}\rangle}{\langle{0}|U(t,-t)|{0}\rangle}\,.

(188)

This is completely time-ordered and would have also held for $y^{0}>x^{0}$ . We can therefore write it as a time-ordered product and use (151)

\displaystyle\langle{\Omega}|T\big{\{}\phi(x)\phi(y)\big{\}}|{\Omega}\rangle=% \lim_{t\to\infty(1-i\delta)}

\displaystyle\frac{\langle{0}|T\Bigg{\{}\phi_{I}(x)\phi_{I}(y)\exp\bigg{(}-i{% \displaystyle\int}_{-t}^{t}{\rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})\bigg{)% }\Bigg{\}}|{0}\rangle}{\langle{0}|T\Bigg{\{}\exp\bigg{(}-i{\displaystyle\int}_% {-t}^{t}{\rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})\bigg{)}\Bigg{\}}|{0}% \rangle}\,.

(189)

This result is known as the Gell-Mann and Low theorem and it allows us to slightly formalise what we have been doing (which amounts to ignoring the denominator).

Haag’s theorem

For the above discussion, we have assumed that both $|{0}\rangle$ and $|{\Omega}\rangle$ exist in the same space and that their overlap $\langle{0}|{\Omega}\rangle\neq 0$ . In practice, this is not true, making the construction invalid. Further, the operator $U(-\infty,t)$ , that we have used to relate the free states to the interacting ones, does not exist either. This result is known as Haag’s theorem and it seriously jeopardises the construction of any QFT. Luckily for us, there are a number of ways to, if not rescue the proof, at least stabilise it enough to be used in calculations. This problem is one of the many issues facing a truly axiomatic construction of QFT.

3.5 Feynman diagrams

To study our first non-trivial Feynman let us consider $\langle{\Omega}|T\big{\{}\phi(x)\phi(y)\big{\}}|{\Omega}\rangle$ , i.e. the propagator of the interacting theory. We will assume implicitly that these operators are in the interaction picture even though they are not contracted just yet. For this we use the Gell-Mann-Low theorem (189) but will ignore the denominator for now. We have

$\displaystyle\langle{\Omega}\|T\big{\{}\phi(x)\phi(y)\big{\}}\|{\Omega}\rangle\sim$
$\displaystyle\langle{0}\|T\Bigg{\{}\phi(x)\phi(y)$	$\displaystyle-i\phi(x)\phi(y)\int{\rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})-% \frac{1}{2!}\phi(x)\phi(y)\int{\rm d}t^{\prime}{\rm d}t^{\prime\prime}\ \tilde% {H}_{I}(t^{\prime})H_{I}(t^{\prime\prime})+\cdots\Bigg{\}}\|{0}\rangle$	(190)
$\displaystyle=\langle{0}\|T\Bigg{\{}\phi(x)\phi(y)$	$\displaystyle+i\phi(x)\phi(y)\int{\rm d}^{4}z\ \mathcal{L}_{I}(z)-\frac{1}{2!}% \phi(x)\phi(y)\int{\rm d}^{4}z{\rm d}^{4}w\ \mathcal{L}_{I}(z)\mathcal{L}_{I}(% w)+\cdots\Bigg{\}}\|{0}\rangle\,.$	(191)

Here we have substituted in $H_{I}=\int{\rm d}^{3}z\ \mathcal{L}_{I}$ and combine the $t^{\prime}$ integration with the $z$ integration. The first term just corresponds to $D_{F}(x-y)$ . For the second, we write

\displaystyle\langle{0}|T\Bigg{\{}i\phi(x)\phi(y)\int{\rm d}^{4}z\ \mathcal{L}% _{I}(z)\Bigg{\}}|{0}\rangle=-\frac{i\lambda}{4!}\langle{0}|T\Bigg{\{}\int{\rm d% }^{4}z\ \phi(x)\phi(y)\ \phi(z)\phi(z)\phi(z)\phi(z)\Bigg{\}}|{0}\rangle\,.

(192)

We can now apply Wick’s theorem to calculate this vev.

Suggested Exercise

Write down all 15 contractions explicitly to show the following

	$\displaystyle\langle{0}\|T\Bigg{\{}i\phi(x)\phi(y)\int{\rm d}^{4}z\ \mathcal{L}% _{I}(z)\Bigg{\}}\|{0}\rangle=-\frac{i\lambda}{4!}\bigg{(}$	$\displaystyle 3\times D_{F}(x-y)\int{\rm d}^{4}z\ D_{F}(z-z)D_{F}(z-z)$
		$\displaystyle+12\times\int{\rm d}^{4}z\ D_{F}(x-z)D_{F}(y-z)D_{F}(z-z)\bigg{)}\,.$		(193)

We can visualise this using Feynman diagrams

\displaystyle\langle{0}|T\Bigg{\{}i\phi(x)\phi(y)\int{\rm d}^{4}z\ \mathcal{L}% _{I}(z)\Bigg{\}}|{0}\rangle=\int{\rm d}^{4}z\Bigg{(}\includegraphics{bmlimages% /notes-28.svg}\bml@image@depth{28}\bmlDescription{A Feynman diagram with a % propagator from x to y, as well as a bowtie diagram connected z to itself % twice}+\includegraphics{bmlimages/notes-29.svg}\bml@image@depth{29}% \bmlDescription{A Feynman diagram with a propagator from x to z to y. There is% a tadpole connecting z to itself}\Bigg{)}\,.

).\displaystyle\langle{0}|T\Bigg{\{}i\phi(x)\phi(y)\int{\rm d}^{4}z\ \mathcal{L}% _{I}(z)\Bigg{\}}|{0}\rangle=\int{\rm d}^{4}z\Bigg{(}\includegraphics{bmlimages% /notes-28.svg}\bml@image@depth{28}\bmlDescription{A Feynman diagram with a % propagator from x to y, as well as a bowtie diagram connected z to itself % twice}+\includegraphics{bmlimages/notes-29.svg}\bml@image@depth{29}% \bmlDescription{A Feynman diagram with a propagator from x to z to y. There is% a tadpole connecting z to itself}\Bigg{)}\,.

(194)

We usually define the symmetry factor, i.e. the 3 and 12 to be included in the diagram. In a Feynman diagram like this, we have propagators (the lines of the diagram) and vertices (points where four lines meet). The vertices are located at positions in spacetime that depend either on the process ( $x$ and $y$ ) or are integrated over ( $z$ ). The number of contractions that contribute can grow quite quickly.

Suggested Exercise

Repeat the above for the $\lambda^{2}$ term to show that

$\displaystyle\frac{1}{2!}\langle{0}\|T\Bigg{\{}\phi(x)\phi(y)\int$	$\displaystyle{\rm d}^{4}z{\rm d}^{4}w\ \mathcal{L}_{I}(z)\mathcal{L}_{I}(w)% \Bigg{\}}\|{0}\rangle=\frac{\lambda^{2}}{2!(4!)^{2}}\ \int{\rm d}^{4}z{\rm d}^{% 4}w\bigg{[}\bigg{(}4!\ D_{F}(x-y)\times D_{F}(z-w)^{4}$
	$\displaystyle+(4\cdot 3/2)^{2}\cdot 2\ D_{F}(x-y)\times D_{F}(w-w)D_{F}(z-z)D_% {F}(w-z)^{2}$
	$\displaystyle+3\cdot 3\ D_{F}(x-y)\times D_{F}(z-z)^{2}D_{F}(w-w)^{2}\bigg{)}$
	$\displaystyle+\bigg{(}4\cdot 3\cdot 3\ D_{F}(x-z)D_{F}(y-z)\times D_{F}(w-w)^{% 2}D_{F}(z-z)$
	$\displaystyle+(4\cdot 3)^{2}\ D_{F}(x-z)D_{F}(y-z)\ D_{F}(z-w)^{2}D_{F}(w-w)$
	$\displaystyle+4\cdot 4\cdot 3\cdot 3\ D_{F}(y-z)D_{F}(z-z)\ D_{F}(z-w)\ D_{F}(% w-w)D_{F}(w-x)$
	$\displaystyle+4\cdot 2\cdot 3\cdot 4\ D_{F}(y-z)\ D_{F}(z-w)^{3}\ D_{F}(w-x)+(% w\leftrightarrow z)\bigg{)}\bigg{]}\,.$	(195)

We can express this using Feynman diagrams as

$\displaystyle\frac{1}{2!}\langle{0}\|T\Bigg{\{}\phi(x)\phi(y)\int$	$\displaystyle{\rm d}^{4}z{\rm d}^{4}w\ \mathcal{L}_{I}(z)\mathcal{L}_{I}(w)% \Bigg{\}}\|{0}\rangle=\lambda^{2}\int{\rm d}^{4}z{\rm d}^{4}w\ \Bigg{(}$
	$\displaystyle\frac{1}{48}\ \includegraphics{bmlimages/notes-30.svg}% \bml@image@depth{30}\bmlDescription{A Feynman diagram with a lone propagator % and a free-floating melon diagram where two vertices are connected with four % lines}+\frac{1}{16}\ \includegraphics{bmlimages/notes-31.svg}\bml@image@depth{% 31}\bmlDescription{A Feynman diagram with a lone propagator and a free-% floating snowman diagram where two vertices, that have a tadpole each, are % connected with two lines}+\frac{1}{128}\ \includegraphics{bmlimages/notes-32.% svg}\bml@image@depth{32}\bmlDescription{A Feynman diagram with a lone % propagator and two free-floating bowties}+\frac{1}{16}\ \includegraphics{% bmlimages/notes-33.svg}\bml@image@depth{33}\bmlDescription{A Feynman diagram % with a propagator that has a tadpole and one free-floating bowtie}$ +116⁢ +1128⁢ +116⁢ \displaystyle\frac{1}{48}\ \includegraphics{bmlimages/notes-30.svg}% \bml@image@depth{30}\bmlDescription{A Feynman diagram with a lone propagator % and a free-floating melon diagram where two vertices are connected with four % lines}+\frac{1}{16}\ \includegraphics{bmlimages/notes-31.svg}\bml@image@depth{% 31}\bmlDescription{A Feynman diagram with a lone propagator and a free-% floating snowman diagram where two vertices, that have a tadpole each, are % connected with two lines}+\frac{1}{128}\ \includegraphics{bmlimages/notes-32.% svg}\bml@image@depth{32}\bmlDescription{A Feynman diagram with a lone % propagator and two free-floating bowties}+\frac{1}{16}\ \includegraphics{% bmlimages/notes-33.svg}\bml@image@depth{33}\bmlDescription{A Feynman diagram % with a propagator that has a tadpole and one free-floating bowtie}
	$\displaystyle+\frac{1}{4}\ \includegraphics{bmlimages/notes-34.svg}% \bml@image@depth{34}\bmlDescription{A Feynman diagram with a propagator that % has a two tadpoles stack on top of each other}+\frac{1}{4}\ \includegraphics{% bmlimages/notes-35.svg}\bml@image@depth{35}\bmlDescription{A Feynman diagram % with a propagator that has a two tadpoles next to each other}+\frac{1}{6}\ % \includegraphics{bmlimages/notes-36.svg}\bml@image@depth{36}\bmlDescription{A % Feynman diagram with a propagator and a sunset, i.e. three lines connecting % two vertices}\Bigg{)}\,.$ +14 +16 ).\displaystyle+\frac{1}{4}\ \includegraphics{bmlimages/notes-34.svg}% \bml@image@depth{34}\bmlDescription{A Feynman diagram with a propagator that % has a two tadpoles stack on top of each other}+\frac{1}{4}\ \includegraphics{% bmlimages/notes-35.svg}\bml@image@depth{35}\bmlDescription{A Feynman diagram % with a propagator that has a two tadpoles next to each other}+\frac{1}{6}\ % \includegraphics{bmlimages/notes-36.svg}\bml@image@depth{36}\bmlDescription{A % Feynman diagram with a propagator and a sunset, i.e. three lines connecting % two vertices}\Bigg{)}\,.	(196)

Here we have written the symmetry factors explicitly even though we previously stated that we consider them part of the diagram. This is for illustration purposes.

Disconnected pieces and the Gell-Mann-Low denominator

In (194) we encountered two types of Feynman diagrams: a connected diagram where all lines and dots are connected; a disconnected diagram where a diagram was free-floating. Let us focus on the disconnected diagram contribution which evaluated to

\displaystyle\includegraphics{bmlimages/notes-37.svg}\bml@image@depth{37}% \bmlDescription{A free-floating bowtie at position z}=\int{\rm d}^{4}z\ D_{F}(% z-z)D_{F}(z-z)\sim\int{\rm d}^{4}z\ (\text{const.})\sim\text{volume of space % time}\,.

=∫d4⁢z⁢DF⁢(z−z)⁢DF⁢(z−z)∼∫d4⁢z⁢(const.)∼volume of space time.\displaystyle\includegraphics{bmlimages/notes-37.svg}\bml@image@depth{37}% \bmlDescription{A free-floating bowtie at position z}=\int{\rm d}^{4}z\ D_{F}(% z-z)D_{F}(z-z)\sim\int{\rm d}^{4}z\ (\text{const.})\sim\text{volume of space % time}\,.

(197)

Once again we find ourselves in a situation where our prediction includes the volume of space time which is obviously infinite. For now let us just call this first disconnected piece $V_{1}$ , the next more complicated disconnected piece is $V_{2}$ etc. We will not really worry what these evaluate to because we will soon see that they cancel.

To see what happens, we start with the full series in $\lambda$

$\displaystyle\langle{\Omega}\|T\{\phi(x)\phi(y)\}\|{\Omega}\rangle\sim$	$\displaystyle\Bigg{(}\includegraphics{bmlimages/notes-38.svg}\bml@image@depth{% 38}\bmlDescription{A lone propagator}\Bigg{)}+\Bigg{(}\includegraphics{% bmlimages/notes-39.svg}\bml@image@depth{39}\bmlDescription{A propagator with a% tadpole}+\includegraphics{bmlimages/notes-40.svg}\bml@image@depth{40}% \bmlDescription{A lone propagator}\includegraphics{bmlimages/notes-41.svg}% \bml@image@depth{41}\bmlDescription{A free floating bow-tie diagram}\Bigg{)}$ )+( + )\displaystyle\Bigg{(}\includegraphics{bmlimages/notes-38.svg}\bml@image@depth{% 38}\bmlDescription{A lone propagator}\Bigg{)}+\Bigg{(}\includegraphics{% bmlimages/notes-39.svg}\bml@image@depth{39}\bmlDescription{A propagator with a% tadpole}+\includegraphics{bmlimages/notes-40.svg}\bml@image@depth{40}% \bmlDescription{A lone propagator}\includegraphics{bmlimages/notes-41.svg}% \bml@image@depth{41}\bmlDescription{A free floating bow-tie diagram}\Bigg{)}
	$\displaystyle+\Bigg{(}\includegraphics{bmlimages/notes-42.svg}\bml@image@depth% {42}\bmlDescription{A propagator with a sunset}+\includegraphics{bmlimages/not% es-43.svg}\bml@image@depth{43}\bmlDescription{A propagator with a tadpole}% \includegraphics{bmlimages/notes-44.svg}\bml@image@depth{44}\bmlDescription{A % free floating bow-tie diagram}+\includegraphics{bmlimages/notes-45.svg}% \bml@image@depth{45}\bmlDescription{A lone propagator}\includegraphics{% bmlimages/notes-46.svg}\bml@image@depth{46}\bmlDescription{A free floating bow% -tie diagram}\includegraphics{bmlimages/notes-47.svg}\bml@image@depth{47}% \bmlDescription{A free floating bow-tie diagram}+\includegraphics{bmlimages/no% tes-48.svg}\bml@image@depth{48}\bmlDescription{A lone propagator}% \includegraphics{bmlimages/notes-49.svg}\bml@image@depth{49}\bmlDescription{A % free floating snowman diagram}+\cdots\Bigg{)}$ + + + +⋯)\displaystyle+\Bigg{(}\includegraphics{bmlimages/notes-42.svg}\bml@image@depth% {42}\bmlDescription{A propagator with a sunset}+\includegraphics{bmlimages/not% es-43.svg}\bml@image@depth{43}\bmlDescription{A propagator with a tadpole}% \includegraphics{bmlimages/notes-44.svg}\bml@image@depth{44}\bmlDescription{A % free floating bow-tie diagram}+\includegraphics{bmlimages/notes-45.svg}% \bml@image@depth{45}\bmlDescription{A lone propagator}\includegraphics{% bmlimages/notes-46.svg}\bml@image@depth{46}\bmlDescription{A free floating bow% -tie diagram}\includegraphics{bmlimages/notes-47.svg}\bml@image@depth{47}% \bmlDescription{A free floating bow-tie diagram}+\includegraphics{bmlimages/no% tes-48.svg}\bml@image@depth{48}\bmlDescription{A lone propagator}% \includegraphics{bmlimages/notes-49.svg}\bml@image@depth{49}\bmlDescription{A % free floating snowman diagram}+\cdots\Bigg{)}
	$\displaystyle+\Bigg{(}\includegraphics{bmlimages/notes-50.svg}\bml@image@depth% {50}\bmlDescription{A propagator with a sunset which itself has a tadpole}+% \includegraphics{bmlimages/notes-51.svg}\bml@image@depth{51}\bmlDescription{A % propagator with a sunset}\includegraphics{bmlimages/notes-52.svg}% \bml@image@depth{52}\bmlDescription{A free floating bow-tie diagram}+% \includegraphics{bmlimages/notes-53.svg}\bml@image@depth{53}\bmlDescription{A % propagator with a tadpole}\includegraphics{bmlimages/notes-54.svg}% \bml@image@depth{54}\bmlDescription{A free floating snowman diagram}+% \includegraphics{bmlimages/notes-55.svg}\bml@image@depth{55}\bmlDescription{A % propagator with a tadpole}\includegraphics{bmlimages/notes-56.svg}% \bml@image@depth{56}\bmlDescription{A free floating bow-tie diagram}% \includegraphics{bmlimages/notes-57.svg}\bml@image@depth{57}\bmlDescription{A % free floating bow-tie diagram}$ + + + \displaystyle+\Bigg{(}\includegraphics{bmlimages/notes-50.svg}\bml@image@depth% {50}\bmlDescription{A propagator with a sunset which itself has a tadpole}+% \includegraphics{bmlimages/notes-51.svg}\bml@image@depth{51}\bmlDescription{A % propagator with a sunset}\includegraphics{bmlimages/notes-52.svg}% \bml@image@depth{52}\bmlDescription{A free floating bow-tie diagram}+% \includegraphics{bmlimages/notes-53.svg}\bml@image@depth{53}\bmlDescription{A % propagator with a tadpole}\includegraphics{bmlimages/notes-54.svg}% \bml@image@depth{54}\bmlDescription{A free floating snowman diagram}+% \includegraphics{bmlimages/notes-55.svg}\bml@image@depth{55}\bmlDescription{A % propagator with a tadpole}\includegraphics{bmlimages/notes-56.svg}% \bml@image@depth{56}\bmlDescription{A free floating bow-tie diagram}% \includegraphics{bmlimages/notes-57.svg}\bml@image@depth{57}\bmlDescription{A % free floating bow-tie diagram}
	$\displaystyle\qquad+\includegraphics{bmlimages/notes-58.svg}\bml@image@depth{5% 8}\bmlDescription{A lone propagator}\includegraphics{bmlimages/notes-59.svg}% \bml@image@depth{59}\bmlDescription{A free floating melon with a tadpole}+% \includegraphics{bmlimages/notes-60.svg}\bml@image@depth{60}\bmlDescription{A % lone propagator}\includegraphics{bmlimages/notes-61.svg}\bml@image@depth{61}% \bmlDescription{A free floating snowman diagram}\includegraphics{bmlimages/not% es-62.svg}\bml@image@depth{62}\bmlDescription{A free floating bow-tie diagram}% +\includegraphics{bmlimages/notes-63.svg}\bml@image@depth{63}\bmlDescription{A% lone propagator}\includegraphics{bmlimages/notes-64.svg}\bml@image@depth{64}% \bmlDescription{A free floating bow-tie diagram}\includegraphics{bmlimages/not% es-65.svg}\bml@image@depth{65}\bmlDescription{A free floating bow-tie diagram}% \includegraphics{bmlimages/notes-66.svg}\bml@image@depth{66}\bmlDescription{A % free floating bow-tie diagram}+\cdots\Bigg{)}+\cdots$ + + +⋯)+⋯\displaystyle\qquad+\includegraphics{bmlimages/notes-58.svg}\bml@image@depth{5% 8}\bmlDescription{A lone propagator}\includegraphics{bmlimages/notes-59.svg}% \bml@image@depth{59}\bmlDescription{A free floating melon with a tadpole}+% \includegraphics{bmlimages/notes-60.svg}\bml@image@depth{60}\bmlDescription{A % lone propagator}\includegraphics{bmlimages/notes-61.svg}\bml@image@depth{61}% \bmlDescription{A free floating snowman diagram}\includegraphics{bmlimages/not% es-62.svg}\bml@image@depth{62}\bmlDescription{A free floating bow-tie diagram}% +\includegraphics{bmlimages/notes-63.svg}\bml@image@depth{63}\bmlDescription{A% lone propagator}\includegraphics{bmlimages/notes-64.svg}\bml@image@depth{64}% \bmlDescription{A free floating bow-tie diagram}\includegraphics{bmlimages/not% es-65.svg}\bml@image@depth{65}\bmlDescription{A free floating bow-tie diagram}% \includegraphics{bmlimages/notes-66.svg}\bml@image@depth{66}\bmlDescription{A % free floating bow-tie diagram}+\cdots\Bigg{)}+\cdots	(198)

Note how in this expression we keep finding the same pieces, both for the connected part and the disconnected ones. We will now try and exploit this structure by rearranging this infinite series by collecting terms not by their power in $\lambda$ but by their diagrammatic topology. This will lead to us factoring our the connected pieces

$\displaystyle\langle{\Omega}\|T\{\phi(x)\phi(y)\}\|{\Omega}\rangle$	$\displaystyle\sim\Bigg{(}\includegraphics{bmlimages/notes-67.svg}% \bml@image@depth{67}\bmlDescription{A lone propagator}+\includegraphics{% bmlimages/notes-68.svg}\bml@image@depth{68}\bmlDescription{A propagator with a% tadpole}+\includegraphics{bmlimages/notes-69.svg}\bml@image@depth{69}% \bmlDescription{A propagator with a sunset}+\includegraphics{bmlimages/notes-7% 0.svg}\bml@image@depth{70}\bmlDescription{A propagator with a sunset which % itself has a tadpole}+\cdots\Bigg{)}$ + + + +⋯)\displaystyle\sim\Bigg{(}\includegraphics{bmlimages/notes-67.svg}% \bml@image@depth{67}\bmlDescription{A lone propagator}+\includegraphics{% bmlimages/notes-68.svg}\bml@image@depth{68}\bmlDescription{A propagator with a% tadpole}+\includegraphics{bmlimages/notes-69.svg}\bml@image@depth{69}% \bmlDescription{A propagator with a sunset}+\includegraphics{bmlimages/notes-7% 0.svg}\bml@image@depth{70}\bmlDescription{A propagator with a sunset which % itself has a tadpole}+\cdots\Bigg{)}
	$\displaystyle\qquad\times\Bigg{(}1+\underbrace{\includegraphics{bmlimages/note% s-71.svg}\bml@image@depth{71}\bmlDescription{A free floating bow-tie diagram}}% _{V_{1}}+\underbrace{\includegraphics{bmlimages/notes-72.svg}\bml@image@depth{% 72}\bmlDescription{A free floating snowman diagram}}_{V_{2}}+\underbrace{% \includegraphics{bmlimages/notes-73.svg}\bml@image@depth{73}\bmlDescription{A % free floating melon with a tadpole}}_{V_{3}}+\cdots$ ⏟V1+ ⏟V2+ ⏟V3+⋯\displaystyle\qquad\times\Bigg{(}1+\underbrace{\includegraphics{bmlimages/note% s-71.svg}\bml@image@depth{71}\bmlDescription{A free floating bow-tie diagram}}% _{V_{1}}+\underbrace{\includegraphics{bmlimages/notes-72.svg}\bml@image@depth{% 72}\bmlDescription{A free floating snowman diagram}}_{V_{2}}+\underbrace{% \includegraphics{bmlimages/notes-73.svg}\bml@image@depth{73}\bmlDescription{A % free floating melon with a tadpole}}_{V_{3}}+\cdots
	$\displaystyle\qquad+\frac{1}{2!}\bigg{[}\includegraphics{bmlimages/notes-74.% svg}\bml@image@depth{74}\bmlDescription{A free floating bow-tie diagram}\bigg{% ]}^{2}+\frac{1}{2!}\bigg{[}\includegraphics{bmlimages/notes-75.svg}% \bml@image@depth{75}\bmlDescription{A free floating snowman diagram}\bigg{]}^{% 2}+\bigg{[}\includegraphics{bmlimages/notes-76.svg}\bml@image@depth{76}% \bmlDescription{A free floating snowman diagram}\bigg{]}\bigg{[}% \includegraphics{bmlimages/notes-77.svg}\bml@image@depth{77}\bmlDescription{A % free floating snowman diagram}\bigg{]}+\bigg{[}\includegraphics{bmlimages/note% s-78.svg}\bml@image@depth{78}\bmlDescription{A free floating bow-tie diagram}% \bigg{]}\bigg{[}\includegraphics{bmlimages/notes-79.svg}\bml@image@depth{79}% \bmlDescription{A free floating melon with a tadpole}\bigg{]}+\cdots$ ]2+12!⁢[ ]2+[ ]⁢[ ]+[ ]⁢[ ]+⋯\displaystyle\qquad+\frac{1}{2!}\bigg{[}\includegraphics{bmlimages/notes-74.% svg}\bml@image@depth{74}\bmlDescription{A free floating bow-tie diagram}\bigg{% ]}^{2}+\frac{1}{2!}\bigg{[}\includegraphics{bmlimages/notes-75.svg}% \bml@image@depth{75}\bmlDescription{A free floating snowman diagram}\bigg{]}^{% 2}+\bigg{[}\includegraphics{bmlimages/notes-76.svg}\bml@image@depth{76}% \bmlDescription{A free floating snowman diagram}\bigg{]}\bigg{[}% \includegraphics{bmlimages/notes-77.svg}\bml@image@depth{77}\bmlDescription{A % free floating snowman diagram}\bigg{]}+\bigg{[}\includegraphics{bmlimages/note% s-78.svg}\bml@image@depth{78}\bmlDescription{A free floating bow-tie diagram}% \bigg{]}\bigg{[}\includegraphics{bmlimages/notes-79.svg}\bml@image@depth{79}% \bmlDescription{A free floating melon with a tadpole}\bigg{]}+\cdots
	$\displaystyle\qquad+\frac{1}{3!}\bigg{[}\includegraphics{bmlimages/notes-80.% svg}\bml@image@depth{80}\bmlDescription{A free floating bow-tie diagram}\bigg{% ]}^{3}+\frac{1}{3!}\bigg{[}\includegraphics{bmlimages/notes-81.svg}% \bml@image@depth{81}\bmlDescription{A free floating snowman diagram}\bigg{]}^{% 3}+\frac{1}{(1!)(2!)}\bigg{[}\includegraphics{bmlimages/notes-82.svg}% \bml@image@depth{82}\bmlDescription{A free floating bow-tie diagram}\bigg{]}% \bigg{[}\includegraphics{bmlimages/notes-83.svg}\bml@image@depth{83}% \bmlDescription{A free floating snowman diagram}\bigg{]}^{2}+\cdots\Bigg{)}$ ]3+13![ ]3+1(1!)⁢(2!)[ ][ ]2+⋯)\displaystyle\qquad+\frac{1}{3!}\bigg{[}\includegraphics{bmlimages/notes-80.% svg}\bml@image@depth{80}\bmlDescription{A free floating bow-tie diagram}\bigg{% ]}^{3}+\frac{1}{3!}\bigg{[}\includegraphics{bmlimages/notes-81.svg}% \bml@image@depth{81}\bmlDescription{A free floating snowman diagram}\bigg{]}^{% 3}+\frac{1}{(1!)(2!)}\bigg{[}\includegraphics{bmlimages/notes-82.svg}% \bml@image@depth{82}\bmlDescription{A free floating bow-tie diagram}\bigg{]}% \bigg{[}\includegraphics{bmlimages/notes-83.svg}\bml@image@depth{83}% \bmlDescription{A free floating snowman diagram}\bigg{]}^{2}+\cdots\Bigg{)}	(199)
	$\displaystyle=\Big{(}\sum\text{connected}\Big{)}\times\bigg{(}1+V_{1}+V_{2}+V_% {3}+\cdots+\frac{1}{2!}V_{1}^{2}+\frac{1}{2!}V_{2}^{2}+V_{1}V_{2}+V_{2}V_{3}+\cdots$
	$\displaystyle\qquad+\frac{1}{3!}V_{1}^{3}+\frac{1}{3!}V_{2}^{3}+\frac{1}{(1!)(% 2!)}V_{1}V_{2}^{2}+\cdots\bigg{)}$	(200)
	$\displaystyle=\Big{(}\sum\text{connected}\Big{)}\times\sum_{\text{all}\ \{n_{i% }\}}\Big{(}\prod_{i}\frac{1}{n_{i}!}V_{i}^{n_{i}}\Big{)}\,.$	(201)

We can do one more step of rearranging

$\displaystyle\langle{\Omega}\|T\{\phi(x)\phi(y)\}\|{\Omega}\rangle$	$\displaystyle\sim\Bigg{(}\includegraphics{bmlimages/notes-84.svg}% \bml@image@depth{84}\bmlDescription{A lone propagator}+\includegraphics{% bmlimages/notes-85.svg}\bml@image@depth{85}\bmlDescription{A propagator with a% tadpole}+\includegraphics{bmlimages/notes-86.svg}\bml@image@depth{86}% \bmlDescription{A propagator with a sunset}+\includegraphics{bmlimages/notes-8% 7.svg}\bml@image@depth{87}\bmlDescription{A propagator with a sunset which % itself has a tadpole}+\cdots\Bigg{)}$ + + + +⋯)\displaystyle\sim\Bigg{(}\includegraphics{bmlimages/notes-84.svg}% \bml@image@depth{84}\bmlDescription{A lone propagator}+\includegraphics{% bmlimages/notes-85.svg}\bml@image@depth{85}\bmlDescription{A propagator with a% tadpole}+\includegraphics{bmlimages/notes-86.svg}\bml@image@depth{86}% \bmlDescription{A propagator with a sunset}+\includegraphics{bmlimages/notes-8% 7.svg}\bml@image@depth{87}\bmlDescription{A propagator with a sunset which % itself has a tadpole}+\cdots\Bigg{)}
	$\displaystyle\qquad\times\Bigg{(}1+\includegraphics{bmlimages/notes-88.svg}% \bml@image@depth{88}\bmlDescription{A free floating bow-tie diagram}+\frac{1}{% 2!}\bigg{[}\includegraphics{bmlimages/notes-89.svg}\bml@image@depth{89}% \bmlDescription{A free floating bow-tie diagram}\bigg{]}^{2}+\frac{1}{3!}\bigg% {[}\includegraphics{bmlimages/notes-90.svg}\bml@image@depth{90}\bmlDescription% {A free floating bow-tie diagram}\bigg{]}^{3}+\cdots\Bigg{)}$ +12!⁢[ ]2+13!⁢[ ]3+⋯)\displaystyle\qquad\times\Bigg{(}1+\includegraphics{bmlimages/notes-88.svg}% \bml@image@depth{88}\bmlDescription{A free floating bow-tie diagram}+\frac{1}{% 2!}\bigg{[}\includegraphics{bmlimages/notes-89.svg}\bml@image@depth{89}% \bmlDescription{A free floating bow-tie diagram}\bigg{]}^{2}+\frac{1}{3!}\bigg% {[}\includegraphics{bmlimages/notes-90.svg}\bml@image@depth{90}\bmlDescription% {A free floating bow-tie diagram}\bigg{]}^{3}+\cdots\Bigg{)}
	$\displaystyle\qquad\times\Bigg{(}1+\includegraphics{bmlimages/notes-91.svg}% \bml@image@depth{91}\bmlDescription{A free floating snowman diagram}+\frac{1}{% 2!}\bigg{[}\includegraphics{bmlimages/notes-92.svg}\bml@image@depth{92}% \bmlDescription{A free floating snowman diagram}\bigg{]}^{2}+\frac{1}{3!}\bigg% {[}\includegraphics{bmlimages/notes-93.svg}\bml@image@depth{93}\bmlDescription% {A free floating snowman diagram}\bigg{]}^{3}+\cdots\Bigg{)}$ +12!⁢[ ]2+13!⁢[ ]3+⋯)\displaystyle\qquad\times\Bigg{(}1+\includegraphics{bmlimages/notes-91.svg}% \bml@image@depth{91}\bmlDescription{A free floating snowman diagram}+\frac{1}{% 2!}\bigg{[}\includegraphics{bmlimages/notes-92.svg}\bml@image@depth{92}% \bmlDescription{A free floating snowman diagram}\bigg{]}^{2}+\frac{1}{3!}\bigg% {[}\includegraphics{bmlimages/notes-93.svg}\bml@image@depth{93}\bmlDescription% {A free floating snowman diagram}\bigg{]}^{3}+\cdots\Bigg{)}
	$\displaystyle\qquad\times\Bigg{(}1+\includegraphics{bmlimages/notes-94.svg}% \bml@image@depth{94}\bmlDescription{A free floating melon with a tadpole}+% \frac{1}{2!}\bigg{[}\includegraphics{bmlimages/notes-95.svg}\bml@image@depth{9% 5}\bmlDescription{A free floating melon with a tadpole}\bigg{]}^{2}+\frac{1}{3% !}\bigg{[}\includegraphics{bmlimages/notes-96.svg}\bml@image@depth{96}% \bmlDescription{A free floating melon with a tadpole}\bigg{]}^{3}+\cdots\Bigg{)}$ +12!⁢[ ]2+13!⁢[ ]3+⋯)\displaystyle\qquad\times\Bigg{(}1+\includegraphics{bmlimages/notes-94.svg}% \bml@image@depth{94}\bmlDescription{A free floating melon with a tadpole}+% \frac{1}{2!}\bigg{[}\includegraphics{bmlimages/notes-95.svg}\bml@image@depth{9% 5}\bmlDescription{A free floating melon with a tadpole}\bigg{]}^{2}+\frac{1}{3% !}\bigg{[}\includegraphics{bmlimages/notes-96.svg}\bml@image@depth{96}% \bmlDescription{A free floating melon with a tadpole}\bigg{]}^{3}+\cdots\Bigg{)}	(202)
	$\displaystyle=\Big{(}\sum\text{connected}\Big{)}\times\Big{(}1+V_{1}+\frac{1}{% 2!}V_{1}^{2}+\cdots\Big{)}\times\Big{(}1+V_{2}+\frac{1}{2!}V_{2}^{2}+\cdots% \Big{)}\times\cdots$	(203)
	$\displaystyle=\Big{(}\sum\text{connected}\Big{)}\times\prod_{i}\Big{(}\sum_{n_% {i}}\frac{1}{n_{i}!}V_{i}^{n_{i}}\Big{)}\,.$	(204)

This can now be written in terms of an exponential, summing all disconnected diagrams to all orders in $\lambda$

\displaystyle\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle\sim\Big{(}% \sum\text{connected}\Big{)}\times\prod_{i}\exp(V_{i})=\Big{(}\sum\text{% connected}\Big{)}\times\exp\Big{(}\sum_{i}V_{i}\Big{)}\,.

(205)

What we have calculated here is the numerator of the Gell-Mann-Low theorem

	$\displaystyle\langle{0}\|T\Bigg{\{}\phi_{I}(x)\phi_{I}(y)\exp\bigg{(}-i{% \displaystyle\int}_{-t}^{t}{\rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})\bigg{)% }\Bigg{\}}\|{0}\rangle=\Bigg{(}$	$\displaystyle\includegraphics{bmlimages/notes-97.svg}\bml@image@depth{97}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-98.svg}% \bml@image@depth{98}\bmlDescription{A propagator with a tadpole}+% \includegraphics{bmlimages/notes-99.svg}\bml@image@depth{99}\bmlDescription{A % propagator with a sunset}+\cdots\Bigg{)}$ + + +⋯)\displaystyle\includegraphics{bmlimages/notes-97.svg}\bml@image@depth{97}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-98.svg}% \bml@image@depth{98}\bmlDescription{A propagator with a tadpole}+% \includegraphics{bmlimages/notes-99.svg}\bml@image@depth{99}\bmlDescription{A % propagator with a sunset}+\cdots\Bigg{)}
		$\displaystyle\times\exp\Bigg{(}\includegraphics{bmlimages/notes-100.svg}% \bml@image@depth{100}\bmlDescription{A free floating bow-tie diagram}+% \includegraphics{bmlimages/notes-101.svg}\bml@image@depth{101}\bmlDescription{% A free floating snowman diagram}+\includegraphics{bmlimages/notes-102.svg}% \bml@image@depth{102}\bmlDescription{A free floating melon with a tadpole}+% \cdots\Bigg{)}\,.$ + + +⋯).\displaystyle\times\exp\Bigg{(}\includegraphics{bmlimages/notes-100.svg}% \bml@image@depth{100}\bmlDescription{A free floating bow-tie diagram}+% \includegraphics{bmlimages/notes-101.svg}\bml@image@depth{101}\bmlDescription{% A free floating snowman diagram}+\includegraphics{bmlimages/notes-102.svg}% \bml@image@depth{102}\bmlDescription{A free floating melon with a tadpole}+% \cdots\Bigg{)}\,.		(206)

We can now consider the denominator which has the same structure but no $\phi(x)$ and $\phi(y)$ . We can use the same logic to show that

\displaystyle\langle{0}|T\Bigg{\{}\exp\bigg{(}-i{\displaystyle\int}_{-t}^{t}{% \rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})\bigg{)}\Bigg{\}}|{0}\rangle=\exp% \Bigg{(}\includegraphics{bmlimages/notes-103.svg}\bml@image@depth{103}% \bmlDescription{A free floating bow-tie diagram}+\includegraphics{bmlimages/no% tes-104.svg}\bml@image@depth{104}\bmlDescription{A free floating snowman % diagram}+\includegraphics{bmlimages/notes-105.svg}\bml@image@depth{105}% \bmlDescription{A free floating melon with a tadpole}+\cdots\Bigg{)}\,.

+⋯).\displaystyle\langle{0}|T\Bigg{\{}\exp\bigg{(}-i{\displaystyle\int}_{-t}^{t}{% \rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})\bigg{)}\Bigg{\}}|{0}\rangle=\exp% \Bigg{(}\includegraphics{bmlimages/notes-103.svg}\bml@image@depth{103}% \bmlDescription{A free floating bow-tie diagram}+\includegraphics{bmlimages/no% tes-104.svg}\bml@image@depth{104}\bmlDescription{A free floating snowman % diagram}+\includegraphics{bmlimages/notes-105.svg}\bml@image@depth{105}% \bmlDescription{A free floating melon with a tadpole}+\cdots\Bigg{)}\,.

(207)

Therefore, the disconnected contributions cancel and we are left with

\displaystyle\langle{\Omega}|T\big{\{}\phi(x)\phi(y)\big{\}}|{\Omega}\rangle=% \text{sum of all connected diagrams with two external points}\,.

(208)

Note that the diagrams we drew in (174) are not disconnected because they are still connected to some external points and therefore do not factor out. This is why the disconnected diagrams are sometimes called vacuum bubbles or vacuum-to-vacuum transitions.

Deriving Feynman diagrams like this is not very efficient. Instead, we usually go the other way around and draw all possible diagrams and then use Feynman rules in position space


For each internal line	$\displaystyle=D_{F}(x-y)\,,$	(209a)
For each vertex	$\displaystyle=-i\lambda\int{\rm d}^{4}z\,,$	(209b)
For each external line	$\displaystyle=1\,,$	(209c)
Divide by the symmetry factor.		(209d)

We obtain a factor $1/n!$ from the Taylor expansion which cancels with the $n!$ ways of arranging the $n$ vertices. Further, there are $n!$ ways to arrange the lines going into a vertex which cancels with the $n!$ in $\mathcal{L}_{I}=\lambda/4!\phi^{4}$ so that our vertex rule is just $\lambda$ . After these factors are accounted for we usually still overcounted the diagram. To avoid this, we add the diagram’s symmetry factor $S$ which are for example explicitly written in (196) Formally, the symmetry factor is $S=|G|$ the order of the symmetry group $G$ of the diagram that keeps the external lines fixed. A more practical set of rules that will cover almost all use cases is

•

lines that start and end in the same vertex, add a factor of 2
•

$n$ propagators connecting the same two vertices, add a factor of $n!$
•

if two vertices are equivalent, add another factor of 2

The rules as formulated above are valid in position space. Often it is more suitable to have them in momentum space where we write the propagator $D_{F}$ as a Fourier transform (cf. (123))

\displaystyle D_{F}(x-y)

\displaystyle=\int\frac{{\rm d}^{4}p}{(2\pi)^{4}}\frac{1}{p^{2}-m^{2}+i% \epsilon}{\rm e}^{-ip\cdot(x-y)}\,.

(210)

We now assign a momentum $p$ to a propagator and split the factor ${\rm e}^{-ip\cdot(x-y)}$ into both ends of the line. This means that for internal vertices we now have

\displaystyle\includegraphics{bmlimages/notes-109.svg}\bml@image@depth{109}% \bmlDescription{A vertex with four momenta going out: p1, p2, p3, p4}\quad\sim% \int{\rm d}^{4}z{\rm e}^{ip_{1}\cdot z}{\rm e}^{ip_{2}\cdot z}{\rm e}^{ip_{3}% \cdot z}{\rm e}^{ip_{4}\cdot z}=(2\pi)^{4}\delta^{(4)}(-p_{1}-p_{2}-p_{3}-p_{4% })\,.

A vertex with four momenta going out: p1, p2, p3, p4

∼∫d4zei⁢p1⋅zei⁢p2⋅zei⁢p3⋅zei⁢p4⋅z=(2π)4δ(4)(−p1−p2−p3−p4).\displaystyle\includegraphics{bmlimages/notes-109.svg}\bml@image@depth{109}% \bmlDescription{A vertex with four momenta going out: p1, p2, p3, p4}\quad\sim% \int{\rm d}^{4}z{\rm e}^{ip_{1}\cdot z}{\rm e}^{ip_{2}\cdot z}{\rm e}^{ip_{3}% \cdot z}{\rm e}^{ip_{4}\cdot z}=(2\pi)^{4}\delta^{(4)}(-p_{1}-p_{2}-p_{3}-p_{4% })\,.

(211)

In other words, momentum is conserved at each vertex. With this, we now have our momentum-space Feynman rules


For each internal line	$\displaystyle=\frac{i}{p^{2}-m^{2}+i\epsilon}\,,$	(212a)
For each vertex	$\displaystyle=-i\lambda\,,$	(212b)
For each external line	$\displaystyle={\rm e}^{-ip\cdot x}\,,$	(212c)
Impose momentum conservation,		(212d)
Integrate over unconstrained momenta	$\displaystyle\int\frac{{\rm d}^{4}p}{(2\pi)^{4}}\,,$	(212e)
Divide by the symmetry factor.		(212f)

Suggested Exercise

Calculate the symmetry factors of the following diagrams

$\displaystyle S$	$\displaystyle=2$	(213)
$\displaystyle S$	$\displaystyle=16$	(214)
$\displaystyle S$	$\displaystyle=6$	(215)
$\displaystyle S$	$\displaystyle=12$	(216)
$\displaystyle S$	$\displaystyle=8$	(217)

Suggested Exercise

Write down all diagrams that contribute to the four-point function $\langle{\Omega}|T\{\phi_{1}\phi_{2}\phi_{3}\phi_{4}\}|{\Omega}\rangle$ up to $\lambda^{2}$ and calculate the amplitude for $\phi(p_{1})\phi(p_{2})\to\phi(p_{3})\phi(p_{4})$ .

At $\lambda^{1}$ we have a single diagram

\displaystyle\includegraphics{bmlimages/notes-118.svg}\bml@image@depth{118}% \bmlDescription{External lines from two positions, x1 and x2, flowing into a % vertex and two lines flowing out to positions x3 and x4}\quad=-i\lambda\times{% \rm e}^{ip_{1}x_{1}}{\rm e}^{ip_{2}x_{2}}{\rm e}^{-ip_{3}x_{3}}{\rm e}^{-ip_{4% }x_{4}}\delta^{(4)}(p_{1}+p_{2}-p_{3}-p_{4})

=−iλ×ei⁢p1⁢x1ei⁢p2⁢x2e−i⁢p3⁢x3e−i⁢p4⁢x4δ(4)(p1+p2−p3−p4)\displaystyle\includegraphics{bmlimages/notes-118.svg}\bml@image@depth{118}% \bmlDescription{External lines from two positions, x1 and x2, flowing into a % vertex and two lines flowing out to positions x3 and x4}\quad=-i\lambda\times{% \rm e}^{ip_{1}x_{1}}{\rm e}^{ip_{2}x_{2}}{\rm e}^{-ip_{3}x_{3}}{\rm e}^{-ip_{4% }x_{4}}\delta^{(4)}(p_{1}+p_{2}-p_{3}-p_{4})

(218)

At $\lambda^{2}$ , we have four interesting diagrams

		$\displaystyle=(-i\lambda)^{2}\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}% -m^{2}+i\epsilon}\frac{1}{(k+p_{1}+p_{2})^{2}-m^{2}+i\epsilon}\times\text{% delta \& exp}\,,$		(219)
		$\displaystyle=(-i\lambda)^{2}\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}% -m^{2}+i\epsilon}\frac{1}{(k+p_{1}-p_{3})^{2}-m^{2}+i\epsilon}\times\text{% delta \& exp}\,,$		(220)
		$\displaystyle=(-i\lambda)^{2}\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}% -m^{2}+i\epsilon}\frac{1}{(k+p_{1}-p_{4})^{2}-m^{2}+i\epsilon}\times\text{% delta \& exp}\,.$		(221)

There are also diagrams where a correction is applied to the external line

\displaystyle\includegraphics{bmlimages/notes-122.svg}\bml@image@depth{122}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the top-right % leg}+\includegraphics{bmlimages/notes-123.svg}\bml@image@depth{123}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the bottom-right% leg}+\includegraphics{bmlimages/notes-124.svg}\bml@image@depth{124}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the bottom-left % leg}+\includegraphics{bmlimages/notes-125.svg}\bml@image@depth{125}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the top-left leg% }\,.

.\displaystyle\includegraphics{bmlimages/notes-122.svg}\bml@image@depth{122}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the top-right % leg}+\includegraphics{bmlimages/notes-123.svg}\bml@image@depth{123}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the bottom-right% leg}+\includegraphics{bmlimages/notes-124.svg}\bml@image@depth{124}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the bottom-left % leg}+\includegraphics{bmlimages/notes-125.svg}\bml@image@depth{125}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the top-left leg% }\,.

(222)

We will discuss these momentarily.

Suggested Exercise

Consider the $\phi^{3}$ theory

\displaystyle\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-\frac{1}{2}m^{2}% \mu^{2}\phi^{2}-\frac{\lambda}{3!}\phi^{3}\,.

(223)

Calculate the symmetry factor of the following diagram through the Wick theorem

\displaystyle\includegraphics{bmlimages/notes-126.svg}\bml@image@depth{126}% \bmlDescription{A propagator with a bubble on it, i.e. two lines}\,.

.\displaystyle\includegraphics{bmlimages/notes-126.svg}\bml@image@depth{126}% \bmlDescription{A propagator with a bubble on it, i.e. two lines}\,.

(224)

Convince yourself that its Feynman rules are the same as in $\phi^{4}$ except for the vertex

For each vertex

\displaystyle=-i\lambda\,,

(225)

Calculate $\langle{\Omega}|T\{\phi_{1}\phi_{2}\phi_{3}\phi_{4}\}|{\Omega}\rangle$ up to $\lambda^{1}$ .

3.6 Returning to the $\mathcal{S}$ matrix

To use our formalism to calculate $\mathcal{S}$ matrix elements we would have to repeat the derivation of the Gell-Mann-Low theorem for states other than $|{\Omega}\rangle$ . However, we had to use the fact that the vacuum is the state with the lowest energy which will not be true for any state that contains particles we would like to scatter. It is possible to rescue this argument but how to do this goes well beyond the scope of this course. Instead, the following construction should motivate why we might think that we can calculate $\mathcal{S}$ matrix elements using Feynman diagrams.

Consider the $\mathcal{S}$ matrix element (cf. (142)) between an outgoing state $f$ (composed of $n$ particles with momenta $\vec{p}_{i}$ ) and an incoming state $i$ (composed of $m$ particles with momenta $\vec{q}_{i}$ )

\displaystyle{}_{i}\langle{f}|\mathcal{S}|{i}\rangle_{i}=\langle{\vec{p}_{1}% \cdots\vec{p}_{n}}|\mathcal{S}|{\vec{q}_{1}\cdots\vec{q}_{m}}\rangle\propto% \langle{\vec{p}_{1}\cdots\vec{p}_{n}}|T\Bigg{\{}\exp\bigg{(}-i{\displaystyle% \int}_{-t}^{t}{\rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})\bigg{)}\Bigg{\}}|{% \vec{q}_{1}\cdots\vec{q}_{m}}\rangle\,.

(226)

The $\propto$ indicates that we have ignored the equivalent denominator of the Gell-Mann-Low formula (189) in that hopes that it will cancel again if we only consider connected diagrams. Since the external states can be expressed through field operators and the vacuum, we can make them part of the Wick contractions that defined the Feynman diagrams.

However, this will lead to plenty of diagrams where nothing of interest happens like the ones in (174). To avoid this, we define the interesting part of the $\mathcal{S}$ matrix

\displaystyle\mathcal{S}=1+i(2\pi)^{4}\delta^{(4)}(P_{i}-P_{f})\mathcal{T}\,,

(227)

where $\mathcal{T}$ contains only the connected diagrams and the $1$ all the cases where no interaction takes place. Since our Feynman rules imply momentum conservation, we have made this explicit already here⁶⁶ 6 Sometimes you will see this factor to be defined as part of the $\mathcal{T}$ matrix instead..

There is one more restriction on the types of diagrams that enter $\mathcal{T}$ . Consider the following diagram which is fully connected

\displaystyle\includegraphics{bmlimages/notes-128.svg}\bml@image@depth{128}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the top-right % leg}\propto\frac{1}{p^{\prime 2}-m^{2}}\delta^{(4)}(p^{\prime}-p_{3})=\frac{1}% {p_{3}^{2}-m^{2}}=\frac{1}{0}\,,

Diagram as at λ^1 but with a tadpole on the top-right leg

∝1p′⁣2−m2⁢δ(4)⁢(p′−p3)=1p32−m2=10,\displaystyle\includegraphics{bmlimages/notes-128.svg}\bml@image@depth{128}% \bmlDescription{Diagram as at \lambda^1 but with a tadpole on the top-right % leg}\propto\frac{1}{p^{\prime 2}-m^{2}}\delta^{(4)}(p^{\prime}-p_{3})=\frac{1}% {p_{3}^{2}-m^{2}}=\frac{1}{0}\,,

(228)

with the intermediary momentum $p^{\prime}$ . Momentum conservation forces $p^{\prime}=p_{3}$ and because we want $p_{3}$ to be on-shell, i.e. $p_{3}^{2}=m^{2}$ , we now have a singularity. This is quite a big problem. $\mathcal{S}$ will only make sense if we exclude this type of diagram where a loop is attached to an external leg like this. One can show that these types of diagrams are similar to the vacuum bubbles we have already excluded in (208). Diagrams that do not have this problem are called amputated.

Let us summarise our achievement

\displaystyle\langle{\vec{p}_{1}\cdots\vec{p}_{n}}|\mathcal{T}|{\vec{q}_{1}% \cdots\vec{q}_{m}}\rangle=\sum\Big{(}\text{amputated \& connected Feynman % diagrams}\Big{)}\,.

(229)

There is one final subtly related to this that we will revisit later in Section A.

	$\displaystyle\phi^{(+)}(x)\|{p}\rangle$	$\displaystyle=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}{\rm e}^{-ik\cdot x}\sqrt{% \frac{2E_{\vec{p}}}{2E_{\vec{k}}}}a_{\rm in}(\vec{k})a_{\rm in}^{\dagger}(p)\|{% 0}\rangle=\int\frac{{\rm d}^{3}k}{(2\pi)^{3}}{\rm e}^{-ik\cdot x}\sqrt{\frac{2% E_{\vec{p}}}{2E_{\vec{k}}}}[a_{\rm in}(\vec{k}),a_{\rm in}^{\dagger}(p)]\|{0}\rangle$		(177)
		$\displaystyle={\rm e}^{-ip\cdot x}\|{0}\rangle\,,$		(178)

	$\displaystyle\|{\Omega}\rangle$	$\displaystyle=\lim_{t\to\infty(1-i\delta)}\Big{(}{\rm e}^{-iE_{0}(t+t^{\prime}% )}\langle{\Omega}\|{0}\rangle\Big{)}^{-1}{\rm e}^{-iH(t+t^{\prime})}\|{0}\rangle$		(182)
		$\displaystyle=\lim_{t\to\infty(1-i\delta)}\Big{(}{\rm e}^{-iE_{0}(t^{\prime}-(% -t))}\langle{\Omega}\|{0}\rangle\Big{)}^{-1}\underbrace{{\rm e}^{-iH(t^{\prime}% -(-t))}{\rm e}^{iH_{0}(t+t^{\prime})}}_{U(t^{\prime},-t)}\|{0}\rangle\,.$		(183)

$\displaystyle\langle{\Omega}\|T\big{\{}\phi(x)\phi(y)\big{\}}\|{\Omega}\rangle\sim$
$\displaystyle\langle{0}\|T\Bigg{\{}\phi(x)\phi(y)$	$\displaystyle-i\phi(x)\phi(y)\int{\rm d}t^{\prime}\ \tilde{H}_{I}(t^{\prime})-% \frac{1}{2!}\phi(x)\phi(y)\int{\rm d}t^{\prime}{\rm d}t^{\prime\prime}\ \tilde% {H}_{I}(t^{\prime})H_{I}(t^{\prime\prime})+\cdots\Bigg{\}}\|{0}\rangle$	(190)
$\displaystyle=\langle{0}\|T\Bigg{\{}\phi(x)\phi(y)$	$\displaystyle+i\phi(x)\phi(y)\int{\rm d}^{4}z\ \mathcal{L}_{I}(z)-\frac{1}{2!}% \phi(x)\phi(y)\int{\rm d}^{4}z{\rm d}^{4}w\ \mathcal{L}_{I}(z)\mathcal{L}_{I}(% w)+\cdots\Bigg{\}}\|{0}\rangle\,.$	(191)

3 Interacting scalar field

3.1 Interaction picture

3.2 The 𝒮\mathcal{S} matrix

3.3 Wick theorem

3.4 Asymptotic states and the interacting vacuum

3.5 Feynman diagrams

3.6 Returning to the 𝒮\mathcal{S} matrix

3.2 The $\mathcal{S}$ matrix

3.6 Returning to the $\mathcal{S}$ matrix