4  Types of Fields and Their Particles

So far we have focussed on the QFT of a real scalar field \(\varphi\) with quartic interactions. The aim was to go into detail about the calculational techniques of QFT, which are largely the same for all QFTs, without being distracted by nuances of particular field types. In this chapter we will look at some of the features that vary between different field types.

4.1 Complex scalar fields

4.1.1 Recap: complex conjugates in quantum mechanics

Consider the single-particle Schroedinger equation

\[i\frac{\partial\psi}{\partial t}=\hat{H}\psi.\]

Assume the Hamiltonian is real: this is always possible in the absence of a magnetic field. We can take the complex conjugate to give

\[\begin{aligned} -i\frac{\partial\psi^{*}}{\partial t} & =\hat{H}\psi^{*} \end{aligned}\]

or equivalently

\[i\frac{\partial\psi^{*}}{\partial\left(-t\right)}=\hat{H}\psi^{*}.\]

This tells us a general property of quantum theories: complex conjugation is equivalent to time reversal. If \(\psi\) evolves according to the TDSE, then \(\psi^{*}\) evolves according to the same equation with time reversed. Now consider the current density associated with a particle of charge \(q\)

\[\boldsymbol{j}\left(\boldsymbol{x},t\right)=\frac{q}{2m}\left\{ \psi^{*}\hat{\boldsymbol{p}}\psi-\psi\hat{\boldsymbol{p}}\psi^{*}\right\}\]

which obeys the continuity equation

\[\dot{\rho}=-\nabla\cdot\boldsymbol{j}\]

which says that if the charge density \(\rho\) changes in a region it must be due to a flow of current into or out of the region. Under complex conjugation \(\boldsymbol{j}\) switches sign. Hence, a flow of \(\psi\) particles into the region is equivalent to a flow of \(\psi^{*}\) particles out of the region, and vice versa. Equivalently, we can force \(\boldsymbol{j}\) to keep the same sign under time reversal by also switching charge \(q\rightarrow-q\).

This reasoning suggests that we can think of \(\psi^{*}\) as being like the antiparticle to \(\psi\), in the sense that it is associated to the opposite charge but same mass. Similarly, a flow of particles forwards in time is equivalent to a flow of antiparticles backwards in time. This is not too mysterious: it just says that a flow of positive charge in one direction is equivalent to a flow of negative charge in the opposite direction.

NB this is again really only an analogy in single-particle quantum mechanics. Antiparticles only become really meaningful in QFT.

Let’s see how this works in QFT.

4.1.2 Complex Klein Gordon action

We can define a Klein Gordon theory over complex scalar fields. Now \(\varphi\) and \(\varphi^{*}\) are treated as independent fields. The action is

\[S^{\mathbb{C}}_{\text{KG}}\left[\varphi^{*},\varphi\right]=\int\text{d}^{4}x\left(\partial^{\mu}\varphi^{*}\partial_{\mu}\varphi-m^{2}\varphi^{*}\varphi\right).\]

Note that it is conventional to multiply by two relative to Eq 2.11. Now we can find two Euler Lagrange equations, by taking functional derivatives with respect to our two different fields \(\varphi\) and \(\varphi^{*}\). We start by extremising the action with respect to \(\varphi^{*}\).

We can do this explicitly, as in Section 2.4.1. But for the sake of variety let’s instead use the functional differentiation method.

\[\begin{aligned} \frac{\delta S^{\mathbb{C}}_{\text{KG}}\left[\varphi^{*},\varphi\right]}{\delta\varphi^{*}\left(y\right)} & =\frac{\delta}{\delta\varphi^{*}\left(y\right)}\int\text{d}^{4}x\left(\partial^{\mu}\varphi^{*}\left(x\right)\partial_{\mu}\varphi\left(x\right)-m^{2}\varphi^{*}\left(x\right)\varphi\left(x\right)\right)\\ & =\frac{\delta}{\delta\varphi^{*}\left(y\right)}\int\text{d}^{4}x\left(-\varphi^{*}\left(x\right)\partial^{2}\varphi\left(x\right)-m^{2}\varphi^{*}\left(x\right)\varphi\left(x\right)\right)\\ & =\int\text{d}^{4}x\left(-\frac{\delta\varphi^{*}\left(x\right)}{\delta\varphi^{*}\left(y\right)}\partial^{2}\varphi\left(x\right)-m^{2}\frac{\delta\left(\partial^{\mu}\varphi^{*}\left(x\right)\right)}{\delta\varphi^{*}\left(y\right)}\varphi\left(x\right)\right) \end{aligned}\]

now use

\[\begin{aligned} \frac{\delta\varphi\left(x\right)}{\delta\varphi\left(y\right)} & =\frac{\delta\varphi^{*}\left(x\right)}{\delta\varphi^{*}\left(y\right)}=\delta^{4}\left(x-y\right)\\ \frac{\delta\varphi^{*}\left(x\right)}{\delta\varphi\left(y\right)} & =\frac{\delta\varphi\left(x\right)}{\delta\varphi^{*}\left(y\right)}=0 \end{aligned}\]

to give

\[\begin{aligned} \frac{\delta S^{\mathbb{C}}_{\text{KG}}\left[\varphi^{*},\varphi\right]}{\delta\varphi^{*}\left(y\right)} & =\int\text{d}^{4}x\delta^{4}\left(x-y\right)\left(-\partial^{2}\varphi\left(x\right)-m^{2}\varphi\left(x\right)\right)\\ & =-\partial^{2}\varphi\left(y\right)-m^{2}\varphi\left(y\right) \end{aligned}\]

where \(\partial_{\mu}\) now means \(\partial/\partial y^{\mu}\). Setting both sides equal to zero and relabelling \(y\rightarrow x\) gives the Klein Gordon equation, as before:

\[\left(\partial^{2}+m^{2}\right)\varphi=0.\]

Instead taking the functional derivative with respect to \(\varphi\) shows that the complex conjugate field also obeys the Klein Gordon equation:

\[\frac{\delta S^{\mathbb{C}}_{\text{KG}}\left[\varphi^{*},\varphi\right]}{\delta\varphi\left(x\right)}=0\implies\left(\partial^{2}+m^{2}\right)\varphi^{*}=0.\]

The fact that the same mass enters the the equation for the particle and antiparticle is a general feature of QFTs. It is not generally true that the antiparticles obey the same equation of motion as the particles: in general, the equations are complex conjugates.

4.1.3 Creation and Annihilation Operators

Since the field and its complex conjugate are independent fields in the path integral formalism, it follows that the field operator and its hermitian conjugate are independent operators in the canonically quantized formalism. This requires a separate creation operator for particles, \(\hat{a}^{\dagger}_{+}\), and for antiparticles \(\hat{a}^{\dagger}_{-}\). Each has its own annihilation operator. The field operators can then be written

\[\begin{aligned} \hat{\varphi}_{x} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left(\hat{a}_{+,\boldsymbol{k}}\exp\left(-ik\cdot x\right)+\hat{a}^{\dagger}_{-,\boldsymbol{k}}\exp\left(ik\cdot x\right)\right)\\ \hat{\varphi}^{\dagger}_{x} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left(\hat{a}^{\dagger}_{+,\boldsymbol{k}}\exp\left(ik\cdot x\right)+\hat{a}_{-,\boldsymbol{k}}\exp\left(-ik\cdot x\right)\right). \end{aligned}\]

Here, \(\hat{\varphi}^{\dagger}_{y}\) creates a particle, or equivalently annihilates an antiparticle, at \(y\). Similarly, \(\hat{\varphi}_{x}\) annihilates a particle/creates an antiparticle at \(x\). In terms of these operators, the Hamiltonian is

\[\hat{H}=\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\omega_{\boldsymbol{k}}\left(\hat{a}^{\dagger}_{+,\boldsymbol{k}}\hat{a}^{\phantom{\dagger}}_{+,\boldsymbol{k}}+\hat{a}^{\dagger}_{-,\boldsymbol{k}}\hat{a}^{\phantom{\dagger}}_{-,\boldsymbol{k}}\right).\]

This commutes with another operator

\[\hat{Q}=\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\left(\hat{a}^{\dagger}_{+,\boldsymbol{k}}\hat{a}^{\phantom{\dagger}}_{+,\boldsymbol{k}}-\hat{a}^{\dagger}_{-,\boldsymbol{k}}\hat{a}^{\phantom{\dagger}}_{-,\boldsymbol{k}}\right).\]

Recall from QM that if an operator commutes with the Hamiltonian then it corresponds to a conserved quantity. Here, the quantity is the electric charge. This makes sense when we remember that \(\hat{a}^{\dagger}\hat{a}^{\phantom{\dagger}}\) is the number operator: the conserved charge simply counts the difference between the number of particles and antiparticles. The propagator for complex Klein Gordon theory is then given by

\[\begin{aligned} G_{xy} & =\langle\Omega|\mathcal{T}\hat{\varphi}^{\phantom{\dagger}}_{\boldsymbol{y}}\hat{\varphi}^{\dagger}_{\boldsymbol{x}}|\Omega\rangle. \end{aligned} \tag{4.1}\]

If we take the complex conjugate (recalling that \(\langle a|\hat{C}\hat{D}|b\rangle^{*}=\langle b|\hat{D}^{\dagger}\hat{C}^{\dagger}|a\rangle\)) we find:

\[\boxed{G^{*}_{xy}=G_{yx}.}\]

Hence, propagating a particle from \(x\rightarrow y\) is equivalent to propagating an antiparticle from \(y\rightarrow x\). This is the relativistic version of what we saw in single particle quantum mechanics, where the ‘antiparticle’ \(\psi^{*}\) is the time-reverse of the particle \(\psi\).

4.1.4 Wick’s Theorem in complex field theories

Let’s take a closer look at Wick’s theorem. It actually provides a neat link between time ordering and normal ordering. Since we have

\[\hat{\varphi}_{x}=\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left(\hat{a}_{+,\boldsymbol{k}}\exp\left(-ik\cdot x\right)+\hat{a}^{\dagger}_{-,\boldsymbol{k}}\exp\left(ik\cdot x\right)\right)\]

let’s define

\[\begin{aligned} \hat{\varphi}^{+}_{x} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\hat{a}_{+,\boldsymbol{k}}\exp\left(-ik\cdot x\right)\\ \hat{\varphi}^{-}_{x} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\hat{a}^{\dagger}_{-,\boldsymbol{k}}\exp\left(ik\cdot x\right) \end{aligned}\]

so that

\[\hat{\varphi}_{x}=\hat{\varphi}^{+}_{x}+\hat{\varphi}^{-}_{x}.\]

Note this implies

\[\begin{aligned} \hat{\varphi}^{+\dagger}_{x} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\hat{a}^{\dagger}_{+,\boldsymbol{k}}\exp\left(ik\cdot x\right)\\ \hat{\varphi}^{-\dagger}_{x}= & \int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\hat{a}_{-,\boldsymbol{k}}\exp\left(-ik\cdot x\right). \end{aligned}\]

Consider the time-ordered product of two fields:

\[\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}=\Theta\left(x^{0}-y^{0}\right)\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}+\Theta\left(y^{0}-x^{0}\right)\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x} \tag{4.2}\]

where, as before, \(\Theta\) is the step function. Let’s focus for now on the first term:

\[\begin{aligned} \hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y} & =\left(\hat{\varphi}^{+}_{x}+\hat{\varphi}^{-}_{x}\right)\left(\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-\dagger}_{y}\right)\\ & =\hat{\varphi}^{+}_{x}\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{-\dagger}_{y}+\hat{\varphi}^{+}_{x}\hat{\varphi}^{-\dagger}_{y}. \end{aligned}\]

To relate this to normal ordering, we need to get all creation operators to the left of all annihilation operators. The only term of the four which is not already normal ordered is the first term. So rewrite this using the commutator:

\[\begin{aligned} \hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y} & =\left[\hat{\varphi}^{+}_{x},\hat{\varphi}^{+\dagger}_{y}\right]+\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{+}_{x}+\hat{\varphi}^{+}_{x}\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{-\dagger}_{y}+\hat{\varphi}^{+}_{x}\hat{\varphi}^{-\dagger}_{y}\\ & =\left[\hat{\varphi}^{+}_{x},\hat{\varphi}^{+\dagger}_{y}\right]+:\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{+}_{x}+\hat{\varphi}^{+}_{x}\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{-\dagger}_{y}+\hat{\varphi}^{+}_{x}\hat{\varphi}^{-\dagger}_{y}:\\ & =\left[\hat{\varphi}^{+}_{x},\hat{\varphi}^{+\dagger}_{y}\right]+:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:. \end{aligned}\]

Now, the trick is to realise that the commutator is simply some function multiplying the identity operator (you can work it out if you like, but it’s not important to the argument). Hence we can re-express it as its own expectation value, multiplied by the identity:

\[\left[\hat{\varphi}^{+}_{x},\hat{\varphi}^{+\dagger}_{y}\right]=\langle\Omega|\left[\hat{\varphi}^{+}_{x},\hat{\varphi}^{+\dagger}_{y}\right]|\Omega\rangle\hat{\mathbb{I}}.\]

Next, note that of the two terms in the commutator, one annihilates the vacuum, giving

\[\begin{aligned} \left[\hat{\varphi}^{+}_{x},\hat{\varphi}^{+\dagger}_{y}\right] & =\langle\Omega|\left(\hat{\varphi}^{+}_{x}\hat{\varphi}^{+\dagger}_{y}-\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{+}_{x}\right)|\Omega\rangle\hat{\mathbb{I}}\\ & =\langle\Omega|\hat{\varphi}^{+}_{x}\hat{\varphi}^{+\dagger}_{y}|\Omega\rangle\hat{\mathbb{I}}. \end{aligned}\]

Finally, note that

\[\begin{aligned} \hat{\varphi}^{+\dagger}_{y}|\Omega\rangle & =\hat{\varphi}^{\dagger}_{y}|\Omega\rangle\\ \langle\Omega|\hat{\varphi}^{+}_{x} & =\langle\Omega|\hat{\varphi}_{x} \end{aligned}\]

since the other term contributing to \(\hat{\varphi}^{\dagger}\) or \(\hat{\varphi}\) annihilates the vacuum in each case. Hence we have the result

\[\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}=:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:+\langle\Omega|\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}|\Omega\rangle\hat{\mathbb{I}}.\]

Doing the same with the other term in Eq 4.2 gives

\[\begin{aligned} \hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x} & =\left(\hat{\varphi}^{+\dagger}_{y}+\hat{\varphi}^{-\dagger}_{y}\right)\left(\hat{\varphi}^{+}_{x}+\hat{\varphi}^{-}_{x}\right)\\ & =\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{+}_{x}+\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{-}_{x}+\hat{\varphi}^{-\dagger}_{y}\hat{\varphi}^{+}_{x}+\hat{\varphi}^{-\dagger}_{y}\hat{\varphi}^{-}_{x}\\ & =\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{+}_{x}+\hat{\varphi}^{+\dagger}_{y}\hat{\varphi}^{-}_{x}+\hat{\varphi}^{-\dagger}_{y}\hat{\varphi}^{+}_{x}+\hat{\varphi}^{-}_{x}\hat{\varphi}^{-\dagger}_{y}+\left[\hat{\varphi}^{-\dagger}_{y},\hat{\varphi}^{-}_{x}\right]\\ & =:\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}:+\langle\Omega|\left[\hat{\varphi}^{-\dagger}_{y},\hat{\varphi}^{-}_{x}\right]|\Omega\rangle\hat{\mathbb{I}}\\ & =:\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}:+\langle\Omega|\hat{\varphi}^{-\dagger}_{y}\hat{\varphi}^{-}_{x}|\Omega\rangle\hat{\mathbb{I}}\\ & =:\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}:+\langle\Omega|\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}|\Omega\rangle\hat{\mathbb{I}} \end{aligned}\]

Putting these expressions together in Eq 4.2 gives

\[\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}=\Theta\left(x^{0}-y^{0}\right)\left(:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:+\langle\Omega|\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}|\Omega\rangle\hat{\mathbb{I}}\right)+\Theta\left(y^{0}-x^{0}\right)\left(:\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}:+\langle\Omega|\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}|\Omega\rangle\hat{\mathbb{I}}\right).\]

Finally, note that \(\Theta\left(a\right)+\Theta\left(-a\right)=1\), and that \(:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:=:\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}:\). Hence,

\[\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}=:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:+\Theta\left(x^{0}-y^{0}\right)\langle\Omega|\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}|\Omega\rangle\hat{\mathbb{I}}+\Theta\left(y^{0}-x^{0}\right)\langle\Omega|\hat{\varphi}^{\dagger}_{y}\hat{\varphi}_{x}|\Omega\rangle\hat{\mathbb{I}}\]

which is simply

\[\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}=:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:+\langle\Omega|\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}|\Omega\rangle\hat{\mathbb{I}}.\]

The last term is the definition of the propagator! Therefore \[\boxed{\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}=:\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}:+G_{xy}\hat{\mathbb{I}}}.\]

This is a simple relation between time ordering and normal ordering. You can see that this is consistent by taking the expectation value of both sides: the expectation value of a normal ordered product is zero, since it has annihilation operators on the right. That is,

\[\langle\Omega|\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}^{\dagger}_{y}|\Omega\rangle=G_{xy}.\]

If you redo the arugments above, you can check that

\[\begin{aligned} \mathcal{T}\hat{\varphi}_{x}\hat{\varphi}_{y} & =:\hat{\varphi}_{x}\hat{\varphi}_{y}:\\ \mathcal{T}\hat{\varphi}^{\dagger}_{x}\hat{\varphi}^{\dagger}_{y} & =:\hat{\varphi}^{\dagger}_{x}\hat{\varphi}^{\dagger}_{y}: \end{aligned}\]

and therefore

\[\langle\Omega|\mathcal{T}\hat{\varphi}_{x}\hat{\varphi}_{y}|\Omega\rangle=\langle\Omega|\mathcal{T}\hat{\varphi}^{\dagger}_{x}\hat{\varphi}^{\dagger}_{y}|\Omega\rangle=0.\]

Continuing the argument by induction to \(N\)-point functions gives

Wick’s theorem (complex fields):

The \(N\)-point function of any complex Gaussian theory decomposes into all possible products of pairings of 2-point functions, where pairs consist of a field and a hermitian conjugate field.

Switching to path integral notation, we have that e.g. the non-interacting 4-point function

\[\langle\varphi_{4}\varphi_{3}\varphi_{2}\varphi_{1}\rangle=0\]

as all the fields are particle fields \(\varphi\) and none are \(\varphi^{*}\). However, the non-interacting 4-point function

\[\begin{aligned} \langle\varphi_{4}\varphi_{3}\varphi^{*}_{2}\varphi^{*}_{1}\rangle & =\langle\varphi_{4}\varphi^{*}_{2}\rangle\langle\varphi_{3}\varphi^{*}_{1}\rangle+\langle\varphi_{4}\varphi^{*}_{1}\rangle\langle\varphi_{3}\varphi^{*}_{2}\rangle\\ & =G_{24}G_{13}+G_{14}G_{23}. \end{aligned}\]

All other terms vanish, e.g. \(\langle\varphi_{4}\varphi_{3}\rangle=0\).

4.1.5 \(U\left(1\right)\) charge conservation

The previous equations can be interpreted another way. They say that pairs of particles or antiparticles cannot be created from the vacuum (since the expectation value of a pair of creation or annihilation operators is zero). Similarly we can interpret the Green’s function as a finite amplitude to create a particle-antiparticle pair from the vacuum.

Together, this tells us that we have a conservation law: the number of particles minus the number of antiparticles is constant in any process. But recall that an antiparticle has opposite charge to a particle. So this conservation law is simply the conservation of electric charge! Of course, our scalar field is not exactly the field of an electron, so you might ask whether it is really the electric charge we are familiar with. In general it is what we call a \(U\left(1\right)\) charge, of which electric charge is an example.

To see where the name comes from, note that charge conservation here follows from the fact that only \(\varphi^{*}\varphi\) terms appear in the action, rather than \(\varphi^{2}\) or \(\varphi^{*2}\). Hence, the field can be rescaled by an arbitrary position-dependent complex phase without changing the action:

\[\begin{aligned} \varphi_{x} & \rightarrow\varphi_{x}\exp\left(i\alpha\left(x\right)\right)\\ \varphi^{*}_{x} & \rightarrow\varphi^{*}_{x}\exp\left(-i\alpha\left(x\right)\right)\\ S^{\mathbb{C}}_{\text{KG}} & \rightarrow S^{\mathbb{C}}_{\text{KG}} \end{aligned}\]

(check this yourself: the kinetic terms seem to change, but only by a total derivative which therefore vanishes under integration). The terms \(\exp\left(i\alpha\right)\) for real \(\alpha\) describe the set of points on the unit circle; under multiplication this describes the group \(U\left(1\right)\).

4.1.6 Feynman diagrams and \(U\left(1\right)\) charge conservation

To keep track of conserved \(U\left(1\right)\) charge in Feynman diagrams, we add an arrow on the propagator indicating the forwards-in-time propagation of a particle (equivalently, the backwards-in-time propagation of an antiparticle). This reduces the symmetry factors assigned to diagrams, since there is no longer a symmetry of reversing the path direction.

4.1.7 Complex \(\varphi^{4}\) theory

We can add interactions as with the real scalar field theory. For example, complex \(\varphi^{4}\) theory is described by the interaction term

\[S_{\text{int}}=\frac{\lambda}{4}\int\left|\varphi_{x}\right|^{4}\text{d}^{D+1}x=\frac{\lambda}{4}\int\left(\varphi^{*}_{x}\varphi_{x}\right)^{2}\text{d}^{D+1}x.\]

Note that the prefactor is \(\lambda/4\), not \(\lambda/4!\) as it was in the real scalar field theory. The reason is that the symmetry factor of the diagrams is different, owing to the \(U\left(1\right)\) conservation arrows, or equivalently owing to the fact that we only contract particle-antiparticle pairs in Wick’s theorem. For the real field we had

\[\langle\mathcal{T}\varphi_{1}\varphi_{2}\varphi_{3}\varphi_{4}\rangle^{\text{connected}}_{1}=i\frac{\lambda}{4!}\int\text{d}^{D+1}x\left\langle \mathcal{T}\varphi^{4}_{x}\varphi_{1}\varphi_{2}\varphi_{3}\varphi_{4}\right\rangle _{0}=i\lambda\int\text{d}^{D+1}xG_{1x}G_{2x}G_{3x}G_{4x}\]

where the \(1/4!\) prefactor was introduced to deal with the fact that there are 4 ways to pair \(\varphi_{1}\) with a \(\varphi_{x}\), three ways to pair \(\varphi_{2}\), and so on. Now we have

\[\langle\mathcal{T}\varphi^{\dagger}_{1}\varphi^{\dagger}_{2}\varphi^{\phantom{\dagger}}_{3}\varphi^{\phantom{\dagger}}_{4}\rangle^{\mathbb{C},\text{ connected}}_{1}=i\frac{\lambda}{4}\int\text{d}^{D+1}x\left\langle \mathcal{T}\varphi^{\dagger}_{1}\varphi^{\dagger}_{2}\varphi^{\phantom{\dagger}}_{3}\varphi^{\phantom{\dagger}}_{4}\varphi^{\dagger 2}_{x}\varphi^{2}_{x}\right\rangle _{0}=i\lambda\int\text{d}^{D+1}xG_{1x}G_{2x}G_{x3}G_{x4}.\]

Here, \(\varphi^{\dagger}_{1}\) only has a choice of two \(\varphi_{x}\), and \(\varphi_{3}\) has a choice of two \(\varphi^{\dagger}_{x}\). The remaining terms are fixed. With this normalisation it has symmetry factor 1, as required (since all 4 legs are external).

4.2 Noether’s Theorem and Conservation Laws

The \(U\left(1\right)\) charge conservation in the previous section can be understood as a special case of one of the most profound results in theoretical physics:

Noether’s Theorem: every continuous symmetry has a corresponding conservation law, and vice versa.

4.2.1 The Euler Lagrange Equations (Relativistic Notation)

We can derive the Euler Lagrange equations in relativistic notation in full generality. Using a real scalar field for simplicity (the argument extends to all other field types), the trick is to notice that the Lagrange density is a function of \(\varphi\) and \(\partial_{\mu}\varphi\), and so any change in the action can be written in terms of changes of these quantities via the chain rule. In particular:

\[\begin{aligned} S\left[\varphi\right] & =\int\text{d}^{D}x\mathcal{L}\left(\varphi,\partial_{\mu}\varphi\right)\\ S\left[\varphi+\lambda\epsilon\right] & =\int\text{d}^{D}x\mathcal{L}\left(\varphi+\lambda\epsilon,\partial_{\mu}\varphi+\lambda\partial_{\mu}\epsilon\right)\\ & \Downarrow\nonumber \\ \frac{\partial S\left[\varphi+\lambda\epsilon\right]}{\partial\lambda} & =\int\text{d}^{D}x\frac{\partial\mathcal{L}}{\partial\left(\varphi+\lambda\epsilon\right)}\frac{\partial\left(\varphi+\lambda\epsilon\right)}{\partial\lambda}+\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi+\lambda\partial_{\mu}\epsilon\right)}\frac{\partial\left(\partial_{\mu}\varphi+\lambda\partial_{\mu}\epsilon\right)}{\partial\lambda}\\ & =\int\text{d}^{D}x\frac{\partial\mathcal{L}}{\partial\left(\varphi+\lambda\epsilon\right)}\epsilon+\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi+\lambda\partial_{\mu}\epsilon\right)}\partial_{\mu}\epsilon\\ & =\int\text{d}^{D}x\left(\frac{\partial\mathcal{L}}{\partial\left(\varphi+\lambda\epsilon\right)}-\partial_{\mu}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi+\lambda\partial_{\mu}\epsilon\right)}\right)\epsilon \end{aligned}\]

and the principle of least action requires

\[\left.\frac{\partial S\left[\varphi+\lambda\epsilon\right]}{\partial\lambda}\right|_{\lambda=0}=0=\int\text{d}^{D}x\left(\frac{\partial\mathcal{L}}{\partial\varphi}-\partial_{\mu}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi\right)}\right)\epsilon.\]

Since this is true for all \(\epsilon\left(x\right)\) matching the boundary conditions, we require the Euler Lagrange equations:

\[\frac{\partial\mathcal{L}}{\partial\varphi}-\partial_{\mu}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi\right)}=0.\]

4.2.2 Noether’s theorem

Now consider a transformation of the fields which leaves the action invariant (the definition of a symmetry). The Lagrange density is free to change, but since the spacetime integral of the Lagrange density is the action, the Lagrange density can only change by a total derivative:

\[\begin{aligned} \varphi & \rightarrow\varphi+\delta\varphi\\ \mathcal{L} & \rightarrow\mathcal{L}+\delta\mathcal{L}\triangleq\mathcal{L}+\partial_{\mu}\Lambda^{\mu}\\ S & \rightarrow S. \end{aligned}\]

Then Noether’s theorem states that the current

\[j^{\mu}=\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\delta\varphi-\Lambda^{\mu}\]

is locally conserved:

\[\partial_{\mu}j^{\mu}=0.\]

This is easily checked:

\[\begin{aligned} \partial_{\mu}j^{\mu} & =\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\delta\varphi\right)-\partial_{\mu}\Lambda^{\mu}\\ & =\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\right)\delta\varphi+\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\partial_{\mu}\delta\varphi-\partial_{\mu}\Lambda^{\mu}\\ & =\frac{\partial\mathcal{L}}{\partial\varphi}\delta\varphi+\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\partial_{\mu}\delta\varphi-\partial_{\mu}\Lambda^{\mu} \end{aligned}\]

where we used the Euler Lagrange equation to rewrite the first term. We notice that since

\[\mathcal{L}=\mathcal{L}\left(\varphi,\partial_{\mu}\varphi\right)\]

the first two terms are simply the chain rule expression for \(\delta\mathcal{L}\), and so

\[\partial_{\mu}j^{\mu}=\delta\mathcal{L}-\partial_{\mu}\Lambda^{\mu}.\]

Hence the divergence vanishes provided \(\delta\mathcal{L}=\partial_{\mu}\Lambda^{\mu}\). The proof of Noether’s theorem is simply this verification run in reverse. We define

\[\delta\mathcal{L}\triangleq\partial_{\mu}\Lambda^{\mu}\]

and we then use the chain rule

\[\delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\varphi}\delta\varphi+\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\left(\partial_{\mu}\delta\varphi\right)\]

and the Euler Lagrange equation on the first term to give

\[\delta\mathcal{L}=\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\right)\delta\varphi+\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\left(\partial_{\mu}\delta\varphi\right)\]

which can be rewritten

\[\delta\mathcal{L}=\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\delta\varphi\right).\]

Equating the two expressions for \(\delta\mathcal{L}\) gives the result

\[\partial_{\mu}\Lambda^{\mu}=\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\delta\varphi\right)\]

or

\[\partial_{\mu}j^{\mu}=0\]

where

\[j^{\mu}=\frac{\partial\mathcal{L}}{\partial\partial_{\mu}\varphi}\delta\varphi-\Lambda^{\mu}.\]

Every Noether current has a corresponding conserved Noether charge:

\[Q\triangleq\int\text{d}^{D}\boldsymbol{x}j^{0}\left(x\right).\] To see that this charge is indeed conserved, note that conservation means \(\dot{Q}=0\):

\[\dot{Q}=\int\text{d}^{D}\boldsymbol{x}\partial_{0}j^{0}\left(x\right)\]

and since Noether’s theorem states

\[\partial_{\mu}j^{\mu}=\partial_{0}j^{0}-\nabla\cdot\boldsymbol{j}=0\]

we can rewrite this as

\[\dot{Q}=\int\text{d}^{D}\boldsymbol{x}\nabla\cdot\boldsymbol{j}\left(x\right).\]

But this is again a total derivative, which vanishes for appropriate boundary conditions.

Note that the derivation of Noether’s theorem used the Euler Lagrange equations, which govern classical trajectories. Hence, Noether’s theorem is classical. The quantum version of Noether’s theorem can be found by promoting fields to field operators, giving a conserved current density operator \(\hat{j}^{\mu}\) and conserved charge operator \(\hat{Q}\) whose eigenvalues are the conserved charges. The result is called the Ward Takahashi identity. Typically we expect that a classically conserved quantity is also conserved quantum mechanically; when this is not the case, we call the result an anomaly.

4.2.3 Example: \(U\left(1\right)\) charge

The motivating example was the \(U\left(1\right)\) symmetry of complex scalar fields:

\[\begin{aligned} \varphi\left(x\right) & \rightarrow\exp\left(i\alpha\right)\varphi\left(x\right)\\ \varphi^{*}\left(x\right) & \rightarrow\exp\left(-i\alpha\right)\varphi^{*}\left(x\right) \end{aligned}\]

with \(\alpha\in\mathbb{R}\). To get this in the right form we can Taylor expand:

\[\varphi\rightarrow\exp\left(i\alpha\right)\varphi\approx\varphi+i\alpha\varphi\]

and so

\[\begin{aligned} \delta\varphi & =i\alpha\varphi\\ \delta\varphi^{*} & =-i\alpha\varphi^{*}. \end{aligned}\]

Under this change our Lagrange density

\[\mathcal{L}=\partial^{\mu}\varphi^{*}\partial_{\mu}\varphi-m^{2}\varphi^{*}\varphi-V\left(\varphi^{*}\varphi\right)\]

is totally invariant, so

\[\Lambda^{\mu}=0.\]

We need the complex version of Noether’s theorem (simply adjust the above argument to account for the fact that \(\varphi\) and \(\varphi^{*}\) are independent fields):

\[j^{\mu}=\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi\right)}\delta\varphi\left(x\right)+\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi^{*}\right)}\delta\varphi^{*}\left(x\right)-\Lambda^{\mu}\left(x\right)\]

to give

\[j^{\mu}=i\left(\varphi\partial^{\mu}\varphi^{*}-\varphi^{*}\partial^{\mu}\varphi\right)\]

(\(\alpha\) played the role of \(\epsilon\)) which is the relativistic version of the probability current density familiar from non-relativistic quantum mechanics. The associated Noether charge is

\[Q=i\int\text{d}^{D}\boldsymbol{x}\left(\varphi\partial^{0}\varphi^{*}-\varphi^{*}\partial^{0}\varphi\right).\]

To make sense of this, look instead at the quantum charge operator:

\[:\hat{Q}:=i\int\text{d}^{D}\boldsymbol{x}:\hat{\varphi}\partial^{0}\hat{\varphi}^{\dagger}-\hat{\varphi}^{\dagger}\partial^{0}\hat{\varphi}:\]

where we have specified normal ordering to avoid the usual ambiguity of operator order when quantizing. We can write the field in terms of creation and annihilation operators:

\[\begin{aligned} \hat{\varphi}_{\boldsymbol{x}} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left(\hat{a}_{+,\boldsymbol{k}}\exp\left(-ik\cdot x\right)+\hat{a}^{\dagger}_{-,\boldsymbol{k}}\exp\left(ik\cdot x\right)\right)\\ \hat{\varphi}^{\dagger}_{\boldsymbol{x}} & =\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\frac{1}{\sqrt{2\omega_{\boldsymbol{k}}}}\left(\hat{a}^{\dagger}_{\boldsymbol{+,k}}\exp\left(ik\cdot x\right)+\hat{a}_{-,\boldsymbol{k}}\exp\left(-ik\cdot x\right)\right). \end{aligned}\]

Inserting into the expression for the charge operator gives

\[:\hat{Q}:=\int\frac{\text{d}^{D}\boldsymbol{k}}{\left(2\pi\right)^{D}}\left(\hat{a}^{\dagger}_{+,\boldsymbol{k}}\hat{a}_{+,\boldsymbol{k}}-\hat{a}^{\dagger}_{-,\boldsymbol{k}}\hat{a}_{-,\boldsymbol{k}}\right).\]

Hence we see that the conserved charge operator is simply the difference in the number operators of the particles and antiparticles. Hence, its eigenvalues, the conserved charges, are the number of particles minus the number of antiparticles.

4.2.4 Example: the Stress Energy Tensor

One symmetry we always require of our relativistic field theories is invariance under spacetime translations:

\[\varphi\left(x\right)\rightarrow\varphi\left(x+\epsilon\right)\approx\varphi\left(x\right)+\epsilon^{\nu}\partial_{\nu}\varphi\left(x\right).\]

This time our variation has an index:

\[\delta\varphi_{\nu}=\partial_{\nu}\varphi.\]

If we allow for an explicit spacetime dependence in the Lagrangian density (the most general case, which we have previously neglected for simplicity):

\[\mathcal{L}\left(\varphi,\partial_{\mu}\varphi,x^{\mu}\right)\rightarrow\mathcal{L}+\epsilon^{\nu}\partial_{\nu}\mathcal{L}=\mathcal{L}+\epsilon^{\nu}\partial_{\mu}\left(\delta^{\mu}_{\nu}\mathcal{L}\right)\]

where the rightmost expression is written to make the total derivative clear:

\[\Lambda^{\mu}_{\nu}=\delta^{\mu}_{\nu}\mathcal{L}.\]

Hence the Noether current is

\[j^{\mu}_{\phantom{\mu}\nu}\triangleq\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi\right)}\partial_{\nu}\varphi-\delta^{\mu}_{\nu}\mathcal{L}.\]

This is conventionally labelled \(T^{\mu}_{\phantom{\mu}\nu}\), the energy momentum tensor (also called the stress energy tensor). Its conserved charges

\[Q_{\nu}=\int\text{d}^{D}\boldsymbol{x}T^{0}_{\phantom{0}\nu}\]

are the energy \(\nu=0\) and momentum. For example,

\[Q_{0}=\int\text{d}^{D}\boldsymbol{x}\left(\frac{\partial\mathcal{L}}{\partial\dot{\varphi}}\dot{\varphi}-\mathcal{L}\right)=\int\text{d}^{D}\boldsymbol{x}\mathcal{H}=H\]

is simply the Hamiltonian.

4.2.5 General results

If this is the first time you have seen Noether’s theorem, it is worth thinking about its profundity. Every conserved quantity has an associated continuous symmetry, and vice versa. This works whether the theory is quantum or classical, relativistic or not. Some symmetries and their conservation laws are given in Table 1.

Symmetries and their corresponding conserved charges according to Noether’s theorem. Each charge is locally conserved, meaning that if it changes within a region, there must be a flow of the corresponding current density into or out of that region.

symmetry	conserved charge
translation	momentum
time translation	energy
\(U\left(1\right)\)	probability (QM) / particle number (QFT)
rotation	angular momentum

Initially this seems quite shocking: energy conservation is equivalent to time translation symmetry: the statement that the laws of physics are the same from instant to instant. But looked at from another perspective this makes perfect sense; after all, what we mean by ‘symmetry’ is that we change something but the system remains unchanged. Not changing is the definition of being conserved!

4.3 Bosons and Fermions

Recall the general argument for the possible behaviours of particles under exchange. Define the exchange operator

\[\hat{P}|1,2\rangle\triangleq|2,1\rangle.\]

Since

\[\hat{P}^{2}|1,2\rangle=|1,2\rangle\]

it follows that

\[\hat{P}=\pm\hat{\mathbb{I}}.\]

Hence, particles are either symmetric or antisymmetric under exchange. We call the former bosons, and the latter fermions.

4.3.1 Bosons

The particles we have considered until now are all bosons, excitations of bosonic fields. To see this, recall that we started from the harmonic oscillator, whose \(n^{\text{th}}\) excited state we interpreted in second quantized language as \(n\) particles. Similarly, we considered it acceptable to raise a given mode in our Fock space as many times as we liked, giving e.g. \(|24300\ldots\rangle\). Only bosons have the property that multiple particles can exist in the same quantum state (with the same set of quantum numbers). Implicit in this statement is the relation that

bosonic ladder operators obey:

\[\begin{aligned} \left[\hat{a}^{\dagger}_{k},\hat{a}^{\dagger}_{l}\right]=\left[\hat{a}_{k},\hat{a}_{l}\right] & =0\\ \left[\hat{a}^{\phantom{\dagger}}_{k},\hat{a}^{\dagger}_{l}\right] & =\delta_{kl}. \end{aligned}\]

For example, Fock states like \(|24300\ldots\rangle\) can only be meaningful if it doesn’t matter in which order we applied the creation operators:

\[|243\rangle=\left(\hat{a}^{\dagger}_{k=1}\right)^{2}\left(\hat{a}^{\dagger}_{k=2}\right)^{4}\left(\hat{a}^{\dagger}_{k=3}\right)^{3}|\Omega\rangle=\left(\hat{a}^{\dagger}_{k=3}\right)^{3}\left(\hat{a}^{\dagger}_{k=1}\right)^{2}\left(\hat{a}^{\dagger}_{k=2}\right)^{4}|\Omega\rangle=\text{\emph{etc}.}\]

This is the same argument behind quantum numbers in single particle QM. For example, you can’t label a ket with both position and momentum, since the operators don’t commute so they cannot be simultaneously well defined.

4.3.2 Fermions

For fermions we must instead define a new operator called

the anticommutator:

\[\left\{ A,B\right\} \triangleq AB+BA.\]

Using this, we have that

fermionic creation and annihilation operators obey:

\[\begin{aligned} \left\{ \hat{c}^{\dagger}_{k},\hat{c}^{\dagger}_{l}\right\} =\left\{ \hat{c}_{k},\hat{c}_{l}\right\} & =0\\ \left\{ \hat{c}^{\phantom{\dagger}}_{k},\hat{c}^{\dagger}_{l}\right\} & =\delta_{kl}. \end{aligned}\]

Defining

\[\hat{c}^{\dagger}|\Omega\rangle=|1\rangle\]

it follows immediately that we can only have 0 or 1 particles in a given quantum state, since

\[\begin{aligned} \left\{ \hat{c}^{\dagger},\hat{c}^{\dagger}\right\} & =0\\ & \Downarrow\nonumber \\ \hat{c}^{\dagger}\hat{c}^{\dagger} & =-\hat{c}^{\dagger}\hat{c}^{\dagger} \end{aligned}\]

and so if we try to define

\[|2\rangle\stackrel{?}{\triangleq}\hat{c}^{\dagger}\hat{c}^{\dagger}|\Omega\rangle\]

we find

\[\begin{aligned} \hat{c}^{\dagger}\hat{c}^{\dagger}|\Omega\rangle & =-\hat{c}^{\dagger}\hat{c}^{\dagger}|\Omega\rangle\\ |2\rangle & =-|2\rangle \end{aligned}\]

which implies that \(|2\rangle\equiv0\), where this is really \(0\) and not the vacuum. Hence, trying to raise a fermionic state a second time annihilates it. This is the Pauli exclusion principle.

4.3.3 Grassman variables & integrals [Peskin & Schroeder Sec 9.5]

Recall that when we move from the canonical quantization picture to the path integral picture, we replace non-commuting field operators \(\hat{\varphi}_{\boldsymbol{x}}\) with classical commuting field variables \(\varphi_{x}\), which assign a real number \(\varphi_{x}\) at each spacetime position \(x\). Quantum non-commutation then arises from the fractal nature of the field configurations being functionally integrated over.

All of this carries over to fermionic fields, with commutators replaced with anti-commutators. This is quite strange: fermionic fields \(\psi_{x}\) in the path integral picture assign a number to each point in space. But two of these numbers must now anti-commute! This wasn’t a problem with bosonic fields, since both real and complex numbers commute.

We call anti-commuting numbers Grassman variables. They obey \(\theta\eta=-\eta\theta\). An immediate corollary is that \(\theta^{2}=\theta^{n>1}=0\). As a result of this, Grassman algebra is actually very simple. For example, any function can be defined by its Taylor series (as with real functions), but now that Taylor series must terminate at the linear term! For \(a\), \(b\in\mathbb{C}\) we have:

\[f\left(\theta\right)\equiv a+b\theta\]

where \(\equiv\) indicates that this is an exact equality. Of particular interest is the function

\[\exp\left(-\overline{\theta}A\theta\right)=1-\overline{\theta}A\theta\]

where \(\overline{\theta}\) is the complex conjugate of \(\theta\) (an independent variable, so that \(\theta\overline{\theta}=-\overline{\theta}\theta\ne0\)). Grassman integrals¹ have some freedom in their definition, but the standard choice leads to:

\[\begin{aligned} \int\text{d}\theta & =0\\ \int\theta\text{d}\theta & =1. \end{aligned}\]

As a result, the Gaussian Grassman integral takes a simple form:

\[\begin{aligned} \int\text{d}\overline{\theta}\int\text{d}\theta\exp\left(-\overline{\theta}A\theta\right) & =\int\text{d}\overline{\theta}\int\text{d}\theta\left(1-\overline{\theta}A\theta\right)\\ & =\int\text{d}\overline{\theta}\int\text{d}\theta-\int\text{d}\overline{\theta}\int\text{d}\theta\overline{\theta}A\theta\\ & =\int\text{d}\overline{\theta}\int\text{d}\theta+A\int\overline{\theta}\text{d}\overline{\theta}\int\theta\text{d}\theta\\ & =A. \end{aligned}\]

Similarly, we have the Grassman Gaussian functional integral:

\[\lim_{N\rightarrow\infty}\prod^{N}_{i=1}\prod^{N}_{j=1}\int\text{d}\overline{\theta}_{i}\text{d}\theta_{j}\exp\left(-\overline{\theta}_{i}A_{ij}\theta_{j}\right)=\det\left(A\right)\]

perhaps written more familiarly over fields \(\psi_{x}\) as

\[\int\mathscr{D}\overline{\psi}\mathscr{D}\psi\exp\left(-\int\text{d}^{D+1}x\int\text{d}^{D+1}y\overline{\psi}_{x}\hat{A}_{xy}\psi_{y}\right)=\det\left(\hat{A}\right).\]

This is proportional to the inverse of the the bosonic Gaussian functional integral

\[\lim_{N\rightarrow\infty}\prod^{N}_{i=1}\prod^{N}_{j=1}\int\text{d}\overline{\varphi}_{i}\text{d}\varphi_{j}\exp\left(-\overline{\varphi}_{i}A_{ij}\varphi_{j}\right)=\frac{\left(2\pi\right)^{N}}{\det\left(A\right)}.\]

The Grassman result is so simple that you will often find theories in which determinants play a key role, otherwise unrelated to QFT, rewritten in ‘free fermion’ form. That is, you rewrite the determinant as a Grassman Gaussian functional integral, and interpret the physical theory as a QFT of non-interacting fermions! As with bosonic integrals we have \[\begin{aligned} \lim_{N\rightarrow\infty}\prod^{N}_{i=1}\prod^{N}_{j=1}\int\text{d}\overline{\theta}_{i}\text{d}\theta_{j}\theta_{k}\overline{\theta}_{l}\exp\left(-\overline{\theta}_{i}A_{ij}\theta_{j}\right) & =\det\left(A\right)A^{-1}_{kl}\\ \lim_{N\rightarrow\infty}\prod^{N}_{i=1}\prod^{N}_{j=1}\int\text{d}\overline{\theta}_{i}\text{d}\theta_{j}\theta_{k}\overline{\theta}_{l}\theta_{m}\overline{\theta}_{n}\exp\left(-\overline{\theta}_{i}A_{ij}\theta_{j}\right) & =\det\left(A\right)\left(A^{-1}_{kl}A^{-1}_{mn}+A^{-1}_{kn}A^{-1}_{ml}\right) \end{aligned}\]

and so on: Wick’s theorem applies as before, with the only change being that it is \(\det A\) rather than \(\det A^{-1}\) which appears as the prefactor.

4.3.4 Quantum statistics of identical non-interacting particles

Using the exchange properties of bosons and fermions, we can immediately derive their respective exchange statistics. Working in the grand canonical ensemble, the grand partition function is

\[\mathscr{Z}\left(\mu,V,T\right)=\sum^{\infty}_{n=0}\exp\left(-\beta\left(E_{n}-N_{n}\mu\right)\right)\]

where \(\beta=1/T\), \(N_{n}\) is the number of particles in microstate \(n\), and \(E_{n}\) is the energy of that (\(V\) is the volume, which will be held fixed). The mean particle number is then given by

\[\langle N\rangle=\beta^{-1}\left(\frac{\partial\ln\mathscr{Z}}{\partial\mu}\right)_{V,T}.\]

For non-interacting identical bosons the energy to add each particle is the same: \(\epsilon-\mu\). Hence

\[\mathscr{Z}_{\text{boson}}=\sum^{\infty}_{n=0}\exp\left(-n\beta\left(\epsilon-\mu\right)\right).\]

Recognising the geometric series, we have

\[\mathscr{Z}_{\text{boson}}=\left(1-\exp\left(-\beta\left(\epsilon-\mu\right)\right)\right)^{-1}\]

and the mean particle number \(\langle N\rangle\) is given by

the Bose Einstein distribution function:

\[n_{B}\left(\mu,\epsilon,T\right)=\frac{1}{\exp\left(\beta\left(\epsilon-\mu\right)\right)-1}.\]

These are Bose-Einstein statistics. At low energies \(\beta\left(\epsilon-\mu\right)\rightarrow0^{+}\) this function diverges: the ground state of a set of identical non-interacting bosons has space for an infinite number of particles. At low energies the particles condense into the ground state, simply by their statistics. The result is a Bose Einstein Condensate (BEC). Examples include superfluids (e.g. helium-4 cooled below 2.17K: helium-4 is a composite spin-0 boson) and superconductors (in which fermionic electrons bind into bosonic pairs, which can then condense). Some refer to Bose Einstein condensation as a macroscopic quantum phenomenon.

For fermions we can only have 0 or 1 particle, and so \[\begin{aligned} \mathscr{Z}_{\text{fermion}} & =\sum^{1}_{n=0}\exp\left(-n\beta\left(\epsilon-\mu\right)\right)\\ & =1+\exp\left(-\beta\left(\epsilon-\mu\right)\right) \end{aligned}\]

and the mean particle number is given by

the Fermi Dirac distribution function:

\[f\left(\mu,\epsilon,T\right)=\frac{1}{\exp\left(\beta\left(\epsilon-\mu\right)\right)+1}.\]

These are Fermi-Dirac statistics. Adding fermions to a system, each quantum state can only contain one particle. At zero temperature, the first particle goes into the lowest energy state, the second particle into the second state, and so on. The result is the ‘Fermi sea’: a set of filled states at low energy, with a sharp cut-off to a set of empty states at higher energies. At higher temperatures, the cut-off broadens, and some of the highest-energy particles move to higher energy states, leaving the low energy states occupied. This situation describes the electrons in a metal, for example.

4.3.5 Bogoliubov Transformations

Sometimes the most natural way to describe a physical system involves writing down operators that do not conserve particle number. For example, we might find a Hamiltonian like

\[\hat{H}_{\text{boson}}=\epsilon\left(\hat{a}^{\dagger}_{1}\hat{a}^{\phantom{\dagger}}_{1}+\hat{a}^{\dagger}_{2}\hat{a}^{\phantom{\dagger}}_{2}\right)+\lambda\left(\hat{a}^{\dagger}_{1}\hat{a}^{\dagger}_{2}+\hat{a}^{\phantom{\dagger}}_{1}\hat{a}^{\phantom{\dagger}}_{2}\right).\]

Acting on a state of \(n_{i}\) particles of type \(i\) (\(i=1,2\)) gives

\[\hat{H}_{\text{boson}}|n_{1},n_{2}\rangle=\epsilon\left(n_{1}+n_{2}\right)|n_{1},n_{2}\rangle+\lambda\left(|n_{1}+1,n_{2}+1\rangle+|n_{1}-1,n_{2}-1\rangle\right).\]

Hamiltonians of this form can always be ‘diagonalised’ by rewriting in terms of new creation and annihilation operators which the commutation relations but which only involve number conserving terms. This is called a Bogoliubov transformation.

First note that we can use the bosonic commutation relations

\[\begin{aligned} \left[\hat{a}^{\phantom{\dagger}}_{i},\hat{a}^{\phantom{\dagger}}_{j}\right] & =\left[\hat{a}^{\dagger}_{i},\hat{a}^{\dagger}_{j}\right]=0\\ \left[\hat{a}^{\phantom{\dagger}}_{i},\hat{a}^{\dagger}_{j}\right] & =\delta_{ij} \end{aligned}\]

to rewrite the Hamiltonian as

\[\begin{aligned} \hat{H}_{\text{boson}} & =\frac{\epsilon}{2}\left(\hat{a}^{\dagger}_{1}\hat{a}^{\phantom{\dagger}}_{1}+\hat{a}^{\phantom{\dagger}}_{1}\hat{a}^{\dagger}_{1}+\hat{a}^{\dagger}_{2}\hat{a}^{\phantom{\dagger}}_{2}+\hat{a}^{\phantom{\dagger}}_{2}\hat{a}^{\dagger}_{2}\right)-\epsilon\\ & +\frac{\lambda}{2}\left(\hat{a}^{\dagger}_{1}\hat{a}^{\dagger}_{2}+\hat{a}^{\dagger}_{2}\hat{a}^{\dagger}_{1}+\hat{a}^{\phantom{\dagger}}_{1}\hat{a}^{\phantom{\dagger}}_{2}+\hat{a}^{\phantom{\dagger}}_{2}\hat{a}^{\phantom{\dagger}}_{1}\right) \end{aligned}\]

and can then put this into matrix form:

\[2\hat{H}_{\text{boson}}+2\epsilon=\left(\hat{a}^{\dagger}_{1},\hat{a}^{\phantom{\dagger}}_{2},\hat{a}^{\dagger}_{2},\hat{a}^{\phantom{\dagger}}_{1}\right)\left(\begin{array}{cccc} \epsilon & \lambda & 0 & 0\\ \lambda & \epsilon & 0 & 0\\ 0 & 0 & \epsilon & \lambda\\ 0 & 0 & \lambda & \epsilon \end{array}\right)\left(\begin{array}{c} \hat{a}^{\phantom{\dagger}}_{1}\\ \hat{a}^{\dagger}_{2}\\ \hat{a}^{\phantom{\dagger}}_{2}\\ \hat{a}^{\dagger}_{1} \end{array}\right).\]

Focus on the top \(2\times2\) block:

\[2\hat{H}^{2\times2}_{\text{boson}}+2\epsilon=\left(\hat{a}^{\dagger}_{1},\hat{a}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \epsilon & \lambda\\ \lambda & \epsilon \end{array}\right)\left(\begin{array}{c} \hat{a}^{\phantom{\dagger}}_{1}\\ \hat{a}^{\dagger}_{2} \end{array}\right).\]

We can diagonalise this using:

\[\left(\begin{array}{c} \hat{a}^{\phantom{\dagger}}_{1}\\ \hat{a}^{\dagger}_{2} \end{array}\right)=\left(\begin{array}{cc} \alpha & \beta\\ \gamma & \delta \end{array}\right)\left(\begin{array}{c} \hat{b}^{\phantom{\dagger}}_{1}\\ \hat{b}^{\dagger}_{2} \end{array}\right).\]

But we must maintain the bosonic commutation relations. This requires:

\[\begin{aligned} \left[\hat{a}^{\phantom{\dagger}}_{1},\hat{a}^{\dagger}_{1}\right] & =1\\ & \downarrow\\ \left[\alpha\hat{b}^{\phantom{\dagger}}_{1}+\beta\hat{b}^{\dagger}_{2},\alpha^{*}\hat{b}^{\dagger}_{1}+\beta^{*}\hat{b}^{\phantom{\dagger}}_{2}\right] & =1\\ \left[\alpha\hat{b}^{\phantom{\dagger}}_{1},\alpha^{*}\hat{b}^{\dagger}_{1}\right]+\left[\beta\hat{b}^{\dagger}_{2},\beta^{*}\hat{b}^{\phantom{\dagger}}_{2}\right] & =1\\ \left|\alpha\right|^{2}-\left|\beta\right|^{2} & =1 \end{aligned}\]

and similarly

\[\left[\hat{a}^{\phantom{\dagger}}_{2},\hat{a}^{\dagger}_{2}\right]=1\implies\left|\delta\right|^{2}-\left|\gamma\right|^{2}=1.\]

A convenient parameterisation is therefore

\[\left(\begin{array}{c} \hat{a}^{\phantom{\dagger}}_{1}\\ \hat{a}^{\dagger}_{2} \end{array}\right)=\left(\begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array}\right)\left(\begin{array}{c} \hat{b}^{\phantom{\dagger}}_{1}\\ \hat{b}^{\dagger}_{2} \end{array}\right).\]

Hence we have

\[\begin{aligned} 2\hat{H}^{2\times2}_{\text{boson}}+2\epsilon & =\left(\hat{b}^{\dagger}_{1},\hat{b}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array}\right)\left(\begin{array}{cc} \epsilon & \lambda\\ \lambda & \epsilon \end{array}\right)\left(\begin{array}{cc} \cosh\theta & \sinh\theta\\ \sinh\theta & \cosh\theta \end{array}\right)\left(\begin{array}{c} \hat{b}^{\phantom{\dagger}}_{1}\\ \hat{b}^{\dagger}_{2} \end{array}\right)\\ & =\left(\hat{b}^{\dagger}_{1},\hat{b}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \epsilon\cosh2\theta+\lambda\sinh2\theta & \epsilon\sinh2\theta+\lambda\cosh2\theta\\ \epsilon\sinh2\theta+\lambda\cosh2\theta & \epsilon\cosh2\theta+\lambda\sinh2\theta \end{array}\right)\left(\begin{array}{c} \hat{b}^{\phantom{\dagger}}_{1}\\ \hat{b}^{\dagger}_{2} \end{array}\right). \end{aligned}\]

Choosing

\[\tanh2\theta=-\lambda/\epsilon\]

gives

\[\begin{aligned} \hat{H}^{2\times2}_{\text{boson}} & =\frac{1}{2}\left(\hat{b}^{\dagger}_{1},\hat{b}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \sqrt{\epsilon^{2}-\lambda^{2}} & 0\\ 0 & \sqrt{\epsilon^{2}-\lambda^{2}} \end{array}\right)\left(\begin{array}{c} \hat{b}^{\phantom{\dagger}}_{1}\\ \hat{b}^{\dagger}_{2} \end{array}\right)-\epsilon \end{aligned}\]

or

\[\hat{H}^{2\times2}_{\text{boson}}=\frac{1}{2}\sqrt{\epsilon^{2}-\lambda^{2}}\left(\hat{b}^{\dagger}_{1}\hat{b}^{\phantom{\dagger}}_{1}+\hat{b}^{\dagger}_{2}\hat{b}^{\phantom{\dagger}}_{2}\right)-\epsilon+\frac{1}{2}\sqrt{\epsilon^{2}-\lambda^{2}}.\]

Hence, in terms of the new \(\hat{b}^{\dagger}_{i}\) quasiparticles, the Hamiltonian is diagonal.

We can do the same for Fermions by switching commutators for anti-commutators. Defining

\[\begin{aligned} \hat{H}_{\text{fermion}} & =\epsilon\left(\hat{c}^{\dagger}_{1}\hat{c}^{\phantom{\dagger}}_{1}+\hat{c}^{\dagger}_{2}\hat{c}^{\phantom{\dagger}}_{2}\right)+\lambda\left(\hat{c}^{\dagger}_{1}\hat{c}^{\dagger}_{2}+\hat{c}^{\phantom{\dagger}}_{2}\hat{c}^{\phantom{\dagger}}_{1}\right) \end{aligned}\]

(note the switch of order in the final term, required for Hermiticity) and using the anti-commutation relations

\[\begin{aligned} \left\{ \hat{c}^{\phantom{\dagger}}_{i},\hat{c}^{\phantom{\dagger}}_{j}\right\} & =\left\{ \hat{c}^{\dagger}_{i},\hat{c}^{\dagger}_{j}\right\} =0\\ \left\{ \hat{c}^{\phantom{\dagger}}_{i},\hat{c}^{\dagger}_{j}\right\} & =\delta_{ij} \end{aligned}\]

to rewrite as

\[\begin{aligned} \hat{H}_{\text{fermion}} & =\frac{\epsilon}{2}\left(\hat{c}^{\dagger}_{1}\hat{c}^{\phantom{\dagger}}_{1}-\hat{c}^{\phantom{\dagger}}_{1}\hat{c}^{\dagger}_{1}+\hat{c}^{\dagger}_{2}\hat{c}^{\phantom{\dagger}}_{2}-\hat{c}^{\phantom{\dagger}}_{2}\hat{c}^{\dagger}_{2}\right)+\epsilon+\frac{\lambda}{2}\left(\hat{c}^{\dagger}_{1}\hat{c}^{\dagger}_{2}-\hat{c}^{\dagger}_{2}\hat{c}^{\dagger}_{1}+\hat{c}^{\phantom{\dagger}}_{2}\hat{c}^{\phantom{\dagger}}_{1}-\hat{c}^{\phantom{\dagger}}_{1}\hat{c}^{\phantom{\dagger}}_{2}\right) \end{aligned}\]

and

\[2\hat{H}_{\text{fermion}}-2\epsilon=\left(\hat{c}^{\dagger}_{1},\hat{c}^{\phantom{\dagger}}_{2},\hat{c}^{\dagger}_{2},\hat{c}^{\phantom{\dagger}}_{1}\right)\left(\begin{array}{cccc} \epsilon & \lambda & 0 & 0\\ \lambda & -\epsilon & 0 & 0\\ 0 & 0 & \epsilon & -\lambda\\ 0 & 0 & -\lambda & -\epsilon \end{array}\right)\left(\begin{array}{c} \hat{c}^{\phantom{\dagger}}_{1}\\ \hat{c}^{\dagger}_{2}\\ \hat{c}^{\phantom{\dagger}}_{2}\\ \hat{c}^{\dagger}_{1} \end{array}\right).\]

Let’s again consider the upper \(2\times2\) block (the other now takes a slightly different form):

\[2\hat{H}^{2\times2}_{\text{fermion}}-2\epsilon=\left(\hat{c}^{\dagger}_{1},\hat{c}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \epsilon & \lambda\\ \lambda & -\epsilon \end{array}\right)\left(\begin{array}{c} \hat{c}^{\phantom{\dagger}}_{1}\\ \hat{c}^{\dagger}_{2} \end{array}\right).\]

This time it turns out (using the same reasoning as before) that to maintain anticommutation relations requires

\[\left(\begin{array}{c} \hat{c}^{\phantom{\dagger}}_{1}\\ \hat{c}^{\dagger}_{2} \end{array}\right)=\left(\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right)\left(\begin{array}{c} \hat{d}^{\phantom{\dagger}}_{1}\\ \hat{d}^{\dagger}_{2} \end{array}\right).\]

This gives

\[2\hat{H}^{2\times2}_{\text{fermion}}-2\epsilon=\cos2\theta\left(\hat{d}^{\dagger}_{1},\hat{d}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \epsilon-\lambda\tan2\theta & \epsilon\tan2\theta+\lambda\\ \epsilon\tan2\theta+\lambda & \lambda\tan2\theta-\epsilon \end{array}\right)\left(\begin{array}{c} \hat{d}^{\phantom{\dagger}}_{1}\\ \hat{d}^{\dagger}_{2} \end{array}\right)\]

and choosing

\[\tan2\theta=-\lambda/\epsilon\]

gives

\[2\hat{H}^{2\times2}_{\text{fermion}}-2\epsilon=\left(\hat{d}^{\dagger}_{1},\hat{d}^{\phantom{\dagger}}_{2}\right)\left(\begin{array}{cc} \sqrt{\epsilon^{2}+\lambda^{2}} & 0\\ 0 & -\sqrt{\epsilon^{2}+\lambda^{2}} \end{array}\right)\left(\begin{array}{c} \hat{d}^{\phantom{\dagger}}_{1}\\ \hat{d}^{\dagger}_{2} \end{array}\right).\]

This Fermionic case occurs in superconductors. Here, the two species of fermion are electrons with opposite crystal momentum. These pair up to form bosonic ‘Cooper pairs’, created by \(\hat{c}^{\dagger}_{k}\hat{c}^{\dagger}_{-k}\), which then Bose-condense in their ground state. There is an energy gap separating the ground state from the first excited state, which is the energy required to break one of these Cooper pairs back into fermions. Interestingly, however, the Bogoliubov transformation shows that the fermions formed by breaking a Cooper pair are not electrons! They are superpositions of electrons and holes:

\[\hat{d}^{\dagger}_{1}=\cos\theta\hat{c}^{\phantom{\dagger}}_{k}-\sin\theta\hat{c}^{\dagger}_{-k}.\]

But electrons and holes have opposite charge (the are antiparticles of one another, in a condensed matter setting). Hence these particles, called Bogoliubons, have an electric charge that is in a quantum superposition! When measured, they are found to have any charge from \(e\) to \(-e\).

4.3.6 The spin-statistics theorem

The exchange operator argument explains that there must be exactly two types of particle in 3D: those that are symmetric or antisymmetric under exchange. We call these bosons and fermions. As we saw in Section X, their exchange properties lead to different statistics: Bose-Einstein for bosons, Fermi-Dirac for fermions.

However, bosons and fermions are also classified by their spin: bosons have integer spin, while fermions have odd-half-integer spin. This perfect correspondence between spin and statistics was for a long time called the spin-statistics connection. In fact it has now been proven, using axiomatic QFT, to be a consequence of CPT symmetry and Lorentz invariance, but the proof is very complicated. The result is the spin-statistics theorem.

Many suspect there should be a simpler explanation for the spin-statistics connection. In a sense, a single spin\(-S\) particle has to be rotated through \(2\pi/S\) before it returns to its original orientation. Hence, the connection says something about the relationship between the symmetry of a single particle under rotation, versus rotating a pair of particles into one another. So far, no simple explanation of spin-statistics has been identified.

The spin of a particle is deeply connected to the symmetries of spacetime. Specifically, spin labels which irreducible representation of the Poincare group a particle lives in. You will see more of this in the second part of the course. For now, we will proceed to list some basic properties of the different irreps. Table 2 provides a summary.

Particles labelled by spin. If we succeed in quantizing gravity, we know that the particle excitations will be spin-2.

spin	statistics	field	examples
0	boson	scalar	Higgs boson
1/2	fermion	spinor	electron, proton
1	boson	vector	\(W^{\pm}\), \(Z\) bosons
2	boson	tensor	graviton?

4.4 A Field Guide to Fields

4.4.1 Spin-0: Scalar fields

Spin-0 fields are Lorentz scalars:

\[\varphi\left(x\right)\rightarrow\varphi'\left(x'\right)=\varphi\left(\Lambda^{-1}x\right)=\varphi\left(\Lambda^{\nu}_{\phantom{}\mu}x^{\mu}\right)\]

(i.e. the field itself does not transform under the Lorentz transformation). They have been our main focus until now. The non-interacting action is the Klein Gordon action. For a real field this is:

\[\mathcal{L}_{\text{spin}-0}=-\frac{1}{2}\partial^{\mu}\varphi\partial_{\mu}\varphi-\frac{1}{2}m^{2}\varphi^{2}.\]

Its Euler Lagrange equation is the Klein Gordon equation, which governs the motion of a non-interacting massive spin-0 particle:

\[\left(\partial^{2}+m^{2}\right)\varphi=0.\]

The only spin-0 particle in the standard model is the Higgs boson, which obeys the Klein Gordon equation with an added potential (and also interacts with other particles).

4.4.2 Spin-1: Vector fields

Spin-1 fields transform as vectors under Lorentz transformations:

\[A^{\mu}\left(x\right)\rightarrow A'^{\mu}\left(x'\right)=\Lambda^{\mu}_{\phantom{}\nu}A^{\nu}\left(\Lambda^{-1}x\right).\]

The corresponding Lagrange density is

\[\mathcal{L}_{\text{spin-1}}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^{2}A^{\mu}A_{\mu}\]

where we have used the field strength:

\[F^{\mu\nu}\triangleq\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}.\]

The corresponding Euler-Lagrange equation is the Proca equation

\[\partial_{\nu}F^{\mu\nu}+m^{2}A^{\mu}=0.\]

This describes massive spin-1 particles, such as the \(W^{+}\), \(W^{-}\), and \(Z\) bosons.

4.4.3 Abelian Gauge fields (spin-1)

With \(m=0\), the Proca equation becomes Maxwell’s equations:

\[\begin{aligned} \mathcal{L}_{\text{Maxwell}} & =-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}\\ & \downarrow\nonumber \\ \partial_{\nu}F^{\mu\nu} & =0. \end{aligned}\]

These describe a massless spin-1 particle: the photon. In this case (only) the action features a gauge symmetry: the action is invariant under a gauge transformation \(A^{\mu}\left(x\right)\rightarrow A^{\mu}\left(x\right)+\partial^{\mu}\lambda\left(x\right)\). In electromagnetism this has a simple interpretation. We have

\[A^{\mu}=\left(\begin{array}{c} \phi\\ \boldsymbol{A} \end{array}\right)^{\mu}\]

where \(\phi\) is the voltage and \(\boldsymbol{A}\) is the magnetic vector potential. That is:

\[\begin{aligned} \boldsymbol{B} & =\nabla\times\boldsymbol{A}\\ \boldsymbol{E} & =-\nabla V-\dot{\boldsymbol{A}}. \end{aligned}\]

Hence the gauge transformation is equivalent to

\[\left(\begin{array}{c} \phi\\ \boldsymbol{A} \end{array}\right)\rightarrow\left(\begin{array}{c} \phi+\partial^{0}\lambda\\ \boldsymbol{A}+\nabla\lambda \end{array}\right)\]

which leaves \(\boldsymbol{B}\) and \(\boldsymbol{E}\) unchanged. The invariance of the action to the gauge transformation encodes the fact that the absolute value of voltage has no meaning: only potential differences are physical. Hence we can shift all voltages in the universe by an amount \(\partial^{0}\lambda\) and must get the same physical results. Similarly, we can shift the magnetic vector potential by a pure gradient and get the same physical situation.

Maxwell theory, of a massless spin-1 field, is therefore an example of an Abelian gauge theory. Gauge fields are of particular importance in QFT, since it is potentials, rather than forces, which appear in quantum mechanics. For example, the non-relativistic Hamiltonian of a single particle in a magnetic field is

\[\hat{H}=\frac{1}{2m}\left(\hat{p}-i\hat{A}\right)^{2}+\hat{V}.\]

This carries over to QFT.

4.4.4 Non-Abelian gauge fields (spin-1)

We can promote any field to a vector of fields simply by adding an index. For example, for a scalar field:

\[\varphi_{a}\left(x\right)=\left(\begin{array}{c} \varphi_{1}\left(x\right)\\ \varphi_{2}\left(x\right)\\ \vdots\\ \varphi_{N}\left(x\right) \end{array}\right).\]

This index is not a spacetime index, so this does not describe a spin-1 field. It is simply a convenient notation for handling multiple independent fields. You can, for example, rewrite a single complex scalar field as a vector of real fields with an appropriate metric. We can just as easily add a vector index to a spin-1 field, to describe \(N\) independent spin-1 fields:

\[A^{\mu}_{a}\left(x\right)=\left(\begin{array}{c} A^{\mu}_{1}\left(x\right)\\ A^{\mu}_{2}\left(x\right)\\ \vdots\\ A^{\mu}_{N}\left(x\right) \end{array}\right).\]

Such fields appear in the Standard Model. Here, the vector space indexed by \(a\) is the non-Abelian gauge group \(G=SU\left(2\right)\) or \(SU\left(3\right)\); then \(N=\dim G\). The result is described by Yang Mills theory:

\[\mathcal{L}_{\text{Yang-Mills}}=-\frac{1}{4}F^{\mu\nu}_{a}F^{a}_{\mu\nu}\]

where

\[F^{\mu\nu}_{a}\triangleq\partial^{\mu}A^{\nu}_{a}-\partial^{\nu}A^{\mu}_{a}+gf^{abc}A^{\mu}_{b}A^{\nu}_{c}.\]

The corresponding Euler-Lagrange equation is

\[D_{\nu}F^{\mu\nu}=0\]

where

\[D^{\mu}\triangleq\partial^{\mu}-igT^{a}A^{\mu}_{a}.\]

Here \(T^{a}\) are the generators, and \(f^{abc}\) the structure constants, of the Lie algebra of \(G\):

\[\left[T^{a},T^{b}\right]=if^{abc}T^{c}\]

(since these are not spacetime indices they can be raised and lowered without any change). For the simplest case of \(G=SU\left(2\right)\), we can choose as generators the Pauli matrices \(\sigma\). Then we have the familiar relation

\[\left[\sigma^{a},\sigma^{b}\right]=2i\epsilon^{abc}\sigma^{c}\]

where \(\epsilon^{abc}\) is antisymmetric in any pair of indices. Non-Abelian gauge fields describe quarks and gluons in the Standard Model: called quantum chromodynamics, the gauge group is \(SU\left(3\right)\). Additionally, electroweak theory describing \(W\) and \(Z\) particles and the photon starts off as an \(SU\left(2\right)\) non-Abelian gauge theory, although the story here is complicated by the Higgs mechanism.

Note that non-abelian gauge theories always contain self-interactions. Expanding the Yang-Mills Lagrange density reveals \(A^{3}\) and \(A^{4}\) terms, while the equation of motion contains \(A^{2}\) and \(A^{3}\) terms and is therefore nonlinear. It cannot be solved by Fourier transform, as was the case with Klein Gordon theory.

4.4.5 Spin-1/2

The spin-1/2 field is called a spinor field. The Lagrange density is

\[\mathcal{L}_{\text{spin-1/2}}\left(\overline{\psi},\psi\right)=\overline{\psi}_{a}\left(i\gamma^{\mu}_{ab}\partial_{\mu}-m\mathbb{I}_{ab}\right)\psi_{b}\]

where

\[\overline{\psi}_{a}\triangleq\psi^{\dagger}_{ab}\gamma^{0}_{b}\]

and the corresponding Euler Lagrange equation is the Dirac equation

\[\left(i\gamma^{\mu}_{ab}\partial_{\mu}-m\mathbb{I}_{ab}\right)\psi_{b}=0.\]

Here, the gamma matrices are any matrices obeying the anti-commutation relations:

\[\left\{ \gamma^{\mu},\gamma^{\nu}\right\} =2\eta^{\mu\nu}\mathbb{I}\]

which defines the Clifford algebra \(\text{Cl}_{1,3}\left(\mathbb{R}\right)\). The smallest dimension of representation is 4, giving \(4\times4\) matrices. In fact any \(4\times4\) representation of this algebra is unitarily equivalent to any other:

\[\gamma^{\mu'}=S^{\dagger}\gamma^{\mu}S\]

for some unitary \(4\times4\) matrix \(S\). One choice, the Weyl representation, is

\[\begin{aligned} \gamma^{0} & =\sigma^{3}\otimes\mathbb{I}_{2}\\ \gamma^{j} & =i\sigma^{2}\otimes\sigma^{j}. \end{aligned}\]

Dirac derived his equation by taking the ‘square root’ of the Klein Gordon equation, and working out a set of matrices which allowed this to work. Hence, you can square the Dirac equation to return to the Klein Gordon equation:

\[\begin{aligned} \left(i\gamma^{\mu}_{ab}\partial_{\mu}-m\mathbb{I}_{ab}\right)\psi_{b} & =0\\ & \downarrow\nonumber \\ \left(-i\gamma^{\nu}_{ca}\partial_{\nu}-m\mathbb{I}_{ca}\right)\left(i\gamma^{\mu}_{ab}\partial_{\mu}-m\mathbb{I}_{ab}\right)\psi_{b} & =0\\ \left(\gamma^{\nu}_{ca}\gamma^{\mu}_{ab}\partial_{\nu}\partial_{\mu}+m^{2}\mathbb{I}_{cb}\right)\psi_{b} & =0\\ \left(\frac{1}{2}\left\{ \gamma^{\nu}_{ca},\gamma^{\mu}_{ab}\right\} \partial_{\nu}\partial_{\mu}+m^{2}\mathbb{I}_{cb}\right)\psi_{b} & =0\\ \left(\partial^{\mu}\partial_{\mu}+m^{2}\right)\psi_{c} & =0 \end{aligned}\]

where it should be noted that \(\psi_{a}\) has the basic appearance of a 4-component vector. It is not a Lorentz vector; rather it is a vector in ‘spinor space’, although this is linked to Minkowski space. As a result the transformation of spinor fields is a bit more complicated:

\[\psi_{a}\left(x\right)\rightarrow\psi_{a}'\left(x'\right)=S_{ab}\left(\Lambda\right)\psi_{b}\left(\Lambda^{-1}x\right).\]

It is conventional to drop the spinor indices, and to additionally define the ‘Feynman scratch notation’

\[\cancel{a}\triangleq\gamma^{\mu}a_{\mu}\]

to write the Dirac equation in the neater form

\[\left(i\cancel{\partial}-m\right)\psi=0.\]

4.5 Scalar Quantum Electrodynamics

Let’s look again at complex Klein Gordon theory:

\[\mathcal{L}^{\mathbb{C}}_{\text{KG}}\triangleq\partial^{\mu}\varphi^{*}\partial_{\mu}\varphi-m^{2}\varphi^{*}\varphi-V\left(\varphi^{*}\varphi\right).\]

I’ve included a potential term for later convenience. This is invariant to global changes in phase of the field:

\[\varphi\left(x\right)\rightarrow\varphi\left(x\right)\exp\left(ie\lambda\right)\]

for constant \(e\lambda\) in much the same way that the single-particle wavefunction in quantum mechanics has an overall global phase which is unobservable. But now imagine we’d like to upgrade this global \(U\left(1\right)\) symmetry to a local gauge symmetry. That is, we’d like invariance under

\[\varphi\left(x\right)\rightarrow\varphi\left(x\right)\exp\left(ie\lambda\left(x\right)\right).\]

The current theory is not invariant under this local symmetry:

\[\mathcal{L}^{\mathbb{C}}_{\text{KG}}\rightarrow\left(\partial^{\mu}-ie\partial^{\mu}\lambda\right)\varphi^{*}\left(\partial_{\mu}+ie\partial^{\mu}\lambda\right)\varphi-m^{2}\varphi^{*}\varphi-V\left(\varphi^{*}\varphi\right).\]

But we can make a new theory which is locally gauge invariant by introducing a new vector gauge field which can cancel the change in the derivative terms. This is called scalar electrodynamics. It is a simplified version of quantum electrodynamics where the spinor field is replaced with a simpler complex scalar field \(\varphi\), which therefore acts like a ‘scalar electron’ field. This is coupled to an abelian gauge field \(A^{\mu}\) corresponding to a photon:

\[\mathcal{L}_{\text{scalar QED}}=\left(D_{\mu}\varphi\right)^{*}D^{\mu}\varphi-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-m^{2}\varphi^{*}\varphi-V\left(\varphi^{*}\varphi\right).\]

In order to make the theory gauge invariant, we were forced to couple the fields via the covariant derivative:

\[D_{\mu}\triangleq\partial_{\mu}-ieA_{\mu}\]

where we now recognise \(e\) as the charge of the \(\varphi\) field. A gauge transformation now takes the form

\[\begin{aligned} A_{\mu}\left(x\right) & \rightarrow A_{\mu}\left(x\right)+\partial_{\mu}\lambda\left(x\right)\\ \varphi & \left(x\right)\rightarrow\varphi\left(x\right)\exp\left(ie\lambda\left(x\right)\right). \end{aligned}\]

4.6 The Standard Model

The Lagrange density for the Standard Model is

\[\mathcal{L}_{\text{SM}}=\mathcal{L}_{\text{boson}}+\mathcal{L}_{\text{fermion}}+\mathcal{L}_{\text{Higgs}}+\mathcal{L}_{\text{Yukawa}}\]

with²

\[\begin{aligned} \mathcal{L}_{\text{boson}}=-\frac{1}{4}G^{a\mu\nu}G^{a}_{\mu\nu}-\frac{1}{4}W^{a\mu\nu}W^{a}_{\mu\nu}-\frac{1}{4}B^{\mu\nu}B_{\mu\nu} \end{aligned}\]

\[\begin{aligned} \mathcal{L}_{\text{fermion}}=\sum_{\psi\in\text{fermions}}\overline{\psi}i\cancel{D}_{\mu}\psi \end{aligned}\]

and

\[\begin{aligned} \mathcal{L}_{\text{Higgs}} =\left(D_{\mu}H\right)^{\dagger}D^{\mu}H+\mu^{2}H^{\dagger}H-\lambda\left(H^{\dagger}H\right)^{2}(\mu^{2}>0). \end{aligned}\]

Here,

• \(G^{a}_{\mu\nu}\) is the field strength of the non-abelian \(SU\left(3\right)\) group (strong force-carrying bosons: gluons)

• \(W^{a}_{\mu\nu}\) is the field strength of the non-abelian \(SU\left(2\right)\) group (electroweak force-carrying bosons)

• \(B_{\mu\nu}\) is the field strength of the abelian \(U\left(1\right)\) field (hypercharge)

• \(\psi\) includes all fermions: quarks and leptons, left- and right-handed

• \(\mathcal{L}_{\text{Yukawa}}\) describes the Fermion masses.

You now have all the tools to do calculations with these fields and their particles in Part II of this course.

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1. More frequently called Berezin integrals, although this is the only example I know of a Russian getting something named after them when an Englishman invented it first (there are many examples the other way around). The first known reference was David Candlin, 1956, a decade before Berezin.

2. This form is before the Higgs field has induced spontaneous symmetry breaking. You will see more of this in Part II.