1 Classical scalar field

At its heart, physics is the study of the symmetries of nature and their consequences. By studying how systems change or more importantly do not change under transformation, we can identify important properties of the system (cf. Noether theorem). Arguably one of the most important symmetries there is, is Lorentz symmetry, i.e. invariance under spacetime rotations in special relativity. These consists of rotation in 3D space and boosts. The following is merely a brief summary of what is required for this course and should not be viewed as complete.

In the above discussion we often had to write $\hbar$ and $c$. To avoid doing this, it is common to choose a unit system where $\hbar=c=1$ and we only have a single unit (such as ${\rm GeV}$). This is perfectly permissible as we will always know how to convert back to SI units by multiplying with the correct powers of $\hbar$ and $c$. With the 2019 revision of the SI system, both $c$ and $\hbar$ have definite values without uncertainties as they are used to define the meter and kilogram respectively. A helpful value to remember is $(\hbar c)=197.3\,{\rm MeV}\cdot{\rm fm}$.

Suggested Exercise

Convert the following values back to SI units:

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  the total cross section for $pp$ collisions at the LHC is $\sigma\approx 250\,{\rm GeV}^{-2}$. How much is this in ${\rm fb}^{2}$?

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  the muon lifetime is $\tau\approx 3.3\times 10^{18}\,{\rm GeV}^{-1}$. How much is this in $\mu{\rm s}$?

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  the electron mass is $m\approx 0.511\,{\rm MeV}$. How much is this in ${\rm kg}$?

Let us briefly review a classical system with $n$ degrees of freedom such as a collection of $n/3$ particles that can all move independently of each other. This system is completely described by $n$ generalised coordinates $q_{1},...,q_{n}$ and $n$ generalised velocities $\dot{q}_{1},...,\dot{q}_{n}$. To find these, we need the Lagrangian $L$. For a system of particles, this is

To derive the equations of motions for this system, we define the action $S$ functional

and use the variational principle, i.e. we require the action is extremal²² 2 This is sometimes referred to as the principle of least action. However, the action does not need to minimised (even if it often is) as long as it is extremal

From this you can easily derive the Euler-Lagrange equation

Derivation of the Euler-Lagrange for $n=1$

In the $n=1$ case we have only one $q$ so that the action only depends on the function $q(t)$. Fixing the boundary conditions $q(t_{0})=q_{0}$ and $q(t_{1})=q_{1}$, we vary the path by a small $\epsilon\,\omega(t)$ with $\omega(t_{0})=\omega(t_{1})=0$. Then,

	$\displaystyle\frac{{\rm d}}{{\rm d}\epsilon}S[q+\epsilon\,\omega]$	$\displaystyle=\int_{t_{0}}^{t_{1}}{\rm d}t\ \frac{{\rm d}}{{\rm d}\epsilon}L\big{(}q(t)+\epsilon\,\omega(t),\dot{q}(t)+\epsilon\,\dot{\omega}(t),t\big{)}$
		$\displaystyle=\int_{t_{0}}^{t_{1}}{\rm d}t\ \Bigg{[}\omega(t)\frac{\partial L}{\partial q}\big{(}q(t)+\epsilon\,\omega(t),\dot{q}(t)+\epsilon\,\dot{\omega}(t),t\big{)}+\dot{\omega}(t)\frac{\partial L}{\partial\dot{q}}\big{(}q(t)+\epsilon\,\omega(t),\dot{q}(t)+\epsilon\,\dot{\omega}(t),t\big{)}\Bigg{]}\,,$		(37)

since $t$ does not depend on $\epsilon$. For $\epsilon=0$, we have an extremal value

$\displaystyle\frac{{\rm d}S[q]}{{\rm d}\epsilon}$

$\displaystyle=\int_{t_{0}}^{t_{1}}{\rm d}t\ \Bigg{[}\omega(t)\frac{\partial L}{\partial q}\big{(}q(t),\dot{q}(t),t\big{)}+\dot{\omega}(t)\frac{\partial L}{\partial\dot{q}}\big{(}q(t),\dot{q}(t),t\big{)}\Bigg{]}\,.$

(38)

Applying integration-by-parts on the second term with $\omega(t_{0})=\omega(t_{1})=0$ such hat the boundary conditions vanish

$\displaystyle\frac{{\rm d}S[q]}{{\rm d}\epsilon}$

$\displaystyle=\int_{t_{0}}^{t_{1}}{\rm d}t\ \omega(t)\Bigg{[}\frac{\partial L}{\partial q}\big{(}q(t),\dot{q}(t),t\big{)}-\frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial\dot{q}}\big{(}q(t),\dot{q}(t),t\big{)}\Bigg{]}\,.$

(39)

Since $\omega$ is an arbitrary smooth function, the fundamental lemma of the calculus of variations requires that the bracket vanishes.

Because of its prevalence in mechanics, we define the conjugate momentum to $q_{i}$

Next, we can define the Hamiltonian

which in the case of (33) is just the total energy of the system

We can now wonder when energy is conversed by calculating the total derivative of $H$ w.r.t. $t$

where we have left the sum over $j$ implicit. Using (36), we can cancel the remaining terms and are left with

In other words, if the Lagrangian $L$ does not explicitly depend on $t$ (for example through a time-dependent potential $V(t)$), energy is conserved. This is the first example of what is often called a Noether current or Noether charge, that is a conserved quantity that arises because the Lagrangian has a certain symmetry such as time-invariance.

Noether Theorem

In 1918, the German mathematician Emmy Noether proved that if the Lagrangian $L$ is invariant under small perturbations of the time variable and the coordinates $q_{i}$, there exists a conserved quantity for each of the $n$ coordinates. To quantify this, let $T$ be the generator of time evolution and $Q_{i}$ the generator of the symmetry

	$\displaystyle t$	$\displaystyle\to t+\delta t=t+\epsilon T\,,$		(45)
	$\displaystyle q_{i}$	$\displaystyle\to q_{i}+\delta q_{i}=q_{i}+\epsilon Q_{i}\,.$		(46)

If $L$ is invariant under the this transformation,

$\displaystyle\mathbb{O}=H\ T-\pi_{i}Q_{i}$

(47)

is conserved. Examples include

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  $T=1$, $Q_{i}=0$: energy is conserved if the potential is not time-dependent.

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  $T=0$, $Q_{i}=1$: linear momentum is conserved if the potential is shift-invariant.

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  $T=0$, for each particle $r$ $\vec{Q}_{r}=\vec{n}\times\vec{q}_{r}$ with some vector $\vec{n}$: angular momentum along the axis $\vec{n}$ is conserved if the Lagrangian is spherical symmetric.

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  If the $L$ is invariant under Lorentz boost, the centre-of-mass system moves with constant velocity.

Additionally to Lagrangian mechanics, sometimes it is helpful to consider the Hamiltonian EoMs

Suggested Exercise

Derive these using the Euler-Lagrange equation.

Before we can study quantum field theorys, let us review classical field theories. A classical field $\phi(\vec{x},t)$ is a function that can take a value for each point in space and time. From a Lagrangian point of view, this means that we one degree of freedom for each point, requiring us to use integrals rather than finite sums. Since we also want to consider relativity, we further want to avoid talking about space differently from time. Therefore, the Lagrangian functional $L$ is now less useful and we instead use the Lagrangian density $\mathcal{L}$ which confusingly is also often called Lagrangian. The generalised coordinate now becomes the field $\phi(x)$ and the generalised velocity the derivative of the field $\partial^{\mu}\phi$. The action functional is still defined the same way

Similarly to the $n=1$ case above, we can derive the Euler-Lagrange equation by requiring $\delta S=0$

where we have once again defined the conjugate momentum $\pi$.

Suggested Exercise

Do this by adding a small variation $\phi\to\phi+\delta\phi$ and use Gauss’s law to remove the surface terms.

Now let us consider the KG as defining a field $\phi$ instead of a wavefunction $\Psi$ and derive the EoM (31). We first require the Lagrangian $\mathcal{L}$. While we could write down a field theory that depends on $\phi$ and $\partial^{\mu}\phi$ in any way we like, we usually want $\mathcal{L}$ to be polynomial in the fields and derivatives. There are plenty of examples where this is not the case though (such as the Sine-Gordon theory or Higgs Effective Theory). We do require $\mathcal{L}$ to be a Lorentz scalar though, meaning that it cannot have open indices, and that $\mathcal{L}$ has units of ${\rm GeV}^{4}$ so that the action $S$ is dimensionless. The first conditions implies that $\mathcal{L}$ can only be a function of $(\partial_{\mu}\phi)(\partial^{\mu}\phi)$ as there are no other vectors to contract with. It turns out that

Note that the normalisation of $1/2$ in front of the kinetic term does not matter classically as it does not impact the Euler-Lagrange equations. This changes once we start studying quantum fields so we will keep it already canonically normalised now. Let us that this reduces to the KG equation by computing

Therefore, we find for (50)

which is just (31).

For the Hamiltonian formulation, we need the conjugate momentum which is written through the time derivative

which, like $\phi$ itself, is a scalar field in that it assigns a scalar value to every point in spacetime.

When we considered the case where $n$ is finite, our next subject was to show that total energy is conserved. We can do something very similar here by defining the Hamiltonian density $\mathcal{H}$

but it is actually helpful to think broader. Once we start considering relativity, energy and momentum become frame-dependent and it would be nice to have a covariant description that works with four-vectors rather than focussing on energy and momentum separately. Such an object is the energy-momentum tensor

which considers energy and momentum densities but also their fluxes as well as pressure and stress. $T^{\mu\nu}$ is a symmetric tensor which is trivial to see in our specific example,

This is very similar to $\mathcal{H}$ and in fact $T^{00}=\mathcal{H}$ is the energy density.

Since we have four translation symmetries (one time and three spatial ones), we expect four conserved Noether currents

Here we have used the chain rule to expand $\partial_{\nu}\mathcal{L}$ and (50) for the first term in the $\eta^{\mu\nu}$ bracket. We have now shown that the energy-momentum tensor is a conserved quantity in a more differential sense.

Let us pause for a moment to understand what this means by considering the $\mu=0$ case, i.e. $j^{\nu}=T^{0\nu}$ for which we still have the same conservation law

By integrating this over a small region of space $V$, we find using Gauss’s divergence theorem for the second term

Since we have already identified $j^{0}$ as the energy density, this implies that the change in energy in the volume is equal to the flux of energy out of this volume. We will often encounter conservation laws like this, from energy or momentum like here to probability densities in wavefunctions or electric charge in electrodynamics.

Before concluding this chapter, let us derive the general solution of the KG equation. Since this is a wave equation, we begin by making an ansatz in terms of plane wave solutions

where $k^{\mu}$ is a four vector indicating the direction of travel of our plane wave. Since we have considered only the real KG where $\phi=\phi^{*}$, we also need a real solution, i.e. $B=A^{*}$. We can constrain $k$ by substituting the ansatz into the KG equation

Therefore, $k^{2}=m^{2}$ which means that the momentum of the plane wave needs to be on its so-called mass-shell. Of course, the general solution is a linear combination of plane waves

where $k^{2}=(k^{0})^{2}-\vec{k}^{2}=E_{\vec{k}}^{2}-\vec{k}^{2}=m^{2}$. The factor $1/\sqrt{2E}$ does not really matter at this point but it will become convenient later.

Suggested Exercise

Repeat the above discussion for the complex KG field where $\phi$ and $\phi^{*}$ are independent degrees of freedom. Start with the Lagrangian

$\displaystyle\mathcal{L}=(\partial_{\mu}\phi)(\partial^{\mu}\phi^{*})-m^{2}\phi\phi^{*}\,.$

(64)

In a QFT context, $\phi^{*}$ would be the antiparticle to $\phi$’s particle.