1 Classical Field Theory

The most familiar examples of fields from classical physics are the electric and magnetic fields, $\overrightarrow{E}(\overrightarrow{x},t)$ and $\overrightarrow{B}(\overrightarrow{x},t)$. Both of these fields are spatial 3-vectors. In a more sophisticated treatement of electromagnetism, we derive these two 3-vectors from a single 4-component field $A^{\mu }(\overrightarrow{x},t)=(\varphi ,\overrightarrow{A})$ where $\mu =0,1,2,3$ shows that this field is a vector in spacetime. The electric and magnetic fields are given by

$\overrightarrow{E}=-\nabla \varphi -{\displaystyle \frac{\partial \overrightarrow{A}}{\partial t}}\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }\overrightarrow{B}=\nabla \times \overrightarrow{A}$

(1.4)

which ensure that two of Maxwell’s equations, $\nabla \cdot \overrightarrow{B}=0$ and $d\overrightarrow{B}/dt=-\nabla \times \overrightarrow{E}$, hold immediately as identities.

The dynamics of the field is governed by a Lagrangian which is a function of $\varphi (\overrightarrow{x},t)$, $\dot{\varphi }(\overrightarrow{x},t)$ and $\nabla \varphi (\overrightarrow{x},t)$. In all the systems we study in this course, the Lagrangian is of the form,

$L(t)={\displaystyle \int d^{3}xℒ({\varphi }_a,{\partial }_{\mu }{\varphi }_a)}$

(1.5)

where the official name for $ℒ$ is the Lagrangian density, although everyone simply calls it the Lagrangian. The action is,

$S={\displaystyle {\int }_{t_1}^{t_2}}𝑑t{\displaystyle \int d^{3}xℒ}={\displaystyle \int d^{4}xℒ}$

(1.6)

Recall that in particle mechanics $L$ depends on $q$ and $\dot{q}$, but not $\ddot{q}$. In field theory we similarly restrict to Lagrangians $ℒ$ depending on $\varphi$ and $\dot{\varphi }$, and not $\ddot{\varphi }$. In principle, there’s nothing to stop $ℒ$ depending on $\nabla \varphi$, ${\nabla }^2\varphi$, ${\nabla }^3\varphi$, etc. However, with an eye to later Lorentz invariance, we will only consider Lagrangians depending on $\nabla \varphi$ and not higher derivatives. Also we will not consider Lagrangians with explicit dependence on $x^{\mu }$; all such dependence only comes through $\varphi$ and its derivatives.

Consider the Lagrangian for a real scalar field $\varphi (\overrightarrow{x},t)$,

	$ℒ$	$=$	$\mathrm{ }\frac{1}{2}{\eta }^{\mu \nu }{\partial }_{\mu }\varphi {\partial }_{\nu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$
		$=$	$\mathrm{ }\frac{1}{2}{\dot{\varphi }}^2-{\displaystyle \frac{1}{2}}{(\nabla \varphi )}^2-\frac{1}{2}{m}^2{\varphi }^2$

where we are using the Minkowski space metric

(1.10)

Comparing (1.1.1) to the usual expression for the Lagrangian $L=T-V$, we identify the kinetic energy of the field as

$T={\displaystyle \int d^{3}x\frac{1}{2}{\dot{\varphi }}^2}$

(1.11)

and the potential energy of the field as

$V={\displaystyle \int d^{3}x\frac{1}{2}{(\nabla \varphi )}^2}+\frac{1}{2}{m}^2{\varphi }^2$

(1.12)

The first term in this expression is called the gradient energy, while the phrase “potential energy”, or just “potential”, is usually reserved for the last term.

We could also consider a Lagrangian that is linear in time derivatives, rather than quadratic. Take a complex scalar field $\psi$ whose dynamics is defined by the real Lagrangian

$ℒ={\displaystyle \frac{i}{2}}({\psi }^{\star }\dot{\psi }-{\dot{\psi }}^{\star }\psi )-\nabla {\psi }^{\star }\cdot \nabla \psi -m{\psi }^{\star }\psi$

(1.17)

We can determine the equations of motion by treating $\psi$ and ${\psi }^{\star }$ as independent objects, so that

${\displaystyle \frac{\partial ℒ}{\partial {\psi }^{\star }}}={\displaystyle \frac{i}{2}}\dot{\psi }-m\psi \mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }{\displaystyle \frac{\partial ℒ}{\partial {\dot{\psi }}^{\star }}}=-{\displaystyle \frac{i}{2}}\psi \mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }{\displaystyle \frac{\partial ℒ}{\partial \nabla {\psi }^{\star }}}=-\nabla \psi$

(1.18)

This gives us the equation of motion

$i{\displaystyle \frac{\partial \psi }{\partial t}}=-{\nabla }^2\psi +m\psi$

(1.19)

This looks very much like the Schrödinger equation. Except it isn’t! Or, at least, the interpretation of this equation is very different: the field $\psi$ is a classical field with none of the probability interpretation of the wavefunction. We’ll come back to this point in Section 2.8.

We may derive Maxwell’s equations in the vacuum from the Lagrangian,

$ℒ=-\frac{1}{2}({\partial }_{\mu }{A}_{\nu })({\partial }^{\mu }{A}^{\nu })+\frac{1}{2}{({\partial }_{\mu }{A}^{\mu })}^2$

(1.20)

Notice the funny minus signs! This is to ensure that the kinetic terms for $A_i$ are positive using the Minkowski space metric (1.10), so $ℒ\sim \frac{1}{2}{\dot{A}}_i^2$. The Lagrangian (1.20) has no kinetic term ${\dot{A}}_0^2$ for $A_0$. We will see the consequences of this in Section 6. To see that Maxwell’s equations indeed follow from (1.20), we compute

${\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{A}_{\nu })}}=-{\partial }^{\mu }{A}^{\nu }+({\partial }_{\rho }{A}^{\rho }){\eta }^{\mu \nu }$

(1.21)

from which we may derive the equations of motion,

${\partial }_{\mu }\left({\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{A}_{\nu })}}\right)=-{\partial }^2{A}^{\nu }+{\partial }^{\nu }({\partial }_{\rho }{A}^{\rho })=-{\partial }_{\mu }({\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu })\equiv -{\partial }_{\mu }{F}^{\mu \nu }$

(1.22)

where the field strength is defined by $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. You can check using (1.4) that this reproduces the remaining two Maxwell’s equations in a vacuum: $\nabla \cdot \overrightarrow{E}=0$ and $\partial \overrightarrow{E}/\partial t=\nabla \times \overrightarrow{B}$. Using the notation of the field strength, we may rewrite the Maxwell Lagrangian (up to an integration by parts) in the compact form

$ℒ=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }$

(1.23)

In each of the examples above, the Lagrangian is local. This means that there are no terms in the Lagrangian coupling $\varphi (\overrightarrow{x},t)$ directly to $\varphi (\overrightarrow{y},t)$ with $\overrightarrow{x}\ne \overrightarrow{y}$. For example, there are no terms that look like

$L={\displaystyle \int d^{3}x{d}^{3}y\varphi (\overrightarrow{x})\varphi (\overrightarrow{y})}$

(1.24)

A priori, there’s no reason for this. After all, $\overrightarrow{x}$ is merely a label, and we’re quite happy to couple other labels together (for example, the term ${\partial }_3{A}_0{\partial }_0{A}_3$ in the Maxwell Lagrangian couples the $\mu =0$ field to the $\mu =3$ field). But the closest we get for the $\overrightarrow{x}$ label is a coupling between $\varphi (\overrightarrow{x})$ and $\varphi (\overrightarrow{x}+\delta \overrightarrow{x})$ through the gradient term ${(\nabla \varphi )}^2$. This property of locality is, as far as we know, a key feature of all theories of Nature. Indeed, one of the main reasons for introducing field theories in classical physics is to implement locality. In this course, we will only consider local Lagrangians.

For a real scalar field we have $\varphi (x)\to {\varphi }^{\prime }(x)=\varphi ({\mathrm{\Lambda }}^{-1}x)$. The derivative of the scalar field transforms as a vector, meaning

$({\partial }_{\mu }\varphi )(x)\to {({\mathrm{\Lambda }}^{-1})}_{\mu }^{\nu }({\partial }_{\nu }\varphi )(y)$

where $y={\mathrm{\Lambda }}^{-1}x$. This means that the derivative terms in the Lagrangian density transform as

	${ℒ}_{deriv}(x)={\partial }_{\mu }\varphi (x){\partial }_{\nu }\varphi (x){\eta }^{\mu \nu }$	$⟶$	$\mathrm{ }{({\mathrm{\Lambda }}^{-1})}_{\mu }^{\rho }({\partial }_{\rho }\varphi )(y){({\mathrm{\Lambda }}^{-1})}_{\nu }^{\sigma }({\partial }_{\sigma }\varphi )(y){\eta }^{\mu \nu }$		(1.29)
		$=$	$\mathrm{ }({\partial }_{\rho }\varphi )(y)({\partial }_{\sigma }\varphi )(y){\eta }^{\rho \sigma }$
		$=$	$\mathrm{ }{ℒ}_{deriv}(y)$

The potential terms transform in the same way, with ${\varphi }^2(x)\to {\varphi }^2(y)$. Putting this all together, we find that the action is indeed invariant under Lorentz transformations,

$S={\displaystyle \int d^{4}xℒ(x)}⟶{\displaystyle \int d^{4}xℒ(y)}={\displaystyle \int d^{4}yℒ(y)}=S$

(1.30)

where, in the last step, we need the fact that we don’t pick up a Jacobian factor when we change integration variables from $\int d^{4}x$ to $\int d^{4}y$. This follows because $det\mathrm{\Lambda }=1$. (At least for Lorentz transformation connected to the identity which, for now, is all we deal with).

In the first-order Lagrangian (1.17), space and time are not on the same footing. ($ℒ$ is linear in time derivatives, but quadratic in spatial derivatives). The theory is not Lorentz invariant.

Under a Lorentz transformation $A^{\mu }(x)\to {\mathrm{\Lambda }}_{\nu }^{\mu }{A}^{\nu }({\mathrm{\Lambda }}^{-1}x)$. You can check that Maxwell’s Lagrangian (1.23) is indeed invariant. Of course, historically electrodynamics was the first Lorentz invariant theory to be discovered: it was found even before the concept of Lorentz invariance.

Every continuous symmetry of the Lagrangian gives rise to a conserved current $j^{\mu }(x)$ such that the equations of motion imply

${\partial }_{\mu }{j}^{\mu }=0$

(1.31)

or, in other words, $\partial j^{\mathrm{\hspace{0.17em}0}}/\partial t+\nabla \cdot \overrightarrow{j}=0$.

$Q={\displaystyle {\int }_{{𝐑}^3}}{d}^{3}x{j}^{\mathrm{\hspace{0.17em}0}}$

(1.32)

which one can immediately see by taking the time derivative,

${\displaystyle \frac{dQ}{dt}}={\displaystyle {\int }_{{𝐑}^3}}{d}^{3}x{{\displaystyle \frac{\partial j}{\partial t}}}^0=-{\displaystyle {\int }_{{𝐑}^3}}{d}^{3}x\nabla \cdot \overrightarrow{j}=0$

(1.33)

assuming that $\overrightarrow{j}\to 0$ sufficiently quickly as $|\overrightarrow{x}|\to \mathrm{\infty }$. However, the existence of a current is a much stronger statement than the existence of a conserved charge because it implies that charge is conserved locally. To see this, we can define the charge in a finite volume $V$,

$Q_V={\displaystyle {\int }_V}{d}^{3}x{j}^{\mathrm{\hspace{0.17em}0}}$

(1.34)

Repeating the analysis above, we find that

${\displaystyle \frac{d{Q}_V}{dt}}=-{\displaystyle {\int }_V}{d}^{3}x\nabla \cdot \overrightarrow{j}=-{\displaystyle {\int }_A}\overrightarrow{j}\cdot 𝑑\overrightarrow{S}$

(1.35)

where $A$ is the area bounding $V$ and we have used Stokes’ theorem. This equation means that any charge leaving $V$ must be accounted for by a flow of the current 3-vector $\overrightarrow{j}$ out of the volume. This kind of local conservation of charge holds in any local field theory.

Proof of Noether’s Theorem: We’ll prove the theorem by working infinitesimally. We may always do this if we have a continuous symmetry. We say that the transformation

$\delta {\varphi }_a(x)=X_a(\varphi )$

(1.36)

is a symmetry if the Lagrangian changes by a total derivative,

$\delta ℒ={\partial }_{\mu }{F}^{\mu }$

(1.37)

for some set of functions $F^{\mu }(\varphi )$. To derive Noether’s theorem, we first consider making an arbitrary transformation of the fields $\delta {\varphi }_a$. Then

	$\delta ℒ$	$=$	${\displaystyle \frac{\partial ℒ}{\partial {\varphi }_a}}\delta {\varphi }_a+{\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\varphi }_a)}}{\partial }_{\mu }(\delta {\varphi }_a)$		(1.38)
		$=$	$\left[{\displaystyle \frac{\partial ℒ}{\partial {\varphi }_a}}-{\partial }_{\mu }{\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\varphi }_a)}}\right]\delta {\varphi }_a+{\partial }_{\mu }\left({\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\varphi }_a)}}\delta {\varphi }_a\right)$

When the equations of motion are satisfied, the term in square brackets vanishes. So we’re left with

$\delta ℒ={\partial }_{\mu }\left({\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\varphi }_a)}}\delta {\varphi }_a\right)$

(1.39)

But for the symmetry transformation $\delta {\varphi }_a=X_a(\varphi )$, we have by definition $\delta ℒ={\partial }_{\mu }{F}^{\mu }$. Equating this expression with (1.39) gives us the result

${\partial }_{\mu }{j}^{\mu }=0\mathit{\hspace{1em}\hspace{1em} }\mathrm{with}\mathit{\hspace{1em}\hspace{1em} }{j}^{\mu }={\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\varphi }_a)}}{X}_a(\varphi )-F^{\mu }(\varphi )$

(1.40)

Recall that in classical particle mechanics, invariance under spatial translations gives rise to the conservation of momentum, while invariance under time translations is responsible for the conservation of energy. We will now see something similar in field theories. Consider the infinitesimal translation

$x^{\nu }\to x^{\nu }-{ϵ}^{\nu }\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{\varphi }_a(x)\to {\varphi }_a(x)+{ϵ}^{\nu }{\partial }_{\nu }{\varphi }_a(x)$

(1.41)

(where the sign in the field transformation is plus, instead of minus, because we’re doing an active, as opposed to passive, transformation). Similarly, once we substitute a specific field configuration $\varphi (x)$ into the Lagrangian, the Lagrangian itself also transforms as

$ℒ(x)\to ℒ(x)+{ϵ}^{\nu }{\partial }_{\nu }ℒ(x)$

(1.42)

Since the change in the Lagrangian is a total derivative, we may invoke Noether’s theorem which gives us four conserved currents ${(j^{\mu })}_{\nu }$, one for each of the translations ${ϵ}^{\nu }$ with $\nu =0,1,2,3$,

${(j^{\mu })}_{\nu }={\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\varphi }_a)}}{\partial }_{\nu }{\varphi }_a-{\delta }_{\nu }^{\mu }ℒ\mathit{\hspace{1em}}\equiv T_{\nu }^{\mu }$

(1.43)

$T_{\nu }^{\mu }$ is called the energy-momentum tensor. It satisfies

${\partial }_{\mu }{T}_{\nu }^{\mu }=0$

(1.44)

The four conserved quantities are given by

$E={\displaystyle \int d^{3}x{T}^{00}}\mathit{\hspace{1em}\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }{P}^i={\displaystyle \int d^{3}x{T}^{0i}}$

(1.45)

where $E$ is the total energy of the field configuration, while $P^i$ is the total momentum of the field configuration.

Consider the simplest scalar field theory with Lagrangian (1.1.1). From the above discussion, we can compute

$T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }ℒ$

(1.46)

One can verify using the equation of motion for $\varphi$ that this expression indeed satisfies ${\partial }_{\mu }{T}^{\mu \nu }=0$. For this example, the conserved energy and momentum are given by

	$E$	$=$	${\displaystyle \int d^{3}x\frac{1}{2}{\dot{\varphi }}^2}+\frac{1}{2}{(\nabla \varphi )}^2+\frac{1}{2}{m}^2{\varphi }^2$		(1.47)
	$P^i$	$=$	${\displaystyle \int d^{3}x\dot{\varphi }{\partial }^i\varphi }$		(1.48)

Notice that for this example, $T^{\mu \nu }$ came out symmetric, so that $T^{\mu \nu }=T^{\nu \mu }$. This won’t always be the case. Nevertheless, there is typically a way to massage the energy momentum tensor of any theory into a symmetric form by adding an extra term

${\mathrm{\Theta }}^{\mu \nu }=T^{\mu \nu }+{\partial }_{\rho }{\mathrm{\Gamma }}^{\rho \mu \nu }$

(1.49)

where ${\mathrm{\Gamma }}^{\rho \mu \nu }$ is some function of the fields that is anti-symmetric in the first two indices so ${\mathrm{\Gamma }}^{\rho \mu \nu }=-{\mathrm{\Gamma }}^{\mu \rho \nu }$. This guarantees that ${\partial }_{\mu }{\partial }_{\rho }{\mathrm{\Gamma }}^{\rho \mu \nu }=0$ so that the new energy-momentum tensor is also a conserved current.

One reason that you may want a symmetric energy-momentum tensor is to make contact with general relativity: such an object sits on the right-hand side of Einstein’s field equations. In fact this observation provides a quick and easy way to determine a symmetric energy-momentum tensor. Firstly consider coupling the theory to a curved background spacetime, introducing an arbitrary metric $g_{\mu \nu }(x)$ in place of ${\eta }_{\mu \nu }$, and replacing the kinetic terms with suitable covariant derivatives using “minimal coupling”. Then a symmetric energy momentum tensor in the flat space theory is given by

${\mathrm{\Theta }}^{\mu \nu }=-{{\displaystyle \frac{2}{\sqrt{-g}}}{\displaystyle \frac{\partial (\sqrt{-g}ℒ)}{\partial g_{\mu \nu }}}|}_{g_{\mu \nu }={\eta }_{\mu \nu }}$

(1.50)

It should be noted however that this trick requires a little more care when working with spinors.

In classical particle mechanics, rotational invariance gave rise to conservation of angular momentum. What is the analogy in field theory? Moreover, we now have further Lorentz transformations, namely boosts. What conserved quantity do they correspond to? To answer these questions, we first need the infinitesimal form of the Lorentz transformations

${\mathrm{\Lambda }}_{\nu }^{\mu }={\delta }_{\nu }^{\mu }+{\omega }_{\nu }^{\mu }$

(1.51)

where ${\omega }_{\nu }^{\mu }$ is infinitesimal. The condition (1.26) for $\mathrm{\Lambda }$ to be a Lorentz transformation becomes

			$\mathrm{ }\mathit{\hspace{1em}\hspace{1em} }({\delta }_{\sigma }^{\mu }+{\omega }_{\sigma }^{\mu })({\delta }_{\tau }^{\nu }+{\omega }_{\tau }^{\nu }){\eta }^{\sigma \tau }={\eta }^{\mu \nu }$		(1.52)
		$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}\hspace{1em}}{\omega }^{\mu \nu }+{\omega }^{\nu \mu }=0$

So the infinitesimal form ${\omega }^{\mu \nu }$ of the Lorentz transformation must be an anti-symmetric matrix. As a check, the number of different $4\times 4$ anti-symmetric matrices is $4\times 3/2=6$, which agrees with the number of different Lorentz transformations (3 rotations + 3 boosts). Now the transformation on a scalar field is given by

	$\varphi (x)\to {\varphi }^{\prime }(x)$	$=$	$\varphi ({\mathrm{\Lambda }}^{-1}x)$		(1.53)
		$=$	$\varphi (x^{\mu }-{\omega }_{\nu }^{\mu }{x}^{\nu })$
		$=$	$\varphi (x^{\mu })-{\omega }_{\nu }^{\mu }{x}^{\nu }{\partial }_{\mu }\varphi (x)$

from which we see that

$\delta \varphi =-{\omega }_{\nu }^{\mu }{x}^{\nu }{\partial }_{\mu }\varphi$

(1.54)

By the same argument, the Lagrangian density transforms as

$\delta ℒ=-{\omega }_{\nu }^{\mu }{x}^{\nu }{\partial }_{\mu }ℒ=-{\partial }_{\mu }({\omega }_{\nu }^{\mu }{x}^{\nu }ℒ)$

(1.55)

where the last equality follows because ${\omega }_{\mu }^{\mu }=0$ due to anti-symmetry. Once again, the Lagrangian changes by a total derivative so we may apply Noether’s theorem (now with $F^{\mu }=-{\omega }_{\nu }^{\mu }{x}^{\nu }ℒ$) to find the conserved current

	$j^{\mu }$	$=$	$-{\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }\varphi )}}{\omega }_{\nu }^{\rho }{x}^{\nu }{\partial }_{\rho }\varphi +{\omega }_{\nu }^{\mu }{x}^{\nu }ℒ$		(1.56)
		$=$	$-{\omega }_{\nu }^{\rho }\left[{\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }\varphi )}}{x}^{\nu }{\partial }_{\rho }\varphi -{\delta }_{\rho }^{\mu }{x}^{\nu }ℒ\right]=-{\omega }_{\nu }^{\rho }{T}_{\rho }^{\mu }{x}^{\nu }$

Unlike in the previous example, I’ve left the infinitesimal choice of ${\omega }_{\nu }^{\mu }$ in the expression for this current. But really, we should strip it out to give six different currents, i.e. one for each choice of ${\omega }_{\nu }^{\mu }$. We can write them as

${({𝒥}^{\mu })}^{\rho \sigma }=x^{\rho }{T}^{\mu \sigma }-x^{\sigma }{T}^{\mu \rho }$

(1.57)

which satisfy ${\partial }_{\mu }{({𝒥}^{\mu })}^{\rho \sigma }=0$ and give rise to 6 conserved charges. For $\rho ,\sigma =1,2,3$, the Lorentz transformation is a rotation and the three conserved charges give the total angular momentum of the field.

$Q^{ij}={\displaystyle \int d^{3}x(x^i{T}^{0j}-x^j{T}^{0i})}$

(1.58)

But what about the boosts? In this case, the conserved charges are

$Q^{0i}={\displaystyle \int d^{3}x(x^0{T}^{0i}-x^i{T}^{00})}$

(1.59)

The fact that these are conserved tells us that

	$0={\displaystyle \frac{d{Q}^{0i}}{dt}}$	$=$	${\displaystyle \int d^{3}x{T}^{0i}}+t{\displaystyle \int d^{3}x\frac{\partial T^{0i}}{\partial t}}-{\displaystyle \frac{d}{dt}}{\displaystyle \int d^{3}x{x}^i{T}^{00}}$		(1.60)
		$=$	$P^i+t{\displaystyle \frac{d{P}^i}{dt}}-{\displaystyle \frac{d}{dt}}{\displaystyle \int d^{3}x{x}^i{T}^{00}}$

But we know that $P^i$ is conserved, so $d{P}^i/dt=0$, leaving us with the following consequence of invariance under boosts:

${\displaystyle \frac{d}{dt}}{\displaystyle \int d^{3}x{x}^i{T}^{00}}=\mathrm{constant}$

(1.61)

This is the statement that the center of energy of the field travels with a constant velocity. It’s kind of like a field theoretic version of Newton’s first law but, rather surprisingly, appearing here as a conservation law.

The above two examples involved transformations of spacetime, as well as transformations of the field. An internal symmetry is one that only involves a transformation of the fields and acts the same at every point in spacetime. The simplest example occurs for a complex scalar field $\psi (x)=({\varphi }_1(x)+i{\varphi }_2(x))/\sqrt{2}$. We can build a real Lagrangian by

$ℒ={\partial }_{\mu }{\psi }^{\star }{\partial }^{\mu }\psi -V({|\psi |}^2)$

(1.62)

where the potential is a general polynomial in ${|\psi |}^2={\psi }^{\star }\psi$. To find the equations of motion, we could expand $\psi$ in terms of ${\varphi }_1$ and ${\varphi }_2$ and work as before. However, it’s easier (and equivalent) to treat $\psi$ and ${\psi }^{\star }$ as independent variables and vary the action with respect to both of them. For example, varying with respect to ${\psi }^{\star }$ leads to the equation of motion

${\partial }_{\mu }{\partial }^{\mu }\psi +{\displaystyle \frac{\partial V({\psi }^{\star }\psi )}{\partial {\psi }^{\star }}}=0$

(1.63)

The Lagrangian has a continuous symmetry which rotates ${\varphi }_1$ and ${\varphi }_2$ or, equivalently, rotates the phase of $\psi$:

$\psi \to e^{i\alpha }\psi \mathit{\hspace{1em}\hspace{1em} }\mathrm{or}\mathit{\hspace{1em}\hspace{1em}}\delta \psi =i\alpha \psi$

(1.64)

where the latter equation holds with $\alpha$ infinitesimal. The Lagrangian remains invariant under this change: $\delta ℒ=0$. The associated conserved current is

$j^{\mu }=i({\partial }^{\mu }{\psi }^{\star })\psi -i{\psi }^{\star }({\partial }^{\mu }\psi )$

(1.65)

We will later see that the conserved charges arising from currents of this type have the interpretation of electric charge or particle number (for example, baryon or lepton number).

Consider a theory involving $N$ scalar fields ${\varphi }_a$, all with the same mass and the Lagrangian

$ℒ={\displaystyle \frac{1}{2}}{\displaystyle \sum _{a=1}^N}{\partial }_{\mu }{\varphi }_a{\partial }^{\mu }{\varphi }_a-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{a=1}^N}{m}^2{\varphi }_a^2-g{\left({\displaystyle \sum _{a=1}^N}{\varphi }_a^2\right)}^2$

(1.66)

In this case the Lagrangian is invariant under the non-Abelian symmetry group $G=SO(N)$. (Actually $O(N)$ in this case). One can construct theories from complex fields in a similar manner that are invariant under an $SU(N)$ symmetry group. Non-Abelian symmetries of this type are often referred to as global symmetries to distinguish them from the “local gauge” symmetries that you will meet later. Isospin is an example of such a symmetry, albeit realized only approximately in Nature.

There is a quick method to determine the conserved current associated to an internal symmetry $\delta \varphi =\alpha \varphi$ for which the Lagrangian is invariant. Here, $\alpha$ is a constant real number. (We may generalize the discussion easily to a non-Abelian internal symmetry for which $\alpha$ becomes a matrix). Now consider performing the transformation but where $\alpha$ depends on spacetime: $\alpha =\alpha (x)$. The action is no longer invariant. However, the change must be of the form

$\delta ℒ=({\partial }_{\mu }\alpha )h^{\mu }(\varphi )$

(1.67)

since we know that $\delta ℒ=0$ when $\alpha$ is constant. The change in the action is therefore

$\delta S={\displaystyle \int d^{4}x\delta ℒ}=-{\displaystyle \int d^{4}x\alpha (x){\partial }_{\mu }{h}^{\mu }}$

(1.68)

which means that when the equations of motion are satisfied (so $\delta S=0$ for all variations, including $\delta \varphi =\alpha (x)\varphi$) we have

${\partial }_{\mu }{h}^{\mu }=0$

(1.69)

We see that we can identify the function $h^{\mu }=j^{\mu }$ as the conserved current. This way of viewing things emphasizes that it is the derivative terms, not the potential terms, in the action that contribute to the current. (The potential terms are invariant even when $\alpha =\alpha (x)$).

For the Lagrangian

$ℒ=\frac{1}{2}{\dot{\varphi }}^2-\frac{1}{2}{(\nabla \varphi )}^2-V(\varphi )$

(1.73)

the momentum is given by $\pi =\dot{\varphi }$, which gives us the Hamiltonian,

$H={\displaystyle \int d^{3}x\frac{1}{2}{\pi }^2}+\frac{1}{2}{(\nabla \varphi )}^2+V(\varphi )$

(1.74)

Notice that the Hamiltonian agrees with the definition of the total energy (1.47) that we get from applying Noether’s theorem for time translation invariance.