Classical Field Theory and the Lagrangian: From a Variational Principle for Infinitely Many Degrees of Freedom to Noether's Theorem

0. Key points

• A field is a quantity that takes a value at every point of space at every instant of time. Mechanically it is a system with infinitely many degrees of freedom, labelled by the points of space; taking the lattice spacing of a chain of coupled oscillators to zero makes this picture concrete.
• The action of a field can always be written as $S[\phi] = \int d^4x\, \mathcal{L}(\phi_a, \partial_\mu \phi_a)$. If the integrand — the Lagrangian density $\mathcal{L}$ — is a Lorentz scalar, the equations of motion that follow from it are automatically Lorentz covariant. This is the single strongest reason to start from the Lagrangian formalism.
• The variational principle $\delta S = 0$ yields the Euler–Lagrange equations for fields, $\partial_\mu\big(\partial\mathcal{L}/\partial(\partial_\mu\phi_a)\big) - \partial\mathcal{L}/\partial\phi_a = 0$ (Theorem 4.3).
• The simplest Lagrangian density allowed for a real scalar field is that of the Klein–Gordon field, and the plane-wave solutions of its equation of motion satisfy $E^2 = \boldsymbol{p}^2 + m^2$. This is why $m$ is called a mass.
• Every continuous symmetry of the action corresponds to one conserved current, $\partial_\mu j^\mu = 0$ (Noether’s theorem, Theorem 6.2). Spacetime translations give the energy–momentum tensor; a global $U(1)$ phase rotation gives a conserved charge.
• The classical field theory assembled here is the starting point for canonical quantization and path-integral quantization in the chapters that follow. Quantization means laying quantum rules on top of the space of classical field configurations.

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1. Motivation: what is it that we quantize?

In the previous chapter we saw that making relativity and quantum mechanics coexist forces the particle number to change, so that the fundamental object must be a field rather than a particle wave function. But before we can speak of quantizing a field, we need the classical field theory that is to be quantized. Building that foundation is the purpose of this chapter.

For a system with finitely many degrees of freedom, classical mechanics offers three formulations: Newton’s equations, the Lagrangian formalism and the Hamiltonian formalism. All three are available in field theory as well, but as a starting point the Lagrangian formalism is overwhelmingly the most convenient. There are four reasons.

First, relativistic invariance stays manifest. The Hamiltonian is the generator of time evolution, so it cannot help but single out the time coordinate from the three spatial ones. The action $S = \int d^4x\,\mathcal{L}$, by contrast, is an integral over four-dimensional spacetime, and if $\mathcal{L}$ is chosen to be a Lorentz scalar then $S$ itself is Lorentz invariant. Since the equations of motion are fixed as the stationarity condition of $S$, Lorentz covariance is guaranteed the moment the expression is written down. For the requirements of special relativity see Lorentz transformations.

Second, it is the natural language for symmetries. Internal symmetries and gauge symmetries alike can be stated in a single line as the invariance of $\mathcal{L}$. And by Noether’s theorem, described below, conservation laws can then be read off mechanically from the symmetries.

Third, the path integral uses the action itself. Transition amplitudes in quantum theory take the form $\int \mathcal{D}\phi\, e^{iS[\phi]}$ (path-integral quantization). What appears here is the action, not the Hamiltonian.

Fourth, it serves as the blueprint of a theory. The modern recipe for building a field theory is: specify the field content and the symmetries, then list every term compatible with them in order of increasing mass dimension. Rather than postulating interactions out of thin air, one narrows the candidates down by symmetry and dimensional counting. This methodology works only because all the information in the theory is concentrated in a single function $\mathcal{L}$.

Historically, too, the Lagrangian formalism for fields is older than quantum theory. Nineteenth-century elasticity already treated the vibrations of continuous media in Lagrangian form. The theory of the electromagnetic field, begun with Faraday’s lines of force and completed in Maxwell’s equations, had by the early twentieth century arrived at the recognition that the electromagnetic field is an independent mechanical degree of freedom carrying energy and momentum. Noether’s 1918 paper established the relation between symmetries and conservation laws in general form, and from the late 1920s Dirac, Jordan, Heisenberg and Pauli began attempting to quantize these classical fields. What we are about to retrace is that first step.

Note

This article extends the Euler–Lagrange equations（Theorem 3.4）［Lagrangian Mechanics］ of Lagrangian mechanics and Noether's theorem（Theorem 4.1）［対称性と保存則］ of symmetries and conservation laws from finitely to infinitely many degrees of freedom. If the finite-dimensional versions feel shaky, read those first.

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2. Preliminaries: notation and conventions

Throughout we use natural units $\hbar = c = 1$. In these units length, time and inverse mass all carry the same dimension, so the dimension of any quantity can be expressed by a single number, its mass dimension. The notational conventions are collected in the table below.

Symbol	Meaning and convention
$\eta_{\mu\nu} = \mathrm{diag}(1,-1,-1,-1)$	Minkowski metric. $\eta^{\mu\nu}$ is its inverse, with the same components
$x^\mu = (x^0, x^1, x^2, x^3) = (t, \boldsymbol{x})$	Spacetime coordinates. Greek indices run over $0,1,2,3$, Latin indices $i,j$ over $1,2,3$
$\partial_\mu = \partial/\partial x^\mu$	$\partial^\mu = \eta^{\mu\nu}\partial_\nu = (\partial_t, -\nabla)$
$\Box = \partial_\mu \partial^\mu = \partial_t^2 - \nabla^2$	d’Alembert operator
$a, b$	Indices distinguishing the components (internal degrees of freedom) of a field. Repeated indices are always summed
$d^4x = dt\, d^3x$	Spacetime volume element. Invariant under Lorentz transformations (their determinant is $1$)

Write $[\,\cdot\,]$ for mass dimension. For $e^{iS}$ to make sense, $S$ must be dimensionless, and since $[d^4x] = -4$ we need

$$[\mathcal{L}] = 4 .$$

This one line is the yardstick we shall use later to classify interaction terms.

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3. The continuum limit: from point masses to fields

Let us verify the statement that a field is a mechanical system with infinitely many degrees of freedom on a concrete example. Consider a one-dimensional chain of point masses joined by springs and let the lattice spacing go to zero. This example has been standard since Goldstein’s textbook, but it is worth once following the mechanical ancestry of the concept of a field with one’s own hands.

In the limit where the lattice spacing a of the discrete spring-and-mass system (top) goes to zero, a continuum field emerges. In the discrete system the degrees of freedom number as many as the point masses; in the limit there is one degree of freedom for each point of space.

Example 3.1（Continuum limit of a spring chain and the wave equation）

Let $N$ point masses of mass $m$ be arranged along a line with spacing $a$, neighbouring masses joined by springs of spring constant $k$. Writing $y_i(t)$ for the displacement of the $i$-th mass from its equilibrium position, the Lagrangian is

$$L = \sum_{i} \left[ \frac{1}{2} m\, \dot{y}_i^{\,2} - \frac{1}{2} k\, (y_{i+1} - y_i)^2 \right] .$$

Apply the finite-dimensional Lagrange equation $\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{y}_i} = \dfrac{\partial L}{\partial y_i}$. The left-hand side is $m\ddot{y}_i$; the right-hand side receives contributions from the two terms containing $y_i$, namely $-\frac{1}{2}k(y_i - y_{i-1})^2$ and $-\frac{1}{2}k(y_{i+1} - y_i)^2$, giving

$$\frac{\partial L}{\partial y_i} = -k(y_i - y_{i-1}) + k(y_{i+1} - y_i) = k\,(y_{i+1} - 2y_i + y_{i-1}) .$$

The equation of motion is therefore $m\ddot{y}_i = k(y_{i+1} - 2y_i + y_{i-1})$.

Now divide both sides by $a$ and take $a \to 0$, $N \to \infty$ (with the total length $Na$ held fixed) while keeping the linear density $\mu := m/a$ and the Young modulus $Y := ka$ fixed. Writing the position of the $i$-th mass as $x = ia$ and reading $y_i(t) \to \phi(x,t)$, we get

$$\frac{k(y_{i+1} - 2y_i + y_{i-1})}{a} = (ka)\,\frac{y_{i+1} - 2y_i + y_{i-1}}{a^2} \longrightarrow Y\,\frac{\partial^2 \phi}{\partial x^2}$$

(using the convergence of the central difference quotient to the second derivative). Hence

$$\mu\, \frac{\partial^2 \phi}{\partial t^2} = Y\, \frac{\partial^2 \phi}{\partial x^2} ,$$

a wave equation with wave speed $v = \sqrt{Y/\mu}$.

Taking the same limit in the Lagrangian gives

$$L = \sum_i a \left[ \frac{1}{2}\frac{m}{a}\dot{y}_i^{\,2} - \frac{1}{2}(ka)\left(\frac{y_{i+1}-y_i}{a}\right)^2 \right] \longrightarrow \int dx\, \underbrace{\left[ \frac{1}{2}\mu\,(\partial_t \phi)^2 - \frac{1}{2}Y (\partial_x \phi)^2 \right]}_{=:\ \mathcal{L}} .$$

The sum $\sum_i a$ has become an integral $\int dx$, and the integrand that appears is the Lagrangian density $\mathcal{L}$.

From this computation we can read off a dictionary between finitely many degrees of freedom and field theory.

Finite-dimensional system	Field theory
Index $i = 1,\dots,N$	Spatial coordinate $\boldsymbol{x} \in \mathbb{R}^3$ (a continuous “index”)
Generalized coordinate $q_i(t)$	Field $\phi(\boldsymbol{x}, t)$
$\sum_i$	$\int d^3x$
Lagrangian $L(q_i, \dot{q}_i)$	$L = \int d^3x\, \mathcal{L}(\phi, \partial_\mu\phi)$
Action $S = \int dt\, L$	Action $S = \int d^4x\, \mathcal{L}$
$\partial L/\partial q_i$	Functional derivative $\delta S/\delta \phi(x)$

Remark 3.2（A field need not be the displacement of anything）

In the example above, $\phi$ had the concrete meaning of the displacement of an elastic medium. The fields of field theory, however, are not necessarily displacements of any medium. Nineteenth-century physicists tried to understand the electromagnetic field as a strain in the ether, but the Michelson–Morley experiment and special relativity denied that picture. The electromagnetic field, and the scalar fields treated from the next section onward, are themselves fundamental mechanical degrees of freedom; there is no vibrating medium behind them.

The continuum-limit argument supplies a mathematical template for handling systems with infinitely many degrees of freedom; it does not supply an ontology of fields. Note in particular that we adjusted $\mu$ and $Y$ precisely so that $\mathcal{L}$ would remain finite in the limit $a \to 0$. This “tuning of how the limit is taken” is the prototype of an operation that reappears later in renormalization theory (introduction to renormalization).

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4. The variational principle for fields and the Euler–Lagrange equations

4.1. The action functional

Definition 4.1（Action functional of a field and Lagrangian density）

Let $\Omega \subset \mathbb{R}^4$ be a bounded region of spacetime with smooth boundary $\partial\Omega$, and let $\phi_a : \Omega \to \mathbb{R}$ ($a = 1,\dots,n$) be a collection of fields of class $C^2$.

Let $\mathcal{L} = \mathcal{L}(\phi_a, \partial_\mu \phi_a)$ be a $C^2$ function of the $n$ real variables $\phi_a$ and the $4n$ real variables $\partial_\mu\phi_a$; we call it the Lagrangian density. Then

$$S[\phi] := \int_\Omega d^4x\ \mathcal{L}\big(\phi_a(x), \partial_\mu \phi_a(x)\big)$$

is called the action functional.

We are assuming here that $\mathcal{L}$ contains no derivatives of $\phi_a$ beyond the first, and that it does not depend explicitly on the coordinate $x$. The former is required in order to keep the equations of motion second order; the latter expresses translation invariance of spacetime, that is, the absence of any externally imposed background.

Before stating the variational principle we need the fundamental lemma of the calculus of variations. It says that if the integral of $\delta S$ vanishes for every test function, then the integrand vanishes; without it the equations of motion would emerge only in integrated form.

Lemma 4.2（Fundamental lemma of the calculus of variations）

Let $\Omega \subset \mathbb{R}^4$ be a nonempty open set and $f : \Omega \to \mathbb{R}$ a continuous function. If

$$\int_\Omega f(x)\, \eta(x)\, d^4x = 0$$

holds for every $C^\infty$ function $\eta$ with compact support contained in $\Omega$, then $f \equiv 0$ on $\Omega$.

Proof（Lemma 4.2）

We prove the contrapositive. Suppose $f(x_0) \ne 0$ at some point $x_0 \in \Omega$; without loss of generality $c := f(x_0) > 0$ (if $f(x_0) < 0$, replace $f$ by $-f$).

Since $f$ is continuous, taking an open ball $B := \{ x : |x - x_0| < r \}$ contained in $\Omega$ small enough (here $|\cdot|$ is the Euclidean norm on the coordinates of $\mathbb{R}^4$ — a purely topological tool, unrelated to the metric $\eta_{\mu\nu}$), we have $f(x) > c/2$ on $B$.

As a test function supported on this $B$ take the standard bump function

$$\eta(x) := \begin{cases} \exp\!\left(\dfrac{-1}{r^2 - |x - x_0|^2}\right) & (|x - x_0| < r) \\[2mm] 0 & (|x-x_0| \ge r) . \end{cases}$$

It is of class $C^\infty$, positive on $B$ and zero outside $B$. Therefore

$$\int_\Omega f \eta\, d^4x = \int_B f \eta\, d^4x > \frac{c}{2}\int_B \eta\, d^4x > 0 ,$$

contradicting the hypothesis. Hence $f = 0$ at every point of $\Omega$.

∎

4.2. The Euler–Lagrange equations

Theorem 4.3（Euler–Lagrange equations for fields）

In the setting of Definition 4.1, suppose a field configuration $\phi = (\phi_a)$ is a stationary point of $S$ in the following sense: for every collection $\eta = (\eta_a)$ of $C^\infty$ functions with compact support contained in $\Omega$,

$$\left.\frac{d}{d\varepsilon}\right|_{\varepsilon = 0} S[\phi + \varepsilon \eta] = 0 .$$

Then on $\Omega$

$$\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)} \right) - \frac{\partial \mathcal{L}}{\partial \phi_a} = 0 \qquad (a = 1, \dots, n) .$$

Conversely, any $\phi$ satisfying this system is a stationary point in the above sense. Here $\partial_\mu$ denotes the total derivative with respect to $x$, including the dependence through $\phi_a(x)$.

Proof（Theorem 4.3）

Put $\phi_a^\varepsilon := \phi_a + \varepsilon \eta_a$. Since the support of $\eta_a$ is compact and contained in $\Omega$, we have $\eta_a = 0$, and hence $\partial_\mu \eta_a = 0$, in a neighbourhood of $\partial\Omega$. As $\mathcal{L}$ is of class $C^2$ we may interchange the $\varepsilon$-derivative with the integral, and the chain rule gives

$$\left.\frac{d}{d\varepsilon}\right|_{0} S[\phi^\varepsilon] = \int_\Omega d^4x \left[ \frac{\partial \mathcal{L}}{\partial \phi_a}\, \eta_a + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}\, \partial_\mu \eta_a \right]$$

(where we used $\partial_\mu(\phi_a + \varepsilon\eta_a) = \partial_\mu\phi_a + \varepsilon\,\partial_\mu\eta_a$).

In the second term use the Leibniz rule to move the derivative off $\eta_a$:

$$\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}\, \partial_\mu \eta_a = \partial_\mu\!\left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}\, \eta_a \right) - \partial_\mu\!\left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)} \right) \eta_a .$$

The first term is a divergence, so by Gauss’s theorem it becomes a boundary integral:

$$\int_\Omega d^4x\ \partial_\mu\!\left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}\, \eta_a \right) = \oint_{\partial\Omega} d\Sigma_\mu\ \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}\, \eta_a = 0 .$$

The last equality holds because $\eta_a = 0$ on $\partial\Omega$. The boundary term drops out only thanks to this assumption; a variation with no boundary condition imposed yields no equations of motion. Therefore

$$\left.\frac{d}{d\varepsilon}\right|_{0} S[\phi^\varepsilon] = \int_\Omega d^4x\ \left[ \frac{\partial \mathcal{L}}{\partial \phi_a} - \partial_\mu\!\left( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)} \right) \right] \eta_a .$$

By hypothesis this vanishes for every $\eta_a$, so for each $a$ separately (take the $\eta$ of the other components to be $0$) we may apply Lemma 4.2 to the expression in square brackets and obtain the conclusion. Since $\phi_a$ and $\mathcal{L}$ are of class $C^2$, that expression is continuous and the hypothesis of the lemma is met.

The converse follows immediately by reading the identity above from right to left.

∎

This is a system of $n$ coupled second-order partial differential equations: the equations of motion of the field. Compared with the finite-dimensional Lagrange equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0$, the only changes are that the time derivative $d/dt$ has become the spacetime derivative $\partial_\mu$, and the one equation per index $i$ has become one equation per field component $a$. Checking it on the elastic string of Example 3.1: for $\mathcal{L} = \frac{1}{2}\mu(\partial_t\phi)^2 - \frac{1}{2}Y(\partial_x\phi)^2$ we have $\partial \mathcal{L}/\partial(\partial_t\phi) = \mu\,\partial_t\phi$, $\partial\mathcal{L}/\partial(\partial_x \phi) = -Y \partial_x\phi$ and $\partial\mathcal{L}/\partial\phi = 0$, so $\partial_t(\mu\partial_t\phi) + \partial_x(-Y\partial_x\phi) = 0$, which is exactly the wave equation obtained above by taking the continuum limit.

Remark 4.4（Why it is good for the Lagrangian density to be a scalar）

Suppose $\mathcal{L}$ is a Lorentz scalar, that is, $\mathcal{L}'(x') = \mathcal{L}(x)$ under a Lorentz transformation $x \to \Lambda x$. Since $d^4x$ is invariant ($|\det\Lambda| = 1$), $S$ is invariant too. The equations of motion are determined by stationarity of $S$, so whatever is a solution in one inertial frame remains a solution when transported to another. In other words, Lorentz covariance of the equations of motion is guaranteed the instant $\mathcal{L}$ is written down.

This is an advantage the Hamiltonian formalism does not offer. The Hamiltonian $H = \int d^3x\,\mathcal{H}$ is an integral over a spatial slice at a fixed time, so its behaviour under Lorentz transformations is not manifest.

4.3. The ambiguity in the Lagrangian density

The equations of motion do not determine $\mathcal{L}$ uniquely. The ambiguity by a total derivative（Proposition 4.4）［Lagrangian Mechanics］ seen in the finite-dimensional case carries over verbatim to field theory. The following proposition is what will later justify extending Noether’s theorem to quasi-invariance.

Proposition 4.5（A total-derivative term does not change the equations of motion）

Let $K^\mu = K^\mu(\phi_a, x)$ be a $C^2$ function of the field values and the coordinates (with no dependence on derivatives of the fields). Then the equations of motion obtained from

$$\mathcal{L}' := \mathcal{L} + \partial_\mu K^\mu$$

via Theorem 4.3 are identical to those obtained from $\mathcal{L}$.

Proof（Proposition 4.5）

We have $S'[\phi] = S[\phi] + \int_\Omega d^4x\, \partial_\mu K^\mu$. Applying Gauss’s theorem to the second term,

$$\int_\Omega d^4x\ \partial_\mu K^\mu = \oint_{\partial\Omega} d\Sigma_\mu\ K^\mu\big(\phi_a(x), x\big) ,$$

which is determined by the values of $\phi_a$ on the boundary $\partial\Omega$ alone.

In the variation of Theorem 4.3, $\eta_a$ vanishes in a neighbourhood of $\partial\Omega$, so $\phi_a^\varepsilon = \phi_a$ on $\partial\Omega$ and this boundary integral does not depend on $\varepsilon$. Hence

$$\left.\frac{d}{d\varepsilon}\right|_0 S'[\phi^\varepsilon] = \left.\frac{d}{d\varepsilon}\right|_0 S[\phi^\varepsilon] ,$$

so the stationarity conditions coincide exactly, and so do the equations of motion obtained through Lemma 4.2.

(If $K^\mu$ also depended on derivatives of the fields, the argument would not go through as it stands, since $\partial_\mu\eta_a$ need not vanish on the boundary. Only the above form is needed in what follows.)

∎

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5. The Klein–Gordon field

5.1. Why that form?

Suppose we have a single real-valued field $\phi(x)$ which transforms as a scalar under Lorentz transformations, $\phi'(x') = \phi(x)$. We narrow down the Lagrangian densities allowed for this field by three requirements.

1. $\mathcal{L}$ is a Lorentz scalar (Remark 4.4).
2. $\mathcal{L}$ contains no derivatives of $\phi$ beyond the first (so that the equations of motion are second order).
3. $\mathcal{L}$ is at most quadratic in $\phi$ (we want a free field, for which the superposition principle holds).

Requirements 1 and 2 leave $\partial_\mu\phi\,\partial^\mu\phi$ as the only scalar that can be built with derivatives (no scalar containing exactly one $\partial_\mu\phi$ exists, since there is nothing to contract the index with). Together with requirement 3, the most general form is, with constants $A, B, C, D$,

$$\mathcal{L} = A\, \partial_\mu\phi\,\partial^\mu\phi + B\, \phi^2 + C\,\phi + D .$$

Here $D$ is a constant that does not contribute to the equations of motion, and $C\phi$ can be absorbed into $B\phi^2$ by a shift $\phi \to \phi + \text{const}$ (when $B \ne 0$). The constant $A$ can be normalized to $1/2$ by rescaling $\phi$ by a constant factor. Writing the remaining $B$ as $B = -\frac{1}{2}m^2$, we arrive at

$$\mathcal{L}_{\mathrm{KG}} = \frac{1}{2}\, \partial_\mu\phi\, \partial^\mu\phi - \frac{1}{2}\, m^2 \phi^2 = \frac{1}{2}\dot\phi^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2 ,$$

the Lagrangian density of the Klein–Gordon field.

The choice of signs has physical content. Unless $A > 0$ (positive kinetic term), the energy density fails to be bounded below, as we shall see. Unless $m^2 > 0$, the configuration $\phi = 0$ is a maximum of the potential $V(\phi) = \frac{1}{2}m^2\phi^2$ and is unstable. What happens when $m^2 < 0$ belongs to the subject of spontaneous symmetry breaking and is treated in gauge theory and spontaneous symmetry breaking.

Example 5.1（The Klein–Gordon equation and its dispersion relation）

Apply Theorem 4.3 to $\mathcal{L}_{\mathrm{KG}}$. First,

$$\frac{\partial \mathcal{L}_{\mathrm{KG}}}{\partial \phi} = -m^2\phi .$$

Next the partial derivative with respect to the derivatives. Writing $\partial_\mu\phi\,\partial^\mu\phi = \eta^{\alpha\beta}\,\partial_\alpha\phi\,\partial_\beta\phi$ and differentiating with respect to $\partial_\rho\phi$, both factors contribute and

$$\frac{\partial}{\partial(\partial_\rho\phi)}\Big( \tfrac{1}{2}\eta^{\alpha\beta}\partial_\alpha\phi\,\partial_\beta\phi \Big) = \tfrac{1}{2}\big( \eta^{\rho\beta}\partial_\beta\phi + \eta^{\alpha\rho}\partial_\alpha\phi \big) = \partial^\rho \phi$$

(using the symmetry of $\eta$). The Euler–Lagrange equation is therefore

$$(\Box + m^2)\,\phi = \partial_\mu\partial^\mu\phi + m^2\phi = 0 ,$$

the Klein–Gordon equation.

Substitute a plane wave $\phi(x) = e^{-i k\cdot x}$ (with $k\cdot x = k_\mu x^\mu = k^0 t - \boldsymbol{k}\cdot\boldsymbol{x}$). Since $\partial_\mu e^{-ik\cdot x} = -i k_\mu e^{-ik\cdot x}$, we have $\Box e^{-ik\cdot x} = (-ik_\mu)(-ik^\mu) e^{-ik\cdot x} = -k_\mu k^\mu\, e^{-ik\cdot x}$, so

$$(-k_\mu k^\mu + m^2)\, e^{-ik\cdot x} = 0 \iff k_\mu k^\mu = m^2 \iff (k^0)^2 = |\boldsymbol{k}|^2 + m^2 .$$

Reading $k^0 = E$ and $\boldsymbol{k} = \boldsymbol{p}$ in units $\hbar = c = 1$, this is nothing but the relativistic energy–momentum relation

$$E^2 = \boldsymbol{p}^2 + m^2$$

(see the energy–momentum relation（Theorem 4.4）［Relativistic Mechanics］ in relativistic mechanics). This is why the parameter $m$ is called a mass. Since $\phi$ is real, the general solution can be written, with $\omega_{\boldsymbol{k}} := \sqrt{|\boldsymbol{k}|^2 + m^2}$, as

$$\phi(x) = \int \frac{d^3k}{(2\pi)^3\, 2\omega_{\boldsymbol{k}}} \left[ \alpha(\boldsymbol{k})\, e^{-ik\cdot x} + \alpha(\boldsymbol{k})^{*}\, e^{ik\cdot x} \right]_{k^0 = \omega_{\boldsymbol{k}}} .$$

Canonical quantization is precisely the step in which $\alpha(\boldsymbol{k})$ turns into an operator $\hat{a}(\boldsymbol{k})$ (canonical quantization of the scalar field).

5.2. A static source and the Yukawa potential

Let us confirm from another angle that $m$ really acts as a mass in the physics. Placing a static point source in a Klein–Gordon field makes the range of the resulting force $1/m$. This is exactly the calculation by which Yukawa predicted the meson in 1935.

Example 5.2（The field of a point source and the range of the nuclear force）

Represent a point source at rest at the origin, coupled to the field, by

$$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2 + g\,\phi\,\delta^3(\boldsymbol{x}) .$$

By Theorem 4.3, with $\partial\mathcal{L}/\partial\phi = -m^2\phi + g\delta^3(\boldsymbol{x})$ and $\partial_\mu(\partial\mathcal{L}/\partial(\partial_\mu\phi)) = \Box\phi$,

$$(\Box + m^2)\phi = g\,\delta^3(\boldsymbol{x}) .$$

We look for a static solution, so $\partial_t\phi = 0$ and $\Box\phi = -\nabla^2\phi$, giving

$$(-\nabla^2 + m^2)\,\phi(\boldsymbol{x}) = g\,\delta^3(\boldsymbol{x}) .$$

Inserting the Fourier transform $\phi(\boldsymbol{x}) = \int \frac{d^3k}{(2\pi)^3} \tilde\phi(\boldsymbol{k}) e^{i\boldsymbol{k}\cdot\boldsymbol{x}}$ gives $(|\boldsymbol{k}|^2 + m^2)\tilde\phi = g$, that is, $\tilde\phi(\boldsymbol{k}) = g/(|\boldsymbol{k}|^2+m^2)$. Carry out the inverse transform in polar coordinates: with $r := |\boldsymbol{x}|$, $k := |\boldsymbol{k}|$ and $\boldsymbol{x}$ taken as polar axis,

$$\begin{aligned} \phi(r) &= \frac{g}{(2\pi)^3}\int_0^{2\pi}\! d\varphi \int_0^\pi\! d\theta\,\sin\theta \int_0^\infty\! dk\ \frac{k^2\, e^{ikr\cos\theta}}{k^2+m^2} \\ &= \frac{g}{(2\pi)^2}\int_0^\infty dk\ \frac{k^2}{k^2+m^2}\cdot \frac{2\sin(kr)}{kr} = \frac{g}{2\pi^2 r}\int_0^\infty dk\ \frac{k\,\sin(kr)}{k^2+m^2} . \end{aligned}$$

In the second line we used $\int_0^\pi d\theta \sin\theta\, e^{ikr\cos\theta} = \big[ e^{ikr\cos\theta}/(-ikr) \big]_{\theta=0}^{\pi} = 2\sin(kr)/(kr)$.

Call the remaining integral $I$. The integrand is even in $k$, so $I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{k\sin(kr)}{k^2+m^2}dk$. Moreover $k\cos(kr)/(k^2+m^2)$ is odd and integrates to $0$, whence

$$\int_{-\infty}^{\infty} \frac{k\,e^{ikr}}{k^2+m^2}\,dk = 0 + i\cdot 2I .$$

Evaluate the left-hand side by residues. Since $r > 0$, close the contour with a semicircle in the upper half plane; the only pole of the integrand there is $k = im$, with residue $\dfrac{im\, e^{i(im)r}}{2im} = \dfrac{e^{-mr}}{2}$. The left-hand side is thus $2\pi i \cdot \frac{e^{-mr}}{2} = i\pi e^{-mr}$, giving $I = \frac{\pi}{2}e^{-mr}$. Therefore

$$\phi(r) = \frac{g}{2\pi^2 r}\cdot\frac{\pi}{2}e^{-mr} = \frac{g}{4\pi}\,\frac{e^{-mr}}{r} .$$

This is the Yukawa potential. As $m \to 0$ it becomes the Coulomb-type long-range force $g/(4\pi r)$, while for $m \ne 0$ the factor $e^{-mr}$ makes it decay rapidly over a distance of order $1/m$.

Let us put in numbers. The range of the nuclear force is about $1.4\ \mathrm{fm}$. Using $\hbar c = 197.3\ \mathrm{MeV\cdot fm}$,

$$m \approx \frac{\hbar c}{1.4\ \mathrm{fm}} = \frac{197.3\ \mathrm{MeV\cdot fm}}{1.4\ \mathrm{fm}} \approx 1.4 \times 10^2\ \mathrm{MeV} .$$

The pion, actually discovered in 1947, has a mass of about $135$–$140\ \mathrm{MeV}$, in good agreement with this estimate.

5.3. Interaction terms and mass dimension

Nothing happens in a free field alone: scattering and decay require interactions. Drop requirement 3 (quadratic in $\phi$) and add potential terms:

$$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2 - \sum_{n \ge 3}\frac{\lambda_n}{n!}\phi^n .$$

Remark 5.3（Mass dimension classifies interactions）

In §2 we established $[\mathcal{L}] = 4$. The kinetic term $\partial_\mu\phi\,\partial^\mu\phi$ gives $2 + 2[\phi] = 4$, that is,

$$[\phi] = 1 .$$

Hence $[\lambda_n] = 4 - n$, and the classification is as follows.

Term	Mass dimension of the coupling	Name
$\phi^3$	$[\lambda_3] = 1$	super-renormalizable
$\phi^4$	$[\lambda_4] = 0$	renormalizable
$\phi^5, \phi^6, \dots$	$[\lambda_n] < 0$	non-renormalizable (meaningful only as an effective theory)

The convention that “the first interaction to write down for a real scalar field in four-dimensional spacetime is $\phi^4$” comes from this dimensional counting. Why a negative mass dimension is problematic is treated in the classification of renormalizability（Definition 5.4）［くりこみ理論入門］ in introduction to renormalization. What matters here is that writing down $\mathcal{L}$ is not a mere formality: it is simultaneously an enumeration of the interactions the theory is allowed to have.

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6. Noether’s theorem (field-theoretic version)

The theorem Emmy Noether proved in 1918 states that every continuous symmetry of the action corresponds to one conserved quantity. In the finite-dimensional version (symmetries and conservation laws) the conserved object was a single function of time; in field theory we obtain the stronger statement of a local conservation law, the continuity equation $\partial_\mu j^\mu = 0$. This asserts that charge cannot disappear here and reappear there, and it is consistent with relativistic causality.

The logical structure of this chapter. A single function, the Lagrangian density, yields both the equations of motion and the conservation laws.

6.1. Defining a symmetry

The definition of a symmetry requires care. Demanding invariance of the action alone would throw away the freedom of Proposition 4.5, so we adopt quasi-invariance, allowing $\mathcal{L}$ to shift by a total derivative.

Definition 6.1（Continuous symmetry of the action (quasi-invariance)）

Consider a smooth one-parameter family of transformations with real parameter $\varepsilon$,

$$x^\mu \longmapsto x'^\mu = x^\mu + \varepsilon\, X^\mu(x) + O(\varepsilon^2), \qquad \phi_a(x) \longmapsto \phi'_a(x') = \phi_a(x) + \varepsilon\, F_a(x) + O(\varepsilon^2) ,$$

where $X^\mu$ is a smooth function of $x$ and $F_a$ is a smooth function of $\phi_b(x)$, $\partial_\mu\phi_b(x)$ and $x$.

If there exists a $K^\mu$ (a smooth function of $\phi_b$, $\partial_\mu\phi_b$ and $x$) such that the identity

$$\mathcal{L}\big(\phi'(x'),\, \partial' \phi'(x')\big)\ \det\!\left( \frac{\partial x'}{\partial x} \right) = \mathcal{L}\big(\phi(x),\, \partial\phi(x)\big) + \varepsilon\, \partial_\mu K^\mu + O(\varepsilon^2)$$

holds for every field configuration, whether or not it satisfies the equations of motion, then the family is called a continuous symmetry of the action. The case $K^\mu = 0$ is called strict invariance, the general case quasi-invariance.

The factor $\det(\partial x'/\partial x)$ on the left represents the change of the integration measure, $d^4x' = \det(\partial x'/\partial x)\, d^4x$; including it makes the left-hand side “the action density of the transformed theory measured in the original coordinates”.

The essential point is that the condition holds identically, independently of the equations of motion. An identity valid only on solutions of the equations of motion yields no new information.

6.2. Statement and proof

Theorem 6.2（Noether's theorem）

Let $\mathcal{L} = \mathcal{L}(\phi_a, \partial_\mu\phi_a)$ have no explicit dependence on the coordinate $x$, and let a continuous symmetry in the sense of Definition 6.1 be given. Put

$$\pi^\mu{}_a := \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi_a)}$$

and define the Noether current by

$$j^\mu := \pi^\mu{}_a \big( F_a - X^\nu \partial_\nu \phi_a \big) + \mathcal{L}\, X^\mu - K^\mu .$$

Then on any solution of the Euler–Lagrange equations (Theorem 4.3),

$$\partial_\mu j^\mu = 0 .$$

Proof（Theorem 6.2）

Step 1: write the symmetry condition to order $\varepsilon$.

The Jacobian matrix is $\partial x'^\mu/\partial x^\nu = \delta^\mu_\nu + \varepsilon\,\partial_\nu X^\mu + O(\varepsilon^2)$, so its determinant is

$$\det\!\left( \frac{\partial x'}{\partial x} \right) = 1 + \varepsilon\, \partial_\mu X^\mu + O(\varepsilon^2)$$

(using $\det(I + \varepsilon M) = 1 + \varepsilon\,\mathrm{tr}\,M + O(\varepsilon^2)$).

Next the transformation of derivatives. The inverse map gives $\partial x^\nu/\partial x'^\mu = \delta^\nu_\mu - \varepsilon\,\partial_\mu X^\nu + O(\varepsilon^2)$, so by the chain rule

$$\partial'_\mu \phi'_a(x') = \frac{\partial x^\nu}{\partial x'^\mu}\,\partial_\nu\big[\phi_a(x) + \varepsilon F_a(x)\big] = \partial_\mu \phi_a + \varepsilon\big( \partial_\mu F_a - (\partial_\mu X^\nu)\, \partial_\nu\phi_a \big) + O(\varepsilon^2) .$$

Since $\mathcal{L}$ has no explicit $x$-dependence, the chain rule gives

$$\mathcal{L}\big(\phi'(x'), \partial'\phi'(x')\big) = \mathcal{L} + \varepsilon\left[ \frac{\partial\mathcal{L}}{\partial\phi_a}F_a + \pi^\mu{}_a\big( \partial_\mu F_a - (\partial_\mu X^\nu)\partial_\nu\phi_a \big) \right] + O(\varepsilon^2) .$$

Multiplying by the determinant and comparing with the right-hand side of Definition 6.1 at order $\varepsilon$, we obtain the identity

$$\mathcal{L}\,\partial_\mu X^\mu + \frac{\partial\mathcal{L}}{\partial\phi_a}F_a + \pi^\mu{}_a\,\partial_\mu F_a - \pi^\mu{}_a\,(\partial_\mu X^\nu)\,\partial_\nu\phi_a = \partial_\mu K^\mu ,$$

which we shall call the symmetry identity. Note that no equation of motion has been used so far, and that it is an identity precisely because Definition 6.1 demanded validity for arbitrary field configurations.

Step 2: use the equations of motion to assemble the left-hand side into a divergence.

From now on let $\phi$ be a solution of the Euler–Lagrange equations, that is, $\dfrac{\partial\mathcal{L}}{\partial\phi_a} = \partial_\mu \pi^\mu{}_a$.

Combine the second and third terms. By the Leibniz rule,

$$\frac{\partial\mathcal{L}}{\partial\phi_a}F_a + \pi^\mu{}_a\partial_\mu F_a = (\partial_\mu \pi^\mu{}_a) F_a + \pi^\mu{}_a \partial_\mu F_a = \partial_\mu\big( \pi^\mu{}_a F_a \big) .$$

Now the first term. Since $\mathcal{L}$ has no explicit $x$-dependence, its total derivative with respect to $x$ is

$$\partial_\mu \mathcal{L} = \frac{\partial\mathcal{L}}{\partial\phi_a}\partial_\mu\phi_a + \pi^\nu{}_a\, \partial_\nu\partial_\mu \phi_a = (\partial_\nu \pi^\nu{}_a)\,\partial_\mu\phi_a + \pi^\nu{}_a\,\partial_\nu\partial_\mu\phi_a = \partial_\nu\big( \pi^\nu{}_a\, \partial_\mu \phi_a \big)$$

(the second equality uses the equations of motion again). Therefore

$$\mathcal{L}\,\partial_\mu X^\mu = \partial_\mu\big(\mathcal{L}X^\mu\big) - X^\mu\,\partial_\mu\mathcal{L} = \partial_\mu\big(\mathcal{L}X^\mu\big) - X^\mu\, \partial_\nu\big( \pi^\nu{}_a \partial_\mu\phi_a \big) .$$

Split the fourth term likewise by the Leibniz rule:

$$-\pi^\mu{}_a (\partial_\mu X^\nu)\partial_\nu\phi_a = -\partial_\mu\big( X^\nu \pi^\mu{}_a \partial_\nu\phi_a \big) + X^\nu \partial_\mu\big( \pi^\mu{}_a \partial_\nu\phi_a \big) .$$

Step 3: the extra terms containing $X$ cancel.

The terms carrying $X$ that appeared in the last two equations are

$$- X^\mu \partial_\nu\big( \pi^\nu{}_a \partial_\mu\phi_a \big) + X^\nu \partial_\mu\big( \pi^\mu{}_a \partial_\nu\phi_a \big) ,$$

and relabelling the summation indices $\mu \leftrightarrow \nu$ in the second term makes it the negative of the first, so the sum is $0$.

Substituting all of this into the symmetry identity gives

$$\partial_\mu\Big[ \mathcal{L}X^\mu + \pi^\mu{}_a F_a - X^\nu \pi^\mu{}_a \partial_\nu\phi_a \Big] = \partial_\mu K^\mu ,$$

that is, $\partial_\mu\big[ \pi^\mu{}_a(F_a - X^\nu\partial_\nu\phi_a) + \mathcal{L}X^\mu - K^\mu \big] = 0$, which is the required identity.

∎

From the local conservation law $\partial_\mu j^\mu = 0$ a globally conserved quantity follows. Note, however, that a physical assumption is needed: the current must fall off fast enough at infinity.

Corollary 6.3（Conservation of the Noether charge）

In the situation of Theorem 6.2, suppose the current $j^\mu$ of a solution $\phi$ satisfies, at each time $t$, on the sphere $S_R$ of radius $R$,

$$\lim_{R\to\infty} \oint_{S_R} j^i\, n^i\, dS = 0$$

($n^i$ the outward unit normal; for instance $|j^i| = o(R^{-2})$ suffices). Suppose also that $j^0$ is integrable over $\mathbb{R}^3$ at each time and that the time derivative may be interchanged with the integral. Then

$$Q(t) := \int_{\mathbb{R}^3} d^3x\ j^0(t, \boldsymbol{x})$$

is a constant independent of time.

Proof（Corollary 6.3）

Since $\partial_\mu j^\mu = \partial_0 j^0 + \partial_i j^i = 0$,

$$\frac{dQ}{dt} = \int_{\mathbb{R}^3} d^3x\ \partial_0 j^0 = -\int_{\mathbb{R}^3} d^3x\ \partial_i j^i = -\lim_{R\to\infty}\int_{|\boldsymbol{x}| \le R} d^3x\ \nabla\cdot \boldsymbol{j} = -\lim_{R\to\infty}\oint_{S_R} j^i n^i\, dS = 0 .$$

The first equality uses the interchange of derivative and integral, the third Gauss’s theorem, and the last the assumption.

∎

Remark 6.4（The current is ambiguous）

Adding $\partial_\nu \Sigma^{\nu\mu}$ to $j^\mu$, where $\Sigma^{\nu\mu} = -\Sigma^{\mu\nu}$ is any antisymmetric tensor, preserves the conservation law, since $\partial_\mu\partial_\nu\Sigma^{\nu\mu} = 0$ (a symmetric pair of derivatives contracted with an antisymmetric tensor). If moreover $\Sigma$ falls off fast enough at infinity, the charge $Q$ is unchanged as well. The Noether current, in other words, is not unique. We shall use this freedom in §6.5.

6.3. Spacetime translations and the energy–momentum tensor

Definition 6.5（Canonical energy–momentum tensor）

When $\mathcal{L}$ has no explicit $x$-dependence,

$$T^\mu{}_\nu := \pi^\mu{}_a\, \partial_\nu \phi_a - \delta^\mu_\nu\, \mathcal{L}, \qquad T^{\mu\nu} = \eta^{\nu\rho}\,T^\mu{}_\rho = \pi^\mu{}_a\, \partial^\nu\phi_a - \eta^{\mu\nu}\mathcal{L}$$

is called the canonical energy–momentum tensor.

Consider a spacetime translation $x'^\mu = x^\mu + \varepsilon\,\xi^\mu$ (with $\xi^\mu$ a constant vector; we write $\xi$ to avoid confusion with the field-component index $a$). The field is merely carried along, $\phi'_a(x') = \phi_a(x)$, so $X^\mu = \xi^\mu$ and $F_a = 0$. The Jacobian matrix is the identity, with $\det = 1$, and since $\mathcal{L}$ has no explicit $x$-dependence the left-hand side of Definition 6.1 is just $\mathcal{L}(\phi(x),\partial\phi(x))$, so we may take $K^\mu = 0$. Applying Theorem 6.2,

$$j^\mu = \pi^\mu{}_a\big( 0 - \xi^\nu \partial_\nu\phi_a \big) + \mathcal{L}\,\xi^\mu = -\xi^\nu\Big( \pi^\mu{}_a \partial_\nu\phi_a - \delta^\mu_\nu \mathcal{L} \Big) = -\xi^\nu\, T^\mu{}_\nu .$$

Since $\xi^\nu$ is arbitrary, we obtain four conservation laws,

$$\partial_\mu T^\mu{}_\nu = 0 \qquad (\nu = 0,1,2,3) .$$

By Corollary 6.3 the corresponding conserved quantities are

$$P^\nu = \int d^3x\ T^{0\nu} ,$$

with $P^0$ the energy and $P^i$ the momentum. Indeed $T^0{}_0 = \pi^0{}_a\dot\phi_a - \mathcal{L}$ has the same form as the finite-dimensional Legendre transform $H = p\dot q - L$ (definition of the Hamiltonian（Definition 3.6）［ハミルトン形式の力学］).

Example 6.6（Energy and momentum of the Klein–Gordon field）

For $\mathcal{L}_{\mathrm{KG}}$ we computed $\pi^\mu = \partial^\mu\phi$ in Example 5.1, so

$$T^{\mu\nu} = \partial^\mu\phi\,\partial^\nu\phi - \eta^{\mu\nu}\left[ \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi - \frac{1}{2}m^2\phi^2 \right] .$$

Noting $\partial_\alpha\phi\,\partial^\alpha\phi = \dot\phi^2 - (\nabla\phi)^2$, compute the $\mu = \nu = 0$ component:

$$T^{00} = \dot\phi^2 - \left[ \frac{1}{2}\dot\phi^2 - \frac{1}{2}(\nabla\phi)^2 - \frac{1}{2}m^2\phi^2 \right] = \frac{1}{2}\dot\phi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 .$$

All three terms are nonnegative, so the energy density is bounded below. This is why we required a positive coefficient for the kinetic term and $m^2 > 0$. The momentum density is

$$T^{0i} = \partial^0\phi\,\partial^i\phi = -\dot\phi\, \frac{\partial\phi}{\partial x^i}, \qquad \boldsymbol{P} = -\int d^3x\ \dot\phi\, \nabla\phi .$$

Let us check the sign. For a wave travelling in the $+x$ direction, $\phi = A\cos(\omega t - k x)$ with $k > 0$, we have $\dot\phi = -A\omega\sin(\omega t - kx)$ and $\partial\phi/\partial x = A k \sin(\omega t - kx)$, so $-\dot\phi\,\partial_x\phi = A^2\omega k \sin^2(\omega t - kx) \ge 0$: the momentum indeed points in the $+x$ direction.

Note also that $T^{\mu\nu}$ is symmetric (both $\partial^\mu\phi\,\partial^\nu\phi$ and $\eta^{\mu\nu}$ are). This will matter in §6.5.

6.4. Internal symmetry and conserved charge

Next come transformations that leave spacetime alone and rotate only within field space. For a complex scalar field $\phi$ (with its complex conjugate $\bar\phi$ regarded as an independent variable), consider

$$\mathcal{L} = \partial_\mu \bar\phi\, \partial^\mu\phi - m^2 \bar\phi \phi .$$

One may equally think of this as two real fields $\phi_1, \phi_2$ packaged as $\phi = (\phi_1 + i\phi_2)/\sqrt{2}$.

Example 6.7（Global U(1) symmetry and its conserved current）

Consider the transformation

$$\phi'(x) = e^{-i\varepsilon}\phi(x) = \phi - i\varepsilon\phi + O(\varepsilon^2), \qquad \bar\phi'(x) = e^{+i\varepsilon}\bar\phi(x) = \bar\phi + i\varepsilon\bar\phi + O(\varepsilon^2) .$$

Since $\varepsilon$ is a constant independent of spacetime, this is a “global” symmetry. In the notation of Definition 6.1, $X^\mu = 0$, $F_\phi = -i\phi$ and $F_{\bar\phi} = +i\bar\phi$. Because $\mathcal{L}$ is built solely out of the combinations $\bar\phi\phi$ and $\partial_\mu\bar\phi\,\partial^\mu\phi$, the phase factors cancel as $e^{+i\varepsilon}e^{-i\varepsilon} = 1$ and $\mathcal{L}$ is strictly invariant. We may thus take $K^\mu = 0$.

The conjugate momenta are

$$\pi^\mu{}_{\phi} = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = \partial^\mu\bar\phi, \qquad \pi^\mu{}_{\bar\phi} = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar\phi)} = \partial^\mu\phi ,$$

so Theorem 6.2 gives

$$j^\mu = \pi^\mu{}_\phi F_\phi + \pi^\mu{}_{\bar\phi} F_{\bar\phi} = \partial^\mu\bar\phi\,(-i\phi) + \partial^\mu\phi\,(i\bar\phi) = i\big( \bar\phi\,\partial^\mu\phi - \phi\,\partial^\mu\bar\phi \big) .$$

Let us verify the conservation directly. The Euler–Lagrange equations of $\mathcal{L}$ are $(\Box + m^2)\phi = 0$ from varying $\bar\phi$, and $(\Box+m^2)\bar\phi = 0$ from varying $\phi$. Hence

$$\partial_\mu j^\mu = i\big( \partial_\mu\bar\phi\,\partial^\mu\phi + \bar\phi\,\Box\phi - \partial_\mu\phi\,\partial^\mu\bar\phi - \phi\,\Box\bar\phi \big) = i\big( \bar\phi(-m^2\phi) - \phi(-m^2\bar\phi) \big) = 0 .$$

The first and third terms cancel, and the equations of motion kill what remains. The conserved charge is

$$Q = \int d^3x\ j^0 = i\int d^3x\ \big( \bar\phi\,\dot\phi - \phi\,\dot{\bar\phi} \big) .$$

Making this $U(1)$ local (letting $\varepsilon$ be a function of $x$) forces a coupling to the electromagnetic field to appear, and $Q$ becomes the electric charge itself (gauge theory and spontaneous symmetry breaking).

Remark 6.8（j⁰ is not a probability density）

The quantity $j^0 = i(\bar\phi\dot\phi - \phi\dot{\bar\phi})$ above has no definite sign. Indeed, for a negative-frequency solution $\phi = e^{+i\omega t}$ one finds $j^0 < 0$. The early attempts to read the Klein–Gordon equation as a relativistic one-particle Schrödinger equation broke down precisely because this $j^0$ could not be interpreted as a probability density (see the Klein–Gordon current and its lack of positivity（Proposition 3.2）［Why We Need Quantum Field Theory］ in why we need quantum field theory).

Within classical field theory this causes no trouble at all. $Q$ is a charge, not a probability, and it is perfectly natural for a charge to take either sign. After quantization one finds that the eigenvalues of $Q$ count the number of particles minus the number of antiparticles.

6.5. Lorentz transformations and angular momentum

Proposition 6.9（Angular momentum tensor of a scalar field）

Let $\mathcal{L}$ be the Lagrangian density of a real scalar field and a Lorentz scalar. As the Noether conserved quantity associated with the infinitesimal Lorentz transformation $x'^\mu = x^\mu + \varepsilon\,\omega^\mu{}_\nu x^\nu$ (with $\omega_{\mu\nu} = -\omega_{\nu\mu}$) together with $\phi'(x') = \phi(x)$, the quantity

$$M^{\mu\alpha\rho} := x^\alpha\, T^{\mu\rho} - x^\rho\, T^{\mu\alpha}$$

satisfies $\partial_\mu M^{\mu\alpha\rho} = 0$. Moreover, it follows from this that the canonical energy–momentum tensor is symmetric, $T^{\alpha\rho} = T^{\rho\alpha}$.

Proof（Proposition 6.9）

In the notation of Definition 6.1, $X^\mu = \omega^\mu{}_\nu x^\nu$ and $F = 0$. The Jacobian is $1 + \varepsilon\,\partial_\mu X^\mu = 1 + \varepsilon\,\omega^\mu{}_\mu$, but $\omega^\mu{}_\mu = \eta^{\mu\alpha}\omega_{\alpha\mu}$ is the contraction of the symmetric tensor $\eta$ with the antisymmetric tensor $\omega$ and therefore vanishes. Since $\mathcal{L}$ is a scalar, $\mathcal{L}(\phi'(x'),\partial'\phi'(x')) = \mathcal{L}(\phi(x),\partial\phi(x))$, and we may take $K^\mu = 0$. Hence

$$j^\mu = -T^\mu{}_\nu X^\nu = -T^\mu{}_\nu\, \omega^\nu{}_\rho\, x^\rho = -T^{\mu\alpha}\,\omega_{\alpha\rho}\, x^\rho$$

(using $T^\mu{}_\nu \eta^{\nu\alpha} = T^{\mu\alpha}$). Antisymmetrizing explicitly with the help of the antisymmetry of $\omega$,

$$j^\mu = -\frac{1}{2}\omega_{\alpha\rho}\big( x^\rho T^{\mu\alpha} - x^\alpha T^{\mu\rho} \big) = \frac{1}{2}\,\omega_{\alpha\rho}\, M^{\mu\alpha\rho} .$$

Since $\omega_{\alpha\rho}$ is an arbitrary constant antisymmetric matrix, the conclusion $\partial_\mu j^\mu = 0$ of Theorem 6.2 means $\partial_\mu M^{\mu\alpha\rho} = 0$ for each pair $(\alpha,\rho)$.

Now the last assertion. From $\partial_\mu x^\alpha = \delta^\alpha_\mu$ and the Leibniz rule,

$$\partial_\mu M^{\mu\alpha\rho} = T^{\alpha\rho} - T^{\rho\alpha} + x^\alpha\,\partial_\mu T^{\mu\rho} - x^\rho\,\partial_\mu T^{\mu\alpha} .$$

The last two terms vanish by $\partial_\mu T^{\mu\nu} = 0$, shown in §6.3, so $\partial_\mu M^{\mu\alpha\rho} = T^{\alpha\rho} - T^{\rho\alpha} = 0$.

∎

Forming $J^i = \frac{1}{2}\epsilon^{ijk}\int d^3x\, M^{0jk}$ from the spatial components of $M^{\mu\alpha\rho}$ gives the ordinary angular momentum. That this is the field-theoretic version of $\boldsymbol{r} \times \boldsymbol{p}$ can be read off from $M^{0jk} = x^j T^{0k} - x^k T^{0j}$.

Remark 6.10（For fields with spin the canonical tensor is not symmetric）

The argument of Proposition 6.9 is specific to scalar fields. For vector or spinor fields the field components mix under Lorentz transformations, so $F_a \ne 0$, and the current acquires a “spin part” $S^{\mu\alpha\rho}$, giving $M^{\mu\alpha\rho} = x^\alpha T^{\mu\rho} - x^\rho T^{\mu\alpha} + S^{\mu\alpha\rho}$. What $\partial_\mu M = 0$ then requires is $T^{\alpha\rho} - T^{\rho\alpha} = -\partial_\mu S^{\mu\alpha\rho}$, not the symmetry of $T$.

The canonical tensor of the electromagnetic field is in fact neither symmetric nor gauge invariant. Using the freedom of Remark 6.4 to absorb the spin part and thereby produce a symmetric, gauge-invariant tensor is the Belinfante–Rosenfeld improvement, and the result coincides with the $T^{\mu\nu}$ that appears as the source of gravity in general relativity. For details see, for example, Weinberg’s textbook.

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7. Toward the Hamiltonian formalism: preparing for quantization

To proceed to canonical quantization we must pass from the Lagrangian to the Hamiltonian formalism. This is the stage at which time is singled out, so Lorentz covariance ceases to be manifest (it is still present in the theory).

Definition 7.1（Conjugate momentum density and Hamiltonian density）

Define the momentum density conjugate to the field $\phi_a$ by

$$\pi_a(x) := \frac{\partial \mathcal{L}}{\partial \dot\phi_a} = \pi^0{}_a(x) .$$

Assuming that $\dot\phi_a$ can be solved for in terms of $\pi_a$ (that is, that the Legendre transform exists), define the Hamiltonian density by

$$\mathcal{H} := \pi_a \dot\phi_a - \mathcal{L}$$

and call $H := \int d^3x\ \mathcal{H}$ the Hamiltonian.

Comparing with Definition 6.5 shows that $\mathcal{H} = T^0{}_0 = T^{00}$ and $H = P^0$. The Hamiltonian is thus nothing but the time component of the Noether charge of spacetime translation symmetry. The statement that energy is conserved because of time-translation symmetry holds in field theory exactly as it does for finitely many degrees of freedom.

For the Klein–Gordon field $\pi = \dot\phi$, so

$$\mathcal{H} = \dot\phi^2 - \mathcal{L}_{\mathrm{KG}} = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2 ,$$

in agreement with the $T^{00}$ of Example 6.6. The $\nabla\phi$ term penalizes differences between the values at neighbouring points; it can be read as the elastic energy of the springs of Example 3.1, surviving intact.

Equal-time Poisson brackets are defined by reading the finite-dimensional $\{q_i, p_j\} = \delta_{ij}$ with a continuous index:

$$\{\phi_a(t,\boldsymbol{x}),\, \pi_b(t,\boldsymbol{y})\} = \delta_{ab}\,\delta^3(\boldsymbol{x} - \boldsymbol{y}), \qquad \{\phi_a(t,\boldsymbol{x}),\, \phi_b(t,\boldsymbol{y})\} = \{\pi_a(t,\boldsymbol{x}),\, \pi_b(t,\boldsymbol{y})\} = 0 .$$

The only change is that the Kronecker delta has become a Dirac delta function (Hamiltonian mechanics, canonical transformations and Poisson brackets).

In the next chapter we replace this bracket by a commutator via $\{\ ,\ \} \to -i[\ ,\ ]$. That is, imposing

$$[\hat\phi(t,\boldsymbol{x}),\, \hat\pi(t,\boldsymbol{y})] = i\,\delta^3(\boldsymbol{x}-\boldsymbol{y})$$

is canonical quantization of the scalar field (canonical quantization of the real scalar field（Definition 3.1）［スカラー場の正準量子化］). The alternative route, which keeps time on the same footing as space and uses the action $S[\phi]$ directly, is path-integral quantization. Whichever road one takes, the starting point is the $\mathcal{L}$ built in this chapter, and the symmetries encoded in it are inherited by the quantum theory — or else broken by quantum effects, which is what an anomaly is.

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8. Exercises

Exercise 8.1Hard

Consider the Lagrangian density of the electromagnetic field,

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - A_\mu J^\mu, \qquad F_{\mu\nu} := \partial_\mu A_\nu - \partial_\nu A_\mu ,$$

where $J^\mu$ is a given external current independent of $A_\mu$. The fundamental fields are the four components $A_\nu$.

1. Apply Theorem 4.3 and derive the equations of motion.
2. Verify that the $\nu = 0$ component of the resulting equation is Gauss’s law.

Solution

1. First, since $F_{\mu\nu}$ contains only derivatives of $A$, the derivative with respect to the field itself is

$$\frac{\partial \mathcal{L}}{\partial A_\nu} = -J^\nu .$$

Next the derivative with respect to $\partial_\mu A_\nu$. From $F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha$,

$$\frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha \delta^\nu_\beta - \delta^\nu_\alpha \delta^\mu_\beta .$$

The chain rule then gives

$$\frac{\partial}{\partial(\partial_\mu A_\nu)}\big( F_{\alpha\beta}F^{\alpha\beta} \big) = 2F^{\alpha\beta}\big( \delta^\mu_\alpha\delta^\nu_\beta - \delta^\nu_\alpha\delta^\mu_\beta \big) = 2\big( F^{\mu\nu} - F^{\nu\mu} \big) = 4F^{\mu\nu}$$

(the last step uses the antisymmetry of $F$). Hence

$$\pi^\mu{}_{(\nu)} = \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\frac{1}{4}\cdot 4 F^{\mu\nu} = -F^{\mu\nu} .$$

Substituting into the Euler–Lagrange equation $\partial_\mu\big(\partial\mathcal{L}/\partial(\partial_\mu A_\nu)\big) - \partial\mathcal{L}/\partial A_\nu = 0$,

$$-\partial_\mu F^{\mu\nu} + J^\nu = 0 \qquad\Longleftrightarrow\qquad \partial_\mu F^{\mu\nu} = J^\nu .$$

These are the two inhomogeneous Maxwell equations (Gauss’s law and the Ampère–Maxwell law). The two homogeneous ones (absence of magnetic flux and Faraday’s law) follow as identities from the definition $F = dA$, so they do not come out of the variational principle.

2. With $A^\mu = (\varphi, \boldsymbol{A})$ we have $A_\mu = (\varphi, -\boldsymbol{A})$, and $F_{0i} = \partial_0 A_i - \partial_i A_0 = -\partial_t A^i - \partial_i\varphi = E^i$ (since $\boldsymbol{E} = -\nabla\varphi - \partial_t\boldsymbol{A}$). Raising indices, $F^{0i} = \eta^{00}\eta^{ii}F_{0i} = -E^i$, that is, $F^{i0} = E^i$. Noting $F^{00} = 0$ by antisymmetry, the $\nu = 0$ component reads

$$\partial_\mu F^{\mu 0} = \partial_0 F^{00} + \partial_i F^{i0} = \nabla\cdot\boldsymbol{E} = J^0 = \rho ,$$

which is Gauss’s law.

Exercise 8.2Standard

Consider a massless real scalar field, $\mathcal{L}_0 = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi$.

1. Verify that the shift $\phi'(x) = \phi(x) + \varepsilon c$ by a constant $c$ (with the spacetime coordinates untouched) is a symmetry in the sense of Definition 6.1, and find the Noether current. Confirm directly from the equation of motion that it is conserved.
2. Show that adding a mass term $-\frac{1}{2}m^2\phi^2$ (with $m \ne 0$) makes this transformation no longer a symmetry.

Solution

1. Here $X^\mu = 0$ and $F = c$. The density $\mathcal{L}_0$ contains no $\phi$ itself, only $\partial_\mu\phi$, and $\partial_\mu(\phi + \varepsilon c) = \partial_\mu\phi$, so $\mathcal{L}_0$ is strictly invariant and we may take $K^\mu = 0$. Since the coordinates are not moved, the Jacobian is $1$.

With $\pi^\mu = \partial^\mu\phi$, Theorem 6.2 gives

$$j^\mu = \pi^\mu F = c\,\partial^\mu\phi .$$

Conservation is checked directly: $\partial_\mu j^\mu = c\,\partial_\mu\partial^\mu\phi = c\,\Box\phi = 0$, the last equality being the Klein–Gordon equation with $m = 0$. The corresponding charge is $Q = c\int d^3x\,\dot\phi$, which states that the “average velocity” of the field does not change.

2. Applying the transformation to $\mathcal{L} = \mathcal{L}_0 - \frac{1}{2}m^2\phi^2$,

$$\mathcal{L}' - \mathcal{L} = -\frac{1}{2}m^2\big[ (\phi + \varepsilon c)^2 - \phi^2 \big] = -\varepsilon\, m^2 c\, \phi + O(\varepsilon^2) .$$

The term at order $\varepsilon$ is $-m^2 c\,\phi$, and this can never equal the total derivative $\partial_\mu K^\mu$ of any $K^\mu(\phi, x)$. Indeed, the chain rule decomposes the total derivative of $K^\mu$ as

$$\partial_\mu K^\mu = \frac{\partial K^\mu}{\partial \phi}\,\partial_\mu\phi + \big( \partial_\mu K^\mu \big)_{\mathrm{expl}} ,$$

the second term differentiating only the explicit $x$-dependence at fixed $\phi$. Suppose the identity $-m^2c\,\phi = \partial_\mu K^\mu$ held for every field configuration. The left-hand side contains no $\partial_\mu\phi$, whereas the first term on the right is proportional to $\partial_\mu\phi$; hence $\partial K^\mu/\partial\phi = 0$, so $K^\mu$ must be a function of $x$ alone. But then the right-hand side does not depend on $\phi$ while the left-hand side is proportional to $\phi$, so $m^2 c = 0$, that is, $m = 0$, on pain of contradiction.

Thus for $m \ne 0$ the shift symmetry is broken. Incidentally, the fact that a massless scalar field enjoys a shift symmetry is deeply related to the reason why Nambu–Goldstone bosons cannot have a mass.

Exercise 8.3Standard

For the energy–momentum tensor of the Klein–Gordon field,

$$T^{\mu\nu} = \partial^\mu\phi\,\partial^\nu\phi - \eta^{\mu\nu}\left[ \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi - \frac{1}{2}m^2\phi^2 \right] ,$$

verify $\partial_\mu T^{\mu\nu} = 0$ by direct computation, without going through Noether’s theorem (take $\phi$ to be a solution of the Klein–Gordon equation).

Solution

Differentiate each term by the Leibniz rule:

$$\partial_\mu\big( \partial^\mu\phi\,\partial^\nu\phi \big) = (\Box\phi)\,\partial^\nu\phi + \partial^\mu\phi\,\partial_\mu\partial^\nu\phi .$$

For the second term, $\eta^{\mu\nu}\partial_\mu = \partial^\nu$, so

$$-\partial_\mu\left( \eta^{\mu\nu}\left[ \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi - \frac{1}{2}m^2\phi^2 \right] \right) = -\partial^\nu\left[ \frac{1}{2}\partial_\alpha\phi\,\partial^\alpha\phi \right] + \frac{1}{2}m^2\,\partial^\nu(\phi^2) = -\partial_\alpha\phi\,\partial^\nu\partial^\alpha\phi + m^2\phi\,\partial^\nu\phi .$$

Now $\partial^\mu\phi\,\partial_\mu\partial^\nu\phi$ and $\partial_\alpha\phi\,\partial^\nu\partial^\alpha\phi$ are the same object: relabel the summation index $\mu \to \alpha$ and use that partial derivatives commute, since $\phi$ is of class $C^2$. These two therefore cancel, leaving

$$\partial_\mu T^{\mu\nu} = (\Box\phi)\,\partial^\nu\phi + m^2\phi\,\partial^\nu\phi = \big( \Box\phi + m^2\phi \big)\,\partial^\nu\phi = 0 .$$

The final equality is the Klein–Gordon equation. Note that the equation of motion was used only in that last line; everything before it was an identity.

Exercise 8.4Easy

Work in natural units in $d$-dimensional spacetime (one time dimension plus $d-1$ spatial dimensions).

1. Find the mass dimension $[\phi]$ of a real scalar field $\phi$.
2. Express the mass dimension $[\lambda_n]$ of the coupling in the interaction term $-\dfrac{\lambda_n}{n!}\phi^n$ in terms of $d$ and $n$, and describe what happens for $d = 4$ and for $d = 2$.

Solution

1. The action $S = \int d^dx\,\mathcal{L}$ is dimensionless and $[d^dx] = -d$, so $[\mathcal{L}] = d$. The kinetic term $\partial_\mu\phi\,\partial^\mu\phi$ has dimension $2[\partial] + 2[\phi] = 2 + 2[\phi]$, hence

$$2 + 2[\phi] = d \iff [\phi] = \frac{d-2}{2} .$$

2. From $[\lambda_n] + n[\phi] = d$,

$$[\lambda_n] = d - n\,\frac{d-2}{2} .$$

For $d = 4$ this is $[\lambda_n] = 4 - n$: the value is $1$ for $n = 3$ (a coupling with the dimension of a mass), $0$ for $n = 4$ (dimensionless), and negative for $n \ge 5$. Couplings of negative dimension grow more important at high energies and are non-renormalizable (Remark 5.3).

For $d = 2$ we get $[\phi] = 0$: the field itself is dimensionless. Consequently $[\lambda_n] = 2$ for every $n$, and an arbitrary function $V(\phi)$ of $\phi$ is allowed as a potential. This is why theories containing non-polynomial functions of $\phi$, such as the sine-Gordon model with $V(\phi) \propto \cos(\beta\phi)$, are studied in two dimensions.

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References

• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Westview Press, 1995 — the first half of Chapter 2, “The Klein-Gordon Field”, gives a concise account of almost exactly the path taken here (Lagrangian density, Euler–Lagrange equations, Noether’s theorem, the Klein–Gordon field).
• H. Goldstein, C. Poole and J. Safko, Classical Mechanics, 3rd edition, Addison-Wesley, 2002 — Chapter 13. Treats the continuum limit of coupled oscillators and the Lagrangian and Hamiltonian formalisms for fields independently of quantum theory. Example 3.1 follows the discussion of that chapter.
• S. Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995 — Chapter 7, “The Canonical Formalism”. Goes further than this article on Noether’s theorem and the improvement (Belinfante–Rosenfeld) of the energy–momentum tensor.
• T. Kugo, Gēji-ba no Ryōshiron I, Baifūkan, 1989 (in Japanese) — Chapter 1. A standard reference giving a careful treatment of classical field theory and Noether’s theorem.
• E. Noether, “Invariante Variationsprobleme”, Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse (1918), 235–257. English translation: M. A. Tavel, “Invariant Variation Problems”, arXiv:physics/0503066.
• H. Yukawa, “On the Interaction of Elementary Particles. I”, Proceedings of the Physico-Mathematical Society of Japan 17 (1935), 48–57. The original source of the calculation in Example 5.2.

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Appendix: Rewriting in terms of functional derivatives

Definition of the functional derivative. In the main text we treated variations as derivatives with respect to $\varepsilon$, but the literature on field theory makes wide use of functional-derivative notation. Given a functional $S[\phi]$, if

$$\left.\frac{d}{d\varepsilon}\right|_{0} S[\phi + \varepsilon\eta] = \int_\Omega d^4x\ \frac{\delta S}{\delta \phi_a(x)}\,\eta_a(x)$$

holds for every smooth $\eta_a$ with compact support contained in $\Omega$, the coefficient $\dfrac{\delta S}{\delta\phi_a(x)}$ in the integrand is called the functional derivative. By Lemma 4.2, this coefficient is uniquely determined (within continuous functions).

Rewriting the Euler–Lagrange equations. In this notation, the expression obtained in the proof of Theorem 4.3 reads

$$\frac{\delta S}{\delta \phi_a(x)} = \frac{\partial\mathcal{L}}{\partial\phi_a} - \partial_\mu\!\left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)} \right) ,$$

so the equations of motion become the single line $\dfrac{\delta S}{\delta\phi_a(x)} = 0$. This is the finite-dimensional $\dfrac{\partial S}{\partial q_i} = 0$ with the index $i$ replaced by the continuous variable $x$, corresponding to the last row of the dictionary in §3.

A basic formula. The definition immediately yields

$$\frac{\delta \phi_a(x)}{\delta \phi_b(y)} = \delta_{ab}\,\delta^4(x - y) .$$

Indeed, applying the defining relation above to $S[\phi] = \phi_a(x)$ (the functional that returns the field value at a fixed $x$), the left-hand side is $\eta_a(x)$ while the right-hand side is $\int d^4y\,\frac{\delta\phi_a(x)}{\delta\phi_b(y)}\eta_b(y)$; for these to agree for every $\eta$, the kernel must be a delta function. This formula is used repeatedly in deriving the Schwinger–Dyson equations from the path integral.