1.8 Gauge theories

At present, physically sensible theories of fundamental interactions are based on gauge theories. The gauge principle—physics should not depend on how we describe it—is in harmony with the principle of general relativity. Here we give a brief summary of classical aspects of gauge theories. For further references, the reader should consult those books listed at the beginning of this chapter.

1.8.1 Abelian gauge theories

The reader should be familiar with Maxwell’s equations:

$$\mathrm{div}\mathbf{B}=0\text{\hspace{1em}(1.241a)}$$

$$\frac{\partial \mathbf{B}}{\partial t}+\mathrm{curl}\mathbf{E}=0\text{\hspace{1em}(1.241b)}$$

$$\mathrm{div}\mathbf{E}=\rho \text{\hspace{1em}(1.241c)}$$

$$\frac{\partial \mathbf{E}}{\partial t}-\mathrm{curl}\mathbf{E}=-\mathbf{j}.\text{\hspace{1em}(1.241d)}$$

The magnetic field $B$ and the electric field $E$ are expressed in terms of the vector potential $A_{\mu }=(\varphi ,A)$ as

$$\mathbf{B}=\mathrm{curl}\mathbf{A}\mathbf{E}=\frac{\partial \mathbf{A}}{\partial t}-\mathrm{grad}\varphi .\text{\hspace{1em}(1.242)}$$

Maxwell’s equations are invariant under the gauge transformation

$$A_{\mu }\to A_{\mu }+{\partial }_{\mu }\chi \text{\hspace{1em}(1.243)}$$

where $\chi$ is a scalar function. This invariance is manifest if we define the electromagnetic field tensor $F_{\mu \nu }$ by

$$F_{\mu v}\equiv {\partial }_{\mu }{A}_v-{\partial }_v{A}_{\mu }=\left(\begin{array}{cccc}0& -E_x& -E_y& -E_z\\ E_x& 0& B_z& -B_y\\ E_y& -B_z& 0& B_z\\ E_z& B_y& -B_x& 0\end{array}\right).\text{\hspace{1em}(1.244)}$$

From the construction, $F$ is invariant under (1.243). The Lagrangian of the electromagnetic fields is given by

$${\mathcal{L}}_{\mathrm{EM}}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }+A_{\mu }{j}^{\mu }\text{\hspace{1em}(1.245)}$$

where $j^{\mu }=(\rho ,j)$

Exercise 1.11

Show that (1.241a) and (1.241b) are written as

$${\partial }_{\xi }{F}_{\mu \nu }+{\partial }_{\mu }{F}_{\nu \xi }+{\partial }_{\nu }{F}_{\xi \mu }=0\text{\hspace{1em}(1.246a)}$$

while (1.241c) and (1.241d) are

$${\partial }_v{F}^{\mu v}=j^{\mu }\text{\hspace{1em}(1.246b)}$$

where the raising and lowering of spacetime indices are carried out with the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ . Verify that (1.246b) is the Euler-Lagrange equation derived from (1.245).

Let $\psi$ be a Dirac field with electric charge $e$ . The free Dirac Lagrangian

$${\mathcal{L}}_0=\bar{\psi }(\mathrm{i}{\gamma }^{\mu }{\partial }_{\mu }+m)\psi \text{\hspace{1em}(1.247)}$$

is clearly invariant under the global gauge transformation

$$\psi \to {\mathrm{e}}^{-\mathrm{i}e\alpha }\psi \bar{\psi }\to \bar{\psi }{\mathrm{e}}^{\mathrm{i}e\alpha }\text{\hspace{1em}(1.248)}$$

where $\alpha \in \mathbb{R}$ is a constant. We elevate this symmetry to invariance under the local gauge transformation,

$$\psi \to {\mathrm{e}}^{-\mathrm{i}e\alpha (x)}\psi \bar{\psi }\to \bar{\psi }{\mathrm{e}}^{\mathrm{i}e\alpha (x)}.\text{\hspace{1em}(1.249)}$$

The Lagrangian transforms under (1.249) as

$$\bar{\psi }\left(\mathrm{i}{\gamma }^{\mu }{\partial }_{\mu }+m\right)\psi \to \bar{\psi }\left(\mathrm{i}{\gamma }^{\mu }{\partial }_{\mu }+e{\gamma }^{\mu }{\partial }_{\mu }\alpha +m\right)\psi .\text{\hspace{1em}(1.250)}$$

Since the extra term $e{\partial }_{\mu }\alpha$ looks like a gauge transformation of the vector potential, we couple the gauge field $A_{\mu }$ with $\psi$ so that the Lagrangian has a local gauge symmetry. We find that

$$\mathcal{L}=\bar{\psi }[\mathrm{i}{\gamma }^{\mu }\left({\partial }_{\mu }-\mathrm{i}e{A}_{\mu }\right)+m]\psi \text{\hspace{1em}(1.251)}$$

is invariant under the combined gauge transformation,

$$\begin{array}{rl}& \psi \to {\psi }^{\prime }={\mathrm{e}}^{-\mathrm{i}e\alpha (x)}\psi \bar{\psi }\to {\bar{\psi }}^{\prime }=\bar{\psi }{\mathrm{e}}^{\mathrm{i}e\alpha (x)}\\ & A_{\mu }\to A_{\mu }^{\prime }=A_{\mu }-{\partial }_{\mu }\alpha (x).\end{array}\text{\hspace{1em}(1.252)}$$

Let us introduce the covariant derivatives,

$${\nabla }_{\mu }\equiv {\partial }_{\mu }-\mathrm{i}e{A}_{\mu }{\nabla }_{\mu }^{\prime }\equiv {\partial }_{\mu }-\mathrm{i}e{A}_{\mu }^{\prime }.\text{\hspace{1em}(1.253)}$$

The reader should verify that ${\nabla }_{\mu }\psi$ transforms in a nice way,

$${\nabla }_{\mu }^{\prime }{\psi }^{\prime }={\mathrm{e}}^{-\mathrm{i}e\alpha (x)}{\nabla }_{\mu }\psi .\text{\hspace{1em}(1.254)}$$

The total quantum electrodynamic (QED) Lagrangian is

$${\mathcal{L}}_{\mathrm{QED}}=-\frac{1}{4}{F}^{\mu \nu }{F}_{\mu \nu }+\bar{\psi }(\mathrm{i}{\gamma }^{\mu }{\nabla }_{\mu }+m)\psi .\text{\hspace{1em}(1.255)}$$

Exercise 1.12

Let $\varphi =({\varphi }_1+\mathrm{i}{\varphi }_2)/\sqrt{2}$ be a complex scalar field with electric charge $e$ . Show that the Lagrangian

$$\mathcal{L}={\eta }^{\mu v}({\nabla }_{\mu }\varphi {)}^{†}({\nabla }_v\varphi )+m^2{\varphi }^{†}\varphi \text{\hspace{1em}(1.256)}$$

is invariant under the gauge transformation

$$\varphi \to {\mathrm{e}}^{-\mathrm{i}e\alpha (x)}\varphi {\varphi }^{†}\to {\varphi }^{†}{\mathrm{e}}^{\mathrm{i}e\alpha (x)}{A}_{\mu }\to A_{\mu }-{\partial }_{\mu }\alpha (x).\text{\hspace{1em}(1.257)}$$

1.8.2 Non-Abelian gauge theories

The gauge transformation just described is a member of a U(1) group, that is a complex number of modulus 1, which happens to be an Abelian group. A few decades ago, Yang and Mills (1954) introduced non-Abelian gauge transformations. At that time, non-Abelian gauge theories were studied from curiosity. Nowadays, they play a central role in elementary particle physics.

Let $G$ be a compact semi-simple Lie group such as $\mathrm{SO}(N)$ or $\mathrm{SU}(N)$ . The anti-Hermitian generators $\{T_{\alpha }\}$ satisfy the commutation relations

$$\left[T_{\alpha },T_{\beta }\right]=f_{\alpha \beta }^{\gamma }{T}_{\gamma }\text{\hspace{1em}(1.258)}$$

where the numbers $f_{\alpha \beta }^{\gamma }$ are called the structure constants of $G$ . An element $U$ of $G$ near the unit element can be expressed as

$$U=\exp \left(-{\theta }^{\alpha }{T}_{\alpha }\right).\text{\hspace{1em}(1.259)}$$

We suppose a Dirac field $\psi$ transforms under $U\in G$ as

$$\psi \to U\psi \bar{\psi }\to \bar{\psi }{U}^{†}.\text{\hspace{1em}(1.260)}$$

[Remark: Strictly speaking, we have to specify the representation of $G$ to which $\psi$ belongs. If readers feel uneasy about (1.260), they may consider $\psi$ is in the fundamental representation, for example.]

Consider the Lagrangian

$$\mathcal{L}=\bar{\psi }\left[\mathrm{i}{\gamma }^{\mu }\left({\partial }_{\mu }+g{\mathcal{A}}_{\mu }\right)+m\right]\psi \text{\hspace{1em}(1.261)}$$

where the Yang-Mills gauge field ${\mathcal{A}}_{\mu }$ takes its values in the Lie algebra of $G$ , that is, ${\mathcal{A}}_{\mu }$ can be expanded in terms of $T_{\alpha }$ as ${\mathcal{A}}_{\mu }=A_{\mu }^{\alpha }{T}_{\alpha }$ . (Script fields are anti-Hermitian.) The constant $g$ is the coupling constant which controls the strength of the coupling between the Dirac field and the gauge field. It is easily verified that $\mathcal{L}$ is invariant under

$$\psi \to {\psi }^{\prime }=U\psi \bar{\psi }\to {\bar{\psi }}^{\prime }=\bar{\psi }{U}^{†}\text{\hspace{1em}(1.262)}$$

$${\mathcal{A}}_{\mu }\to {\mathcal{A}}_{\mu }^{\prime }=U{\mathcal{A}}_{\mu }{U}^{†}+g^{-1}U{\partial }_{\mu }{U}^{†}.$$

The covariant derivative is defined by ${\nabla }_{\mu }={\partial }_{\mu }+g{\mathcal{A}}_{\mu }$ as before. The covariant derivative ${\nabla }_{\mu }\psi$ transforms covariantly under the gauge transformation

$${\nabla }_{\mu }^{\prime }{\psi }^{\prime }=U{\nabla }_{\mu }\psi .\text{\hspace{1em}(1.263)}$$

The Yang-Mills field tensor is

$${\mathcal{F}}_{\mu v}\equiv {\partial }_{\mu }{\mathcal{A}}_v-{\partial }_v{\mathcal{A}}_{\mu }+g[{\mathcal{A}}_{\mu },{\mathcal{A}}_v].\text{\hspace{1em}(1.264)}$$

The component $F_{\mu \nu }^{\alpha }$ is

$$F_{\mu v}^{\alpha }={\partial }_{\mu }{A}_v^{\alpha }-{\partial }_v{A}_{\mu }^{\alpha }+g{f}_{\beta \gamma }^{\alpha }{A}_{\mu }^{\beta }{A}_v^{\gamma }.\text{\hspace{1em}(1.265)}$$

If we define the dual field tensor $\ast {\mathcal{F}}_{\mu \nu }\equiv \frac{1}{2}{\epsilon }_{\mu \nu \kappa \lambda }{\mathcal{F}}^{\kappa \lambda }$ , it satisfies the Bianchi identity,

$${\mathcal{D}}_{\mu }\ast {\mathcal{F}}^{\mu v}\equiv {\partial }_{\mu }\ast {\mathcal{F}}^{\mu v}+g[{\mathcal{A}}_{\mu },\ast {\mathcal{F}}^{\mu v}]=0.\text{\hspace{1em}(1.266)}$$

Exercise 1.13

Show that ${\mathcal{F}}_{\mu \nu }$ transforms under (1.262) as

$${\mathcal{F}}_{\mu v}\to U{\mathcal{F}}_{\mu v}{U}^{†}.\text{\hspace{1em}(1.267)}$$

From this exercise, we find a gauge-invariant action

$${\mathcal{L}}_{\mathrm{YM}}=-\frac{1}{2}\mathrm{tr}\left({\mathcal{F}}^{\mu v}{\mathcal{F}}_{\mu v}\right)\text{\hspace{1em}(1.268a)}$$

where the trace is over the group matrix. The component form is

$${\mathcal{L}}_{\mathrm{YM}}=-\frac{1}{2}{F}^{\mu v\alpha }{F}_{\mu v}^{\beta }\mathrm{tr}\left(T_{\alpha }{T}_{\beta }\right)=\frac{1}{4}{F}^{\mu v\alpha }{F}_{\mu v\alpha }\text{\hspace{1em}(1.268b)}$$

where we have normalized $\{T_{\alpha }\}$ so that $\mathrm{tr}(T_{\alpha }{T}_{\beta })=-\frac{1}{2}{\delta }_{\alpha \beta }$ . The field equation derived from (1.268) is

$${\mathcal{D}}_{\mu }{\mathcal{F}}_{\mu v}={\partial }_{\mu }{\mathcal{F}}_{\mu v}+g\left[{\mathcal{A}}_{\mu },{\mathcal{F}}_{\mu v}\right]=0.\text{\hspace{1em}(1.269)}$$

1.8.3 Higgs fields

If the gauge symmetry is manifest in our world, there would be many observable massless vector fields. The absence of such fields, except for the electromagnetic field, forces us to break the gauge symmetry. The theory is left renormalizable if the symmetry is broken spontaneously.

Let us consider a U(1) gauge field coupled to a complex scalar field $\varphi$ , whose Lagrangian is given by

$$\mathcal{L}=-\frac{1}{4}{F}^{\mu v}{F}_{\mu v}+({\nabla }_{\mu }\varphi {)}^{†}({\nabla }_{\mu }\varphi )-\lambda {\left({\varphi }^{†}\varphi -v^2\right)}^2.\text{\hspace{1em}(1.270)}$$

The potential $V(\varphi )=\lambda ({\varphi }^{†}\varphi -v^2{)}^2$ has minima $V=0$ at $|\varphi |=v$ . The Lagrangian (1.270) is invariant under the local gauge transformation

$$A_{\mu }\to A_{\mu }-{\partial }_{\mu }\alpha \varphi \to {\mathrm{e}}^{-\mathrm{i}e\alpha }\varphi {\varphi }^{†}\to {\mathrm{e}}^{\mathrm{i}e\alpha }{\varphi }^{†}.\text{\hspace{1em}(1.271)}$$

This symmetry is spontaneously broken due to the vacuum expectation value (VEV) $⟨\varphi ⟩$ of the Higgs field $\varphi$ . We expand $\varphi$ as

$$\varphi =\frac{1}{\sqrt{2}}[v+\rho (x)]{\mathrm{e}}^{\mathrm{i}\alpha (x)/v}\sim \frac{1}{\sqrt{2}}[v+\rho (x)+\mathrm{i}\alpha (x)]$$

assuming $v\ne 0$ . If $v\ne 0$ , we may take the unitary gauge in which the phase of $\varphi$ is ‘gauged away’ so that $\varphi$ has only the real part,

$$\varphi (x)=\frac{1}{\sqrt{2}}(v+\rho (x)).\text{\hspace{1em}(1.272)}$$

If we substitute (1.272) into (1.270) and expand in $\rho$ , we have

$$\begin{array}{rl}& \mathcal{L}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }+\frac{1}{2}{\partial }_{\mu }\rho {\partial }^{\mu }\rho +\frac{1}{2}{e}^2{A}_{\mu }{A}^{\mu }(v^2+2v\rho +{\rho }^2)\\ & -\frac{1}{4}\lambda \left(4v^2{\rho }^2+4v{\rho }^3+{\rho }^4\right).\end{array}\text{\hspace{1em}(1.273)}$$

The equations of motion for $A_{\mu }$ and $\rho$ derived from the free parts are

$${\partial }^{\nu }{F}_{\nu \mu }+2e^2{v}^2{A}_{\mu }=0{\partial }_{\mu }{\partial }^{\mu }\rho +2\lambda v^2\rho =0.\text{\hspace{1em}(1.274)}$$

From the first equation, we find $A_{\mu }$ must satisfy the Lorentz condition ${\partial }_{\mu }{A}^{\mu }=0$ . The apparent degrees of freedom of (1.270) are 2(photon) + 2(complex scalar) = 4. If VEV ≠ 0, we have 3(massive vector) + 1(real scalar) = 4. The field $A_0$ has a mass term with the wrong sign and so cannot be a physical degree of freedom. The creation of massive fields out of a gauge field is called the Higgs mechanism.

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：99f24b0f0425ab0defc0680a34ed5993adccb5253b280224f14e25daeafb9c78