Topological objects in field theory

The 1+1d 'sine-Gordon' equation of motion is

$$\begin{eqnarray}&&\left({\partial }_{t}^{2}-{\partial }_{x}^{2}\right)\phi +\alpha \,\sin \,\beta \phi =0,\end{eqnarray} \tag{ 8.1 }$$

which describes a scalar field in one space and time dimension in a periodic potential. The corresponding Lagrangian is

$$\begin{eqnarray}&&{ \mathcal L }=\frac{1}{2}{\partial }_{\mu }\phi {\partial }^{\mu }\phi -V(\phi ),\end{eqnarray} \tag{ 8.2 }$$

and the Hamiltonian density is

$$\begin{eqnarray}&&{ \mathcal H }=\frac{1}{2}{\left(\frac{\partial \phi }{\partial t}\right)}^{2}+\frac{1}{2}{\left(\frac{\partial \phi }{\partial x}\right)}^{2}+V(\phi ),\end{eqnarray} \tag{ 8.3 }$$

with

$$\begin{eqnarray}&&V(\phi )=\frac{\alpha }{\beta }(1-\cos \,\beta \phi ).\end{eqnarray} \tag{ 8.4 }$$

We have plotted this potential in figure 8.1. As we can see from this figure, there are an infinite number of stable degenerate 'vacuum' solutions given by

$$\begin{eqnarray}&&{\phi }_{n}=\frac{2\pi n}{\beta }\,{\rm{with}}\,n\in {\mathbb{Z}}.\end{eqnarray} \tag{ 8.5 }$$

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Such trivial static solutions have zero energy $({ \mathcal H }=0)$ and all non-trivial solutions we obtain below will only be defined up to a shift by ${\phi }_{n}$. As we will see, there are solutions to these equations that allow us to connect the different degenerate minima.

Note that, if we were to treat this problem perturbatively, we would expand equation (8.4)

$$\begin{eqnarray}&&V(\phi )=-\frac{\alpha }{\beta }\sum _{n=1}^{\infty }\frac{{(-1)}^{n}{(\beta \phi )}^{2n}}{(2n)!}\simeq \frac{1}{2}\mathop{\underbrace{\alpha \beta }}\limits_{\equiv {m}^{2}}{\phi }^{2}-\frac{1}{24}\mathop{\underbrace{\alpha {\beta }^{3}}}\limits_{\equiv \lambda }{\phi }^{4}+\cdots ,\end{eqnarray} \tag{ 8.6 }$$

which looks like a potential for a particle with mass ${m}^{2}=\alpha \beta$ and coupling $\lambda =\alpha {\beta }^{3}$.

We begin by looking for static solutions with ${\partial }_{t}\phi =0$, and hence need to solve

$$\begin{eqnarray}&&{\partial }_{x}^{2}\phi =\alpha \,\sin \,\beta \phi .\end{eqnarray} \tag{ 8.7 }$$

Multiplying left and right by ${\partial }_{x}\phi$ and simplifying gives

$$\begin{eqnarray}&&\frac{1}{2}{\partial }_{x}{({\partial }_{x}\phi )}^{2}=\alpha ({\partial }_{x}\phi )\sin \,\beta \phi .\end{eqnarray} \tag{ 8.8 }$$

Integrating, gives

$$\begin{eqnarray}&&\frac{1}{2}{({\partial }_{x}\phi )}^{2}=-\frac{\alpha }{\beta }\,\cos \,\beta \phi +{c}_{1}.\end{eqnarray} \tag{ 8.9 }$$

We will search for solutions that have ${\partial }_{x}\phi \to 0$ at both $x\to \pm \infty$ and $\phi \to 0$ and $\phi \to 0\,{\rm{mod}}\,2\pi /\beta$ for $x\to -\infty$ and $x\to +\infty$, respectively. This implies that ${c}_{1}=\alpha /\beta$, giving

$$\begin{eqnarray}&&\frac{1}{2}{({\partial }_{x}\phi )}^{2}=\frac{\alpha }{\beta }(1-\cos \,\beta \phi )=V(\phi )=\frac{2\alpha }{\beta }{\sin }^{2}\,\left(\frac{\beta \phi }{2}\right),\end{eqnarray} \tag{ 8.10 }$$

and hence

$$\begin{eqnarray}&&\frac{d\phi }{dx}=\pm 2\sqrt{\frac{\alpha }{\beta }}\,\sin \,\left(\frac{\beta \phi }{2}\right).\end{eqnarray} \tag{ 8.11 }$$

Separating and integrating, we obtain

$$\begin{eqnarray}&&{\int }_{{\phi }_{0}}^{{\phi }_{f}}\frac{d\phi }{\sin (\beta \phi /2)}=\pm 2\sqrt{\frac{\alpha }{\beta }}{\int }_{{x}_{0}}^{{x}_{f}}dx.\end{eqnarray} \tag{ 8.12 }$$

This gives

$$\begin{eqnarray}&&\frac{2}{\beta }\,\mathrm{log}\,\left[\tan \,\left(\frac{\beta \phi (x)}{4}\right)\right]=\pm 2\sqrt{\frac{\alpha }{\beta }}(x-{x}_{0}),\end{eqnarray} \tag{ 8.13 }$$

with ${x}_{0}$ being an integration constant. Solving for $\phi$, we obtain

$$\begin{eqnarray}&&\phi (x)=\frac{4}{\beta }\,\arctan \,\left({{\rm{e}}}^{\pm \sqrt{\alpha \beta }(x-{x}_{0})}\right).\end{eqnarray} \tag{ 8.14 }$$

The kink (+) and anti-kink (−) solutions are plotted figure 8.2 left panel. In the right panel of the same figure we show a visualization of field rotation associated with the kink solution. As can be seen from this figure, the kink solution moves from a vacuum state with $\phi =0$ to one with $\phi =2\pi /\beta$. The kink is topological in the sense that one cannot continuously deform the right panel of figure 8.2 while keeping the final vectors the same and get rid of the kink (rotation in the phase vector). The anti-kink is similar but instead moves one from $\phi =2\pi /\beta$ to $\phi =0$. Classically, these solutions are ones in which the system starts asymptotically at one minimum and reaches the other minimum at large times.

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In the non-stationary case, we search for solutions of the form $\phi (t,x)=f(x-vt)$ with $v\lt 1$. Plugging this ansatz into the sine-Gordon equation gives

$$\begin{eqnarray}&&f^{\prime\prime} (u)={\gamma }^{2}\alpha \,\sin (\beta f(u)),\end{eqnarray} \tag{ 8.15 }$$

with $\gamma ={(1-{v}^{2})}^{-\mathrm{1/2}}$ and $u=x-vt$. This looks exactly like equation (8.7), so we can take the solution directly over to obtain

$$\begin{eqnarray}&&\phi (t,x)=\frac{4}{\beta }\,\arctan \,\left({{\rm{e}}}^{\pm \gamma \sqrt{\alpha \beta }(x-vt-{x}_{0})}\right).\end{eqnarray} \tag{ 8.16 }$$

This solution represents a kink (or anti-kink) moving at finite velocity. It has a finite energy, which can be computed from the integral of the Hamiltonian density. Since adding the time dependence just gives a moving kink, it suffices to consider the energy of a static kink

$$\begin{eqnarray}\begin{array}{rcl}E & = & \int dx{ \mathcal H }=\int dx\left[\frac{1}{2}{\left(\frac{\partial \phi }{\partial x}\right)}^{2}+V(\phi )\right]=\int dx2V(\phi )\\ & = & {\int }_{0}^{\frac{2\pi }{\beta }}d\phi \sqrt{2V(\phi )}=\sqrt{\frac{2\alpha }{\beta }{\int }_{0}^{\frac{2\pi }{\beta }}d\phi {(1-\cos \beta \phi )}^{\mathrm{1/2}}}\\ & = & 8\sqrt{\frac{\alpha }{{\beta }^{3}}}=\frac{8{m}^{3}}{\lambda },\end{array}\end{eqnarray} \tag{ 8.17 }$$

where we used equations (8.10) and (8.11) and, in the last line, we re-expressed the energy in terms of the mass and coupling constant introduced in equation (8.6). This shows that the energy of the kink solution is finite and, importantly, that the energy goes inversely with the coupling $\lambda$ meaning that, in the strong coupling limit, the energy associated with such configurations can become small.

We note that similar non-trivial solutions can be obtained by assuming that $\phi (t,x)$ can be expressed in the form [1]

$$\begin{eqnarray}&&\phi (t,x)=\frac{4}{\beta }\,\arctan \,\left(\frac{F(x)}{G(t)}\right).\end{eqnarray} \tag{ 8.18 }$$

For example, for $\alpha =\beta =1$, there is a two-kink solution of the form

$$\begin{eqnarray}&&\phi (t,x)=4\,\arctan \,\left(\frac{v\,\sinh (\gamma x)}{\cosh (v\gamma t)}\right).\end{eqnarray} \tag{ 8.19 }$$

This solution is plotted and visualized in figure 8.3 for the case $v=0.6$ at $t=0$.

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In fact, a family of solutions with an arbitrary number of kinks + anti-kinks can be constructed in this manner. Each of these solutions has a conserved number associated with the number of kinks minus anti-kinks. One can construct a conserved current of the form

$$\begin{eqnarray}&&{J}_{\mu }=\frac{\beta }{2\pi }{\varepsilon }^{\mu \nu }{\partial }_{\nu }\phi ,\end{eqnarray} \tag{ 8.20 }$$

with $\mu ,\nu \in \mathrm{\{0,1\}}$ and ${\varepsilon }^{\mu \nu }$ being the 1+1 anti-symmetric tensor with ${\varepsilon }^{01}=1$. This current is conserved, that is ${\partial }_{\mu }{J}^{\mu }=0$, and the conserved charge associated with it is

$$\begin{eqnarray}&&Q=\int dx\,{J}^{0}=\frac{\beta }{2\pi }\int dx{\partial }_{x}\phi =\frac{\beta }{2\pi }\left[\phi (\infty )-\phi (-\infty )\right]=N,\end{eqnarray} \tag{ 8.21 }$$

where $N$ can be identified as the difference of the number of kinks minus anti-kinks. From this we see the important role played by the boundary conditions at infinity. The (anti)-kink and two-kink solutions presented earlier in this section have charge $\pm 1$ and 2, respectively. Note that the solitonic current is not a Noether current associated with a global symmetry, it instead comes from the conservation of 'winding number', which is a topological invariant.

Next we consider a complex scalar field (charged scalar) in two spatial dimensions. We take the boundary of space to be the circle at infinity, ${S}^{1}$. We will impose a boundary condition at infinity

$$\begin{eqnarray}&&\psi =a{{\rm{e}}}^{in\theta }\,(r\to \infty ),\end{eqnarray} \tag{ 8.22 }$$

where $r$ and $\theta$ are polar coordinates in the plane, $a$ is an arbitrary amplitude, and $n$ is an integer in order to guarantee that $\psi$ is single valued.

The Lagrangian and Hamiltonian densities are

$$\begin{eqnarray}\begin{array}{rcl}{ \mathcal L } & = & \frac{1}{2}{\partial }_{t}{\psi }^{\dagger }{\partial }_{t}\psi -\frac{1}{2}\nabla {\psi }^{\dagger }\cdot \nabla \psi -V(| \psi {| }^{2}),\\ { \mathcal H } & = & \frac{1}{2}{\partial }_{t}{\psi }^{\dagger }{\partial }_{t}\psi +\frac{1}{2}\nabla {\psi }^{\dagger }\cdot \nabla \psi +V(| \psi {| }^{2}).\end{array}\end{eqnarray} \tag{ 8.23 }$$

As our example, let us take $V(| \psi {| }^{2})={({a}^{2}-| \psi {| }^{2})}^{2}$ such that $V\to 0$ at the boundary.

To start with, let us consider a static configuration. In this case, the Hamiltonian density at the boundary is

$$\begin{eqnarray}&&\mathop{\mathrm{lim}}\limits_{r\to \infty }\,{ \mathcal H }=\frac{1}{2}| \nabla \psi {| }^{2}=\frac{1}{2}{\left|\nabla (a{{\rm{e}}}^{in\theta })\right|}^{2}=\frac{1}{2}{\left|\frac{ina}{r}{{\rm{e}}}^{in\theta }\right|}^{2}=\frac{{n}^{2}{a}^{2}}{2{r}^{2}}.\end{eqnarray} \tag{ 8.24 }$$

Since the Hamiltonian density only falls off like ${r}^{-2}$, this implies that the energy of such a configuration is infinite. This means that there is no finite-energy generalization of the 1d kink solution for a 2d complex scalar.

Next, let us consider what happens if we couple our complex scalar field to an abelian gauge field by taking ${\partial }_{\mu }\to {D}_{\mu }$ with

$$\begin{eqnarray}&&{D}_{\mu }\psi ={\partial }_{\mu }\psi +i{\rm{e}}{A}_{\mu }\psi ,\end{eqnarray} \tag{ 8.25 }$$

and require that the vector potential has the boundary condition

$$\begin{eqnarray}&&{\bf{A}}=\frac{1}{{\rm{e}}}\nabla (n\theta )\,(r\to \infty ).\end{eqnarray} \tag{ 8.26 }$$

As we demonstrate below, by adding a gauge field with these non-trivial boundary conditions, one can construct a discrete set of finite-energy configurations of a complex scalar coupled to an abelian gauge field. Including the gauge-field contribution to the Lagrangian density, we have

$$\begin{eqnarray}&&{ \mathcal L }=\frac{1}{2}{D}^{\mu }{\psi }^{\dagger }{D}_{\mu }\psi -V(| \psi {| }^{2})-\frac{1}{4}{F}^{\mu \nu }{F}_{\mu \nu }.\end{eqnarray} \tag{ 8.27 }$$

To see that the energy of a configuration satisfying equations (8.22) and (8.26) is finite, we first note that, given the boundary condition for ${A}^{\mu }$ above, one has

$$\begin{eqnarray}\begin{array}{rcl}{A}_{r} & \to & 0,\\ {A}_{\theta } & \to & -\,\frac{n}{{\rm{e}}r},\end{array}\end{eqnarray} \tag{ 8.28 }$$

which implies that, at asymptotically large distances, one has

$$\begin{eqnarray}\begin{array}{rcl}{D}_{r}\psi & \to & 0,\\ {D}_{\theta }\psi & \to & \frac{1}{r}\left(\frac{\partial \psi }{\partial \theta }\right)+i{\rm{e}}{A}_{\theta }\psi =0.\end{array}\end{eqnarray} \tag{ 8.29 }$$

This means ${D}_{\theta }\psi$ falls faster than $1/r$ at infinity and the kinetic energy contribution will be UV finite. Additionally, since equation (8.26) can be expressed in the form ${A}_{\mu }={\partial }_{\mu }\chi$ with $\chi =n\theta /{\rm{e}}$, this gauge field is a pure gauge transform and carries zero energy. As a result, all terms in equation (8.27) result in a finite contribution to the total energy.

The gauge field configuration possesses a quantized magnetic flux. To see this, consider the magnetic flux ${\rm{\Phi }}$ generated by equation (8.28) through a disk of radius $R$ with boundary $C$

$$\begin{eqnarray}&&{\rm{\Phi }}=\int {\bf{B}}\cdot d{\bf{S}}={\oint }_{C}{\bf{A}}\cdot d{\bf{l}}={\int }_{0}^{2\pi }Rd\theta {A}_{\theta }(R)=-\frac{2\pi n}{{\rm{e}}},\end{eqnarray} \tag{ 8.30 }$$

which shows that the flux is quantized. We have demonstrated that it is possible to construct a 2d configuration consisting of a charged scalar field plus a gauge field that carries a quantized magnetic flux.

This 2d solution can be extended to 3d by simplying requiring cylindrical symmetry of the 3d system. In this case, the quantized flux lines are the 'Abrikosov flux lines' which appear in the theory of type II superconductors. The model we analyzed (8.23) corresponds to scalar electrodynamics with spontaneous symmetry breaking (Higgs model). This is a relativistic generalization of the condensed matter field theory, with the field $\psi$ corresponding to the Bardeen-Cooper-Schrieffer condensate. In a type II superconducting medium, the magnetic field normally cannot penetrate the material and if it does, it can only do so through quantized flux lines called Abrikosov flux lines. For a more extensive discussion of topological solutions in the context of condensed matter see ref. [2].

The next question one naturally asks is whether it is possible to construct analogous classical Yang–Mills topological solutions. To determine whether such non-abelian topological solutions are possible, we ask if it is possible to construct finite-energy classical solutions with non-trivial boundary conditions. We specialize to the case of pure-gauge fields, in which case the canonical energy momentum tensor is expressible solely in terms of the field-strength tensor (see volume 1)

$$\begin{eqnarray}&&{T}^{\mu \nu }={F}_{a}^{\mu \lambda }{{F}_{a\lambda }}^{\nu }+\frac{1}{4}{\eta }^{\mu \nu }{F}_{a}^{\alpha \beta }{F}_{a\alpha \beta },\end{eqnarray} \tag{ 8.31 }$$

which obeys

$$\begin{eqnarray}&&{\partial }_{\mu }{T}^{\mu \nu }=0.\end{eqnarray} \tag{ 8.32 }$$

The energy-momentum tensor is gauge-invariant. From equation (8.31) we have

$$\begin{eqnarray}&&{{T}^{\mu }}_{\mu }={F}_{a}^{\mu \lambda }{F}_{a\lambda \mu }+\frac{1}{4}(d+1){F}_{a}^{\alpha \beta }{F}_{a\alpha \beta }=\frac{1}{4}(d-3){F}_{a}^{\alpha \beta }{F}_{a\alpha \beta },\end{eqnarray} \tag{ 8.33 }$$

where $d$ is the number of spatial dimensions.

8.3.1. Static solutions

Topological solutions can be time-dependent or time-independent (static). A static solution is one where ${A}^{\mu }$ can be made time-independent using a continuous gauge transformation. For such static solutions, the general time evolution of ${A}^{\mu }$ can be obtained from a continuous gauge transformation

$$\begin{eqnarray}&&{\dot{A}}^{\mu }(x)=\frac{1}{g}{\partial }^{\mu }\,\dot{{\rm{\Omega }}}(x)-i[\dot{{\rm{\Omega }}}(x),\,{A}^{\mu }({\bf{x}},t)],\end{eqnarray} \tag{ 8.34 }$$

where ${\rm{\Omega }}(x)$ is an arbitrary continuous function. We can make the right-hand side vanish if

$$\begin{eqnarray}&&\frac{1}{g}{\partial }^{\mu }\,\dot{{\rm{\Omega }}}(x)=i[\dot{{\rm{\Omega }}}(x),\,{A}^{\mu }({\bf{x}},t)],\end{eqnarray} \tag{ 8.35 }$$

which is solved by

$$\begin{eqnarray}&&\dot{{\rm{\Omega }}}(x)\,=\,L(P)\,{\rm{\Omega }}\,({x}_{0}){L}^{-1}(P),\end{eqnarray} \tag{ 8.36 }$$

with $L(P)$ being the gauge-link, which is the exponential of the integral of the gauge potential along the path $P$

$$\begin{eqnarray}&&L(P)\equiv {P}\,\exp \,\left(-ig{\int }_{P}d{x}^{\mu }{A}_{\mu }(x)\right).\end{eqnarray} \tag{ 8.37 }$$

In the definition of $L$, ${P}$ implies the path-ordering operator, which orders $A$ similar to the time-ordering operator, but now for a path in space-time.

With ${\rm{\Omega }}$ determined in this way, we have ${\dot{A}}^{\mu }(x)=0$, that is the field configuration is time-independent. The gauge in which our static field is time-independent is called the static gauge. Static solutions with finite-energy are either the vacuum or a static solution. For pure Yang–Mills theory it turns out that there are no static topological solutions unless the number of spatial dimensions is four [3]. We will now review the proof of this statement by considering Yang–Mills in $(d+1)-$ dimensional Minkowski space.

The requirement of finite-energy implies that

$$\begin{eqnarray}&&\int {d}^{n}x{T}^{00}=\int {d}^{n}x\left({F}_{a}^{0\lambda }{F}_{a\lambda }^{0}+\frac{1}{4}{F}_{a}^{\alpha \beta }{F}_{a\alpha \beta }\right)\lt \infty .\end{eqnarray} \tag{ 8.38 }$$

One can show that, for a static solution, ${F}_{a}^{i0}=0$ with i = 1, 2, 3 (homework), and hence one has

$$\begin{eqnarray}&&\int {d}^{n}x\frac{1}{4}{F}_{a}^{ij}{F}_{aij}\lt \infty .\end{eqnarray} \tag{ 8.39 }$$

As a result, ${\mathrm{lim}}_{r\to \infty }{r}^{n-1}| {F}_{a}^{ij}{| }^{2}\lt C{r}^{-1-\varepsilon }$ with $\varepsilon \gt 0$ and ${r}^{2}\equiv {{\rm{\Sigma }}}_{i=1}^{n}{x}_{i}^{2}$ and one must have ${\mathrm{lim}}_{r\to \infty }{F}_{a}^{ij}\sim { \mathcal O }({r}^{-n/2-\varepsilon })$ for convergence in the UV.

Next, consider

$$\begin{eqnarray}&&{\partial }_{i}({x}^{j}{T}^{ij})={{T}^{i}}_{i}+{x}^{j}{\partial }_{i}{T}^{ij}.\end{eqnarray} \tag{ 8.40 }$$

Since ${\partial }_{\mu }{T}^{\mu \nu }=0$, one has ${\partial }_{i}{T}^{ij}=-{\partial }_{t}{T}^{0j}$, giving

$$\begin{eqnarray}&&{\partial }_{i}({x}^{j}{T}^{ij})={{T}^{i}}_{i}-{x}^{j}{\partial }_{t}{T}^{0j}.\end{eqnarray} \tag{ 8.41 }$$

For a static solution, the second term vanishes, leaving

$$\begin{eqnarray}&&{\partial }_{i}({x}^{j}{T}^{ij})={{T}^{i}}_{i}.\end{eqnarray} \tag{ 8.42 }$$

Integrating on the left and right we see that, since ${T}^{ij}$ vanishes at infinity, the integral of the left-hand side has to vanish and, therefore, so does the integral over the right-hand side

$$\begin{eqnarray}&&\int {d}^{n}x{T}_{i}^{i}=0.\end{eqnarray} \tag{ 8.43 }$$

Using ${T}_{i}^{i}={T}_{\mu }^{\mu }-{T}^{00}$, equation (8.33), and ${T}^{00}=(\mathrm{1/4}){F}_{a}^{\alpha \beta }{F}_{a\alpha \beta }$ we obtain

$$\begin{eqnarray}&&(n-4)\int {d}^{n}x{F}_{a}^{ij}{F}_{a}^{ij}=0.\end{eqnarray} \tag{ 8.44 }$$

Therefore, unless $d=4$ one must have ${F}_{a}^{ij}=0$ everywhere for this integral to be zero (integrand is positive-definite). We thus conclude that for $d\ne 4$ there cannot be static pure-gauge topological solutions. For $d=4$ we can find a solution, which is called the instanton solution. We will construct such a solution next.

A static topological solution in 4 + 1 Minkowski space is a time-independent solution that only depends on the four-dimensional Euclidean space coordinates. Since we can formulate the path integral in four-dimensional Euclidean or Minkowski spaces, a four-dimensional Euclidean solution might be of some interest. As an example, let us search for a solution that is invariant under $SU\mathrm{(2)}$ gauge transformations with ${f}^{abc}={\varepsilon }^{abc}$. What we are looking for is a mapping from the $SU\mathrm{(2)}$ group space ${S}^{3}$ to the boundary of the physical space, which is also ${S}^{3}$.

We will label Euclidean space with spatial coordinates $({x}_{1},{x}_{2},{x}_{3},{x}_{4})$. The Euclidean field tensor ${F}_{\mu \nu }^{a}$ is defined in the same way as the Minkowski tensor

$$\begin{eqnarray}&&{F}_{\mu \nu }^{a}={\partial }_{\mu }{A}_{\nu }^{a}-{\partial }_{\nu }{A}_{\mu }^{a}+g{\varepsilon }^{abc}{A}_{\mu }^{b}{A}_{\nu }^{c}.\end{eqnarray} \tag{ 8.45 }$$

Defining ${A}_{\mu }={t}^{a}{A}_{\mu }^{a}$ and ${F}_{\mu \nu }={t}^{a}{F}_{\mu \nu }^{a}$ we can write this compactly as ${F}_{\mu \nu }={\partial }_{\mu }{A}_{\nu }\,-{\partial }_{\nu }{A}_{\mu }-ig[{A}_{\mu },{A}_{\nu }]$. We can introduce the dual of ${F}_{\mu \nu }$ as ${\tilde{F}}_{\mu \nu }$ which is defined by

$$\begin{eqnarray}&&{\tilde{F}}_{\mu \nu }\equiv \frac{1}{2}{\varepsilon }_{\mu \nu \rho \sigma }{F}_{\rho \sigma },\end{eqnarray} \tag{ 8.46 }$$

where, since we are in Euclidean space, we do not have to distinguish up and down indices.

We can express

$$\begin{eqnarray}&&\frac{1}{4}{\rm{Tr}}{\tilde{F}}_{\mu \nu }{F}_{\mu \nu }={\partial }_{\mu }{K}_{\mu },\end{eqnarray} \tag{ 8.47 }$$

with

$$\begin{eqnarray}\begin{array}{rcl}{K}_{\mu } & \equiv & \frac{1}{4}{\varepsilon }_{\mu \nu \kappa \lambda }\left({A}_{\nu }^{a}{\partial }_{\kappa }{A}_{\lambda }^{a}+\frac{g}{3}{\varepsilon }_{abc}{A}_{\nu }^{a}{A}_{\kappa }^{b}{A}_{\lambda }^{c}\right)\\ & = & {\varepsilon }_{\mu \nu \kappa \lambda }{\rm{Tr}}\left(\frac{1}{2}{A}_{\nu }{\partial }_{\kappa }{A}_{\lambda }-\frac{ig}{3}{A}_{\nu }{A}_{\kappa }{A}_{\lambda }\right),\end{array}\end{eqnarray} \tag{ 8.48 }$$

being the Chern–Simons current. Since $(\mathrm{1/4}){\rm{Tr}}{\tilde{F}}_{\mu \nu }{F}_{\mu \nu }$ can be expressed as a total derivative of the Chern–Simons current, its integral can only depend on the boundary conditions for the current. Considering a four-dimensional Euclidean volume ${V}^{4}$ with boundary $\partial {V}^{4}={S}^{3}$. Suppose that on the boundary we have a pure vacuum solution with ${A}_{\mu }=0$, ${F}_{\mu \nu }=0$, and, hence, ${K}_{\mu }=0$. In the absence of matter, the equation of motion for the field in the entire volume ${V}^{4}$ is

$$\begin{eqnarray}&&{D}_{\mu }{F}^{\mu \nu }=0.\end{eqnarray} \tag{ 8.49 }$$

Additionally, the dual satisfies

$$\begin{eqnarray}&&{D}_{\mu }{\tilde{F}}^{\mu \nu }=0,\end{eqnarray} \tag{ 8.50 }$$

which follows from the Bianchi identity. Using the current ${K}_{\mu }$, we can write

$$\begin{eqnarray}&&{\int }_{{V}^{4}}{d}^{4}x{\rm{Tr}}{\tilde{F}}_{\mu \nu }{F}_{\mu \nu }=4{\int }_{{V}^{4}}{d}^{4}x{\partial }_{\mu }{K}_{\mu }=4{\int }_{{S}^{3}}{d}^{3}x{K}_{\perp },\end{eqnarray} \tag{ 8.51 }$$

where ${K}_{\perp }$ is the projection of $K$ that is perpendicular to the surface ${S}^{3}$. Note that this is trivially satisfied if the solution is pure vacuum everywhere in ${V}^{4}$.

To proceed, we imagine performing a space-time dependent gauge transformation on the boundary ${S}^{3}$

$$\begin{eqnarray}&&{A}_{\mu }\to -\frac{i}{g}({\partial }_{\mu }{\rm{\Omega }}){{\rm{\Omega }}}^{-1}.\end{eqnarray} \tag{ 8.52 }$$

Since the result is a pure-gauge field, one still has ${F}^{\mu \nu }=0$ on the boundary, but it may be possible to have ${K}_{\mu }\ne 0$.

Choosing

$$\begin{eqnarray}&&{\rm{\Omega }}=\frac{1}{{x}^{2}}({x}_{4}+i{\bf{x}}\cdot {\boldsymbol{\sigma }}),\end{eqnarray} \tag{ 8.53 }$$

with ${x}^{2}\equiv {{\rm{\Sigma }}}_{i\mathrm{=1}}^{4}{x}_{i}^{2}$ gives

$$\begin{eqnarray}&&{A}_{i}=\frac{i}{g{x}^{2}}[{x}_{i}-{\sigma }_{i}({\bf{x}}\cdot {\boldsymbol{\sigma }}+i{x}_{4})],\end{eqnarray} \tag{ 8.54 }$$

$$\begin{eqnarray}&&{A}_{4}=-\frac{1}{g{x}^{2}}{\bf{x}}\cdot {\boldsymbol{\sigma }},\end{eqnarray} \tag{ 8.55 }$$

and

$$\begin{eqnarray}&&{K}_{\mu }=\frac{2{x}_{\mu }}{{g}^{2}{x}^{4}}.\end{eqnarray} \tag{ 8.56 }$$

Using this, we obtain

$$\begin{eqnarray}&&{\int }_{{V}^{4}}{d}^{4}x{\rm{Tr}}{\tilde{F}}_{\mu \nu }{F}_{\mu \nu }=4{\int }_{{S}^{3}}{d}^{3}x{K}_{\perp }=\frac{8}{{g}^{2}{x}^{3}}\mathop{\underbrace{{\int }_{{S}^{3}}{d}^{3}x}}\limits_{2{\pi }^{2}{x}^{3}}=\frac{16{\pi }^{2}}{{g}^{2}}.\end{eqnarray} \tag{ 8.57 }$$

We see from this expression that although ${F}_{\mu \nu }$ vanishes on the boundary at infinity, it cannot be zero everywhere within ${V}^{4}$ since the integral above is finite. This means that the solution cannot be a pure gauge everywhere in ${V}^{4}$. We will return to this issue shortly.

Exercise 8.1 Show that equation (8.48) is correct.

Exercise 8.2 Show that equation (8.50) is correct.

Exercise 8.3 Show that equations (8.54) and (8.55) are correct.

The quantity introduced above $\int {d}^{4}x{\rm{Tr}}{\tilde{F}}_{\mu \nu }{F}_{\mu \nu }$ is related to the Potryagin or topological index, which is denoted as $q$,

$$\begin{eqnarray}&&q\equiv \frac{{g}^{2}}{16{\pi }^{2}}\int {d}^{4}x{\rm{Tr}}{\tilde{F}}_{\mu \nu }{F}_{\mu \nu }.\end{eqnarray} \tag{ 8.58 }$$

The vacuum has $q=0$. Equation (8.53) gives $q=1$ according to equation (8.57).

We will now show that $q$ is the degree of the mapping of the group space from the spatial coordinate boundary (number of times it covers the group space). In this case we have a mapping of ${S}^{3}$, which is the group space of $SU\mathrm{(2)}$, onto a ${S}^{3}$ in coordinate space. Computing ${K}_{\mu }$ using equations (8.48) and (8.52) one obtains

$$\begin{eqnarray}&&{K}_{\mu }=\frac{1}{6{g}^{2}}{\varepsilon }^{\mu \nu \kappa \lambda }{\rm{Tr}}\left[\left({{\rm{\Omega }}}^{-1}{\partial }_{\nu }{\rm{\Omega }}\right)\left({{\rm{\Omega }}}^{-1}{\partial }_{\lambda }{\rm{\Omega }}\right)\left({{\rm{\Omega }}}^{-1}{\partial }_{\kappa }{\rm{\Omega }}\right)\right].\end{eqnarray} \tag{ 8.59 }$$

This allows us to express $q$ as

$$\begin{eqnarray}\begin{array}{rcl}q & = & \frac{1}{24{\pi }^{2}}{\int }_{{S}^{3}}{d}^{3}x{\hat{n}}^{\mu }{\varepsilon }^{\mu \nu \kappa \lambda }{\rm{Tr}}\left[({{\rm{\Omega }}}^{-1}{\partial }_{\nu }{\rm{\Omega }})\left({{\rm{\Omega }}}^{-1}{\partial }_{\lambda }{\rm{\Omega }}\right)\left({{\rm{\Omega }}}^{-1}{\partial }_{\kappa }{\rm{\Omega }}\right)\right]\\ & = & \frac{1}{24{\pi }^{2}}{\int }_{{S}^{3}}{d}^{3}x{\varepsilon }^{ijk}{\rm{Tr}}[({{\rm{\Omega }}}^{-1}{\partial }_{i}{\rm{\Omega }})({{\rm{\Omega }}}^{-1}{\partial }_{j}{\rm{\Omega }})({{\rm{\Omega }}}^{-1}{\partial }_{k}{\rm{\Omega }})],\end{array}\end{eqnarray} \tag{ 8.60 }$$

where ${\hat{n}}^{\mu }$ is an outward pointing unit vector. To proceed, let us consider a parameterization of the ${\rm{\Omega }}$ using parameters ${a}_{i}$. With this we have

$$\begin{eqnarray}\begin{array}{rcl}q & = & \frac{1}{24{\pi }^{2}}{\int }_{{S}^{3}}{d}^{3}x{\rm{Tr}}\left[\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{{\ell }}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{m}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{n}}\right)\right]{\varepsilon }^{ijk}\frac{\partial {a}_{{\ell }}}{\partial {x}_{i}}\frac{\partial {a}_{m}}{\partial {x}_{j}}\frac{\partial {a}_{n}}{\partial {x}_{k}}\\ & = & \frac{1}{24{\pi }^{2}}{\int }_{{S}^{3}}{d}^{3}x{\rm{Tr}}\left[\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{{\ell }}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{m}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{n}}\right)\right]\frac{\partial \left({a}_{{\ell }},{a}_{m},{a}_{n}\right)}{\partial ({x}_{1},{x}_{2},{x}_{3})}\\ & = & \frac{1}{4{\pi }^{2}}{\int }_{{S}^{3}}{d}^{3}x{\rm{Tr}}\left[\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{1}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{2}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{3}}\right)\right]\frac{\partial ({a}_{1},{a}_{2},{a}_{3})}{\partial ({x}_{1},{x}_{2},{x}_{3})}\\ & = & \frac{1}{4{\pi }^{2}}{\int }_{x\in {S}^{3}}d{a}_{1}d{a}_{2}d{a}_{3}{\rm{Tr}}\left[\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{1}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{2}}\right)\left({{\rm{\Omega }}}^{-1}\frac{\partial {\rm{\Omega }}}{\partial {a}_{3}}\right)\right]\\ & = & {\int }_{x\in {S}^{3}}dU.\end{array}\end{eqnarray} \tag{ 8.61 }$$

From the last line, we learn that $q$ is the number of times that the group space is covered by the map the coordinate ${S}^{3}$ to the group ${S}^{3}$. This is sometimes called the 'winding number' of the map.

As a simpler example of this, let us look at a map from ${S}^{1}$ to ${S}^{1}$ using a $U\mathrm{(1)}$ configuration of the form

$$\begin{eqnarray}&&{\rm{\Omega }}(x)={e}^{i2\pi \alpha (x)},\end{eqnarray} \tag{ 8.62 }$$

where $x={S}^{1}$ with $0\leqslant x\lt 2\pi$ covering the circle. Requiring that ${\rm{\Omega }}$ be singled-valued for $x\in {S}^{1}$ we see that $\alpha (x+2\pi )=\alpha (x)+n$ where $n\in {\mathbb{Z}}$. The corresponding pure-gauge vector potential constructed from this is

$$\begin{eqnarray}&&{{\rm{\Omega }}}^{-1}{\partial }_{\mu }{\rm{\Omega }}=2\pi i{\partial }_{\mu }\alpha (x).\end{eqnarray} \tag{ 8.63 }$$

One, therefore, has

$$\begin{eqnarray}&&\frac{1}{2\pi i}{\int }_{0}^{2\pi }d\alpha {{\rm{\Omega }}}^{-1}{\partial }_{\mu }{\rm{\Omega }}={\int }_{0}^{2\pi }d\alpha {\partial }_{\mu }\alpha =\alpha (2\pi )-\alpha (0)=n,\end{eqnarray} \tag{ 8.64 }$$

where $n$ is the winding number, which counts the number of times ${\rm{\Omega }}$ goes around the $U\mathrm{(1)}$ circle when $x$ runs around the ${S}^{1}$ circle once $(x=0\to 2\pi )$. This is similar to the phase we accumulated in the sine-Gordon model when we made a transition from one vacuum to another. The situation is similar in pure-gauge QCD, the mapping of ${S}^{3}\to {S}^{3}$ generates a topologically conserved number, which is the Potryagin index of the field configuration.

Exercise 8.4 Derive equation (8.59).

As mentioned previously, the field cannot be a pure-gauge configuration everywhere if $q$ is non-zero. The instanton solution in all of ${V}^{4}$ is ¹

$$\begin{eqnarray}&&{A}_{\mu }(x)=-\frac{i}{g}\frac{{x}^{2}}{{x}^{2}+{\rho }^{2}}({\partial }_{\mu }{\rm{\Omega }}){{\rm{\Omega }}}^{-1},\end{eqnarray} \tag{ 8.65 }$$

where ${x}^{2}={{\bf{x}}}^{2}+{x}_{4}^{2}$ and $\rho$ is an arbitrary real constant which sets the size of the instanton solution. As ${x}_{4}^{2}\to \pm \infty$ this solution reduces to the pure-gauge form given by equation (8.52). Generally, one has

$$\begin{eqnarray}&&{A}_{i}=\frac{i}{g}\frac{1}{{x}^{2}+{\rho }^{2}}[{x}_{i}-{\sigma }_{i}({\bf{x}}\cdot {\boldsymbol{\sigma }}+i{x}_{4})],\end{eqnarray} \tag{ 8.66 }$$

$$\begin{eqnarray}&&{A}_{4}=-\frac{1}{g}\frac{1}{{x}^{2}+{\rho }^{2}}{\bf{x}}\cdot {\boldsymbol{\sigma }}.\end{eqnarray} \tag{ 8.67 }$$

The asymptotic form of this solution can be obtained from gauge transformations of the type

$$\begin{eqnarray}&&{g}_{n}={({g}_{1})}^{n},\end{eqnarray} \tag{ 8.68 }$$

with

$$\begin{eqnarray}&&{g}_{1}\equiv \exp \,\left(\frac{-i\pi {\bf{x}}\cdot {\boldsymbol{\sigma }}}{\sqrt{{x}^{2}+{\rho }^{2}}}\right),\end{eqnarray} \tag{ 8.69 }$$

and

$$\begin{eqnarray}\begin{array}{rcl}\mathop{\mathrm{lim}}\limits_{{x}_{4}\to \infty }\,{A}_{i} & = & i{({g}_{n})}^{-1}({\partial }_{i}{g}_{n}),\\ \mathop{\mathrm{lim}}\limits_{{x}_{4}\to -\infty }\,{A}_{i} & = & i{({g}_{n-1})}^{-1}({\partial }_{i}{g}_{n-1}).\end{array}\end{eqnarray} \tag{ 8.70 }$$

The gauge transform ${g}_{n}$ is an element of $SU\mathrm{(2)}$ but ${g}_{n}$ and ${g}_{m}$ for $n\ne m$ are not homotopic, that is they have a different topology and cannot be continuously deformed into one another. The $q=1$ instanton configuration describes a solution of the gauge-field equations in which, as ${x}_{4}$ goes from $-\infty$ to $\infty$, a vacuum belonging to homotopy class $n-1$ evolves into another vacuum with homotopy class $n$. The energy density of the pure-gauge field at the end-point is vanishing, however, the full configuration has a positive field energy. As a result, we see that the Yang–Mills vacuum is infinitely degenerate with an infinite number of homotopically non-equivalent vacua. The instanton represents a transition from one vacuum class to another. Due to the finite energy of the instanton, classically there can be no transition between the degenerate vacua, however, quantum mechanically we have the possibility of quantum tunneling.

The barrier potential amplitude is given $\exp (-{S}_{E})$ where ${S}_{E}$ is the Euclidean action. To see this, consider the motion of a particle with energy $E$ in a one-dimensional potential $V$ in the WKB approximation. If $V\gt E$, the transition is classically forbidden, and the quantum tunneling amplitude is

$$\begin{eqnarray}&&\exp \,\left(-{\int }_{a}^{b}dx\sqrt{2m(V-E)}\right)\equiv \exp (-{S}_{E}),\end{eqnarray} \tag{ 8.71 }$$

where we have identified the Euclidean action ${S}_{E}$ with the integral appearing on the left. Let us see why this is true. If $E\gt V$, the transition is classically allowed and the wave function oscillates with the number of oscillations given by

$$\begin{eqnarray}&&{\int }_{a}^{b}dxp={\int }_{a}^{b}dx\sqrt{2m(E-V)}.\end{eqnarray} \tag{ 8.72 }$$

Alternatively, we can express the integral of $p$ as

$$\begin{eqnarray}&&\int dxp=\int dtp\dot{x}=\int dt(H+L)=\int dt(E+L).\end{eqnarray} \tag{ 8.73 }$$

If the total energy is normalized to zero which it always can be, then

$$\begin{eqnarray}&&{\int }_{a}^{b}dxp={\int }_{{t}_{0}}^{{t}_{f}}dtL=S.\end{eqnarray} \tag{ 8.74 }$$

This is the total action induced transversing from $a$ to $b$.

The only difference between the classically forbidden case and the allowed case just considered, is the sign of $E-V$ and, since we have normalized the energy such that $E=0$, we see that we simply pick up a relative factor of $i$ from the square root. The sign of $V$ in the classical equation of motion

$$\begin{eqnarray}&&m\ddot{x}=\frac{\partial V}{\partial x}\end{eqnarray} \tag{ 8.75 }$$

is reversed if we take $t\to it$. As a result, ${S}_{E}$ is the action for imaginary times. This justifies calling $\exp (-{S}_{E})$ the tunneling amplitude.

Now we need to determine the action for our instanton. For this we need to be able to evaluate

$$\begin{eqnarray}&&{S}_{E}=\frac{1}{4}{\int }_{x}{F}_{\mu \nu }^{a}{F}_{\mu \nu }^{a}.\end{eqnarray} \tag{ 8.76 }$$

The $q=1$ solution given in equation (8.65) is special because it is self-dual meaning that

$$\begin{eqnarray}&&{\tilde{F}}_{\mu \nu }={F}_{\mu \nu }.\end{eqnarray} \tag{ 8.77 }$$

As a result,

$$\begin{eqnarray}&&{S}_{E}=\frac{1}{4}{\int }_{x}{\tilde{F}}_{\mu \nu }^{a}{F}_{\mu \nu }^{a}=\frac{8{\pi }^{2}}{{g}^{2}}.\end{eqnarray} \tag{ 8.78 }$$

For the $q=1$ instanton constructed herein, this gives a transition probability of $\exp (-8{\pi }^{2}/{g}^{2})$ and in general, one finds, $\exp (-8{\pi }^{2}| q| /{g}^{2})$. For small $g$, such transitions are extremely strongly suppressed. Regardless of the magnitude of $g$, however, the fact that these tunneling solutions exist means that all of the degenerate vacua of Yang–Mills are coupled by instantons.

We can label each of the generate vacuum by the topological index which must be an integer, so we have a basis of states of the form $\left|n\right.\rangle$ with $n\in {\mathbb{Z}}$. Positive $n$ map to multi-instanton solutions with more instanton than anti-instanton solutions and vice-verse for negative $n$ and $n=0$ is the instanton-free vacuum. The true wave function of the QCD vacuum is therefore a linear superposition of states, however, since instantons couple the various vacuum states, we must diagonalize the Hamiltonian to obtain the true ground state. The end result is that the QCD vacuum will have the form

$$\begin{eqnarray}&&\left|\theta \right.\rangle \equiv \sum _{n}{{\rm{e}}}^{in\theta }\left|n\right.\rangle ,\end{eqnarray} \tag{ 8.79 }$$

where $\theta$ is an arbitrary constant. This is vacuum is called the theta vacuum. The form of the coefficients in this expansion guarantee that the $\theta$-vacuum is invariant under gauge transformations of the type ${g}_{1}$. Acting with ${g}_{1}$ we find

$$\begin{eqnarray}&&\left|n\right.\rangle \mathop{\to }\limits^{{g}_{1}}\left|n+1\right.\rangle ,\end{eqnarray} \tag{ 8.80 }$$

and hence the theta vacuum simply picks up a phase

$$\begin{eqnarray}&&\left|\theta \right.\rangle \mathop{\to }\limits^{{g}_{1}}{{\rm{e}}}^{-i\theta }\left|\theta \right.\rangle .\end{eqnarray} \tag{ 8.81 }$$

Although one can construct stationary states for any value of $\theta$, they are not excitations of the $\theta =0$ vacuum, because in QCD the value of $\theta$ cannot be changed. As far as the strong interaction is concerned, different values of $\theta$ correspond to different worlds. We can fix the value of $\theta$ by adding an additional term of the form

$$\begin{eqnarray}&&{ \mathcal L }=-\frac{{g}^{2}\theta }{32{\pi }^{2}}{F}_{\mu \nu }^{a}{\tilde{F}}_{\mu \nu }^{a},\end{eqnarray} \tag{ 8.82 }$$

to the QCD Lagrangian. Does physics depend on the value of $\theta$? The interaction above violates both T and CP invariance. On the other hand, it is a surface term and it might be possible that confinement somehow screens the effects of the $\theta$-term. A similar phenomenon is known to occur in three-dimensional compact electrodynamics. In QCD, however, one can show that, if the $U{\mathrm{(1)}}_{A}$ problem is solved (there is no massless $\eta ^{\prime}$ state in the chiral limit) and none of the quarks are massless, a non-zero value of $\theta$ implies that CP symmetry is broken [5–8]. The most severe limits on CP violation in the strong interaction come from the electric dipole of the neutron. Current experiments imply that $\theta \lesssim {10}^{-10}$ [9, 10]. The question of why $\theta$ is so small is known as the strong CP problem.

Exercise 8.5 Show that equation (8.77) is obeyed by a $q=1$ instanton solution.

It is possible that classical symmetries of a system do not survive quantization, in which case it is said that the theory possesses a quantum anomaly. If this is case, the Noether current associated with the classical symmetry is no longer conserved after quantization and the current conservation law is said to receive an anomalous contribution. This is relevant for this course because QCD has a chiral anomaly, which is associated with the non-conservation of the chiral current in the limit of vanishing light quark masses. This will be the focus of the remainder of the chapter.

Historically, the reaction ${\pi }^{0}\to \gamma \gamma$ is the best example of a process that proceeds primarily via the chiral anomaly. The original calculation of this anomalous decay was performed in 1969 by Bell and Jackiw and, independently, Adler. As a result, it is sometimes referred to as the Adler-Bell-Jackiw (ABJ) anomaly [11–13]. Earlier calculations of the ${\pi }^{0}\to \gamma \gamma$ decay width, which did not take into chiral anomaly, resulted in a decay lifetime on the order of 10⁻³³ s, which was approximately three orders of magnitude longer than the experimentally observed pion lifetime. As of the 2015 PDG listings, the pion lifetime is $\tau =\hslash /{{\rm{\Gamma }}}_{{\rm{tot}}}({\pi }^{0})=(8.52\pm 0.18)\times {10}^{-17}\,{\rm{s}}$. The branching ratio for the $2\gamma$ decay channel is ${\rm{BR}}({\pi }^{0}\to \gamma \gamma )\simeq 0.9882$ and hence it dominates the total lifetime calculation for the pion. Taking into account the chiral anomaly, ABJ obtained a decay width of ${\rm{\Gamma }}({\pi }^{0}\to \gamma \gamma )=7.76\,{\rm{eV}}$. This maps to a total pion lifetime of approximately $8.38\times {10}^{-17}\,{\rm{s}}$, which is in the right ballpark and, within the modern experimental error bars ² .

In perturbation theory, one can understand the emergence of the chiral anomaly through the consideration of triangle graphs of the form

where A stands for 'axial' and V stands for 'vector'. Such graphs naturally arise in the calculation of the pion decay rate. There are also VVV graphs and other configurations that occur at higher orders, however, it turns out that once we understand the anomaly in the AVV graph it is automatically handled in all of the other graphs. Before proceeding to this technical calculation, however, I would first like to discuss the physics of the anomaly.

8.8.1. The chiral anomaly in the Schwinger model

The Schwinger model is simply 1+1d massless QED. The Lagrangian density is

$$\begin{eqnarray}&&{ \mathcal L }=\bar{\psi }iD\psi -\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu },\end{eqnarray} \tag{ 8.84 }$$

where, as usual, ${D}_{\mu }\equiv {\partial }_{\mu }-i{\rm{e}}{A}_{\mu }$ is the covariant derivative and the $2\times 2$ Dirac matrices can be written in terms of the Pauli matrices

$$\begin{eqnarray}\begin{array}{rcl}{\gamma }^{0} & = & {\sigma }_{1},\\ {\gamma }^{1} & = & i{\sigma }_{2}.\end{array}\end{eqnarray} \tag{ 8.85 }$$

The resulting classical equations of motion are

$$\begin{eqnarray}&&iD\psi =0,\end{eqnarray} \tag{ 8.86 }$$

$$\begin{eqnarray}&&\square {A}_{\mu }=-{\rm{e}}{j}_{\mu },\end{eqnarray} \tag{ 8.87 }$$

where

$$\begin{eqnarray}&&{j}_{\mu }\equiv \bar{\psi }{\gamma }_{\mu }\psi ,\end{eqnarray} \tag{ 8.88 }$$

is the conserved vector current, ${\partial }_{\mu }\,{j}^{\mu }=0$. This conserved current results from the local gauge invariance of the Schwinger model and QED in general.

There is also an axial current or chiral current

$$\begin{eqnarray}&&{j}_{\mu }^{5}\equiv \bar{\psi }{\gamma }_{\mu }{\gamma }_{5}\psi ,\end{eqnarray} \tag{ 8.89 }$$

where ${\gamma }_{5}\equiv -{\gamma }^{0}{\gamma }^{1}={\sigma }_{3}$. This chiral current is classically conserved ${\partial }^{\mu }{j}_{\mu }^{5}=0$ and this conservation law is associated with classical invariance of QED under global chiral transformations of the fermionic fields

$$\begin{eqnarray}&&\psi \to {{\rm{e}}}^{i\alpha {\gamma }_{5}}\psi ={{\rm{e}}}^{i\alpha {\sigma }_{3}}\psi .\end{eqnarray} \tag{ 8.90 }$$

We can identify the two components of the spinor as 'left'- and 'right'-handed

$$\begin{eqnarray}&&\psi =\left(\begin{array}{c}{\psi }_{L}\\ {\psi }_{R}\end{array}\right),\end{eqnarray} \tag{ 8.91 }$$

where ${\psi }_{L,R}=(\mathrm{1/2})(1\pm {\gamma }^{5})\psi \equiv {P}_{L,R}\psi$ with ${P}_{L,R}$ being a left/right projector. Under a chiral transformation the left and right fields pick up opposite phases

$$\begin{eqnarray}\begin{array}{rcl}{\psi }_{L} & \to & {{\rm{e}}}^{i\alpha }{\psi }_{L},\\ {\psi }_{R} & \to & {{\rm{e}}}^{-i\alpha }{\psi }_{R}.\end{array}\end{eqnarray} \tag{ 8.92 }$$

Expanding out the free fermionic contribution to the Lagrangian in terms of ${\psi }_{L,R}$, we obtain

$$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal L }}_{{\rm{free}}{\rm{fermionic}}} & = & i{\psi }^{\dagger }{\gamma }^{0}\left({\gamma }^{0}{\partial }_{t}+{\gamma }^{1}{\partial }_{x}\right)\psi \\ & = & i{\psi }^{\dagger }({\partial }_{t}-{\gamma }_{5}{\partial }_{x})\psi \\ & = & i{\psi }_{L}^{\dagger }({\partial }_{t}-{\partial }_{x}){\psi }_{L}+i{\psi }_{R}^{\dagger }({\partial }_{t}+{\partial }_{x}){\psi }_{R},\end{array}\end{eqnarray} \tag{ 8.93 }$$

and the general solution to the equations of motion in this case will be of the form

$$\begin{eqnarray*}\begin{array}{ll}{\psi }_{L}(t,x)={\psi }_{L}(x+t) & \,{\rm{left}}\,{\rm{mover}},\\ {\psi }_{R}(t,x)={\psi }_{R}(x-t) & \,{\rm{right}}\,{\rm{mover}}.\end{array}\end{eqnarray*}$$

This demonstrates that, in the Schwinger model, left- and right-handed particles are quite literally left- and right-moving particles. This is different than the 3+1d case where chirality is related to the alignment of the particle's spin with its momentum (helicity). In 1+1d, there is no spin and the handedness is related to particles propagating either to the left and right. This only makes sense in the massless case where particles move at the speed of light and the direction of propagation is the same in all reference frames. In addition, in this case we see that a parity transformation $x\to -x$ transforms left-movers into right-movers. This will be become important since, if the symmetry between left- and right-movers is broken, then we might break parity symmetry.

We also see that the free part of the Lagrangian is invariant under chiral transformations (8.92) since left- and right-handed fields only couple to themselves. The same holds for the interaction part, where one finds

$$\begin{eqnarray}&&{{ \mathcal L }}_{{\rm{interaction}}}=e\bar{\psi }A\psi =e{\psi }_{L}^{\dagger }({A}_{0}+{A}_{1}){\psi }_{L}+e{\psi }_{R}^{\dagger }({A}_{0}-{A}_{1}){\psi }_{R}.\end{eqnarray} \tag{ 8.94 }$$

Note that, in the classical theory, the number of left- and right-handed fermions, ${N}_{L}$ and ${N}_{R}$, are independent constants of the motion. This will remain true if the fermions are coupled to a gauge field.

Exercise 8.6 Derive equations (8.86) and (8.87).

8.8.2. Understanding the anomaly

Before proceeding more formally, let us consider what would happen if we apply an external electric field ${\bf{E}}=E\hat{{\bf{x}}}$ to a system of positively charged particles for a short amount of time ${\rm{\Delta }}t$. In this case, the right-movers will gain energy

$$\begin{eqnarray}&&{\rm{\Delta }}{ \mathcal E }={\rm{e}}E{\rm{\Delta }}t,\end{eqnarray} \tag{ 8.95 }$$

and left-movers will lose the same amount of energy. If the initial state before the electric field was turned on is the Dirac vacuum, with all negative energy levels filled, then after the time interval ${\rm{\Delta }}t$, the right-handed 'Fermi-level' has increased by ${\rm{\Delta }}{ \mathcal E }$ and the left-handed Fermi level has decreased by ${\rm{\Delta }}{ \mathcal E }$. As a result, right-handed fermions are created along with left-handed anti-fermions (holes). This is sketched in figure 8.4.

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Since the one-dimensional density of states is $dp/(2\pi )=d{ \mathcal E }/(2\pi )$, the number density per unit length of left- and right-handed fermions become

$$\begin{eqnarray}\begin{array}{rcl}{\rho }_{L} & = & -\,\frac{{\rm{e}}E}{2\pi }{\rm{\Delta }}t,\\ {\rho }_{R} & = & \frac{{\rm{e}}E}{2\pi }{\rm{\Delta }}t.\end{array}\end{eqnarray} \tag{ 8.96 }$$

Therefore, the 'vector' fermion number associated with charge conservation is conserved during this process

$$\begin{eqnarray}&&\dot{\rho }{f}_{V}={\dot{\rho }}_{R}+{\dot{\rho }}_{L}=0,\end{eqnarray} \tag{ 8.97 }$$

however, the 'axial' fermion number, which is also conserved at the classical level, changes according to

$$\begin{eqnarray}&&{\dot{\rho }}_{A}\equiv {\dot{\rho }}_{R}-{\dot{\rho }}_{L}=\frac{{\rm{e}}E}{\pi }.\end{eqnarray} \tag{ 8.98 }$$

The fact that the right hand side is non-vanishing is the chiral anomaly. The chiral anomaly corresponds to a kind of dielectric breakdown of the vacuum where we create particle-hole pairs with a chiral imbalance.

At this point, you should be scratching your head in confusion since everything we just did was classical; however, the subtlety in the argument is the existence of an infinitely occupied Dirac sea in the first place. Imagine that instead of an infinite number of negative energy states, there were a finite number of them, regulated by a cutoff ${\rm{\Lambda }}$ on the lowest possible negative energy state before turning on the electric field. In that case, figure 8.4 would look instead like figure 8.5 and we would always have the same number of left- and right-moving states even in the presence of an external electric field. As a result, we see that the physics of the anomaly will be tied intimately with the regularization of the theory. If we regularize and do not remove the regulator, we might even miss it! Note that, if you hear condensed matter theorists discussing the chiral anomaly, the figure they use to illustrate the concept will look more like figure 8.6.

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8.8.3. The chiral anomaly in 3+1d

The natural followup question is, of course, if this kind of logic can be extended to 3+1d. As it turns out, the precise discussion we just had applies in 3+1d to massless fermions in a magnetic field. This is because fermions in a background magnetic field are restricted to Landau levels labeled by an integer $n$ [17]. For massless fermions, the Landau levels are

$$\begin{eqnarray}&&{E}_{n,{s}_{z}}^{2}={k}_{z}^{2}+(2n+1)| q| B-2qB{s}_{z},\end{eqnarray} \tag{ 8.99 }$$

where $n\geqslant 0$ is an integer, ${\bf{B}}=B\hat{{\bf{z}}}$, and ${s}_{z}=\pm \mathrm{1/2}$. The fermionic motion in the $x-y$ plane is quantized circular motion and the fermions effectively propagate along the z-axis like 1+1d fermions with mass

$$\begin{eqnarray}&&{m}_{{\rm{eff}}}^{2}=(2n+1)| q| B-2qB{s}_{z}.\end{eqnarray} \tag{ 8.100 }$$

For a positively charged particle, $q={\rm{e}}$, with ${s}_{z}=\mathrm{1/2}$, we see that the lowest Landau level $n=0$ the effective mass vanishes and, hence, these particles will behave like massless 1+1d fermions. For a negatively charged particle, we have the same behavior for the ${s}_{z}=-\mathrm{1/2}$.

For massless 3+1d fermions, chirality is identified with the helicity of the state. The right-handed fermion with ${s}_{z}=\mathrm{1/2}$ is a right-mover along the z-axis and the left-handed fermion is a left-mover. So, by adding a background field, at least some subset of the allowed states are effectively 1+1 and our previously setup can be applied. We now imagine applying an electric field along the $z$ direction, ${\bf{E}}=E\hat{{\bf{z}}}$, and our earlier discussion applies. The density of states per unit area in a Landau level is

$$\begin{eqnarray}&&{g}_{n}=\frac{{\rm{e}}B}{2\pi },\end{eqnarray} \tag{ 8.101 }$$

therefore, axial charge is created at a rate of

$$\begin{eqnarray}&&{\dot{\rho }}_{A}=\frac{{\rm{e}}B}{2\pi }\frac{{\rm{e}}E}{\pi }=\frac{{{\rm{e}}}^{2}}{2{\pi }^{2}}{\bf{E}}\cdot {\bf{B}},\end{eqnarray} \tag{ 8.102 }$$

where now the right-hand side indicates the presence of a 3+1d axial anomaly. Stated succinctly, an electric field applied to the vacuum causes pair production and, if a parallel magnetic field is applied, the pairs created are chiral. Positively-charged fermions will align their spins with B, while anti-fermions (negative charge in this case), anti-align their spins with B ³ .

We would like to express ${\rho }_{A}$ in terms of the axial (chiral) charge ${Q}_{5}$

$$\begin{eqnarray}&&{Q}_{5}\equiv \int dx{j}_{0}^{5}=\int dx\bar{\psi }{\gamma }_{0}{\gamma }_{5}\psi =\int dx{\psi }^{\dagger }{\sigma }_{3}\psi =\int dx\left({\psi }_{L}^{\dagger }{\psi }_{L}-{\psi }_{R}^{\dagger }{\psi }_{R}\right)\equiv {\rho }_{L}-{\rho }_{R},\end{eqnarray} \tag{ 8.103 }$$

and the contraction of the field strength tensor and the dual field strength tensor

$$\begin{eqnarray}&&{F}_{\mu \nu }{\tilde{F}}^{\mu \nu }=-4{\bf{E}}\cdot {\bf{B}},\end{eqnarray} \tag{ 8.104 }$$

where we remind you that ${\tilde{F}}_{\mu \nu }\equiv (\mathrm{1/2}){\varepsilon }_{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. This gives

$$\begin{eqnarray}&&\frac{d{Q}_{5}}{dt}=\frac{{{\rm{e}}}^{2}}{8{\pi }^{2}}{F}_{\mu \nu }{\tilde{F}}^{\mu \nu }.\end{eqnarray} \tag{ 8.105 }$$

This is the standard way to present the anomalous contribution, which breaks chiral current conservation.

Exercise 8.7 Derive equation (8.99).

Exercise 8.8 Derive equation (8.104).

Now that we have a kind of intuitive understanding of the phenomenon, let us return to the mathematical development in the context of the 1+1d Schwinger model. To begin, we note that the axial current (8.89) can be expressed as

$$\begin{eqnarray}&&{j}_{\mu }^{5}={\varepsilon }_{\mu \nu }{j}^{\nu },\end{eqnarray} \tag{ 8.106 }$$

where ${\varepsilon }^{\mu \nu }$ is the two-dimensional Levi-Civita tensor. Upon quantization, it can be shown that the Lagrangian can be written as

$$\begin{eqnarray}&&{ \mathcal L }=\bar{\psi }i\partial \psi -\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }-\frac{{{\rm{e}}}^{2}}{2\pi }{A}_{\mu }{A}^{\mu }.\end{eqnarray} \tag{ 8.107 }$$

This is the Lagrangian of a system of free massless spin-1/2 particles and free 'photons' having mass ${m}_{\gamma }^{2}={{\rm{e}}}^{2}/\pi$. Also, in the quantized theory the axial current is no longer conserved. Instead, we have

$$\begin{eqnarray}&&{\partial }^{\mu }{j}_{\mu }^{5}=-\frac{{\rm{e}}}{2\pi }{\varepsilon }^{\mu \nu }{F}_{\mu \nu }.\end{eqnarray} \tag{ 8.108 }$$

As a result, quantization breaks axial current conservation and there exists an anomaly.

To see how this arises physically, returning to our consideration of an electric field applied to the chiral Schwinger model, we see that the applied electric field induces a current density along the x-direction $j\equiv {j}^{1}$ which grows in time with

$$\begin{eqnarray}&&\frac{dj}{dt}={\int }_{-\infty }^{\infty }\frac{dp}{2\pi }\frac{dv}{dt},\end{eqnarray} \tag{ 8.109 }$$

where, using the Lorentz force law, $dp/dt={\rm{e}}E$, we have

$$\begin{eqnarray}&&\frac{dv}{dt}=\frac{d}{dt}\frac{d}{dp}\sqrt{{m}^{2}+{p}^{2}}=\frac{{\rm{e}}E{m}^{2}}{{({p}^{2}+{m}^{2})}^{\mathrm{5/2}}}.\end{eqnarray} \tag{ 8.110 }$$

Integrating equation (8.109), we then obtain

$$\begin{eqnarray}&&\frac{dj}{dt}=\frac{{\rm{e}}E}{\pi },\end{eqnarray} \tag{ 8.111 }$$

which is independent of the mass. Since the vacuum charge density ${\rho }_{vac}$ is independent of the position, we have $d{j}^{0}/dx=0$. Defining ${j}^{\mu }=(\,{j}^{0},j)$ and using equation (8.106), we have

$$\begin{eqnarray}&&{\partial }^{\mu }{j}_{\mu }^{5}={\partial }^{\mu }{\varepsilon }_{\mu \nu }{j}^{\nu }={\rm{e}}{\partial }_{t}\,j=\frac{{\rm{e}}E}{\pi }=-\frac{{\rm{e}}}{2\pi }{\varepsilon }^{\mu \nu }{F}_{\mu \nu },\end{eqnarray} \tag{ 8.112 }$$

where, in the last step, we have used ${\varepsilon }^{\mu \nu }{F}_{\mu \nu }={F}_{01}-{F}_{10}=2{F}_{01}=-2E$.

To see how we obtain a theory with massive photons, we note that the last equation can be written equivalently as

$$\begin{eqnarray}&&{\varepsilon }_{\mu \nu }{\partial }^{\mu }\left({j}^{\nu }+\frac{{\rm{e}}}{\pi }{A}^{\nu }\right)=0.\end{eqnarray} \tag{ 8.113 }$$

In Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, we have (homework)

$$\begin{eqnarray}&&{j}_{\mu }=-\frac{{\rm{e}}}{\pi }{A}_{\mu }.\end{eqnarray} \tag{ 8.114 }$$

Plugging this into the equation of motion (8.87) gives

$$\begin{eqnarray}&&\left(\square +\frac{{{\rm{e}}}^{2}}{\pi }\right){A}_{\mu }=0,\end{eqnarray} \tag{ 8.115 }$$

which shows that the photon has developed an effective mass ${m}_{\gamma }^{2}={{\rm{e}}}^{2}/\pi$. The picture is as before: the anomaly arises from an alteration of the vacuum state of a quantized system in the presence of an applied electric field.

Exercise 8.9 Derive equation (8.107).

Exercise 8.10 Derive equation (8.114).

Perhaps while you were reading the previous section you noticed that the rate of change of the axial (chiral) current (8.105) is proportional to the local topological density ${F}_{\mu \nu }{\tilde{F}}^{\mu \nu }$. Although, the argumentation in the previous section relied on consideration of abelian electric and magnetic fields, the same phenomena occurs in the presence of color electric and magnetic fields ${{\bf{E}}}^{a}$ and ${{\bf{B}}}^{a}$. Hence, if there is region where there is a non-vanishing topological density, there will also be breaking of chiral symmetry. Since instantons are 'topological lumps' that realized precisely this situation, one concludes that the presence of instantons in the QCD vacuum will result in local breaking of chiral symmetry. This leads to, for example, a nonzero amplitude for chirality breaking processes such as

$$\begin{eqnarray}&&{u}_{L}+{d}_{L}\to {u}_{R}+{d}_{R}.\end{eqnarray} \tag{ 8.116 }$$

Our previous discussion of instantons was restricted to pure gauge theory (Yang–Mills); however, in relation to chiral symmetry breaking instantons are important because the Dirac operator has a chiral zero model in the instanton background. These zero modes correspond to localized quark states that can become collective if many instantons and anti-instantons interact. The delocalized state that results corresponds to the wave function of the quark condensate and the instanton zero modes generate an effective four-quark interaction as indicated above. That is all we will say on this point and instead refer you to the literature: see for example refs. [22–25] and references therein.

We will now discuss how to see the emergence of the chiral anomaly in the context of perturbation theory. Unfortunately, since there is no well-defined generalization of ${\gamma }_{5}$ to non-integer dimensions, typically different regularization methods such as Pauli–Villars or Schwinger regularization are used. Here we will think in terms of momentum-space cutoffs with the understanding that, if done using a proper gauge-invariant regulator the same result emerges. The analysis begins by considering the perturbative corrections to the axial-vector vertex between gauge and matter fields. The corresponding diagrams through $2\,({{\rm{e}}}^{5})$ are

where a dashed line is a particle that couples via ${\gamma }_{\mu }{\gamma }_{5}$ to quarks such as a pion field, but in general QFT it could also be the Weinberg–Salam theory of weak interactions as well. All that require is that there is an axial vector coupling between the matter and gauge fields of the form ${g}_{A}{j}_{\mu }^{5}{W}^{\mu }$, where ${W}^{\mu }$ is the particle that couples to the axial current and ${j}_{\mu }^{5}$ is the axial current (8.89). At the classical level, we have from the Dirac equation

$$\begin{eqnarray}&&{\partial }^{\mu }{j}_{\mu }^{5}=2im\bar{\psi }{\gamma }_{5}\psi \equiv 2m{n}_{5},\end{eqnarray} \tag{ 8.118 }$$

where ${n}_{5}$ is the chiral density. For $m\ne 0$ this current is not conserved since axial symmetry is explicitly broken, however, even in this case the relation results in an axial Ward identity. In the case that $m=0$, this current is conserved at the classical level, however, as we have discussed previously, in the massless case this current is no longer conserved when the theory is quantized. The problem remains when $m\ne 0$, so it suffices to consider the chiral limit to understand the problem.

In quantum field theory, conservation laws result from analysis of the vertex functions. Analyzing the graphs in (8.117), one finds that the last graph, which contains a triangle-shaped fermionic closed loop, fails to satisfy the axial Ward identity and gives rise to the chiral anomaly. Let us now focus our attention on the triangle subgraph that couples AVV. As noted in the last lecture, there are other configurations such as AAA-triangle, squares, pentagons, etc. but to understand the basic mechanism of the chiral anomaly it suffices to consider the AVV graph. Since the photons are indistinguishable, there are two graphs that enter. They can be expressed in terms of two permutations of the diagram shown to the right.

The resulting Feynman diagrams can be expressed in terms of the three-tensor

$$\begin{eqnarray}&&{T}^{\kappa \lambda \mu }({q}_{1},{q}_{2})={U}^{\kappa \lambda \mu }({q}_{1},{q}_{2})+{U}^{\lambda \kappa \mu }({q}_{2},{q}_{1}),\end{eqnarray} \tag{ 8.119 }$$

where

$$\begin{eqnarray}&&{U}^{\kappa \lambda \mu }({q}_{1},{q}_{2})\equiv -\frac{i{{\rm{e}}}^{2}{K}_{F}}{2}{\int }_{k}{\rm{Tr}}\left[\frac{1}{\rlap{/}{k}+{\rlap{/}{q}}_{1}}{\gamma }^{\kappa }\frac{1}{\rlap{/}{k}}{\gamma }^{\lambda }\frac{1}{\rlap{/}{k}-{\rlap{/}{q}}_{2}}{\gamma }^{\mu }{\gamma }_{5}\right],\end{eqnarray} \tag{ 8.120 }$$

with ${\int }_{k}=\int {d}^{4}k/{(2\pi )}^{4}$. The factor of ${K}_{F}$ arises from a sum over colored light quark loops and is

$$\begin{eqnarray}&&{K}_{F}={C}_{A}\sum _{u,d}{Q}_{q}^{2}{\sigma }_{3q}=3\left[{\left(\frac{2}{3}\right)}^{2}-{\left(\frac{1}{3}\right)}^{2}\right]=1,\end{eqnarray} \tag{ 8.121 }$$

where ${\sigma }_{3u}={({\sigma }_{3})}_{11}$ and ${\sigma }_{3d}={({\sigma }_{3})}_{22}$.

Generally, the quantity ${T}^{\kappa \lambda \mu }$ is the Fourier transform of the AVV current amplitude

$$\begin{eqnarray}&&{T}^{\kappa \lambda \mu }({q}_{1},{q}_{2})=-i{{\rm{e}}}^{2}{\int }_{x}{\int }_{y}{{\rm{e}}}^{-i{q}_{1}\cdot x-i{q}_{2}\cdot y}\langle 0| T\left[{J}_{{\rm{em}}}^{\kappa }(x){J}_{{\rm{em}}}^{\lambda }(\,y){J}_{5}^{\mu }(x)\right]| 0\rangle .\end{eqnarray} \tag{ 8.122 }$$

The diagrams result from the perturbative expansion of this quantity

We can now check the various conservation laws ${\partial }_{\mu }{J}_{{\rm{em}}}^{\mu }=0$ and ${\partial }_{\mu }{J}_{5}^{\mu }=0$. Based on the last equation, current conservation for the vector and axial-vector currents gives the following conditions

$$\begin{eqnarray}&&{q}_{1}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})={q}_{2}^{\lambda }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})={({q}_{1}+{q}_{2})}^{\mu }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2}).\end{eqnarray} \tag{ 8.123 }$$

Looking at the first one, which expresses current conservation at one of the electromagnetic vertices, we find

$$\begin{eqnarray}\begin{array}{rcl}{q}_{1}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2}) & = & -\,i{{\rm{e}}}^{2}{\int }_{k}{\rm{Tr}}\left[\left(\frac{1}{\rlap{/}{k}}-\frac{1}{\rlap{/}{k}+{\rlap{/}{q}}_{1}}\right){\gamma }_{\lambda }\frac{1}{k-{q}_{2}}{\gamma }_{\mu }{\gamma }_{5}\right.\\ & & \left.+\,\frac{1}{\rlap{/}{k}+{\rlap{/}{q}}_{2}}{\gamma }_{\lambda }\left(\frac{1}{\rlap{/}{k}}-\frac{1}{\rlap{/}{k}-{\rlap{/}{q}}_{1}}\right){\gamma }_{\mu }{\gamma }_{5}\right].\end{array}\end{eqnarray} \tag{ 8.124 }$$

The integrals involving a single factor of photon momentum ${q}_{1}$ or ${q}_{2}$ vanish since the Levi-Civita tensor associated with the trace

$$\begin{eqnarray}&&{\rm{Tr}}[{\gamma }_{\kappa }{\gamma }_{\lambda }{\gamma }_{\alpha }{\gamma }_{\beta }]=4i{\varepsilon }_{\kappa \lambda \alpha \beta },\end{eqnarray} \tag{ 8.125 }$$

requires contraction with two independent momenta in order to be non-vanishing. Defining

$$\begin{eqnarray}&&{W}_{\lambda \mu }(\,p\,)={\rm{Tr}}\left[\frac{1}{\rlap{/}{p}}{\gamma }_{\lambda }\frac{1}{\rlap{/}{p}-{\rlap{/}{q}}_{1}-{\rlap{/}{q}}_{2}}{\gamma }_{\mu }{\gamma }_{5}\right],\end{eqnarray} \tag{ 8.126 }$$

one finds

$$\begin{eqnarray}&&{q}_{1}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=-i{{\rm{e}}}^{2}{\int }_{k}\left[{W}_{\lambda \mu }(k+{q}_{1})-{W}_{\lambda \mu }(k+{q}_{2})\right].\end{eqnarray} \tag{ 8.127 }$$

This integral linearly divergent in the ultraviolet. If the integrals above were convergent, or diverged at worst logarithmically, then we could shift the integration variables by $-{q}_{1}$ in the first term and by $-{q}_{2}$ in the second term and they would exactly cancel. The linear divergence, however, means that this shift would appear in the upper limit of the integration due to the need for regulation, and would break the symmetry between the two terms. In order to carry out the shift more carefully, we Taylor expand

$$\begin{eqnarray}&&{\int }_{k}F(k+a)={\int }_{k}[F(k)+{a}^{\nu }{\partial }_{\nu }F(k)+\cdots ],\end{eqnarray} \tag{ 8.128 }$$

to obtain

$$\begin{eqnarray}&&{q}_{1}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=-i{{\rm{e}}}^{2}{({q}_{2}-{q}_{1})}^{\nu }{\int }_{k}[{\partial }_{\nu }{W}_{\lambda \mu }+\cdots ].\end{eqnarray} \tag{ 8.129 }$$

The omitted higher-order terms in the Taylor series vanish when we remove the cutoff since, in the case at hand, $F(k)\sim {k}^{-3}$ and the integration measure contains a factor of ${k}^{3}$. One can use Gauss' law to evaluate the resulting integral, which gives (homework)

$$\begin{eqnarray}&&{q}_{1}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=-\frac{i{{\rm{e}}}^{2}}{4{\pi }^{2}}{\varepsilon }_{\kappa \lambda \mu \delta }{q}_{1}^{\kappa }{q}_{2}^{\delta },\end{eqnarray} \tag{ 8.130 }$$

where we have used

$$\begin{eqnarray}\begin{array}{rcl}{\rm{Tr}}[{\gamma }_{5}{\gamma }_{\rho }{\gamma }_{\lambda }{\gamma }_{\sigma }{\gamma }_{\kappa }{\gamma }_{\tau }{\gamma }_{\mu }] & = & 4i{\varepsilon }_{\kappa \tau \mu \alpha }\left({\delta }_{\rho }^{\alpha }{\eta }_{\lambda \sigma }-{\delta }_{\lambda }^{\alpha }{\eta }_{\rho \sigma }+{\delta }_{\sigma }^{\alpha }{\eta }_{\rho \lambda }\right)\\ & & -\,4i{\varepsilon }_{\rho \lambda \sigma \alpha }\left({\delta }_{\kappa }^{\alpha }{\eta }_{\tau \mu }-{\delta }_{\tau }^{\alpha }{\eta }_{\kappa \mu }+{\delta }_{\mu }^{\alpha }{\eta }_{\kappa \tau }\right),\end{array}\end{eqnarray} \tag{ 8.131 }$$

which gives

$$\begin{eqnarray}&&{\rm{Tr}}[{\gamma }_{5}\rlap{/}{k}{\gamma }_{\lambda }\rlap{/}{k}{\gamma }_{\kappa }\rlap{/}{k}{\gamma }_{\mu }]=4i{\varepsilon }_{\kappa \lambda \mu \alpha }{k}^{\alpha }{k}^{2}.\end{eqnarray} \tag{ 8.132 }$$

Similarly, one finds

$$\begin{eqnarray}&&{q}_{2}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=\frac{i{{\rm{e}}}^{2}}{4{\pi }^{2}}{\varepsilon }_{\kappa \lambda \mu \delta }{q}_{1}^{\kappa }{q}_{2}^{\delta },\end{eqnarray} \tag{ 8.133 }$$

and

$$\begin{eqnarray}&&{({q}_{1}-{q}_{2})}^{\kappa }{T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=0.\end{eqnarray} \tag{ 8.134 }$$

This implies that current is not conserved and that electromagnetic gauge invariance is broken. Since gauge symmetry is special we can redefine the amplitude for the triangle graph by adding a polynomial in the external momentum, which can be be done without affecting the absorptive component of the amplitude. Thus, defining the physical decay amplitude via

$$\begin{eqnarray}&&{\tilde{T}}_{\kappa \lambda \mu }({q}_{1},{q}_{2})\equiv {T}_{\kappa \lambda \mu }({q}_{1},{q}_{2})-\frac{i{{\rm{e}}}^{2}}{4{\pi }^{2}}{\varepsilon }_{\kappa \lambda \mu \delta }{({q}_{1}-{q}_{2})}^{\delta },\end{eqnarray} \tag{ 8.135 }$$

we can enforce electromagnetic gauge invariance

$$\begin{eqnarray}&&{q}_{1}^{\kappa }{\tilde{T}}_{\kappa \lambda \mu }({q}_{1},{q}_{2})={q}_{2}^{\kappa }{\tilde{T}}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=0,\end{eqnarray} \tag{ 8.136 }$$

at the expense of a non-vanishing axial divergence

$$\begin{eqnarray}&&{({q}_{1}-{q}_{2})}^{\mu }{\tilde{T}}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=\frac{i{{\rm{e}}}^{2}}{2{\pi }^{2}}{\varepsilon }_{\kappa \lambda \mu \delta }{q}_{1}^{\mu }{q}_{2}^{\delta },\end{eqnarray} \tag{ 8.137 }$$

and the axial current is no longer conserved. Using the last expression we have

$$\begin{eqnarray}&&{\tilde{T}}_{\kappa \lambda \mu }({q}_{1},{q}_{2})=\frac{{q}_{1\mu }+{q}_{2\mu }}{{({q}_{1}+{q}_{2})}^{2}}\frac{i{{\rm{e}}}^{2}}{2{\pi }^{2}}{\varepsilon }_{\kappa \lambda \mu \delta }{q}_{1}^{\mu }{q}_{2}^{\delta }.\end{eqnarray} \tag{ 8.138 }$$

Using

$$\begin{eqnarray}\begin{array}{rcl}{F}^{\mu \nu }{\tilde{F}}_{\mu \nu } & = & \frac{1}{2}{\varepsilon }_{\mu \nu \rho \sigma }({\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu })({\partial }^{\rho }{A}^{\sigma }-{\partial }^{\sigma }{A}^{\rho })\\ & = & 2{\varepsilon }_{\mu \nu \rho \sigma }{\partial }^{\mu }{A}^{\nu }{\partial }^{\rho }{A}^{\sigma }\\ & = & {\partial }^{\mu }(2{\varepsilon }_{\mu \nu \rho \sigma }{A}^{\nu }{\partial }^{\rho }{A}^{\sigma }),\end{array}\end{eqnarray} \tag{ 8.139 }$$

this maps to the condition

$$\begin{eqnarray}&&{\partial }_{\mu }{J}_{5}^{\mu }=\frac{{{\rm{e}}}^{2}}{8{\pi }^{2}}{F}^{\mu \nu }{\tilde{F}}_{\mu \nu }.\end{eqnarray} \tag{ 8.140 }$$

This is precisely the same form, we obtain based on more physical argumentation (8.105). If we were to repeat this exercise, without taking the fermion masses to zero in the beginning, we would have found instead

$$\begin{eqnarray}&&{\partial }_{\mu }{J}_{5}^{\mu }=2m{n}_{5}+\frac{{{\rm{e}}}^{2}}{8{\pi }^{2}}{F}^{\mu \nu }{\tilde{F}}_{\mu \nu }.\end{eqnarray} \tag{ 8.141 }$$

Exercise 8.11 Derive equation (8.119).

Exercise 8.12 Derive equation (8.123).

Exercise 8.13 Derive equation (8.130).