# How do instantons cause vacuum decay?

From what I read about on instantons (Zee, QFT in a Nutshell, pg 309-310), an instanton is a vacuum solution that maps $S^3 \rightarrow S^3$ (the boundary of Euclideanized spacetime), which comes from minimizing the Euclidean action for some Lagrangian with a nontrivial vacuum structure. I've also read (for example in Muckanov, Physical Foundations of Cosmology, pg 180-199) about how instantons can mediate quantum tunneling from one vacuum state to another.

My question is: how are these two ideas/definitions of instantons related? All of the simples examples that I have looked at of nontrivial vacuum solutions involve solitons, vortices, or hedgehogs, which as far as I know cannot mediate decay from a metastable vacuum. Solitons etc are defined on spacial infinity so I know (suspect?) that the fact that an instanton lives on the boundary of space$time$ is related to its connection to rate of vacuum decay. I would greatly appreciated some simple examples/links to references as well.

Let us look at the instantons of an ordinary pure Yang-Mills theory for gauge group $G$ in four Euclidean dimensions:

An instanton is a local minimum of the action

$$S_{YM}[A] = \int \mathrm{tr}(F \wedge \star F)$$

which is, on $\mathbb{R}^4$, precisely given by the (anti-)self-dual solutions $F = \pm \star F$. For (anti-)self-dual solutions, $\mathrm{tr}(F \wedge \star F) = \mathrm{tr}(F \wedge F)$. The latter is a topological term known as the second Chern class, and its integral is discrete:

$$\int \mathrm{tr}(F \wedge F) = 8\pi^2 k$$

with integer $k \in \mathbb{Z}$ (don't ask about the $\pi$). For given $k$, one also speaks of the corresponding curvature/gauge field as the $k$-instanton. Now, how does this relate to the things you have asked about?

Instantons as vacua

Since the instanton provides a local minimum of the action, it is a natural start for perturbation theory, where it naturally then represents the vacuum. We have infinitely many vacuua to choose from, since $k$ is arbitrary.

Instantons and the three-sphere

(The motivation here is, that, for the vacuum to have finite energy, $F = 0$ at infinity, so we seek actually a solution for the field equations on $\mathbb{R}^4 \cup \{\infty\} = S^4$ such that $F(\infty) = 0$)

Take two local instanton solutions $A_1,A_2$ (for same Chern class $k$) on some open disks $D_1, D_2$´. Now, glue them together by a gauge transformation $t : D_k \cap D_{k'} \rightarrow G$ as per

$$A_2 = tA_1t^{-1} + t\mathrm{d}t^{-1}$$

(we are essentially defining the principal bundle over $S^4$ here) and observe that $\mathrm{tr}(F_i \wedge F_i) = \mathrm{d}\omega_i$ with $\omega_i$ the Chern-Simons form

$$\omega_i := \mathrm{tr}(F_i \wedge A_i - \frac{1}{3} A_i \wedge A_i \wedge A_i)$$

Take the two disks as being the hemispheres of an $S^4$, overlapping only at the equator. If we now calculate the chern class again, we find:

$$8\pi^2 k = \int_{D_1} \mathrm{d}\omega_1 + \int_{D_2} \mathrm{d}\omega_2 = \int_{\partial D_1} \omega_1 + \int_{\partial D_2} \omega_2 = \int_{S^3} \omega_1 - \int_{S^3} \omega_2$$

due to Stokes' theorem and different orientation of the hemisphere boundary w.r.t. each other. If we examine the RHs further, we find that

$$k = - \frac{1}{24\pi^2} \int_{S^3} \mathrm{tr}(t\mathrm{d}t^{-1} \wedge t\mathrm{d}t^{-1} \wedge t\mathrm{d}t^{-1})$$

so the $k$ is completely determined by the chosen gauge transformation! As all $k$-vacua have the same value in the action, they are not really different. This means we can already classify an $k$-instanton by giving the gauge transformation $t : S^3 \rightarrow G$. The topologist immediately sees that $t$ is therefore given by choosing an element of the third homotopy group $\pi_3(G)$, since homotopic maps integrate to the same things. For simple Lie group, which we always choose our gauge groups to be, $\pi_3(G) = \mathbb{Z}$, which is a nice result: $t$ is (up to homotopy, which is incidentally the same as up to global gauge transformation here) already defined by the $k$-number of the instanton.

Instantons and tunneling

Now we can see what tunneling between an $N$- and an $N + k$-vacuum might mean:

Take a $[-T,T] \times S^3$ spacetime, that is, a "cylinder", and fill it with a $k$-instanton field configuration $A_k$. This is essentially, by usual topological arguments, a propagator between the space of states at the one $S^3$ to the other $S^3$. If you calculate its partition function, you get a tunneling amplitude for the set of states belonging to $\{-T\} \times S^3$ turning into the set of states belonging to $\{T\} \times S^3$.

Calculate again the Chern class (or winding number or Poyntragin invariant - this thing has more names than cats have lives):

$$8\pi^2 k = \int_{[-T,T] \times S^3} \mathrm{d}\omega = \int_{\{T\}\times S^3} \omega(-T) - \int_{\{-T\}\times S^3} \omega(T)$$

If the $S^3$ represent vacua, the field strength vanishes there and $A(-T),A(T)$ are pure gauge, i.e. $A(\pm T) = t_\pm \mathrm{d} t_\pm^{-1}$, so we have the Chern-Simons form reducing to the Cartan-Maurer form $\omega(\pm T) = \frac{1}{3} t_\pm \mathrm{d} t_\pm^{-1} \wedge t_\pm \mathrm{d} t_\pm^{-1} \wedge t_\pm \mathrm{d} t_\pm^{-1}$. But now the two boundary integrals for the winding number are simply determined by the homotopy class of $t_\pm : \{\pm T\} \times S^3 \rightarrow G$, let's call them $k_\pm$. Therefore, we simply have $k = k_+ - k_-$.

So, we have here that a cylinder spacetime with a $k$-instanton configuration indeed is the propagator between the space of states associated with a spatial slice of a $k-$-instanton and the space of states associated with a spatial slice of a $k_+$-instanton, where $k_\pm$ differ exactly by $k$, so you would get the amplitude from the partition function of that cylinder. To actually calculate that is a work for another day (and question) ;)

• Ah, I guess I was just used to seeing $\Lambda$ used instead of $t$ as the element of the gauge group. Point 2 makes more sense now as well. My last question about your (quite helpful) answer is: how is calculating the Chern class related to the partition function (e.g. I don't see how the $k$ is related to your discussion about the partition function of the transition from $\{-T\} \times S^3 \rightarrow \{T\} \times S^3$). – Lloyd Jul 24 '14 at 16:05
• @Lloyd: It's not related to the partition function directly, but it tells what kind of vacua are at the ends of the propagator - namely those whose winding numbers differ by $k$ from each other. Note that I did just put a $k$-instanton on the cylinder without assuming anything other than "vacuum" about the ends, and got that difference from calculating the Chern class of the cylinder. – ACuriousMind Jul 24 '14 at 16:23
• @ACuriousMind There's a step in this argument I've never understood: you start with describing instantons on $\mathbb{R}^4$ compactified to $S^4$, but then consider an instanton in $X = I \times S^3$ where $I$ is an interval. But $X$ is not topologically the same as $S^4$ unless you demand more stringent boundary conditions, i.e. that the gauge field be constant on the two spatial slices as well. But then that would ruin the argument. – knzhou Jul 27 '18 at 17:15
• @ACuriousMind What step am I missing here? Would it be worth asking a separate question about that? – knzhou Jul 27 '18 at 17:15
• @knzhou I never claim that $S^4$ and the 3-cylinder are the same. I'm just saying that if you put a k-instanton on the 3-cylinder then it mediates between two spatial slices of instantons on $S^4$ (spatial slice is $S^3$) that differ by k. – ACuriousMind Jul 27 '18 at 17:22