# 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 $\{-T\} \times S^3 \rightarrow \{T\} \times S^3$). Commented Jul 24, 2014 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. Commented Jul 24, 2014 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. Commented Jul 27, 2018 at 17:15
• @ACuriousMind What step am I missing here? Would it be worth asking a separate question about that? Commented Jul 27, 2018 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. Commented Jul 27, 2018 at 17:22