The vacuum state of QCD is a superposition of different ground states with non-tirvial topological charge $\propto n\in \mathbb{Z}$. We lable this vacuum configurations with $|{n}\rangle$. The transition amplitude is given by:

$$ \langle n|m\rangle = \int \mathcal{DA}_{n-m}\ \ \text{exp} (-iS) $$

where $\mathcal{A}_{n-m}$ denots the gauge field configuration to the topological charge $n-m$.

Maybe I missed something but I don't understand why this transition amplitude depends only on the difference $n-m$? Shouldn't one calculate the the vacuum-to-vacuum transition amplitude by the generating functional (including all configurations)?

  • $\begingroup$ $DA_{n-m}$ is funny (and unclear) notation, but it is clear that the transition amplitude only depends on $n-m$, not on $n,m$ separately. Of course you have to integrate over all configurations that start in $n$ and end in $m$. $\endgroup$
    – Thomas
    Dec 17, 2017 at 0:10

1 Answer 1


The path integral starts from a configuration $A^{(m)}$ of a topological charge $m$ and ends in a configuration $A^{(n)}$ of topological charge $n$.

The reason that it depends only on the difference is that both the QCD Lagrangian and the path integral measure are invariant under large gauge transformation, i.e., gauge transformation with nonvanishing winding number: $$Q = \frac{1}{24 \pi^2} \int \mathrm{Tr}\left((g^{-1}dg)^3\right)$$ If we change the integration variables by performing a gauge transformation $$ g\cdot A = g^{-1}Ag + g^{-1} dg$$ The Lagrangian will not change, while the boundary configurations will become of topological charges $m+Q$ and $n+Q$, respectively. Since a transformation of the integration variables does not change the result. The final value should depend only on the difference $m-n$.


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