In 't Hooft's paper "Computation of the quantum effects of a four-dimensional pseudo-particle" (1976), on page 3434, the author writes the following:

(i) The vacuum-to-vacuum amplitude in the absence of sources must be normalized to $1$, so that the vacuum state has norm $1$. This implies that $W$ must be divided by the same expression with $A^{\textrm{cl}}=0$.

Isn't it standard QFT to divide by the amplitude with the sources set to zero? i.e. with $J_{st}=0$ ? Setting $A^{\textrm{cl}}=0$ doesn't make sense to me, because $A^{\textrm{cl}}$ is set by the classical action.

In standard QFT, the generator of normalized correlation functions is:

$$\frac{Z[J]}{Z[0]}=\sum_{k=0}^{\infty}\,\,\,\sum_{m_1,m_2,\cdots m_k}\frac{1}{m_1!\cdots m_k!}\int d^4x_{1}\cdots d^4x_k\, J(x_1) \cdots J(x_k)\,\langle\Omega|\phi^{m_1}_{1}(x_1)\cdots\phi^{m_k}_{k}(x_k)|\Omega\rangle_{\textrm{normalized}}$$

$$\implies \langle\Omega|\phi^{m_1}_{1}(x_1)\cdots\phi^{m_k}_{k}(x_k)|\Omega\rangle_{\textrm{normalized}}=\frac{\delta^{m_1}}{\delta J(x_1)^{m_1}}\cdots\frac{\delta^{m_k}}{\delta J(x_k)^{m_k}}\frac{Z[J]}{Z[0]}$$

Which correctly gives $\langle \Omega |\Omega\rangle_{\textrm{normalized}}=1$.

I suppose my question can be generalized to, how do you correctly normalize a quantum amplitude in QFT? I really thought that all you do is divide by $\langle \Omega|\Omega\rangle=Z[0]$, i.e. the partition function with $J=0$.

  • $\begingroup$ Link to abstact page? $\endgroup$
    – Qmechanic
    May 1, 2020 at 18:28
  • $\begingroup$ @Qmechanic Oh snap, I thought I had linked it. It's there now. $\endgroup$ May 1, 2020 at 19:11

1 Answer 1


QCD has topological vacua $|n\rangle$ labeled by an integer $n$. 't Hooft is computing (in weak coupling) the transition amplitude $\langle n\pm 1|\exp(-iHt)|n\rangle$, normalized to the vacuum amplitude $\langle n|\exp(-iHt)|n\rangle$. For this purpose he performs a Gaussian integral around the one-instanton configuration, normalized to a Gaussian inergral around the classical vacuum. This is the same strategy that is used in QM when we compute the transition amplitude in the double well potential.

  • $\begingroup$ I was not aware that he was calculating the transition amplitude from $|n\rangle$ to $|n+1\rangle$. I was under the impression that we were simply calculating correlation functions in the Euclidean QFT. Does it even make sense to calculate correlators in the euclidean QFT? $\endgroup$ May 7, 2020 at 16:15
  • $\begingroup$ @ArturodonJuan Yes, you can obviously compute correlation functions in a euclidean QFT and they are related (by analytic continuation) to Minkowski space correlation functions. Also note that in the presence of massless fermions the transition amplitude is a correlation function, of the form $\langle n+1 | det(\bar\psi_L(x)\psi_R(y))|n\rangle$, as also shown by 't Hooft. $\endgroup$
    – Thomas
    May 8, 2020 at 14:15
  • $\begingroup$ Ok I came back to this and now I'm confused about something else. Why is he even calculating the transition amplitude $\langle n\pm 1,t=+\infty | n,t=-\infty\rangle$? What relevance does that have with anything? If we assume our vacuum state is given by $\theta=0$ then our vacuum for all times is $|\Omega\rangle=\sum |n\rangle$. Why don't we use this physical theta-vacuum, and calculate correlators as $\langle \Omega| \psi\bar\psi|\Omega\rangle / \langle \Omega |\Omega\rangle$? $\endgroup$ May 20, 2020 at 13:25
  • $\begingroup$ When we calculate a correlator in QCD, we should be calculating with respect to the true vacuum $|\Omega\rangle=|\theta=0\rangle$. Calculating the transition amplitude $\langle n\pm 1|U(\infty,-\infty )|n\rangle$ is only relevant for a chapter/section on theta vacua in a pedagogical text. I don't see the point of calculating this transition amplitude after the idea of theta vacua has been made clear. $\endgroup$ May 21, 2020 at 13:31
  • $\begingroup$ @ArturodonJuan By the usual arguments (explained by Coleman etc) onservables in theta vacua (incl theta=0) are sensitive to the transition amplitudes. For example, E(theta)=-cos(theta)*tunneling rate. $\endgroup$
    – Thomas
    May 21, 2020 at 20:29

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