# Understanding a simple remark in 't Hooft's "Computation of the quantum effects of four-dimensional pseudo-particle"

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:

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

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

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 $$\langle \Omega|\Omega\rangle=Z$$, i.e. the partition function with $$J=0$$.

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

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.
• 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? May 7, 2020 at 16:15
• @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. May 8, 2020 at 14:15
• 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 $\langle \Omega| \psi\bar\psi|\Omega\rangle / \langle \Omega |\Omega\rangle$? May 20, 2020 at 13:25
• 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. May 21, 2020 at 13:31