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  Doubts about the theta angle and the ground state energy density in Euclidean Yang-Mills theory

+ 3 like - 0 dislike
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I am reading the following notes https://munsal.files.wordpress.com/2014/10/marino-lectures2014.pdf. On section 4.3 the euclidean Yang-Mills theory is considered. It is said that renormalizability and gauge invariance allow the following Lagrangian form
$$
\mathcal{L}_{\theta}=\mathcal{L}_{YM}-i\theta{}q(x)
$$
where
$$
q(x)=\frac{1}{32\pi^2}F_{\mu\nu}^a\tilde{F}^{a\mu\nu}
$$
where we have the usual YM lagrangian and the theta term. In equation (4.54) he defines what he calls the partition function in the presence of the $\theta$ angle as
$$
Z(\theta)=\int[\mathcal{D}A]e^{-\int{}d^4x\mathcal{L}}
$$
he then says that he will write this expression above as an exponential
$$
Z(\theta)=\int[\mathcal{D}A]e^{-\int{}d^4x\mathcal{L}}=e^{-VE_V(\theta)}
$$
where $V$ is the volume of the space. I need more motivation dor this step. Why can we write this as an exponential? He then says that the ground state energy density is given by
$$
\lim_{V\to\infty}E_V(\theta)=E(\theta)
$$
it is completely far from obvious why this should be the ground state energy. What  is behind this?

asked Jun 20, 2016 in Theoretical Physics by anonymous [ no revision ]

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