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  QCD symmetry breaking and pseudoscalar mass

+ 2 like - 0 dislike

Suppose we have action
S  = \int d^{4}x\left(\frac{\theta (x)}{f_{\theta}}G\tilde{G} + L_{SM} + \frac{1}{2}(\partial_{\mu}\theta )^{2} \right), \qquad (1)

where $G$ denotes SM gluon field strength, $\theta$ denotes pseudoscalar field, and $L_{SM}$ denotes Standard model lagrangian.

How to see that $\theta$ field acquires mass due to nontrivial vacuum expectation value of $G\tilde{G}$ during QCD phase transition? I know the standard trick with performing the chiral transformation $q \to e^{i\gamma_{5}C\frac{\theta}{f_{\theta}}}$, but I want to know how to get the mass term directly from the first summand of $(1)$.

Also, I see that for shift transformation $\theta \to \theta + f_{\theta}c$ modification of action is

\Delta S = c\int d^{4}x G\tilde{G}\to  c i\int d^{4}x_{E}G_{E}\tilde{G}_{E} = i\frac{16 \pi n}{\alpha_{s}}c \Rightarrow e^{iS} \to e^{iS}e^{-\frac{16 \pi n c}{\alpha_{s}}},$$

so I don't understand why people claim that the first summand in $(1)$ leaves discrete symmetry $\theta \to \theta + 2\pi kf_{\theta}$. It can be demonstrated directly by computing of pseudoscalar effective potential, but I can't see this from first principles.

asked Dec 4, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Dec 5, 2015 by NAME_XXX

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