Undergrad Why is the QCD Lagrangian often abbreviated?

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The QCD Lagrangian is often abbreviated by omitting the term involving the theta parameter, which is represented as ##\frac{\theta g^2}{32\pi^2}G^{a}_{\mu\nu}\widetilde{G}^{a\mu\nu}##. This omission occurs because observationally, ##\theta = 0## to experimental precision, and including this term would imply CP violation, which has not been observed. The discussion highlights the strong CP problem and mentions the Peccei-Quinn mechanism as a potential solution to this issue.

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The QCD Lagrangian is
##\mathcal{L}=-\frac{1}{4}G^{a}_{\mu\nu}G^{a\mu\nu}+\sum\limits_{j=1}^n \left[\bar{q}_j\gamma^{\mu}iD_{\mu}q_j - (m_jq^{\dagger}_{Lj}q_{Rj}+h.c.)\right]+\frac{\theta g^2}{32\pi^2}G^{a}_{\mu\nu}\widetilde{G}^{a\mu\nu}##
Why is it so often quoted as just
##\mathcal{L}=-\frac{1}{4}G^{a}_{\mu\nu}G^{a\mu\nu}+\sum\limits_{j=1}^n \left[\bar{q}_j\gamma^{\mu}iD_{\mu}q_j - (m_jq^{\dagger}_{Lj}q_{Rj}+h.c.)\right]##?
I've seen both and I'm assuming the longer one is more complete somehow, but in those cases where the short version is being used, there's not even a mention of the missing term.
 
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Observationally, ##\theta = 0## to experimental precision. The presence of the last term would introduce CP violation that is not observed. Why this is the case is known as the strong CP problem. One possible solution is the Peccei-Quinn mechanicsm.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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