# Application of CPT invariance : some trivial algebra

I am having some problem in understanding one step in the following algebra.

Consider an interaction where initial state is defined as $\left|i\right>$ and final state by $\left|f\right>$. Now, $$\left|i\right> = \mathcal{ CPT}\left | \bar{i}\right>$$ $$\left|f\right> = \mathcal{ CPT} \left| \bar{f}\right>$$ Using the CPT invariance condition, $\left(\mathcal{ CPT} \right)T \left(\mathcal{ CPT}\right)^{-1}= T^{\dagger}$, where $T$ is the transition matrix; $$\left<f|T^{\dagger}|i\right> = \left<\bar{f}|T|\bar{i}\right>^{*}$$

Please show explicitly how to derive the last equation.

• Isn't the invariance condition for $T$ that $(\mathcal{CPT})T\left(\mathcal{T^{-1}P^{-1}C^{-1}}\right) = T^\dagger$? – rob Sep 28 '15 at 12:24
• @rob This condition says, that Transition matrix is invariant under $\mathcal{CPT}$. – seeking_infinity Sep 28 '15 at 13:01
• Well if $$\left(\mathcal{ CPT} \right)T \left(\mathcal{ CPT}^{-1}\right)= T^{\dagger}$$ Then the converse $$\left(\mathcal{ CPT} \right)T^{\dagger} \left(\mathcal{ CPT}^{-1}\right)= T$$ is true too right? – Horus Sep 28 '15 at 15:44
• What I am asking is, how to get to the last step. – seeking_infinity Sep 28 '15 at 17:34
• @seeking_infinity Your edit is better. Inverse of a product of operators is the product of the inverses in the reverse order. – rob Sep 28 '15 at 18:46

So let's start from the relations you gave and transform one of them from ket to bra. $$\left|i\right> = \mathcal{ CPT}\left | \bar{i}\right>$$ $$\left<f\right| = \left< \bar{f}\right| (\mathcal{ CPT})^{\dagger}$$ Using the CPT invariance condition, $\left(\mathcal{ CPT} \right)T \left(\mathcal{ CPT}\right)^{-1}= T^{\dagger}$,
It is easy to show that: $$\left<f|T^{\dagger}|i\right> = \left<\bar{f}|(\mathcal{CPT})^{\dagger}(\mathcal{ CPT})T|\bar{i}\right>$$
Then by the anti-linearity of $\mathcal{CPT}$, $\left<\bar{f}|(\mathcal{CPT})^{\dagger}(\mathcal{ CPT})T|\bar{i}\right> = \left<\bar{f}|(\mathcal{CPT})(\mathcal{ CPT})T|\bar{i}\right>^*$
Since $\mathcal{CPT}^2$ = $\mathcal{I}$
$\left<\bar{f}|(\mathcal{CPT})(\mathcal{ CPT})T|\bar{i}\right>^* = \left<\bar{f}|T|\bar{i}\right>^*$