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.

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

They key step is the anti-linearity condition.


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