# How does Dirac show that $\langle B|\bar{\bar{\alpha}}|P\rangle\;=\; \overline{\langle P|{\bar{\alpha}}|B\rangle}\;=\; \langle B|{\alpha}|P\rangle$?

Dirac shows that the conjugate imaginary of $\langle \!P|\alpha$ is $\bar{\alpha} |P\!\rangle$ and then starts with the identity on page 27 in his book:

$$\langle B|\bar{{\alpha}}|P\rangle\;=\; \overline{\langle P|{{\alpha}}|B\rangle}\tag {4}$$

He then says this expression is true for any linear operator $\alpha$ and ket vectors$|P\!\rangle$ and $|B\!\rangle$; so replacing $\alpha$ with $\bar\alpha$ we get

$$\langle B|\bar{\bar{\alpha}}|P\rangle\;=\; \overline{\langle P|{\bar{\alpha}}|B\rangle}\;=\; \langle B|{\alpha}|P\rangle$$

by using (4) again with $|P\!\rangle$ and $|B\!\rangle$ interchanged.

Why should this give the second equality?

If (4) is applied again, I would expect ${\bar\alpha}\rightarrow \bar{\bar\alpha}$ getting back to the LHS expression, yet Dirac has ${\bar\alpha}\rightarrow \alpha$

• It's just hermitian conjugation. It obeys $(ABC\dots Z)^\dagger = Z^\dagger Y^\dagger \dots C^\dagger B^\dagger A^\dagger$. The Hermitian conjugation of a $c$-number $\alpha$ is simply the complex conjugate $\bar \alpha$, and adding the bar (or dagger) twice is like erasing both. Sep 2, 2014 at 17:07

Thus we have $$\langle B \mid (\alpha^\dagger)^\dagger \mid P \rangle = \left(\langle P \mid \alpha^\dagger \mid B \rangle\right)^* = \left(\langle B \mid \alpha \mid P \rangle^*\right)^* = \langle B \mid \alpha \mid P \rangle.$$ Since this holds for any $\lvert B \rangle$, $\lvert P \rangle$, and $\alpha$, this shows in a roundabout way that $\alpha^{\dagger\dagger} = \alpha$ for any $\alpha$. I think what's confusing you is that you assumed such an obvious fact before Dirac did, and you thought (4) meant "switch states, add an overall complex conjugate, and add a hermitian bar over the operator (which may cancel with one already there)."