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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 ${\bar\alpha}\rightarrow \alpha$

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  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 2 '14 at 17:07
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The relation (4) literally switches the states, adds an overall complex conjugate, and removes a hermitian bar over the operator.

(Actually, no one uses bars anymore to denote hermitian conjugates, they use daggers instead. And because stacked bars get ugly, I'll use stars for complex conjugation of plain complex numbers.)

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)."

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  • $\begingroup$ Indeed; I assumed that the adjoint of an adjoint cancelled like the inverse operator. It seems obvious now that the adjoint of an adjoint should initially be assumed to give a different linear operator since its still just a linear operator. $\endgroup$ – Physiks lover Sep 3 '14 at 17:40

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