# Why does Dirac write $\langle\xi'|\overline{f(\xi)} = \overline f(\xi ')\langle\xi'|$?

Starting on page 41 of Dirac's The Principles of Quantum Mechanics, he defines $f(\xi)$ in general to be that linear operator which satisfies

$$f(\xi)|\xi'\rangle = f(\xi')|\xi'\rangle\tag {34}$$

for every eigenket $|\xi'\rangle$ of the real dynamical variable $\xi$, $f(\xi')$ being a number for each eigenvalue $\xi'$

He then defines the conjugate complex $\overline{f(\xi)}$ of ${f(\xi)}$ by the conjugate imaginary equation to (34) as

$$\langle\xi'|\overline{f(\xi)} = \overline f(\xi ')\langle\xi'|$$

holding for any eigenbra $\langle\xi'|$, $\overline f(\xi')$ being the conjugate complex function to $f(\xi')$

Why doesn't he write $\overline f(\xi ')$ as $\overline {f(\xi ')}$?

• Looks to me that $\bar{f}(\xi')$ is the eigenvalue of the operator $\overline{f(\xi)}$ on $\langle\xi'\vert$. The difference in the lengths of the bars signifies the difference between the eigenvalue & the operator. – Kyle Kanos Oct 5 '14 at 16:59
• @KyleKanos Dirac uses $\xi$ for a real linear operator, ' to label objects connected with eigenvalues-- $\xi'$ for an eigenvalue, $\langle\xi'|$ for an eigenbra – Physiks lover Oct 5 '14 at 18:18
• @KyleKanos Why doesn't Dirac just substitute $\xi'$ into $\overline{f(\xi)}$ as for (34)? – Physiks lover Oct 5 '14 at 21:21
• I am not sure the reason, that seems like a question for Dirac ;). – Kyle Kanos Oct 5 '14 at 21:31
• I also agree with the answer of Nikos M. below. If you want to go further into deep & unchartered territories ;-), this link might be useful. As a gist, since $\xi$ is an observable, its eigenvalues $\xi^{'}$ must be real. In that case, its easy to see $\overline{f(\xi^{'})} = \overline{f}(\xi^{'})$ – jayann Oct 6 '14 at 8:52