D.17 In­ner prod­uct for the ex­pec­ta­tion value

To see that $\left\langle\vphantom{A}\Psi\hspace{-\nulldelimiterspace}\hspace{.03em}\right.\!\left\vert\vphantom{\Psi}A\right\rangle $ works for get­ting the ex­pec­ta­tion value, just write $\Psi$ out in terms of the eigen­func­tions $\alpha_n$ of $A$:

\begin{displaymath}
\langle c_1 \alpha_1 + c_2 \alpha_2 + c_3 \alpha_3 + \ldots...
...rt
c_1 \alpha_1 + c_2 \alpha_2 + c_3 \alpha_3 + \ldots\rangle
\end{displaymath}

Now by the de­f­i­n­i­tion of eigen­func­tions $A\alpha_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_n\alpha_n$ for every $n$, so you get

\begin{displaymath}
\langle c_1 \alpha_1 + c_2 \alpha_2 + c_3 \alpha_3 + \ldots...
...\alpha_1 + c_2 a_2 \alpha_2 + c_3 a_3 \alpha_3 + \ldots\rangle
\end{displaymath}

Since eigen­func­tions are or­tho­nor­mal:

\begin{displaymath}
\langle\alpha_1\vert\alpha_1\rangle = 1 \quad
\langle\alph...
... 1\quad
\langle\alpha_3\vert\alpha_3\rangle = 1 \quad
\ldots
\end{displaymath}


\begin{displaymath}
\langle\alpha_1\vert\alpha_2\rangle =\langle\alpha_2\vert\a...
...ha_3\rangle =\langle\alpha_3\vert\alpha_2\rangle =
\ldots = 0
\end{displaymath}

So, mul­ti­ply­ing out pro­duces the de­sired re­sult:

\begin{displaymath}
\langle \Psi\vert A \Psi\rangle =
\vert c_1\vert^2 a_1 + \...
... c_3\vert^2 a_3 + \ldots \equiv
\left\langle{A}\right\rangle
\end{displaymath}