D.17 Inner product for the expectation value

To see that $\big\langle\Psi\big\vert A\big\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 + \ldot...
...
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 + \ldot...
...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\al...
...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\...
..._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 +...
...rt c_3\vert^2 a_3 + \ldots \equiv
\big\langle A\big\rangle
\end{displaymath}