Normalisation and Hermitean conjugates

If you look at the expression {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}f{(y)}^{∗}{\hat{a}}^{†}g(y)\kern 1.66702pt dy and use the explicit form {\hat{a}}^{†} = {1\over \sqrt{2}}\left (y − {d\over dy}\right ), you may guess that we can use partial integration to get the operator acting on f,

\begin{eqnarray} {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}f{(y)}^{∗}{\hat{a}}^{†}g(y)\kern 1.66702pt dy&& %& \\ & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}f{(y)}^{∗} {1\over \sqrt{2}}\left (y − {d\over dy}\right )g(y)\kern 1.66702pt dy%& \\ & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞} {1\over \sqrt{2}}\left (y + {d\over dy}\right )f{(y)}^{∗}g(y)\kern 1.66702pt dy%& \\ & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{[\hat{a}f(y)]}^{∗}g(y)\kern 1.66702pt dy\qquad . %&(9.21) \\ \end{eqnarray}

This is the first example of an operator that is clearly not Hermitean, but we see that \hat{a} and {\hat{a}}^{†} are related by “Hermitean conjugation”. We can actually use this to normalise the wave function! Let us look at

\begin{eqnarray}{ O}_{n}& =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{\left [{\left ({\hat{a}}^{†}\right )}^{n}{e}^{−{y}^{2}∕2 }\right ]}^{∗}{\left ({\hat{a}}^{†}\right )}^{n}{e}^{−{y}^{2}∕2 }\kern 1.66702pt dy %& \\ & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{\left [\hat{a}{\left ({\hat{a}}^{†}\right )}^{n}{e}^{−{y}^{2}∕2 }\right ]}^{∗}{\left ({\hat{a}}^{†}\right )}^{n−1}{e}^{−{y}^{2}∕2 }\kern 1.66702pt dy%&(9.22) \\ \end{eqnarray}

If we now use \hat{a}{\hat{a}}^{†} ={ \hat{a}}^{†}\hat{a} +\hat{ 1} repeatedly until the operator \hat{a} acts on {u}_{0}(y), we find

Question: Show that this agrees with the normalisation proposed in the previous study of the harmonic oscillator!

Question: Show that the states {u}_{n} for different n are orthogonal, using the techniques sketched above.

9.4 Normalisation and Hermitean conjugates