For the wave function in Example 2.2, find the expectation value of H, at time t = 0 , the “old fashioned” way:
\left\langle H\right\rangle =\int \Psi(x, 0)^{*} \hat{H} \Psi(x, 0) d xCompare the result we got in Example 2.3. Note: Because \left\langle H\right\rangle is independent of time, there is no loss of generality in using t=0.
\hat{H} \Psi(x, 0)=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}[A x(a-x)]=-A \frac{\hbar^{2}}{2 m} \frac{\partial}{\partial x}(a-2 x)=A \frac{\hbar^{2}}{m} .
\int \Psi(x, 0)^{*} \hat{H} \Psi(x, 0) d x=A^{2} \frac{\hbar^{2}}{m} \int_{0}^{a} x(a-x) d x=\left.A^{2} \frac{\hbar^{2}}{m}\left(a \frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\right|_{0} ^{a} .
=A^{2} \frac{\hbar^{2}}{m}\left(\frac{a^{3}}{2}-\frac{a^{3}}{3}\right)=\frac{30}{a^{5}} \frac{\hbar^{2}}{m} \frac{a^{3}}{6}= \frac{5 \hbar^{2}}{m a^{2}} .
(same as Example 2.3).