Question 2.3: Check that Equation 2.20 is satisfied, for the wave function...

The starting wave function (Figure 2.3) closely resembles the ground state (Figure 2.2).
This suggests that \left|c_{1}\right|^{2} should dominate,  and in fact

\left|c_{1}\right|^{2}=\left(\frac{8\sqrt{15} }{\pi^{3}} \right)^{2}=0.998555 ……

The rest of the coefficients make up the difference:

\sum\limits_{ n=1}^{\infty}{\left|c_{n} \right|^{2} } =\left(\frac{8\sqrt{15} }{\pi^{3}} \right)^{2} \sum\limits_{n=1,3,5,…}^{\infty } {\frac{1}{n^{6}} =1}.

The most likely outcome of an energy measurement is E_{1}=\pi^{2}\hbar ^{2}/2ma^{2} more than 99.8% of all measurements will yield this value. The expectation value of the energy (Equation 2.21) is

<H>=\sum\limits_{n=1}^{\infty}{|c_n|^2 E_n}         (2.21)

<H>=\sum\limits_{n=1,3,5,…}^{\infty } \left(\frac{8 \sqrt{15} }{n^{3}\pi^{3}} \right)^{2} \frac{n^ {2}\pi^{2}\hbar ^{2}}{2ma^{2}} =\frac{480\hbar ^{2}}{\pi^{4}ma^{2}} \sum\limits_{n=1,3,5,…} ^{\infty } {\frac{1}{n^{4}} }=\frac{5\hbar ^{2}}{ma^{2}}

As one would expect, it is very close to E_{1} (5 in place of \pi^{2}/2\thickapprox 4.935) slightly larger, because of the admixture of excited states.