Expectation value position of sine wave in infinite square well

Question

Given a neutron (mass$\approx$939 MeV/c$^2$) in an infinite square well of size $a$, the value of the expectation value for position should be in the range $[0-a]$. I know that the general form of the expectation value for position is $$\langle X\rangle=\int_{-\infty}^{\infty}\psi^*x\psi dx=\int^a_0\psi^*x\psi dx \, ,$$ My wave function is given by: $$\psi[x]=\sqrt[]{\frac{2}{7a}}\sin{\frac{x\pi}{a}}+\sqrt[]{\frac{4}{7a}}\sin{\frac{2x\pi}{a}}+\sqrt[]{\frac{8}{7a}}\sin{\frac{3x\pi}{a}}\, ,$$ which is a superposition of wavefunctions of the form $\sin(n\pi x / a)$. Because all of these $\sin$ functions are orthogonal, the expectation value can be written: $$\sum_n[p_n\int^a_0\phi_n^*x\phi_ndx]$$ For the probabilities of each $n$ state given by $p_n$. However, for all $n$, the above integrals all evaluate to $\frac{a^2}{4}$, which not only has the wrong units, but for $a>4$ gives a magnitude larger than the size of the well. How can this be?

EDIT: this was poorly worded, and under-explained due to a mixture of pressure and lack of sleep (not that anyone on stack exchange cares). I think I've fixed it.

Answer 1

The problem I was having is that the integral must be taken of the normalized wavefunction, not the orthonormal wavefunction. It's incorrect to multiply by $p_n$, that's not what the expectation value entails. Specifically, for this case: $$\Psi[x]=\sqrt[]{\frac{4}{7a}}\sin[\frac{\pi x}{a}]+\sqrt[]{\frac{2}{7a}}\sin[\frac{2\pi x}{a}]+\sqrt[]{\frac{8}{7a}}\sin[\frac{3\pi x}{a}]\, .$$ Note that the coefficients are neither the probabilities of each state, nor their square, they are simply the normalization coefficients. Because $$\Psi\in\mathbb{R}\, ,$$ The expectation value evaluates to: $$\langle X\rangle=\int_0^ax\Psi^2dx\\ ~\\ =\frac{4}{7}a\left(\frac{7}{8}-\frac{8(54+25\sqrt[]{2})}{255\pi^2}\right)\\ ~\\ \approx0.316054\space a\, ,$$ within the box, more or less what you'd expect.