Non-Degeneracy and Reality of Bound States in 1D

Source: Cambridge Quantum Mechanics, Example Sheet 1, Question 4; accompanies David Tong’s notes.

Consider a particle in one dimension in a potential $U(x)$ that tends to zero rapidly as $x \to \pm\infty$.

Let $\chi_1(x)$ and $\chi_2(x)$ be normalisable energy eigenfunctions of the Hamiltonian with the same energy eigenvalue $E$. By considering $\chi_1 \chi_2' - \chi_2 \chi_1'$, show that $\chi_1$ and $\chi_2$ must be proportional to one another. What does this mean, physically?
Can the wavefunction for a normalised energy eigenstate always be chosen to be real?
Show that if $\chi(x)$ is any normalised energy eigenstate then $\langle \hat{p} \rangle_\chi = 0$.

Consider the Wronskian, \[W(x) = \chi_1\chi_2'-\chi_2\chi_1'.\tag{1.1}\] Note that \[W'(x) = \chi_1\chi_2''-\chi_2\chi_1''.\tag{1.2}\] Since both $\chi_1$ and $\chi_2$ satisfy the Schrödinger equation with the same eigenvalue, $W'(x)$ vanishes. Therefore, we have $W(x) = \text{const}$. Since the wavefunctions are normalisable, we can only have $W(x)=0$. Thus, the two wavefunctions are linearly dependent, which means they are the same eigenfunction to the Hamiltonian.

Given that $\chi$ satisfies \[H\chi=E\chi,\tag{1.3}\] take the complex conjugate on both side, we find \[H\chi^*=E\chi^*.\tag{1.4}\] As we have shown that $\chi$ and $\chi^*$ are linearly dependent, we can always construct a real eigenfunction for the eigenvalue, $E$, namely \[\psi = \chi + \chi^*.\tag{1.5}\]

The expectation value for the momentum operator is given by ($\chi$ taken to be real) \[\langle p\rangle = -i\hbar\int dx \chi\chi' = -\frac{i\hbar}{2}\int dx (\chi^2)' = -\frac{i\hbar}{2}[\chi^2(+\infty) - \chi^2(-\infty)],\tag{1.6}\] which must vanish for normalisable $\chi$.