I'm currently working on a Quantum Mechanics problem and am stuck. The problem I'm working on is 6.7 in Griffiths Introduction to Quantum Mechanics. The problem is:
Consider a particle of mass $m$ that is free to move in a one-dimensional region of length $L$ that closes on itself (for instance, a bead that slides frictionlessly on a circular wire of circumference $L$.
I'm stuck on part C. To save some space, I'll simply write the results of parts A and B, and my work for part C. Part C states.
What are the "good" linear combinations of $\psi_n$ and $\psi_{-n}$ for this problem? (Hint: Use equation 6.22) Show that with these states you get the first-order correction for the energy.
So, I understand how to do the second part of the question. My issue is finding the "good" linear combination. Here's my work so far, I believe I'm pretty close to the answer.
From parts A and B, I have the following equations: $$ E_{\pm} = -\frac{V_0}{L}a\sqrt \pi \bigg( 1 \mp e^{-(2 \pi a^2 /L)^2} \bigg) \\ W_{aa} = -\frac{V_0}{L}a\sqrt \pi \\ W_{ab} = -\frac{V_0}{L}e^{-(2\pi a^2 / L)^2} $$
Equation 6.22 states: $$ \alpha W_{aa} +\beta W_{ab} = \alpha E^1 $$ Where $E^1$ denotes the first order correction to the energy.
So, using these equations, I solved for $\beta$. To save some algebra, I'll skip a few steps here. I simply plugged in all the known equations into equation 6.22 and chose to solve for $\beta$. So, $$ \beta = \alpha \frac{(E^1 - W_{aa})}{W_{ab}} = \mp \alpha $$
Now, for the part I'm confused about. Part A gave me the following wave functions:
$$ \psi_a = \frac{1}{\sqrt L} e^{2 \pi i n x / L} \\ \psi_b = \frac{1}{\sqrt L} e^{-2 \pi i n x / L} $$
So, my goal with the following work is to find the wave function for when $\alpha$ is positive and when it is negative, as we have two results ($\beta = \mp \alpha$). So, starting out with $\beta = -\alpha$,
$$ \begin{align} \psi_+ (x) &= \alpha\psi_a + \beta\psi_b \\ &=\alpha\psi_a - \alpha\psi_b \\ &= \frac{\alpha}{\sqrt L} \bigg( e^{i2\pi n x /L} - e^{-i 2\pi n x /L} \bigg) \\ &=\frac{2i\alpha}{\sqrt L} \sin{\bigg( \frac{2n\pi x}{L} \bigg)} \end{align}$$
Now, the solution to this problem suggests that $\alpha = 1/\sqrt 2$, making the final result $$ \psi_+ (x) =i\sqrt{\frac{2}{L}} \sin{\bigg( \frac{2n\pi x}{L} \bigg)} $$
I'm not sure how to find $\alpha$ here. My final result, for the positive case, is correct barring the value for $\alpha$. Any advice would be appreciated, thank you.
Normalization should do the trick.
$\int^{L}_0|\psi(x)|^2dx=1$