[Physics] Interpreting Schrödinger Equation’s solution for a free particle

Question

Let's say we have the time-independent Schrödinger Equation for a free particle (without potential):
$$-\frac{\hbar}{2m} \frac{d^2 \psi}{dx^2} = E\psi$$
Whose general solution is:
$$\psi(x) = Ae^{ikx} + Be^{-ikx}$$
Where $k = \frac{2\pi}{\lambda}$ (i am supposing just one dimension, hence the use of a scalar quantity). Let's say that i have the following contour values: $\psi(0) = 0$ and $\psi(L) = 0 $ (First Question: What is the meaning of this? I know that this means that there is no probability of finding the particle in either $x=0$ or $x=L$. But how does this relate physically to the system and why can i impose this? A particle with no force shouldn't be an uniform distribution over $[0,L]$? (with me not being allowed to set these contour values). To help solving the contour value, i will put the solution in real form:
$$\psi(x) = A\cos(kx) + B\sin(kx)$$
Using the first condidtion:
$$\psi(0) = 0 \rightarrow A\cos(0) + B\sin(0) = 0 \rightarrow \boxed{A = 0}$$
Now for the second condition:
$$\psi(L) = 0 \rightarrow B\sin(kL) = 0 \rightarrow kL = \Big(n+\frac12\Big)\pi$$
$$ k = \frac{\big(n + \frac12 \big)\pi}{L}$$
Finally, the solution is:
$$\psi(x) = B \sin \Big[\frac{(n+\frac12)\pi}{L}x\Big]$$
Now, the second question is: What is the meaning of $n$? If i choose, for example, $n=1$. What does that mean? Why does the potential and the contour values can't tell us how the system will evolve? I know that $n$ is analogous to the "harmonics" of a wave. But i can't understand how does that relate to a particle in QM.

Answer 1

The difficulty with your question is that, by forcing time-independent conditions such as $\psi(0)=\psi(L)=0$, the solutions become standing waves rather than plane waves. To be explicit, you can restore full time dependence $$ \Psi(x,t)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)} $$ and see that the $e^{i(kx-\omega )t}$ and $e^{-i(kx+\omega t)}$ correspond to two plane waves travelling in different directions, with momentum $p=\hbar k$ in reverse directions. Thus your real form $\psi(x)\sim \sin(kx)$ does not describe a free particle since it does not correspond to a single value of momentum. Since this solution is not for a free particle, it cannot be expected to have a uniform probability density.

In a more mathematical language, the solution in terms of sine is not an eigenfunction of the momentum operator $-i\hbar d/dx$.

Once you have standing waves you have (loosely speaking) quantization of energy values, with only a selected number of energies allowed. In your case, your standing waves must have a wavelength that is a multiple of $L/2$. This implies some conditions on $\vert k\vert$ and, since $p=\hbar k$, some conditions on $p^2$ and thus on the allowed energy.