Hard-wall boundary conditions Hard-wall boundary conditions says that wave function = 0 on walls. And you can't really use running wave, must be stationary. How to prove that these boundary conditions leads to Δki=π/L? Where ki is one of wave vector k dekart components.
 A: The Classical wave equation in 1D is the PDE:
$$u_{tt}=cu_{xx}$$
We solve this by separation of variables by assuming:
$$u(x,t)=X(x)\Gamma(t)$$
Inserting into the PDE:
$$X\Gamma''=c\Gamma X''$$
Dividing both sides by $cX\Gamma$ gives:
$$\frac1c\frac{\Gamma''}{\Gamma}=\frac{X''}{X}=-k^2$$
Where $-k^2$ is a separation constant. We then obtain two ODEs of which the latter is:
$$X''(x)+k^2X(x)=0$$
This has the standard solution:
$$X(x)=c_1\cos kx+c_2\sin kx$$
To have standing waves we need the following boundary conditions:
$$u(0,t)=0\implies X(0)=0$$
$$u(L,t)=0\implies X(L)=0$$
The first one yields:
$$c_1=0$$
And with the second boundary condition:
$$X(x)=c_2\sin kx$$
$$X(L)=c_2\sin kL=0$$
$$\implies k_nL=n\pi$$
$$n=1,2,3,...$$
$$\implies k_n=\frac{n\pi}{L}$$
$$X(x)=c_2\sin\Big(\frac{n\pi x}{L}\Big)$$
Now look at the Schrödinger equation, here in one dimension and with zero potential ($V(x)=0$):
$$i\hbar\frac{\partial\Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(x,t)}{\partial t^2}$$
Like the Classical wave equation above, we can solve this by separation of variables by assuming:
$$\Psi(x,t)=\psi(x)\phi(t)$$
In the case of a particle in a box with infinitely high potential walls, the ODE for $\psi(x)$ has the same type boundary conditions as above:
$$\psi(0)=0,\psi(L)=0$$
Which leads to the solution:
$$\psi_n(x)=A_n\sin\Big(\frac{n\pi x}{L}\Big)$$
And:
$$k_n=\frac{n\pi}{L}$$
For:
$$n=1,2,3,...$$
