Inside the box, the particle is free, so the wavelength is h/p, ?(x) = sin kx, but the probability has to be zero outside the box-a boundary condition!! Prob (x) = |?(x)|2 = 0 at edges of the box, and our practice with strings stands us in good stead here, and we see that sin k0 =0 (no information) and at the other end sin kL = 0 -> kL = n? ->
Inside the box, the particle is free, so the wavelength is h/p, ?(x) = sin kx, but the probability has to be zero outside the box-a boundary condition!! Prob (x) = |?(x)|2 = 0 at edges of the box, and our practice with strings stands us in good stead here, and we see that sin k0 =0 (no information) and at the other end sin kL = 0 -> kL = n? ->
k = n?/L or 2 ?/? = n ?/L or ? = 2L/n
much like the string. But this now gives us the momentum!! From ? = h/p, p = (nh)/(2L)
and there is a smallest momentum h/(2L) which we call the “zero point energy” E=p˛/2m=(n˛h˛)/(8mL˛)