Briefly, the 1-D box problem goes like this: A particle can move between x=0 and x=L. The potential is: V=Infinite ,x=<0 and x>=L. V=0 ,0<x<L. So the wavefunction is zero at x=0 and x=L. Given these boundary conditions we find, by solving the Schroedinger equation, that the wavefunction is: G[x]=sqrt(2/L)*Sin[n*Pi*x/L] We also know that the energy of the particle is quantized: E=hbar^2*Pi^2*n^2/(2mL^2), n=1,2,... And just because E=p^2/2m, it`s momentum will be quantized as well: p=(+/-)n*Pi*hbar/L, n=1,2,... We all know the above. My problem begins now: Let`s take the wavefunction for n=1: G1[x]=sqrt(2/L)*Sin[Pi*x/L] Since the eigenfunctions of momentum constitute a complete system, we can express G1 (and every eigenfunction) as: G1[x]=integral[Cp*F[x], x:-Infinity,...,+Infinity], where Cp are coefficients and F[x]=Exp[i*p*x/hbar]/sqrt[2*Pi*hbar] the normalised eigenfunction of momentum. Using the Fourrier transformation we can solve and find Cp: We`ll find: Cp=sqrt(2*Pi*h)^1/2*Integral[Exp[-i*p*x/h]*sqrt[2/L]*Sin[Pi*x/L]. Doing the math, we`ll find that: |C(p)|^2=4Pi*hbar^3*L*Cos[p*L/2hbar]^2/(p^2*L^2-Pi^2*hbar^2)^2 As we know |C(p)|^2*dp gives as the probability to find momentum between p and p+dp. My question is this: We know that for the wavefunction G1[x] there are 2 possible values for the momentum: p=(+/-)Pi*hbar/L. However! C(p)^2 is a continuous function which means that there is a probability to find ANY value for momentum. There is no restriction to find just the 2 discrete values: p=(+/-)Pi*hbar/L as we found above. Can anyone tell me, why this is happening?