- #1
JK423
Gold Member
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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?
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?