Briefly, the 1-D box problem goes like this:(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# 1-D Box. Paradox?

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