1. Jan 11, 2008

### JK423

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, its momentum will be quantized as well:
p=(+/-)n*Pi*hbar/L, n=1,2,...

We all know the above. My problem begins now:
Lets 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:
Well find:
Cp=sqrt(2*Pi*h)^1/2*Integral[Exp[-i*p*x/h]*sqrt[2/L]*Sin[Pi*x/L].

Doing the math, well 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?

2. Jan 11, 2008

### Mute

So, it's a little hard to tell from the way you've expressed all of your equations in text, but, I think what you're saying is:

You have some wavefunction for the ground state:

$$\psi_1(x) = \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L} \right)$$

which you want to expand in a fourier series

$$\psi (x) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{0}^{\infty} dp \phi (p) e^{i p x / \hbar}$$

where

$$\phi (p) \propto \int_{-\infty}^{\infty} dx \psi(x) e^{-i p x/\hbar}$$

In doing this, however, you've already implicitly assumed that p is a continuous variable by writing $\psi(x)$ as a fourier integral instead of a Fourier Series:

$$\psi(x) = \sum c_n e^{i x p_n/\hbar}$$

where $p_n[/tex] is discrete, as expected. When [itex]\psi$ is just the ground state function, all c_n but c_1 should vanish, and so |c_n|^2 will correspond to the probablilty of finding the system the n = 1 eigenstate.

Last edited: Jan 11, 2008
3. Jan 11, 2008

### George Jones

Staff Emeritus
The wavefunction isn't quite right, i.e., it should be something like

$$\psi_1(x) = \left( H\left(x\right) - H\left(x - L\right)\right)\sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L} \right),$$

where $H$ is the Heaviside step function, so that $\psi_1$ is zero outside the box.

Now integrate by parts, differentiating the step functions to get delta functions.

4. Jan 11, 2008

### JK423

Mute thanks for writing down the equations in more proper way..
Why should i hypothesize that the momentum is discrete? The math should show me that..

Mr Jones, if we replace the wavefunction $$\psi_1(x) = \left( H\left(x\right) - H\left(x - L\right)\right)\sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L} \right),$$
at the equation $$\phi (p) \propto \int_{-\infty}^{\infty} dx \psi(x) e^{-i p x/\hbar}$$ , the limits of the integral will change from [-00,+00] to [0,L] and the Heaviside function will disappear. So we will get $$\phi (p)$$ which will be equal to C(p) i wrote at the first post.
No delta function and the spectrum is still continuous.
Where am i mistaken?

Even if i get the delta function inside the equation still doesnt help...

5. Jan 11, 2008

### George Jones

Staff Emeritus
I wrote this off the top of my head without actually doing a calculation, so I don't know if it works.

I think you might find a mathematically careful treatment in

http://arxiv.org/abs/quant-ph/0103153

6. Jan 11, 2008

### Mute

But you did get the fact that the momentum is discrete from the math! You said it right here:

The fact that the energy is discrete comes from the solution to the time independent Schrodinger Equation and your boundary conditions. The boundary conditions impose the discreteness of E, which inside the box is equal to the particle's kinetic energy $p^2/2m$, and hence the momentum is quantized as well, just like you originally supposed.

This tells you that the momentum is also discrete, so if you are going to use momentum as the basis states to expand the wavefunction, you must use a Fourier Series instead of a Fourier integral, as the Fourier integral requires your basis states be continuous.

What I was saying above was that the moment you wrote down $\psi(x)$ as a Fourier Integral, you were implicitly assuming that momentum is continuous, and hence you of course will find in the end that momentum is continuous. If you had written $\psi(x)$ as a Fourier series instead, you would have had to conclude that momentum was discrete.

So, the fact that momentum was discrete comes from the fact that energy is discrete, like you had already found, and that tells you which kind of Fourier expansion to use.

7. Jan 11, 2008

### JK423

Thanks for the contribution guys.

I found a paper that discusses exactly this problem.
This trully is a paradox afterall, and it has a name: Einstein-Pauli-Yukawa paradox.
The article is very interesting and can be found here:
http://arxiv.org/pdf/quant-ph/9803001