Exploring the Paradox of the 1-D Box: Solving for Eigenfunctions and Momentum

In summary, the 1-D box problem involves a particle moving between x=0 and x=L with a potential that is infinite at x=<0 and x>=L, and zero at 0<x<L. The wavefunction for this problem is found to be G[x]=sqrt(2/L)*Sin[n*Pi*x/L] by solving the Schroedinger equation with the given boundary conditions. It is also known that the energy and momentum of the particle are quantized, with the energy given by E=hbar^2*Pi^2*n^2/(2mL^2) and the momentum given by p=(+/-)n*Pi*hbar/L. However, when attempting to expand the wavefunction in a Fourier
  • #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?
 
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  • #2
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:

[tex]\psi_1(x) = \sqrt{\frac{2}{L}} \sin \left(\frac{\pi x}{L} \right)[/tex]

which you want to expand in a Fourier series

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


where

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

(which is your C_p).

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

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

where [itex]p_n[/tex] is discrete, as expected.

When [itex]\psi[/itex] 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.
 
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  • #3
The wavefunction isn't quite right, i.e., it should be something like

[tex]\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),[/tex]

where [itex]H[/itex] is the Heaviside step function, so that [itex]\psi_1[/itex] is zero outside the box.

Now integrate by parts, differentiating the step functions to get delta functions.
 
  • #4
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 [tex]\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),[/tex]
at the equation [tex]\phi (p) \propto \int_{-\infty}^{\infty} dx \psi(x) e^{-i p x/\hbar}[/tex] , the limits of the integral will change from [-00,+00] to [0,L] and the Heaviside function will disappear. So we will get [tex]\phi (p)[/tex] 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 doesn't help...
 
  • #5
JK423 said:
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 [tex]\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),[/tex]
at the equation [tex]\phi (p) \propto \int_{-\infty}^{\infty} dx \psi(x) e^{-i p x/\hbar}[/tex] , the limits of the integral will change from [-00,+00] to [0,L] and the Heaviside function will disappear. So we will get [tex]\phi (p)[/tex] 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 doesn't help...

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

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,...

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 [itex]p^2/2m[/itex], 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 [itex]\psi(x)[/itex] 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 [itex]\psi(x)[/itex] 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
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
 

1. What is the 1-D box paradox?

The 1-D box paradox is a theoretical concept in quantum mechanics that explores the behavior of a particle confined to a one-dimensional space. It is considered a paradox because it challenges the classical understanding of particle behavior, where particles are thought to have definite positions and momenta at all times.

2. How is the 1-D box paradox solved?

The 1-D box paradox is solved by solving the Schrödinger equation for the particle confined in the one-dimensional space. This involves finding the eigenfunctions and eigenvalues of the system, which represent the possible states and energies of the particle.

3. What are eigenfunctions and eigenvalues?

Eigenfunctions are mathematical functions that represent the allowed states of a system, while eigenvalues are the corresponding energies associated with each eigenfunction. In the context of the 1-D box paradox, the eigenfunctions and eigenvalues represent the possible positions and momenta of the confined particle.

4. How does the 1-D box paradox relate to the uncertainty principle?

The 1-D box paradox is closely related to the uncertainty principle, which states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. In the 1-D box paradox, the confined particle is limited in its possible positions, leading to a spread of possible momenta.

5. What are the practical applications of studying the 1-D box paradox?

Studying the 1-D box paradox can help us better understand the behavior of particles in confined spaces, which has important applications in fields such as nanotechnology and quantum computing. It also allows us to explore the fundamental principles of quantum mechanics and deepen our understanding of the physical world.

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