Eigenvalue of momentum for particle in a box

1. Jan 18, 2009

feynmann

The wave function of "particle in a box" is Asin(kx).
Since potential energy is zero inside the box, so the Hamiltonian is just kinetic energy
In principle, I should be able to find eigenvalue of momentum using momentum operator,
but stymied in solving the equation. Can somebody help me find whereabouts I am going wrong here? Thanks.

Here's the link to Particle in a Box
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html#c1

Last edited: Jan 18, 2009
2. Jan 18, 2009

scorpion990

I didn't think that the particle in a box wavefunctions are eigenfunctions of the momentum operator. Just try it!

Operate on the wavefunction with the momentum operator, and set it equal to a constant * the wavefunction. Is it possible to solve that equation for all x?

3. Jan 19, 2009

Staff: Mentor

The classical relationship between kinetic energy and momentum is

$$K = \frac{p^2}{2m}$$

In one-dimensional motion, a given value of K corresponds to two possible values of momentum.

4. Jan 19, 2009

Staff Emeritus
And that's why the expectation of momentum is 0.

5. May 11, 2010

sweet springs

Hi.
This is rather old QA but I have some uneasiness to the answers of jtbell and Vanadium 50.
About momentum distribution, some people say it is discrete values of ±sqrt(2mE).
Examples in Web
http://itl.chem.ufl.edu/4412_aa/partinbox.html 　　(16)
http://people.ccmr.cornell.edu/~muchomas/P214/Notes/QuantumMechanics/node8.html
Other people say it is continuous and its probability amplitude is derived from Fourier transform of wave function.
Examples in Web
http://www.ecse.rpi.edu/~schubert/C...um mechanics/Ch03 Position&momentum space.pdf 　　　P.21　Figure 3.2
Which is the right answer? I think the latter is logical.
Regards.

Last edited: May 11, 2010
6. May 11, 2010

DrDu

As I laid out in another thread on the same topic some days ago, the momentum operator does not exist in that problem. The momentum operator is defined as the generator of a translation of the wavefunction on which it acts. However in the problem at hand, the wavefunctions have to strictly vanish outside the box. Any shift of the wavefunction will generally violate this condition whence the operator cannot exist.
Independent from this reasoning, it is possible to define the operator hbar/i d/dx, e.g. with periodic boundary conditions. The eigenvalues of the latter operator form a basis for a Fourier transform, however, they do not correspond to momentum in that example.

7. May 11, 2010

sweet springs

Hi.

Thanks DrDu
Let me ask you a question. Energy operator or Hamiltonian H is p^2/2m + V. Even if P does not exist, H which consists of P can exist?
Regards.

8. May 11, 2010

DrDu

Yes, you are completely right. p^2 =-d^2/dx^2 is not the square of the momentum operator. That is a common problem with differential operators on a bounded interval.
Another, perhaps less artificial example, is the radial momentum, i.e. the momentum operator conjugate to the radial distance r. This operator can also shown not to exist. However its square is well defined and contributes to kinetic energy.
The classic text is John von Neumanns "Mathematical foundations of quantum mechanics".

9. May 11, 2010

sweet springs

Hi.

Concerning momentum disritibution of a particle in a box,
a. two discrete values of ±sqrt(2mE) of each 50% probability are observed.
b. continuous values whose probability is calculated from Fourier transform of wave function are observed.
c. the momentum operator does not exist in this system.

Now we have another choice c thanks to DrDu. Is there any way to fix one answer to this system?
Regards.

10. May 12, 2010

DrDu

case a corresponds to an analysis in terms of the eigenfunction of the operator hbar/i d/dx with periodic boundary conditions which is certainly not the momentum. Hence it can be ruled out.
The choice between b and c depends on your definition of the problem. If you consider only the functions defined on the interval 0 to L (the length of the box), then c is correct.
If you consider the full range of x values, x=- infinity to x=infinity, then a is correct.

11. May 12, 2010

sweet springs

Hi, DrDu.
Thank you for your reasonable suggestions. I will restate below.

Concerning Momentum distribution of a particle in a box,
-in case the world is x(- infinity, + infinity):
b. continuous values whose probability is calculated from Fourier transform of wave function are observed.,
-in case the world is x[0,L]:
c. the momentum operator does not exist.
-No Good at all times:
a. two discrete values of ±sqrt(2mE) of each 50% probability are observed.

Many authors have stated "a". Even W.Pauli in his Wave Mechanics (Vol. 5 of Pauli Lectures on Physics) did it.

Regards.

12. May 12, 2010

SpectraCat

Hmmm ... I have a little trouble understanding this in the sense of a limit. I completely understand what you are saying if the potential well is actually infinite, however what happens when it is just really really big compared with the energy level spacing (which occurs for some real problems)? In such cases, a small shift of the wavefunction will lead to an abrupt damping of the parts that are shifted into the classically forbidden regions. I guess in such cases, the Fourier transform version (option b), would be correct?

In other words, is it correct to say that:
$$\lim\limits_{V\rightarrow\infty}\hat{p}$$
strictly speaking does not exist ... however for practical purposes the FT approach should be used, since it gives correct results (to a good approximation) in the case where V is very large but not infinite?

13. May 12, 2010

sweet springs

Hi. SpectraCat

If the world is limited to x[0,L], it is meaningless to say anything about potential V(x) where x does NOT exist. So I think case c actually has nothing to do with potential V, instead the given conditions　ψ(0)=ψ(L)=0 play role.
Regards.

Last edited: May 13, 2010