Probability distribution problem

[danja347]

I have the normalized wavefunction u(x)=1/sqrt(L) when -L/2 < x < L/2 and zero everywhere else.

I want the value of p when the probability of getting a measurement result close to p i maximal/minimal. I seek for maximum/minimum probability density!

I was thinking of Fouriertransform u(x) to get y(p). Then take |y(p)|^2 -- that is the probability distribution in p-space isn't it? Then take d/dp(|y(p)|^2) and get the maxima and the minima of the function. The thing is that in this way i get a strange expression for the derivate and i don't find the maxima and the minima like i want! I can´t evaluate the expression i get!

Can anyone help me in this way or in another way please!?

 

[lalbatros]

You need to find the maxima and minima of the function:

\frac{sin(z)^2}{z^2}


where

z=\frac{kL}{2}

and

p=\hbar k

The maxima and minima of this function cannot be found analitically, except the z=0 maximum.
It would be quite useful to make a drawing of this function.

 

[danja347]

One more question...

Thanks!

I had it right but i just could not see it!

One more question. If you see the wavefunction above as a particle in the box problem! Is it a solution to that problem? A superposition of energy eigenfunctions should be able to build up the wavefunction above? In that case the energy is quantized and never zero? Isn´t it strange that the most probable value of p is zero when the energy is never zero?

Thanks again!

/Daniel

 

[ZapperZ]

Thanks!

I had it right but i just could not see it!

One more question. If you see the wavefunction above as a particle in the box problem! Is it a solution to that problem? A superposition of energy eigenfunctions should be able to build up the wavefunction above? In that case the energy is quantized and never zero? Isn´t it strange that the most probable value of p is zero when the energy is never zero?

Thanks again!

/Daniel

You are forgetting that "p" is a vector, while "energy" isn't. So if the most probable value of the momentum is zero, it means there are equal probability of the particle moving to the "left" as it is moving to the "right" in the potential well.

Zz.

 

[danja347]

Good answer!

Thank you for the fast answer and for clearing that up!

/Daniel

 

[lalbatros]

Daniel,


I had been first wondering if you were talking about the particle-in-a-box problem.
It caused me some trouble before I understood you were talking about a wavepacket.
Your last question now revives my problem:

If you see the wavefunction above as a particle in the box problem! Is it a solution to that problem?

My difficulty was about the boundary conditions that should apply for this problem.
I found a web site describing the particle-in-a-box problem.
There is also an explanation about why the wave function should vanish at the boundaries.
You will also find there the solution of the SE (careful: the x is offset with respect to yours).
As you will see, the boundary conditions produce a discrete set of solutions and energies.
The reflexion of the stationary waves on the boundaries implies that they are not eigenfunctions of the momentum.
The momentum changes direction by reflexion.
Note finally that [H,p]=0 inside the box, but not on the boudary.
On the boundary there is a force which precisely produces the reflexion. This force make H incompatible with p.

 

[danja347]

Thanks again for answer!

The original problem didn't involve the particle in a box problem! I was just curious about if the wave packet was a solution of the particle in a box problem and why the most probable measurement of the momentum would yield zero! But i think i got it all figured out now!

Thanks everyone!