Quantum mechanics: free partical in spherical coordinates

In summary: The observation value for the radial momentum is 0. It has no physical meaning, but you can use it to anticipate results by looking at the wavefunction.
  • #1
liorda
28
0

Homework Statement


My wavefunction is [tex]\psi (r, \theta, \phi )=N cos(\theta) e^{-(r/R_0)^2}[/tex].

I need to calculate [tex]<p_r>[/tex] and [tex]\Delta p_r[/tex] where [tex]p_r[/tex] is the radial momentum.


Homework Equations


I think i know [tex]p_r=\frac{\hbar}{i} \left( \frac{d}{dr}+\frac{1}{r} \right) [/tex].


The Attempt at a Solution



When I try to calculate the observation value I got infinity (the integral does not seem to converge):

[tex]<\psi | p_r | \psi > = \int_0^{2 \pi}d\phi \int_0^\pi d\theta \int_0^{\infty}dr [\psi^\star p_r \psi] [/tex]

Are the limits for the integral correct? What am I doing wrong? :(

thank.
 
Physics news on Phys.org
  • #2
[tex] ( \psi |p_r|\psi) = \frac{\hbar}{i} \left( \frac{d}{dr}+\frac{1}{r} \right) N^2 cos^2( \theta) e^{-2(r/R_0)^2} [/tex]...agreed?

So,

[tex] < \psi |p_r| \psi> = \frac{N^2 \hbar}{i} \int_0^{\pi}cos^2(\theta)d \theta \int_0^{2 \pi}d \phi \int_0^{\infty} \left \frac{d}{dr}- \frac{1}{r} \right e^{-2(r/R_0)^2} dr [/tex]

[tex] = \frac{ N^2 \hbar \pi^2}{i} \int_0^{\infty} \left \frac{d}{dr}- \frac{1}{r} \right e^{-2(r/R_0)^2} dr [/tex]

Since the theta int. goes to pi/2 and the phi int. goes to 2pi

Is this the integral you ended up computing?

Josh
 
  • #3
Hi Josh,
Thanks for answering.

I ended up with this:
[tex]< \psi |p_r| \psi> = \frac{N^2 \hbar}{i} \int_0^{\pi}cos^2(\theta)d \theta \int_0^{2 \pi}d \phi \int_0^{\infty} \left( \frac{d}{dr}+ \frac{1}{r} \right) \right e^{-2(r/R_0)^2} dr [/tex]

My problem is with [tex]\int_0^\infty \frac{1}{r} e^{-2(r/R_0)^2} dr[/tex] which does not converge.
 
  • #4
Ok, we're on the same page then with the equations.

The r integral will separate into two integrals after you distribute the variables in parenthesis... that is, one term is d/dr of the e-fcn, and the other is 1/r times the e-fcn.

Because of this, its ok for the 1/r term to blow up as long as the other one does not (I haven't computed either one of them, so let me know if there is an additional problem..)

Easiest way to integrate these, in case you didnt already know, is with the wolfram online integrator:

http://integrals.wolfram.com/
 
  • #5
kreil said:
Ok, we're on the same page then with the equations.
Because of this, its ok for the 1/r term to blow up as long as the other one does not (I haven't computed either one of them, so let me know if there is an additional problem..)

Are you sure? Can I just disregard the inf in the equation?!?

thanks again for answering.
 
  • #6
Yes, if a term goes to infinity it is called 'blowing up'

This is a common situation when evaluating boundary criterion in, for example, the finite square well, where the psi contains e^x and e^-x for x>0, and since as x-> infty e^x blows up, psi reduces to just the e^-x term.
 
  • #7
the integral you have is incorrect, i believe.
[tex] dV = r^{2} sin\theta d\theta d\phi dr [/tex]

Edit: in your equation you have:
[tex] dV = dr d\phi d\theta [/tex]
 
  • #8
the r^2*sin(t) is the |Jacobian| isn't it? I thought it should only be added when making a transformation between Cartesian and spherical coordinates, i.e [tex]\int\int\int dxdydz = \int\int\int r^2 sin(\theta) d\phi d\theta dr[/tex]...

although the r^2 sin(t) will solve the integral convergence problem, i don't understand yet if it is necessary.
 
Last edited:
  • #9
Well, the volume element I suggested to you is just the general form of a volume integral in spherical coordinates. Check it on a sphere to see what you get. The transformation you give is correct. You are essentially doing a transformation from cartesian to polar spherical here (if you want to look at it like you started in cartesian). Regardless of which, you must use dV = r^2 sin(theta) dphi dtheta dr if you're integrating over phi, theta, and r, in spherical coordinates. You should draw out the element, and you will see why you must do this. dV = dr dtheta dphi doesn't make sense as you will see if you try to draw out the element.
 
  • #10
[tex]
\int\int\int_{D} \Psi(x,y,z)^{*}p\Psi(x,y,z) dxdydz = \int\int\int_{D} \Psi(r,\phi,\theta)^{*}p\Psi(r,\phi,\theta) r^2 sin(\theta) d\phi d\theta dr[/tex]
 
  • #11
kreil and EngageEngage: thanks a lot!

I now get that the observation value for the radial momentum is 0. does it have a physical meaning? is there any way of anticipating that result by looking at the wavefunction? and should I even try to search physical meanings in quantum mechanics questions?

thanks again.
 

1. What is quantum mechanics?

Quantum mechanics is a branch of physics that explains the behavior of particles at the atomic and subatomic level. It uses mathematical equations to describe the behavior of particles, which can behave like both waves and particles.

2. What is a free particle in spherical coordinates?

A free particle in spherical coordinates is a particle that is not subject to any external forces or interactions. It is described by its position in three-dimensional space using spherical coordinates, which include the distance from the origin, the angle from the positive z-axis, and the azimuthal angle.

3. How is quantum mechanics applied to a free particle in spherical coordinates?

In quantum mechanics, the behavior of a free particle in spherical coordinates is described using the Schrödinger equation, which is a mathematical equation that describes the wave function of a particle. The wave function represents the probability of finding the particle at a certain position and time.

4. What are the implications of quantum mechanics for a free particle in spherical coordinates?

The implications of quantum mechanics for a free particle in spherical coordinates include the uncertainty principle, which states that the position and momentum of a particle cannot be known simultaneously. It also shows that particles can exist in multiple states simultaneously, known as superposition, and can become entangled with other particles.

5. What practical applications does quantum mechanics have for a free particle in spherical coordinates?

Quantum mechanics has many practical applications for a free particle in spherical coordinates, including understanding the behavior of atoms and molecules, developing new technologies such as transistors and lasers, and creating new materials with unique properties. It also plays a crucial role in the field of quantum computing.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
878
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
229
  • Introductory Physics Homework Help
Replies
4
Views
748
  • Introductory Physics Homework Help
Replies
21
Views
991
Replies
16
Views
394
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
109
Back
Top