U(0)=0 for real expectation values of momentum

In summary, the conversation was about the position-space representation of the radial component of the momentum operator and how to show that its expectation value must be real by satisfying a specific condition for the radial wave function. The suggestion was to express the expectation value in position space using spherical coordinates and then integrate by parts. The attempt at a solution focused on solving for a general \psi but the integration by parts was not clear. The conversation ended with a request for insight on how to approach the problem.
  • #1
rtellez700
2
0

Homework Statement


The position-space representation of the radial component of the momentum operator is given by
## p_r \rightarrow \frac{\hbar}{i}\left ( \frac{\partial }{\partial r} + \frac{1}{r}\right ) ##

Show that for its expectation value to be real:## \left \langle \psi|p_r|\psi \right \rangle = \left \langle \psi|p_r|\psi \right \rangle ^{*}##, the radial wave function must satisfy the condition ##u(0)=0##. Suggestion: Express the expectation value in position space in spherical coordinates and integrate by parts.

Homework Equations


##u(r)=r*R(r)##

The Attempt at a Solution


I think this can be solved for a general \psi but I'm having a hard time figuring out where the integration by parts would even come into play. Any insight on how to approach this problem would be appreciated.
 
Physics news on Phys.org
  • #2
rtellez700 said:
Suggestion: Express the expectation value in position space in spherical coordinates
Where did you follow that suggestion?
 

Related to U(0)=0 for real expectation values of momentum

1. What does "U(0)=0 for real expectation values of momentum" mean?

This statement refers to the fact that the expected value of momentum (a physical quantity related to an object's mass and velocity) is equal to zero when the momentum is measured at a specific point in time (represented by the symbol "0"). This applies to real scenarios, where the momentum can be physically measured, as opposed to hypothetical or theoretical situations.

2. Why is it important to have a real expectation value of momentum?

Having a real expectation value of momentum is important because it reflects the true physical state of an object. In other words, it accurately describes the momentum of an object at a specific point in time, which is essential for understanding and predicting its behavior and interactions with other objects.

3. How is the expectation value of momentum calculated?

The expectation value of momentum is calculated by taking the integral of the momentum operator (represented by the symbol "p") and the wave function (represented by the symbol "ψ"). This can be expressed as: ⟨p⟩ = ∫ pψ*ψ dx, where the asterisk (*) denotes the complex conjugate of the wave function.

4. What is the significance of U(0) in the equation?

The value of U(0) represents the specific point in time where the momentum is being measured. This point can be chosen arbitrarily, but it is typically chosen to be the initial time or the time when the system is in its ground state. It is important to note that the expectation value of momentum may change at different points in time.

5. How does this concept relate to the uncertainty principle?

The uncertainty principle, also known as the Heisenberg uncertainty principle, states that it is impossible to know both the exact position and momentum of a particle simultaneously. This means that there will always be some uncertainty in the measurement of momentum, and this concept of U(0)=0 for real expectation values of momentum reflects this uncertainty. It shows that the more precisely we know the position of an object, the less precisely we can know its momentum, and vice versa.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
677
  • Advanced Physics Homework Help
Replies
29
Views
278
Replies
16
Views
668
  • Advanced Physics Homework Help
Replies
1
Views
409
Replies
14
Views
2K
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
3K
  • Advanced Physics Homework Help
Replies
4
Views
962
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
5
Views
1K
Back
Top