How Can You Calculate the Expectation Value of Momentum in Quantum Mechanics?

Click For Summary
SUMMARY

The expectation value of momentum for a particle in an infinite square potential well can be calculated using the wave function \(\psi(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n \pi x}{a}\right)\). The correct calculation of the expectation value

results in 0, as derived from the integral \(\frac{2 \hbar k}{\pi} \int^{a}_{0} \sin^{2} \frac{\pi x}{a} dx\). The error in the initial approach stemmed from using an incorrect procedure for calculating

. It is essential to refer to standard textbooks for the proper methodology in calculating expectation values in quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wave functions.
  • Familiarity with the concept of expectation values in quantum mechanics.
  • Knowledge of integral calculus, specifically techniques for evaluating definite integrals.
  • Proficiency in using the Schrödinger equation and its applications in potential wells.
NEXT STEPS
  • Review the calculation of expectation values in quantum mechanics, focusing on operators and wave functions.
  • Study the properties of the infinite square potential well and its implications on particle behavior.
  • Learn about the role of the Schrödinger equation in determining wave functions and their physical interpretations.
  • Explore advanced topics in quantum mechanics, such as perturbation theory and its effect on expectation values.
USEFUL FOR

Students of quantum mechanics, physicists, and anyone involved in theoretical physics or quantum calculations will benefit from this discussion.

kasse
Messages
383
Reaction score
1

Homework Statement



A particle is in a infinite square poteltian well between x=0 and x=a. Find <p> of a particle whose wave function is [tex]\psi(x) = \sqrt{\frac{2}{a}}sin\frac{n \pi x}{a}[/tex] (the ground state).

2. The attempt at a solution

[tex]<p> = \frac{2 \hbar k}{\pi} \int^{a}_{0}sin^{2} \frac{\pi x}{a} dx = \frac{2 \hbar k }{\pi} \int^{\pi}_{0}sin^{2}u du = \hbar k = p[/tex]

which certainly isn't the answer I wanted. The correct answer is 0. Where's my mistake?
 
Physics news on Phys.org
You get the wrong answer because you invented your own procedure to calculate <p>. Check your textbook for how to calculate the expectation value <A> of any operator A, in an arbitrary state psi.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
Replies
9
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K