- #1
laser1
- 170
- 23
- Homework Statement
- Find <p^2>
- Relevant Equations
- NA
psi is given above. I have checked multiple times but can't find my mistake. Thank you!
Why is the expectation value of momentum negative? (QM)
- Thread starter laser1
- Start date
Click For Summary
SUMMARY
The discussion centers on the negative expectation value of momentum in quantum mechanics, particularly in relation to a bound state represented by the wave function psi. Participants emphasize the importance of splitting the integral due to the modulus and highlight the role of the discontinuity in the derivative of psi at x=0. The calculation of the expectation value using the operator p^2 confirms a positive result, indicating that the issue lies in the treatment of the wave function's discontinuity. The fictitious potential is also mentioned as a point of clarification.
PREREQUISITES- Understanding of quantum mechanics principles, particularly bound states.
- Familiarity with wave functions and their properties in quantum systems.
- Knowledge of momentum operators in quantum mechanics.
- Basic calculus, specifically integration and differentiation of functions.
- Study the implications of discontinuities in wave functions on quantum mechanical calculations.
- Research the concept of fictitious potentials in quantum mechanics.
- Learn about the calculation of expectation values using quantum operators.
- Explore the properties of the delta function and its role in quantum mechanics.
Students and professionals in quantum mechanics, physicists analyzing wave functions, and anyone interested in the mathematical foundations of quantum theory.
psi is given above. I have checked multiple times but can't find my mistake. Thank you!
- #2
- 29,472
- 21,217
My first thought is that you have to split the integral because of the modulus.
- #3
hutchphd
Science Advisor
Homework Helper
- 6,951
- 6,035
This is a bound state (can you tell me what fictitious potential?) Does it make sense now?
- #4
vela
Staff Emeritus
Science Advisor
Homework Helper
- 16,212
- 2,871
If you calculate it as ##\langle \psi \rvert p^2 \lvert \psi \rangle = \langle p\psi \vert p\psi \rangle##, you'll get a positive answer. My guess is that it's the discontinuity in ##\psi'## at ##x=0## that's causing your problem.
- #5
pines-demon
Science Advisor
Gold Member
- 1,018
- 860
Distributionally
$$\frac{\mathrm d^2 }{\mathrm d x^2} (e^{-\lambda |x|})= \lambda^2 e^{-\lambda|x|}-2\lambda \delta(x)$$
$$\frac{\mathrm d^2 }{\mathrm d x^2} (e^{-\lambda |x|})= \lambda^2 e^{-\lambda|x|}-2\lambda \delta(x)$$