- #1
laser1
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- 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
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Homework Help Overview
The discussion revolves around the expectation value of momentum in quantum mechanics, specifically addressing a bound state and the implications of the wave function's properties.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants explore the calculation of the expectation value, question the nature of the potential involved, and discuss the implications of discontinuities in the wave function's derivative.
Discussion Status
There are various lines of reasoning being explored, including the need to split integrals and the effects of specific mathematical properties of the wave function. Some participants suggest potential sources of error while others provide insights into the mathematical framework.
Contextual Notes
Participants are considering the implications of a bound state and the characteristics of the wave function, including its behavior at specific points, which may affect the calculations being performed.
psi is given above. I have checked multiple times but can't find my mistake. Thank you!
- #2
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My first thought is that you have to split the integral because of the modulus.
- #3
hutchphd
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This is a bound state (can you tell me what fictitious potential?) Does it make sense now?
- #4
vela
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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
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Distributionally
$$\frac{\mathrm d^2 }{\mathrm d x^2} (e^{-\lambda |x|})= \lambda^2 e^{-\lambda|x|}-2\lambda \delta(x)$$
$$\frac{\mathrm d^2 }{\mathrm d x^2} (e^{-\lambda |x|})= \lambda^2 e^{-\lambda|x|}-2\lambda \delta(x)$$