- #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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The discussion centers on the negative expectation value of momentum in quantum mechanics, with participants examining the implications of the wave function psi. There is an emphasis on the need to split the integral due to the modulus, and the concept of a fictitious potential is raised. The calculation of momentum squared is highlighted, suggesting it yields a positive result. A key point is the discontinuity in the derivative of psi at x=0, which may be contributing to the observed issue. The conversation underscores the importance of understanding distributional derivatives in this context.
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
$$\frac{\mathrm d^2 }{\mathrm d x^2} (e^{-\lambda |x|})= \lambda^2 e^{-\lambda|x|}-2\lambda \delta(x)$$
$$\frac{\mathrm d^2 }{\mathrm d x^2} (e^{-\lambda |x|})= \lambda^2 e^{-\lambda|x|}-2\lambda \delta(x)$$