Derivation of Quantum Momentum

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We go through the mathy steps of the derivation justifying the form of the momentum operator in quantum mechanics.
  1. Dr. Underwood's Physics YouTube Page
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Комментарии
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So clearly explained thank you man! Makes sense.

georgerevell
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Great video, thank you for posting it! I was wondering, though, isn't it a little circular to derive the momentum operator using the Hamiltonian? Because the the kinetic term of the Hamiltonian was derived using (p^2)/2m, it seems circular.

sahandyman
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Why are we "allowed" to use the Schrodinger's equation in this proof? I'm a bit confused...

HidekazuOki
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It is assume that V(x) is real. But in the in the derivation of the expectation value of momentum V(x) multiplies the wave function and also tis conjugate. Are those resulting terms real? If not they do not cancel out.

jg
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Great video! Im just wondering, why in the last integration by parts you're allowed to mov the d/dx onto psi and off of psi*? Is that allowed?

jaredaitken
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I am a certified professor and princeton and realized the immediate mistake at the beginning of the video. You noted both the -infinity and infinity on the integral symbol, which is completely unneccessary.

hardiehero
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What do you mean V is "real"?

samdull