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yukcream

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- Thread starter yukcream
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- #1

yukcream

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- #2

Galileo

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[tex]\frac{d}{dt}\langle Q \rangle = \frac{i}{\hbar}\langle [H,Q] \rangle + \langle \frac{\partial Q}{\partial t}\rangle[/tex]

where Q is an observable?

Try putting Q=p (momentum) and Q=r (position).

- #3

DaTario

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Are you looking for Earnshaw's Theorem?DaTario said:

- #5

DaTario

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hehehehe...thank you a lot... I am getting old, my son...

- #6

yukcream

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Galileo said:

[tex]\frac{d}{dt}\langle Q \rangle = \frac{i}{\hbar}\langle [H,Q] \rangle + \langle \frac{\partial Q}{\partial t}\rangle[/tex]

where Q is an observable?

Try putting Q=p (momentum) and Q=r (position).

It is just the result of the theorem! Any mathematical divide inside?

- #7

Gokul43201

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