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How to take differential of Quantum mean value over hermitian operator (mean or expectation value)? $$d\langle \hat A\rangle$$ remark: or time evolution of mean value over operator $$\frac {d\langle \hat A\rangle}{dt}$$

what is the problem here? ok let me talk a little more special in three steps. 1. In classical phyaics differential of momentum is: $$dp=dmv=mdv$$ 2. In relativistic physics differential of a momentum is: $$dp=dmv=mdv + vdm$$ 3. how about differential of relativistic quantum mean value over momentum operator? $$d\langle \hat P\rangle=?$$

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but i dont have access to my questions!.

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What's the problem here? A differential of anything is always the same thing - applied to different quantities. I don't quite understand how it's possible to know what "differential of something" means so that "differential of the mean value" would still need a clarification. –  Luboš Motl Dec 13 '12 at 8:17
Dear mare, I saw your update. The Leibniz rule $d(xy)=dx\cdot y+x\cdot dy$ still holds even if $x,y$ are mean values or operators or anything of the sort. Also, $d\langle A\rangle = \langle dA\rangle$ and similar identities. For $d\langle\hat P\rangle$ one just adds hats to all the quantities in the unhatted identity. Does it answer your questions? –  Luboš Motl Dec 13 '12 at 9:20
If you want your accounts merged, please fill both profile pages with a "Please merge with <link to profile>", where <link to profile> links to the other profile. –  Manishearth Dec 13 '12 at 12:11

2 Answers 2

In fact, you missed another type of force:

In Lagrangian Mechanics there is a scalar potential field V in which the gradient of V is the force: $$F=-\nabla V$$

and this is exactly what we dealing with in QM.

$$\frac {d \langle \hat P\rangle}{dt}=\langle \hat F\rangle=\langle -\nabla V\rangle$$

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I assume that you only work in the 'wavefunction' approximation then $\langle \hat A \rangle = \langle \Psi | \hat A | \Psi \rangle$

The differential or time derivative follow in the usual way. For instance

$$\frac {d \langle \hat A \rangle}{dt} = \frac {d \langle \Psi |}{dt} \hat A | \Psi \rangle + \langle \Psi | \frac{\partial \hat{A}}{\partial t} | \Psi \rangle + \langle \Psi | \hat A \frac {d | \Psi \rangle}{dt}$$

Using the Schrodinger equation for the 'wavefunction' and working a bit one finally arrives to

$$\frac {d \langle \hat A \rangle}{dt} = \frac{1}{i\hbar} \langle [\hat{A},\hat{H}] \rangle + \left\langle \frac{\partial \hat{A}}{\partial t} \right\rangle$$

A result known as Eherenfest theorem. The differential can be obtained by multiplying everything by $dt$. The momentum operator $\hat{P}$ does not depend explicitly on time and the last term vanishes

$$\frac {d \langle \hat P \rangle}{dt} = \frac{1}{i\hbar} \langle [\hat{P},\hat{H}] \rangle$$

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