How to Derive the Velocity Operator in Quantum Mechanics?

[divB]

Hello!

I have a task to do where I do not know where to start or where to find more information.

At first, this is just the problem statement:

Velocity operator $$\mathbf{\hat{v}}$$ is defined by the following equations:

$$\frac{d}{dt} \mathbf{\bar{r}} = \left< \psi | \mathbf{\hat{v}} | \psi \right>$$

$$\mathbf{\bar{r}} = \left< \psi | \mathbf{\hat{r}} | \psi \right>$$

where $$\mathbf{\hat{r}}$$ is the position operator (just the coordinate $$\mathbf{r}$$ itself). Show that the following relation holds between the operators:

$$\mathbf{\hat{v}} = \frac{\mathbf{\hat{p}}}{m}$$

(momentum operator $$\mathbf{\hat{p}}$$ is given by $$\mathbf{\hat{p}} = -i \hbar \nabla$$)

I do not understand what the momentum has to do with that. I tried around a while with the integral-definitions of the operators but was not successful. At the moment I have absolutely no idea where to start.

Can anybody give me a hint?

Also some literature which might help me would be greatly appreciated!

Thank you very much and Regards,
divB

 

[Hurkyl]

Two operators are equal iff they have the same effect on all kets.

 

[divB]

Hi,

Thanks for your reply.

Anyway I think I found a source where it is derived:

* http://de.wikibooks.org/wiki/Quantenmechanik#Der_Impulsoperator

But - for sure - I do not really understand the derivation :-( Maybe somebody can help me with that?

First question: Why is

$$\frac{d}{dt}\int \psi^* r \psi\,dr = \int \psi^* r \frac{d}{dt}\psi + \psi r \frac{d}{dt}\psi^*\,d r$$

?

(For the sake of simplicity I take only scalars - I hope this does not harm the generalization)

divB

 

[bluebandit26]

That's your old buddy, the chain rule...
Hurky is telling you what to do. Try applying the operators to the psi ket vector to show that they give the same result.

 

[divB]

Hi,

Thank you very much! I just wanted to write that I got it, really dumb easy :-) Thank you anyway.

About Hurky's hint: Aaah, now understand what he meant. But I think this is exactly what I am doing now: I start with the derivation of the $$\mathbf{\hat{r}}$$ (as this is v!) and the result should and will be something proportional to p with factor m. Of course this is true if I replace

$$-i\hbar\nabla = \mathbf{\hat{p}}$$

in the final equation :-) So I have shown that both are the same.

But I think there is just one piece of mathematics missing for me to understand:

How to come from

$$\frac{\hbar}{2mi} \left[-\int d^3r \psi^*(\vec{r},t) \vec{r} \Delta\psi(\vec{r},t)+\psi(\vec{r},t) \vec{r} \Delta\psi^*(\vec{r},t)\right]$$

to

$$\frac{\hbar}{2mi} \left[-\int \vec{f}\left(\psi^*\vec{r}\vec\nabla\psi-\psi\vec{r}\vec\nabla\psi^*\right) + \int d^3r \left(\vec\nabla\psi^*\vec{r}\right)\vec\nabla\psi - \int d^3r\left(\vec\nabla\psi\vec{r}\right)\vec\nabla\psi^* \right]$$

Thank you again,
divB