# Commutator relations for the Ehrenfest Theorem

1. May 21, 2008

### tommy01

Hi there,...

For a derivation of the Ehrenfestequations i found the following commutator relations for the Hamilton-Operator in a book:
$$H = \frac{p_{op}^2}{2m} + V(r,t)$$
and the momentum-operator $$p_{op} = - i \hbar \nabla$$ respectively the position-operator $$r$$ in position space:

$$[H,p_{op}] = -i \hbar (V(r,t) \nabla - \nabla V(r,t)) = i \hbar \nabla V(r,t)$$

and

$$[H,r] = \frac{\hbar^2}{2m} (\Delta r - r \Delta) = - \frac{\hbar^2}{m} \nabla$$

so i don't understand how to get these results. it looks nearly like the use of the product rule but the signs don't match and especially in equation two: how does the laplace-operator become a nabla?

sorry if this problem is obviously but i dont't see it.

thanks and greetings.
tommy

2. May 21, 2008

### StatusX

Working in one dimension for simplicity:

$$[H,p] = [V,p]=-i\hbar(V \frac{d}{dx} - \frac{d}{dx} V)$$

Now, the operator $d/dx V$ corresponds to first applying the operator V (multiplying the wavefunction by V) and then the operator d/dx. It might help to use a test function:

$$\frac{d}{dx} V f = \frac{d}{dx} (Vf) = \frac{dV}{dx} f + V \frac{df}{dx}$$

So that, in terms of operators:

$$\frac{d}{dx} V = \frac{dV}{dx} + V \frac{d}{dx}$$

Note the first term on the RHS is different from the LHS: it just says to multiply the wavefunction by the function dV/dx.

Plugging this in above:

$$[H,p] = -i\hbar(V \frac{d}{dx} - \frac{dV}{dx} - V\frac{d}{dx}) = i \hbar \frac{dV}{dx}$$

You can work out [H,x] similarly. I would recommend using a test function first.

3. May 21, 2008

### tommy01

thanks a lot,... i was blind ;)
greetings