We're breaking new ground! Chapter 4! Let's get to the solution. First of all, we have to work out the
canonical commutation relations. First of all, let us find $[r_i, \ r_j]$, where $r_x$, $r_y$, and $r_z$ correspond
to $x$, $y$, and $z$. Now we have:
$$[r_i, \ r_j] \ = \ r_i r_j \ - \ r_j r_i \ = \ 0$$
And for the components of momentum, we have:
$$[p_i, \ p_j] \ = \ p_i p_j \ - \ p_j p_i \ = \ 0$$
Now, for finding the commutator of position and momentum components. We have:
$$[r_i, \ p_j] \ = \ \frac{\hbar}{i} \ r_i \ \frac{\partial f}{\partial r_j} \ - \ \frac{\hbar}{i} \ \frac{\partial}{\partial r_j} \ (r_i f)
\ = \ \frac{\hbar}{i} \ r_i \ \frac{\partial f}{\partial r_j} \ - \ \frac{\hbar}{i} \ r_i \ \frac{\partial f}{\partial r_j} \ - \ \frac{\hbar}{i} \ f \ \frac{\partial r_i}{\partial r_j}
\ = \ - \ \frac{\hbar}{i} \ f \ \frac{\partial r_i}{\partial r_j} \ = \ i\hbar \ \frac{\partial r_i}{\partial r_j}$$
Ok, now, the derivative of some position componenet with respect to itself will be $1$, while the derivative of some position component with respect to another, different position component, will be $0$.
This means that we have:
$$[r_i, \ p_j] \ = \ i\hbar \delta_{ij}$$
Now, when the we reverse the commutator, we will have $-i\hbar \delta_{ij}$ (this can be very easily checked, but I'm not going to go through the motions in this solution).
The next part of this problem requires us to confirm Ehrenfest's Theorem for three-dimensions. We can do this fairly easily
if we confirm the equation:
$$\frac{\text{d}}{\text{dt}} \ \langle \hat{Q} \rangle \ = \ \frac{i}{\hbar} \ \langle [\hat{H}, \ \hat{Q}] \rangle \ + \ \Big\langle \frac{\partial \hat{Q}}{\partial t} \Big\rangle$$
For three dimensions. This can easily be seen, as the above equation only depends on time and the expectation of operators that can easily be rewritten in three-dimensions. The
expectation value of the derivative of the operator with respect to time is usually $0$, so in three-dimensions, nothing really changes:
$$\frac{\text{d}}{\text{dt}} \ \langle \hat{Q} \rangle \ = \ \frac{i}{\hbar} \ \langle [\hat{H}, \ \hat{Q}] \rangle$$
So we also have:
$$\frac{\text{d}}{\text{dt}} \ \langle \textbf{r} \rangle \ = \ \frac{i}{\hbar} \ \langle [\hat{H}, \ \textbf{r}] \rangle \ = \ \frac{i}{\hbar} \ \Big\langle \Big(\frac{-\hbar^2}{2m} \nabla^2 \ + \ V\Big)\textbf{r}f \ - \ \textbf{r}\Big(\frac{-\hbar^2}{2m} \nabla^2 \ + \ V\Big)f \Big\rangle
\ = \ \frac{i}{\hbar} \ \Big\langle \frac{-\hbar^2}{2m} \nabla\Big( \textbf{r}\nabla f \ + \ f\nabla\textbf{r} \Big) \ + \ V\textbf{r}f \ - \ \textbf{r}\frac{-\hbar^2}{2m} \nabla^2 f \ + \ \textbf{r}Vf \Big\rangle$$
$$\Rightarrow \ \frac{i}{\hbar} \ \Big\langle \frac{-\hbar^2}{2m} \Big( 2\nabla\textbf{r}\nabla f \ + \ f\nabla^2 \textbf{r} \ + \ \textbf{r}\nabla^2 f \Big) \ + \ V\textbf{r}f \ - \ \textbf{r}\frac{-\hbar^2}{2m} \nabla^2 f \ + \ \textbf{r}Vf \Big\rangle \ = \
\frac{i}{\hbar} \ \Big\langle \frac{-\hbar^2}{m} \nabla\textbf{r}\nabla f \Big\rangle \ = \ \frac{i}{\hbar} \ \Big\langle \frac{-\hbar^2}{m} \nabla \Big\rangle \ = \ \frac{1}{m} \langle \hat{p} \rangle$$
And for momentum, we have:
$$\frac{\text{d}}{\text{dt}} \ \langle \hat{p} \rangle \ = \ \frac{i}{\hbar} \ \langle [\hat{H}, \ \hat{p}] \rangle \ \Rightarrow \ [\hat{H}, \ \hat{p}] \ = \ \frac{\hbar}{i}\Big(-\frac{\hbar^2}{2m} \nabla^3 f \ + \ V \nabla f \Big) \ - \ \frac{\hbar}{i}\Big(-\frac{\hbar^2}{2m} \nabla^3 f \ + \ V \nabla f \ + \ f \nabla V \Big) \
\Rightarrow \ \frac{i}{\hbar} \ \langle [\hat{H}, \ \hat{p}] \rangle \ = \ \langle -\nabla V \rangle$$
Finally, we are asked to check the uncertainty principle for three-dimensions. The general uncertainty principle tells us that:
$$\sigma_A^2 \sigma_B^2 \ \geq \ \Big(\frac{1}{2i} \ \langle [\hat{A}, \ \hat{B}] \rangle \Big)^2$$
From the previously proved canonical commutation relations, we have:
$$\sigma_{r_i}^2 \sigma_{p_j}^2 \ \geq \ \Big(\frac{1}{2i} \ \langle [r_i, \ p_j] \rangle \Big)^2 \ = \ \Big(\frac{1}{2i} \ \langle i\hbar \delta_{ij} \rangle \Big)^2 \ = \
\frac{\hbar^2}{4}\delta_{ij}^2 \ \Rightarrow \ \sigma_{r_i} \sigma_{p_j} \ \geq \ \frac{\hbar}{2}\delta_{ij}$$