(a) Work out all of the canonical commutation relations for components of the operator r and p : $\mathbf{[}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{]}\mathbf{,}\mathbf{[}\mathbf{x}\mathbf{,}{\mathbf{p}}_{\mathbf{y}}\mathbf{]}\mathbf{,}\mathbf{[}\mathbf{x}\mathbf{,}{\mathbf{p}}_{\mathbf{x}}\mathbf{]}\mathbf{,}\mathbf{[}{\mathbf{p}}_{\mathbf{y}}\mathbf{,}{\mathbf{p}}_{\mathbf{z}}\mathbf{]}\mathbf{,}$and so on.

(b) Confirm Ehrenfest’s theorem for 3 dimensions

_{$\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{<}\mathbf{r}\mathbf{>}\mathbf{=}\frac{\mathbf{1}}{\mathbf{m}}\mathbf{<}\mathbf{p}\mathbf{>}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{<}\mathbf{p}\mathbf{>}\mathbf{=}\mathbf{<}\mathbf{-}\mathbf{\nabla }\mathbf{v}\mathbf{>}$}

(Each of these, of course, stand for three equations- one for each component.)

(c) Formulate Heisenberg’s uncertainty principle in three dimensions Answer:

_{${\mathit{\sigma }}_{\mathbf{x}}{\mathit{\sigma }}_{\mathbf{p}}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}\mathbf{;}{\mathit{\sigma }}_{\mathbf{y}}{\mathit{\sigma }}_{\mathbf{p}}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}\mathbf{;}{\mathit{\sigma }}_{\mathbf{z}}{\mathit{\sigma }}_{\mathbf{p}}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}}$}

But there is no restriction on, say, ${\mathit{\sigma }}_{\mathbf{x}}{\mathit{\sigma }}_{\mathbf{py}}\mathbf{.}$

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

$\left[\hat{x},\hat{y}\right]=-\left[\hat{y},\hat{x}\right]$From the axioms of quantum mechanics, we get that

$\hat{x}f=xf$ and
_{${\hat{p}}_i=-ih\frac{\partial }{\partial r_i}$}

By definition, we also know that
_{${\delta }_{ij}=1$}if otherwise,
_{${\delta }_{ij}=0$}

This letter of definition leads to the following useful identities

_{$\left[\hat{A},\hat{B}\right]=-\left[\hat{B},\hat{A}\right]$}

_{$$\begin{gathered} \left[\hat{A}\hat{B},\hat{C}\right]=\hat{B}\left[\hat{A},\hat{C}\right]+\hat{A}\left[\hat{B},\hat{C}\right] \\ \left[{\hat{A}}^2,\hat{B}\right]=\hat{A}\left[\hat{A},\hat{B}\right]+\left[\hat{A},\hat{B}\right]\hat{A} \end{gathered}$$}

Let, f be a twice continuously differentiable function,

_{$$\begin{gathered} \left[\hat{x},\hat{y}\right]f=\left(\hat{x}\hat{y}-\hat{y}\hat{x}\right)f \\ \left[\hat{x},\hat{y}\right]f=\left(\hat{x}\hat{y}\right)f-\left(\hat{y}\hat{x}\right)f \\ \left[\hat{x},\hat{y}\right]f=\left(xy\right)f-\left(yx\right)f \\ \left[\hat{x},\hat{y}\right]f=\left(yx\right)f-\left(xy\right)f \\ \left[\hat{x},\hat{y}\right]f=\left(\hat{y}\hat{x}-\hat{x}\hat{y}\right)f \\ \left[\hat{x},\hat{y}\right]f=\left[\hat{y},\hat{x}\right]f \end{gathered}$$}

But since we also know that$\left[\hat{x},\hat{y}\right]=-\left[\hat{y},\hat{x}\right]$, this implies that $\left[\hat{x},\hat{y}\right]=\left[\hat{y},\hat{x}\right]=0$

By symmetry, we then see that for all $i,j$

_{$\left[r_i,r_j\right]=\left[r_j,r_i\right]=0$}

Now consider $\left[{\hat{p}}_y,{\hat{p}}_z\right]$

_{$$\begin{gathered} \left[{\hat{p}}_y,{\hat{p}}_z\right]f=-ih(\frac{{\partial }^2}{\partial y\partial z}-\frac{{\partial }^2}{\partial z\partial y})f \\ \left[{\hat{p}}_y,{\hat{p}}_z\right]f=-ih(\frac{{\partial }^{2}f}{\partial y\partial z}-\frac{{\partial }^{2}f}{\partial z\partial y}) \\ \left[{\hat{p}}_y,{\hat{p}}_z\right]f=0 \end{gathered}$$}

By equality of mixed second partial derivatives,

By symmetry, we then see that for all, i,j

_{$$\begin{gathered} \left[{\hat{p}}_i,{\hat{p}}_j\right]=\left[{\hat{p}}_j,{\hat{p}}_i\right] \\ =0 \end{gathered}$$}

Next, consider $\left[\hat{x}{\hat{p}}_y\right]$applying this operator to the function f , we find that

_{$$\begin{gathered} \left[\hat{x}{\hat{p}}_y\right]f=\hat{x}{\hat{p}}_{y}f-{\hat{p}}_y\hat{x}f \\ \left[\hat{x}{\hat{p}}_y\right]f=-ih(x\frac{\partial f}{\partial y}-\frac{\partial \left(xf\right)}{\partial y}) \\ \left[\hat{x}{\hat{p}}_y\right]f=-ih(x\frac{\partial f}{\partial y}-\frac{\partial f}{\partial y}) \\ \left[\hat{x}{\hat{p}}_y\right]f=0 \end{gathered}$$}

By symmetry, we then see that for all

_{$\left[{\hat{r}}_i{\hat{p}}_j\right]=\left[{\hat{p}}_j{\hat{r}}_i\right]=0$}

But we also want to know what $\left[{\hat{r}}_i{\hat{p}}_j\right]$ is when i=j so we consider$\left[\hat{x}{\hat{p}}_x\right]$

_{$$\begin{gathered} \left[\hat{x}{\hat{p}}_x\right]f=\hat{x}{\hat{p}}_{x}f-{\hat{p}}_x\hat{x}f \\ \left[\hat{x}{\hat{p}}_x\right]f=-ih(x\frac{\partial f}{\partial x}-\frac{\partial \left(xf\right)}{\partial x} \\ \left[\hat{x}{\hat{p}}_x\right]f=-ih(x\frac{\partial f}{\partial x}-f-\frac{\partial f}{\partial x}) \\ \left[\hat{x}{\hat{p}}_x\right]f=ihf \end{gathered}$$}

_{$\left[{\hat{r}}_i{\hat{p}}_j\right]=ih$}for all i follows by symmetry

Together with the above result $i\ne j$we see that for all $i,j$

by the symmetry of$x,y,z$