91影视

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
_{${\hat{p}}_i=-ih\frac{鈭�}{鈭�r_i}$}

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

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

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

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

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

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$\left[\hat{x}{\hat{p}}_x\right]$

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

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

Together with the above result $i鈮�j$we see that for all $i,j$

by the symmetry of$x,y,z$