Apply Equation 3.71 to the following special cases: (a)Q=1; (b)Q=H; (c)Q=x; (d)Q=p. In each case, comment on the result, with particular reference to Equations 1.27,1.33,1.38, and conservation of energy (comments following Equation 2.39).

The rate of change of the expectation value of an operator $\mathbf{Q}$is:

$$\begin{gathered} \frac{\mathbf{d}}{\mathbf{dt}}<\hat{Q}>\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}<\Psi \overline{)\hat{Q}\Psi }> \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{i}}{\sout{\mathbf{h}}}<\left[\hat{H},\hat{Q}\right]>\mathbf{+}<\frac{\partial \hat{Q}}{\partial t}> \end{gathered}$$

Where,localid="1656314946232" $\hat{\mathit{H}}$is the Hamiltonian of the system.

(a)

Any operator commutes with a constant, so $\hat{Q}=1$commutes with everything, that is$\left[\hat{H},\hat{Q}\right]=0$, substitute into (1):

this is the conservation of normalization. which is proved in equation 1.27.

(b)

Any operator commutes with itself, so$\hat{Q}=\hat{H}$commutes with $\hat{H}$, that is role="math" localid="1656315468447" $\left[\hat{H},\hat{H}\right]=0$, substitute into (l):

If the energy has no explicit time dependence, then$\frac{d\left(H\right)}{dt}=0$this result expresses the conservation of energy.

(c)

For $\hat{Q}=\hat{x}$,find the commutator of [H, x], as:

$$\begin{gathered} \left[x,H\right]=\left[x,\frac{p^2}{2m}+V\right] \\ =\frac{1}{2m}\left[x,p^2\right]+\left[x,V\right] \end{gathered}$$

Where:

$$\begin{gathered} \left[x,p^2\right]=x{p}^2-p^{2}x \\ =x{p}^2-pxp+pxp-p^{2}x \\ =\left[x,p\right]p+p\left[x,p\right] \end{gathered}$$

$$\begin{gathered} \mathrm{But}\left[\mathrm{x},{\mathrm{p}}^2\right]=\mathrm{i}\sout{\mathrm{h}},\mathrm{so}: \\ \left[\mathrm{x},{\mathrm{p}}^2\right]=2\mathrm{i}\sout{\mathrm{h}} \\ \mathrm{and}\left[\mathrm{x},\mathrm{V}\right]=0\mathrm{hence}, \\ \left[\mathrm{x},\mathrm{H}\right]=\frac{1}{2\mathrm{m}}2\mathrm{i}\sout{\mathrm{h}}\mathrm{p}=\frac{\mathrm{i}\sout{\mathrm{h}}\mathrm{p}}{\mathrm{m}} \end{gathered}$$

Substitute into (1), to get:

But and are independent variables, then $\partial x/\partial t=0$, substitute to get:

This is the quantum equivalent of the classical equation $p=mv$.

(d)

For $\hat{Q}=\hat{p}$,find the commutator of [H, p] as:

$$\begin{gathered} \left[H,p\right]g=\left[\frac{{\sout{h}}^2}{2m}\frac{{\partial }^2}{\partial x^2}+V,\frac{\sout{h}}{i}\frac{\partial }{\partial x}\right]g \\ =\frac{\sout{h}}{i}\left(V\frac{\partial }{\partial x}-\frac{\partial }{\partial x}\left(Vg\right)\right) \\ =-\frac{\sout{h}}{i}\frac{\partial V}{\partial x}g \end{gathered}$$

The above equation will be,

$\left[\hat{H},\hat{p}\right]=\frac{\sout{h}}{i}\frac{\partial V}{\partial x}$

Substitute into , to get:

But pand tare independent variables, then $\partial \hat{p}/\partial t=0$, substitute to get:

This is Ehrenfest's theorem.