Question

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).

Short answer

a) When $\hat{Q}=1$then .

b) When $\hat{Q}=\hat{H}$then .

c) When $\hat{Q}=\hat{x}$then .

d) When $\hat{Q}=\hat{p}$then .

Step-by-step solution

Expectation value of an operator Q

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.

Apply Equation 3.71 to the cases whenQ^=1.

(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.

Apply Equation 3.71 to the cases when

(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.

Apply Equation 3.71 to the cases when Q^=x^

(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 $p=mv$.

Apply Equation 3.71 to the cases when Q^=P^

(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.