91影视

When time is not explicitly included in the function, it is equal to the total energy of the system. It is used to describe a dynamic system (such as the motion of a particle) in terms of components of momentum and coordinates of space and time, and it is equal to the total energy of the system.

(a)

An exact solution to the time-dependent Schrodinger equation with the moving delta-function well$V\left(x,t\right)=-伪未\left(x-vt\right)$, is given by:

$蠄\left(x,t\right)=\sqrt{\frac{m伪}{\sout{h}}}{e^{-m伪\overline{)x-vtl\sout{h}}}}^2{e}^{-i}\left[\left(E+m{v}^{2}l2\right)t-mvx\right]l\sout{h}$

Prove that this wave function satisfies the Schrodinger equation. The first derivative w.r.t is:

$$\begin{gathered} \frac{鈭�蠄}{鈭�t}=\left[-\frac{m伪}{{\sout{h}}^2}v\left[2胃\left(x-vt\right)-1\right]-i\frac{\left(E+{\displaystyle \frac{1}{2}}m{v}^2\right)}{\sout{h}}\right]蠄 \\ Thus: \\ i\sout{h}\frac{鈭�蠄}{鈭�t}=\left[i\frac{m伪v}{{\sout{h}}^2}\left[2胃\left(x-vt\right)-1\right]+E+\frac{1}{2}m{v}^2\right]蠄 \end{gathered}$$

Where:

$\frac{鈭�}{鈭�t}\left|x-vt\right|=\left\{\begin{array}{cc}-V& ifx-vt>0\\ V,& ifx-vt<0\end{array}\right.\left.\begin{array}{r}\\ \end{array}\right\}$

the function $胃$is defined by equation 2.143 as:

$胃\left(x\right)=\left\{\begin{array}{l}1x<0\\ 0x>0\end{array}\right.$

Using this definition, we can write equation (2) as:

$\frac{鈭�}{鈭�t}\left|x-vt\right|=-v\left[2胃\left(x-vt\right)-1\right]$

Substitute with this equation into (1) to get:

$\frac{鈭�蠄}{鈭�t}=\left[\frac{m伪}{\sout{h^2}}v\left[2胃\left(x-vt\right)\right]-1-i\frac{\left(E+{\displaystyle \frac{1}{2}}m{v}^2\right)}{\sout{h}}\right]蠄$

Thus;

$i\sout{h}\frac{鈭�蠄}{鈭�t}=\left[i\frac{m伪v}{\sout{h}}\left[2胃\left(x-vt\right)-1\right]+E+\frac{1}{2}m{v}^2\right]蠄$

The first derivative w.r.t x is;

$\frac{{鈭�}^2蠄}{鈭�x^2}={\left[-\frac{m伪}{\sout{h^2}}\left[2胃\left(x-vt\right)-1\right]+\frac{imv}{\sout{h}}\right]}^2蠄-\frac{2m伪}{\sout{h}}\left[\frac{鈭�}{鈭�x}胃\left(x-vt\right)\right]蠄$

Note that,

$\frac{鈭�}{鈭�x}胃\left(x-vt\right)=未\left(x-vt\right)$

Thus,

$$\begin{gathered} -\frac{\sout{h^2}}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}=-\frac{\sout{h^2}}{2m}{\left[-\frac{m伪}{\sout{h^2}}\left[2胃\left(x-vt\right)-1\right]+\frac{imv}{\sout{h}}\right]}^2蠄+伪未\left(x-vt\right)蠄 \\ =\left(-\frac{\sout{h^2}}{2m}{\left[-\frac{m伪}{\sout{h^2}}\left[2胃\left(x-vt\right)-1\right]+\frac{imv}{\sout{h}}\right]}^2蠄+伪未\left(x-vt\right)\right)蠄 \\ =-\frac{\sout{h^2}}{2m}-\frac{m^2{伪}^2}{\sout{h^4}}{\left[2胃\left(x-vt\right)-1\right]}^2-\frac{m^2{v}^2}{\sout{h^2}} \\ =-2i\frac{mv}{\sout{h}}\frac{m伪}{\sout{h}}\left[2胃\left(x-vt\right)-1\right]+伪未\left(x-vt\right)蠄 \end{gathered}$$

Now $$\begin{gathered} {\left[2胃\left(x-vt\right)-1\right]}^2=1,thus: \\ -\frac{\sout{h^2}}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}=\left[-\frac{m{伪}^2}{2\sout{h^2}}+\frac{1}{2}m{v}^2+i\frac{mv伪}{\sout{h}}\left[2胃\left(x-vt\right)-1\right]+伪未\left(x-vt\right)\right]蠄 \\ -\frac{\sout{h^2}}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}-伪未\left(x-vt\right)蠄=\left[-\frac{m{伪}^2}{2\sout{h^2}}+\frac{1}{2}m{v}^2+i\frac{mv伪}{\sout{h}}\left[2胃\left(x-vt\right)-1\right]\right]蠄 \end{gathered}$$

$$\begin{gathered} Also,E=\frac{1}{2}m{v}^2,thus: \\ -\frac{{\sout{h}}^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}-伪未\left(x-vt\right)蠄=i\sout{h\frac{鈭�蠄}{鈭�t}} \end{gathered}$$

Therefore, the time-dependent Schr枚dinger equation conforms with the exact solution.

(b)

The expectation value of Hamiltonian is expressed by the equation,

Here,

$H蠄=i\sout{h}\frac{鈭�蠄}{鈭�t}$

which is determined in part (a) and substituted into the following equation in equation (3); we get:

role="math" localid="1658299633201"

Let $y=x-vt$

thus:

from the normalization ${鈭�}_{-鈭�}^{鈭�}{\left|蠄\right|}^2=1$and $\left[2胃\left(x-vt\right)-1\right]$ is an odd function, and the integration of an odd function from $-鈭�$ to $鈭�$is zero,

so:

Thus, the expectation value of the Hamiltonian in this state is .