Consider the moving delta-function well: $\mathbf{V}\left(x,t\right)\mathbf{=}\mathbf{-}\mathbf{\alpha \delta }\left(x-\mathrm{vt}\right)$

where v is the (constant) velocity of the well. (a) Show that the time-dependent Schrödinger equation admits the exact solution $\mathbf{\text{ψ}}\left(x,t\right)\mathbf{=}\sqrt{\frac{\mathbf{m\alpha }}{\sout{\mathbf{h}}}}{\mathbf{e}}^{\mathbf{-}\mathbf{m\alpha }\left|x-\mathrm{vt}\right|}\mathbf{l}{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{i}}\left[\left(E+\left(1/2\right){\mathrm{mv}}^2\right)t-\mathrm{mvx}\right]\mathbf{l}\sout{\mathbf{h}}$where $\mathbf{E}\mathbf{=}\mathbf{-}{\mathbf{m\alpha }}^{\mathbf{2}}\mathbf{}\mathbf{l}\mathbf{}\mathbf{2}{\sout{\mathbf{h}}}^{\mathbf{2}}$ is the bound-state energy of the stationary delta function. Hint: Plug it in and check it! Use the result of Problem 2.24(b). (b) Find the expectation value of the Hamiltonian in this state, and comment on the result.

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$V\left(x,t\right)=-\alpha \delta \left(x-vt\right)$, is given by:

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

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

$$\begin{gathered} \frac{\partial \psi }{\partial t}=\left[-\frac{m\alpha }{{\sout{h}}^2}v\left[2\theta \left(x-vt\right)-1\right]-i\frac{\left(E+{\displaystyle \frac{1}{2}}m{v}^2\right)}{\sout{h}}\right]\psi \\ Thus: \\ i\sout{h}\frac{\partial \psi }{\partial t}=\left[i\frac{m\alpha v}{{\sout{h}}^2}\left[2\theta \left(x-vt\right)-1\right]+E+\frac{1}{2}m{v}^2\right]\psi \end{gathered}$$

Where:

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

the function $\theta$is defined by equation 2.143 as:

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

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

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

Substitute with this equation into (1) to get:

$\frac{\partial \psi }{\partial t}=\left[\frac{m\alpha }{\sout{h^2}}v\left[2\theta \left(x-vt\right)\right]-1-i\frac{\left(E+{\displaystyle \frac{1}{2}}m{v}^2\right)}{\sout{h}}\right]\psi$

Thus;

$i\sout{h}\frac{\partial \psi }{\partial t}=\left[i\frac{m\alpha v}{\sout{h}}\left[2\theta \left(x-vt\right)-1\right]+E+\frac{1}{2}m{v}^2\right]\psi$

The first derivative w.r.t x is;

$\frac{{\partial }^2\psi }{\partial x^2}={\left[-\frac{m\alpha }{\sout{h^2}}\left[2\theta \left(x-vt\right)-1\right]+\frac{imv}{\sout{h}}\right]}^2\psi -\frac{2m\alpha }{\sout{h}}\left[\frac{\partial }{\partial x}\theta \left(x-vt\right)\right]\psi$

Note that,

$\frac{\partial }{\partial x}\theta \left(x-vt\right)=\delta \left(x-vt\right)$

Thus,

$$\begin{gathered} -\frac{\sout{h^2}}{2m}\frac{{\partial }^2\psi }{\partial x^2}=-\frac{\sout{h^2}}{2m}{\left[-\frac{m\alpha }{\sout{h^2}}\left[2\theta \left(x-vt\right)-1\right]+\frac{imv}{\sout{h}}\right]}^2\psi +\alpha \delta \left(x-vt\right)\psi \\ =\left(-\frac{\sout{h^2}}{2m}{\left[-\frac{m\alpha }{\sout{h^2}}\left[2\theta \left(x-vt\right)-1\right]+\frac{imv}{\sout{h}}\right]}^2\psi +\alpha \delta \left(x-vt\right)\right)\psi \\ =-\frac{\sout{h^2}}{2m}-\frac{m^2{\alpha }^2}{\sout{h^4}}{\left[2\theta \left(x-vt\right)-1\right]}^2-\frac{m^2{v}^2}{\sout{h^2}} \\ =-2i\frac{mv}{\sout{h}}\frac{m\alpha }{\sout{h}}\left[2\theta \left(x-vt\right)-1\right]+\alpha \delta \left(x-vt\right)\psi \end{gathered}$$

Now $$\begin{gathered} {\left[2\theta \left(x-vt\right)-1\right]}^2=1,thus: \\ -\frac{\sout{h^2}}{2m}\frac{{\partial }^2\psi }{\partial x^2}=\left[-\frac{m{\alpha }^2}{2\sout{h^2}}+\frac{1}{2}m{v}^2+i\frac{mv\alpha }{\sout{h}}\left[2\theta \left(x-vt\right)-1\right]+\alpha \delta \left(x-vt\right)\right]\psi \\ -\frac{\sout{h^2}}{2m}\frac{{\partial }^2\psi }{\partial x^2}-\alpha \delta \left(x-vt\right)\psi =\left[-\frac{m{\alpha }^2}{2\sout{h^2}}+\frac{1}{2}m{v}^2+i\frac{mv\alpha }{\sout{h}}\left[2\theta \left(x-vt\right)-1\right]\right]\psi \end{gathered}$$

$$\begin{gathered} Also,E=\frac{1}{2}m{v}^2,thus: \\ -\frac{{\sout{h}}^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}-\alpha \delta \left(x-vt\right)\psi =i\sout{h\frac{\partial \psi }{\partial t}} \end{gathered}$$

Therefore, the time-dependent Schrödinger equation conforms with the exact solution.

(b)

The expectation value of Hamiltonian is expressed by the equation,

Here,

$H\psi =i\sout{h}\frac{\partial \psi }{\partial t}$

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 ${\int }_{-\infty }^{\infty }{\left|\psi \right|}^2=1$and $\left[2\theta \left(x-vt\right)-1\right]$ is an odd function, and the integration of an odd function from $-\infty$ to $\infty$is zero,

so:

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