A charged particle, traveling in from $-\infty$ along the x axis, encounters a rectangular potential energy barrier

$\mathbf{U}\left(\mathbf{x}\right)=\left\{\begin{array}{l}{\mathbf{U}}_0\mathbf{,}\mathbf{if}\mathbf{0}<\mathbf{x}<\mathbf{L}\mathbf{,}\\ \mathbf{0}\mathbf{,}\mathbf{otherwise}\end{array}\right\}$

Show that, because of the radiation reaction, it is possible for the particle to tunnel through the barrier-that is, even if the incident kinetic energy is less than${\mathbf{U}}_0$, the particle can pass through ${}^{26}$.[Hint: Your task is to solve the equation

$\mathbf{a}=\tau \dot{\mathbf{a}}+\frac{F}{m}$

Subject to the force

$\mathbf{F}\left(\mathbf{x}\right)={\mathbf{U}}_0[-\delta \left(\mathbf{x}\right)+\delta (\mathbf{x}-\mathbf{L})]$

Refer to Probs. 11.19 and 11.31, but notice that this time the force is a specified function of$\mathbf{x}$, not$\mathbf{t}$. There are three regions to consider: (i)$\mathbf{x}<\mathbf{0}$, (ii) $\mathbf{0}<\mathbf{x}<\mathbf{L}$, (iii)$\mathbf{x}\mathbf{>}\mathbf{L}$. Find the general solution for$\mathbf{a}\left(\mathbf{t}\right)$, $\upsilon \left(\mathbf{t}\right)$, and$\mathbf{x}\left(\mathbf{t}\right)$in each region, exclude the runaway in region (iii), and impose the appropriate boundary conditions at $\mathbf{x}=\mathbf{0}$and$\mathbf{x}=\mathbf{L}$. Show that the final velocity $\left({\upsilon }_f\right)$is related to the time spent traversing the barrier by the equation

,$\mathbf{L}={\upsilon }_f\mathbf{T}-\frac{{\mathbf{U}}_0}{\mathbf{m}{\upsilon }_f}(\tau {\mathbf{e}}^{-\mathbf{T}/\tau }+\mathbf{T}-\tau )$

and the initial velocity (at$\mathbf{x}=-\infty$) is

${\upsilon }_i={\upsilon }_f-\frac{{\mathbf{U}}_0}{\mathbf{m}{\upsilon }_f}[\mathbf{1}-\frac{1}{\mathbf{1}+\frac{{\upsilon }_0}{{\mathbf{mv}}_f^2}({\mathbf{e}}^{-\mathbf{T}/\tau }-\mathbf{1})}]$

To simplify these results (since all we're looking for is a specific example), suppose the final kinetic energy is half the barrier height. Show that in this case

${\upsilon }_i=\frac{{\upsilon }_f}{\mathbf{1}-(\mathbf{L}/{\upsilon }_f\tau )}$

In particular, if you choose $\mathbf{L}={\upsilon }_f\tau /\mathbf{4}$ , then ${\upsilon }_i=(\mathbf{4}/\mathbf{3}){\upsilon }_f$, the initial kinetic energy is $(\mathbf{8}/\mathbf{9}){\mathbf{U}}_0$, and the particle makes it through, even though it didn't have sufficient energy to get over the barrier!]

Consider acharged particle, traveling in from $-\infty$ along the x axis, encounters a rectangular potential energy barrier.

Consider the initial velocity (at$\text{x}=-\infty$).

Consider the following boundary conditions:

(1) $x$ Continuous at$x=0$ and.$x=L$

(2) $\nu$Continuous at $x=0$and.$x=L$

(3) $\Delta a=\pm \frac{U_0}{mt\nu }$ (Plus at $x=0$, minus at $x=L$.

Write the formula ofradiation reaction subject to the force.

$\mathbf{a}=\tau \dot{\mathbf{a}}+\frac{F}{m}$ …… (1)

Here, $\dot{\mathbf{a}}$is radiation of reaction, $\mathbf{F}$is force and $\mathbf{m}$is mass of the particle.

Write the formula ofregion of the general solution of.

$\mathbf{a}\left(\mathbf{t}\right)={\mathbf{Ae}}^{\mathbf{t}/\tau }$…… (2)

Here, $\mathbf{A}$ is runway we pick.

Write the formula ofregion of the general solution of $\nu \left(t\right)$.

. $\nu \left(\mathbf{t}\right)={\mathbf{Ate}}^{\mathbf{t}/\tau }$ …… (3)

Here, $\mathbf{A}$ is runway we pick, $\mathbf{t}$ is time.

Write the formula ofregion of the general solution of $x\left(t\right)$.

$\mathbf{x}\left(\mathbf{t}\right)={\mathbf{At}}^2{\mathbf{e}}^{\mathbf{t}/\tau }$ …… (4)

Here, $\mathbf{A}$ is runway we pick, $\mathbf{t}$ is time.

The third of these follows from integrating the equation of motion:

Determine theradiation reaction subject to the force.

Substitute $U_0[-\delta \left(x\right)+\delta (x-L)]$ for $F$ into equation (1).

$\begin{array}{c}\int \frac{dv}{dt}dt=t\int \frac{da}{dt}dt+\frac{U_0}{m}\int [-\delta \left(x\right)+\delta (x-L)dt]\\ \Delta \nu =t\Delta a+\frac{U_0}{m}\int [-\delta \left(x\right)+\delta (x-L)]\frac{dt}{dx}dx=0\\ \Delta a=\frac{U_0}{m}\int \frac{1}{\nu }[-\delta \left(x\right)+\delta (x-L)]dx\\ \Delta a=\pm \frac{U_0}{mt\nu }\end{array}$

Therefore, the value of radiation reaction subject to the force is $\Delta a=\pm \frac{U_0}{mt\nu }$.

In each of the three regions the force is zero (it acts only at $x=0$and$x=l$).

For.$(x>L)$

Determine the${\text{3}}^{\text{th}}$region the general solution is.$a\left(t\right)$

Substitute $0$ for $A_3$into equation (2).

$a\left(t\right)=0$

Determine the${\text{3}}^{\text{th}}$region the general solution is .

Substitute $0$for $A_3$ into equation (3).

$\nu \left(t\right)=B_3$

Determine the${\text{3}}^{\text{th}}$region the general solution is .

Substitute $0$ for$A_3$ into equation (4).

$x\left(t\right)=B_{3}t+C_3$

Let the final velocity be${\nu }_f(=B_3)$, set the clock so that $t=0$ when the particle is at $x=0$, and let $T$ be the time it takes to transverse the barrier.

So, $x\left(T\right)=L={\nu }_{f}T+C_3$ and hence $C_3=L-{\nu }_{f}T$

Then$a\left(t\right)=0$ ; $\nu \left(t\right)={\nu }_f$,

$x\left(t\right)=L+{\nu }_f(t-T)$

Therefore, the value of ${\text{3}}^{\text{th}}$ region the general solution is$a\left(t\right)=0$ , $\upsilon \left(t\right)=B_3$ and .$x\left(t\right)=B_{3}t+C_3$

For $(0<x<L)$

Determine the ${\text{2}}^{\text{nd}}$region the general solution is $a\left(t\right)$.

Substitute $\frac{U_0}{mt{\nu }_f}{e}^{T/t}$ for $A_2$into equation (2).

$\begin{array}{c}a\left(t\right)=\frac{U_0}{mt{\nu }_f}{e}^{T/\tau }{e}^{t/\tau }\\ =\frac{U_0}{mt{\nu }_f}{e}^{\frac{(t-T)}{\tau }}\end{array}$

Determine the${\text{2}}^{\text{nd}}$ region the general solution is$\nu \left(t\right)$.

Substitute $\frac{U_0}{mt{\nu }_f}{e}^{T/t}$ for $A_2$and ${\nu }_f-\frac{U_0}{m{\nu }_f}$ for $B_2$ into equation (3).

$\begin{array}{c}\nu \left(t\right)={\nu }_f-\frac{U_0}{m{\nu }_f}+\frac{U_0}{mt{\nu }_f}{e}^{T/\tau }\\ ={\nu }_f+\frac{U_0}{m{\nu }_f}[e^{\frac{(t-T)}{\tau }}-1]\end{array}$

Determine the ${\text{2}}^{\text{nd}}$region the general solution is$x\left(t\right)$.

Substitute $\frac{U_0}{mt{\nu }_f}{e}^{T/t}$ for $A_2$ and $L-{\nu }_{f}T+\frac{U_0}{m{\nu }_f}(T-t)$for $C_2$into equation (4).

$x\left(t\right)=L+{\nu }_f(t-T)+\frac{U_0}{m{\nu }_f}[t{e}^{\frac{(t-T)}{\tau }}-t+T-t]$

Therefore, thevalue of ${\text{2}}^{\text{nd}}$ region the general solution is$a\left(t\right)=\frac{U_0}{mt{\nu }_f}{e}^{\frac{(t-T)}{\text{t}}}$, $\nu \left(t\right)={\nu }_f+\frac{U_0}{m{\nu }_f}[e^{\frac{t-T}{t}}-1]$and $x\left(t\right)=L+{\nu }_f(t-T)+\frac{U_0}{m{v}_f}[t{e}^{\frac{(t-T)}{t}}-t+T-t]$

Note:[The particle may reverse course before reaching L if the barrier is sufficiently wide (or high), but we're interested in the case when it actually tunnels through.]

In particular for $t=0$ (When$x=0$)

$0=L-{\nu }_{f}T+\frac{U_0}{m{\nu }_f}\left[t{e}^{\frac{T}{\tau }+T-t}\right]$

Thus $L={\nu }_{f}T-\frac{U_0}{m{\nu }_f}[t{e}^{-T/\tau }+T-t]$

For $(x<0)$

Determine the${\text{1}}^{\text{st}}$ region the general solution is $a\left(t\right)$.

$a\left(t\right)=A_1{e}^{t/\tau }$

Determine the${\text{1}}^{\text{st}}$ region the general solution is .$\nu \left(t\right)$

$\nu \left(t\right)=A_{1}t{e}^{t/\tau }+B_1$

Determine the${\text{1}}^{\text{st}}$ region the general solution is $x\left(t\right)$.

$x\left(t\right)A_1{t}^2{e}^{t/\tau }+B_{1}t+C_1$

Therefore, the

Let ${\nu }_i$ be the incident velocity (at $t\to -\infty$); then $B_1={\nu }_i$condition (3) says

$\frac{U_0}{mt{\nu }_f}{e}^{-T/t}-A=\frac{U_0}{mt{\nu }_0}$

Where, ${\nu }_0$is the speed of the particle as it passes $x=0$. From the solution in region (ii) it follows the ${\nu }_0={\nu }_f+\frac{U_0}{m{\nu }_f}(e^{-T/t}-1)$. But we can also express it in terms of the solution in region (i).

${\nu }_0=A_{1}t+{\nu }_i$

Therefore, the value of ${\text{1}}^{\text{st}}$ region the general solution is $a\left(t\right)=A_1{e}^{l/t}$, $\nu \left(t\right)=A_{1}t{e}^{l/t}+B_1$and .$x\left(t\right)=A_1{t}^2{e}^{l/t}+B_{1}t+C_1$

$\begin{array}{c}{\nu }_i={\nu }_f+\frac{U_0}{m{\nu }_f}(e^{\frac{T}{t}}-1)-A_{1}t={\nu }_f+\frac{U_0}{m{\nu }_f}(e^{\frac{T}{t}-1}-1)+\frac{U_0}{m{\nu }_0}-\frac{U_0}{m{\nu }_f}{e}^{\frac{T}{t}}\\ ={\nu }_f-\frac{U_0}{m{\nu }_f}+\frac{U_0}{m{\nu }_0}\\ ={\nu }_f-\frac{U_0}{m{\nu }_f}(1-\frac{{\nu }_f}{{\nu }_0})\\ ={\nu }_f-\frac{U_0}{m{\nu }_f}\{1-\frac{{\nu }_f}{{\nu }_f+\frac{U_0}{m{\nu }_f}[e^{-T/t}-1]}\}\end{array}$

If $\frac{1}{2}m{\nu }_f^2=\frac{1}{2}{U}_0$, then

Determine the inductance.

$\begin{array}{c}L={\nu }_{f}T-{\nu }_f[t{e}^{-T/t}+T-t]\\ ={\nu }_f[T-t{e}^{-T/t}-T+t]\\ =t{\nu }_f(1-e^{-T/t})\end{array}$

Now incident voltage will be:

$\begin{array}{c}{\nu }_i={\nu }_f-{\nu }_f[1-\frac{1}{1+e^{\frac{T}{t}-1}}]\\ ={\nu }_f(1-1+e^{\frac{T}{t}})\\ ={\nu }_f{e}^{\frac{T}{t}}\end{array}$

Putting these together since, $\frac{L}{t{\nu }_f}=1-e^{\frac{T}{t}}$then $e^{\frac{T}{t}}=1-\frac{L}{t{\nu }_f}$

Since,$e^{T/t}=\frac{1}{1-\left(\frac{L}{t{\nu }_f}\right)}$

Thus ${\nu }_i=\frac{{\nu }_f}{1-\left(\frac{L}{{\nu }_{f}t}\right)}$

In particular, for ,$L=\frac{{\nu }_{f}t}{4}$solve as:

$\begin{array}{c}{\nu }_i=\frac{{\nu }_f}{1-\frac{1}{4}}\\ =\frac{4}{3}{\nu }_f\end{array}$

So incident kinetic energy will be:

$\begin{array}{c}\frac{K{E}_i}{K{E}_f}=\frac{\frac{1}{2}m{v}_i^2}{\frac{1}{2}m{v}_i^2}\\ \frac{K{E}_i}{K{E}_f}={\left(\frac{v_i}{v_f}\right)}^2\\ \frac{K{E}_i}{K{E}_f}=\frac{16}{9}\\ K{E}_i=\frac{16}{9}K{E}_f\end{array}$

Solve further as

$\begin{array}{c}K{E}_i=\frac{16}{9}\frac{1}{2}{U}_0\\ =\frac{8}{9}{U}_0\end{array}$