Question

A particle of mass$\mathit{m}$and energyrole="math" localid="1656064863125" $E$is incident from the left on the potential

$v\left(x\right)=\left\{\begin{array}{c}0,\left(x<-a\right)\\ -V_0,\left(-a\le x\le 0\right)\\ \infty ,\left(x>0\right)\end{array}\right.$

(a) If the incoming wave is$A{e}^{ikx}$(where$k=\sqrt{2mE}l\sout{h}$), find the reflected wave.

(b) Confirm that the reflected wave has the same amplitude as the incident wave.

(c) Find the phase shift$\delta$(Equation 11.40) for a very deep well$\left(E\ll v_0\right)$.

Short answer

(a) The reflected wave is$B=A{e}^{-2ika}\left[\frac{k-i\dot{kcot\left(\dot{k}a\right)}}{k+i\dot{kcot\left(\dot{k}a\right)}}\right]$

(b) The reflected wave has the same amplitude as the incident wave is verified,${\left|B\right|}^2={\left|A\right|}^2$

(c) The phase shift is$\delta =-ka$

Step-by-step solution

Find Reflected wave.

Consider a particle of mass$m$and energy$E$is incident from the left on the potential

$v\left(x\right)=\left\{\begin{array}{c}0,\left(x<-a\right)\\ -V_0,\left(-a\le x\le 0\right)\\ \infty ,\left(x>0\right)\end{array}\right.$……(1)

In the first region we have incoming and reflected parts in the wave function, if the incoming wave is $A{e}^{ikx}$(where $k=\$\$\sqrt{2mE}l\sout{h}$, then the wave function in the first are is:

$\psi \left(x\right)={\mathrm{Ae}}^{ikx}+B^{-ikx}$……(2)

In the second region, the Schrodinger equation is:

$$\begin{gathered} -\frac{h^2}{2m}\frac{d^2\psi }{d{x}^2}-V_0\psi =E\psi \\ \frac{d^2\psi }{d{x}^2}=-k\psi \end{gathered}$$

Where$\dot{k}=\sqrt{2m\left(E+V_0\right)l\sout{h}}$, the general solution of this equation is:

$\psi \left(x\right)=C\sin \left(\dot{k}x\right)+D\cos \left(\dot{k}x\right)$……(3)

In the third region the wave function is zero, since the potential is infinite. So the total wave function is:

$\psi \left(x\right)=\left\{\begin{array}{cc}A{e}^{ikx}+B^{-ikx}& \left(x<-a\right)\\ C\sin \left(\dot{k}x\right)+D\cos \left(\dot{k}x\right)& \left(-a\le x\le 0\right)\\ 0& \begin{array}{c}\left(x>0\right)\end{array}\end{array}\right.$……(4)

Now we need to apply the boundary conditions,

$$\begin{gathered} \psi \left(0\right)=0 \\ D=0 \end{gathered}$$

Thus, the wave function in equation (4) becomes,

$\psi \left(x\right)=\left\{\begin{array}{cc}A{e}^{ikx}+B^{-ikx}& \left(x<-a\right)\\ C\sin \left(k^{t}x\right)& \left(-a\le x\le 0\right)\\ 0& \begin{array}{c}\left(x>0\right)\end{array}\end{array}\right.$

The wave function and its derivative continue at$x=-a$, so we get:

$$\begin{gathered} A{e}^{-ika}+B{e}^{ika}=-C\sin \left(\dot{k}a\right) \\ ikA{e}^{-ika}-ik{B}^{ika}=\dot{kC\cos \left(\dot{k}a\right)} \end{gathered}$$

The reflected wave is$B$so we need to solve for it, divide the second equation by the first one we get:

localid="1656068719836" role="math" $$\begin{gathered} \frac{ikA{e}^{-ika}-ikB{e}^{ika}}{A{e}^{-ika}+B{e}^{ika}}=-\dot{k}cot\left(\dot{k}a\right) \\ ikA{e}^{-ika}-ikB{e}^{ika}=-A{e}^{-ika}\dot{kcot\left(\dot{k}a\right)-B{e}^{ika}\dot{k}cot\left(\dot{k}a\right)} \\ B{e}^{ika}\left[-ik+\dot{k}cot\left(\dot{k}a\right)\right]=A{e}^{-ika}\left[-ik-\dot{k}cot\left(\dot{k}a\right)\right] \\ B=A{e}^{-2ika}\left[\frac{k-i\dot{k}cot\left(\dot{k}a\right)}{k+i\dot{k}cot\left(\dot{k}a\right)}\right] \end{gathered}$$……..(6)

To confirm that the reflected wave has the same amplitude as the incident wave.

We need to show that the amplitude of the incoming wave equals the amplitude of the reflected wave, as:

$$\begin{gathered} {\left|B\right|}^2={\left|A\right|}^2\left[\frac{k-\dot{ik}cot\left(\dot{k}a\right)}{k+\dot{ik}cot\left(\dot{k}a\right)}\right].\left[\frac{k+\dot{ik}cot\left(\dot{k}a\right)}{k-\dot{ik}cot\left(\dot{k}a\right)}\right] \\ {\left|B\right|}^2={\left|A\right|}^2 \end{gathered}$$

Find the phase shift.

Substitute with equation (6) into the wave function for$x<-a$, then we get:

$\psi \left(x\right)=A{e}^{ikx}+A{e}^{-2ika}\left[\frac{k+\dot{ik}cot\left(\dot{k}a\right)}{k-\dot{ik}cot\left(\dot{k}a\right)}\right]e^{-ikx}$

The phase shift$\delta$is defined in equation as:

$\psi \left(x\right)=A\left[e^{ikx}-e^{i\left(2\delta -kx\right)}\right]$

By comparing this equation with the above one we can see that,

$e^{-2ika}\left[\frac{k+\dot{ik}cot\left(\dot{k}a\right)}{k-\dot{ik}cot\left(\dot{k}a\right)}\right]=e^{2i\delta }$……..(7)

Assume that the well is very deep so$E\ll v_0$, and:

$$\begin{gathered} \frac{\sqrt{2mE}}{\sout{h}}\square \frac{\sqrt{2m\left(E+V_0\right)}}{\sout{h}} \\ k\square \dot{k} \end{gathered}$$

So, we can neglect $k$in (6) to get:

$$\begin{gathered} e^{-2ika}\left[\frac{-\dot{ik}cot\left(\dot{k}a\right)}{\dot{ik}cot\left(\dot{k}a\right)}\right]=-e^{i\delta } \\ {e}^{-2ika}=e^{i\delta } \\ -2ika=2i\delta \\ \delta =ka \end{gathered}$$

Thus, the phase shift is $\delta =-ka$.