91Ó°ÊÓ

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

$v\left(x\right)=\left\{\begin{array}{c}0,\left(x<-a\right)\\ -V_0,\left(-a≤x≤0\right)\\ ∞,\left(x>0\right)\end{array}\right.$…â¶Ä¦(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:

$\mathrm{ψ}\left(x\right)={\mathrm{Ae}}^{ikx}+B^{-ikx}$…â¶Ä¦(2)

In the second region, the Schrodinger equation is:

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

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

$\mathrm{ψ}\left(x\right)=C\sin \left(\stackrel{Ë™}{k}x\right)+D\cos \left(\stackrel{Ë™}{k}x\right)$…â¶Ä¦(3)

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

$\mathrm{ψ}\left(x\right)=\left\{\begin{array}{cc}A{e}^{ikx}+B^{-ikx}& \left(x<-a\right)\\ C\sin \left(\stackrel{Ë™}{k}x\right)+D\cos \left(\stackrel{Ë™}{k}x\right)& \left(-a≤x≤0\right)\\ 0& \begin{array}{c}\left(x>0\right)\end{array}\end{array}\right.$…â¶Ä¦(4)

Now we need to apply the boundary conditions,

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

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

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

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

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

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" $$\begin{gathered} \frac{ikA{e}^{-ika}-ikB{e}^{ika}}{A{e}^{-ika}+B{e}^{ika}}=-\stackrel{Ë™}{k}cot\left(\stackrel{Ë™}{k}a\right) \\ ikA{e}^{-ika}-ikB{e}^{ika}=-A{e}^{-ika}\stackrel{Ë™}{kcot\left(\stackrel{Ë™}{k}a\right)-B{e}^{ika}\stackrel{Ë™}{k}cot\left(\stackrel{Ë™}{k}a\right)} \\ B{e}^{ika}\left[-ik+\stackrel{Ë™}{k}cot\left(\stackrel{Ë™}{k}a\right)\right]=A{e}^{-ika}\left[-ik-\stackrel{Ë™}{k}cot\left(\stackrel{Ë™}{k}a\right)\right] \\ B=A{e}^{-2ika}\left[\frac{k-i\stackrel{Ë™}{k}cot\left(\stackrel{Ë™}{k}a\right)}{k+i\stackrel{Ë™}{k}cot\left(\stackrel{Ë™}{k}a\right)}\right] \end{gathered}$$…â¶Ä¦..(6)

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

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

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

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

The phase shift$\mathrm{δ}$is defined in equation as:

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

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

$e^{-2ika}\left[\frac{k+\stackrel{Ë™}{ik}cot\left(\stackrel{Ë™}{k}a\right)}{k-\stackrel{Ë™}{ik}cot\left(\stackrel{Ë™}{k}a\right)}\right]=e^{2i\mathrm{δ}}$…â¶Ä¦..(7)

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

$$\begin{gathered} \frac{\sqrt{2mE}}{\sout{h}}â–¡\frac{\sqrt{2m\left(E+V_0\right)}}{\sout{h}} \\ kâ–¡\stackrel{Ë™}{k} \end{gathered}$$

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

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

Thus, the phase shift is $\mathrm{δ}=-ka$.