-consider the “step” potential:

$\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{cc}\mathbf{0}\mathbf{,}& \mathbf{if}\mathbf{}\mathbf{x}\mathbf{\le }\mathbf{0}\mathbf{,}\\ {\mathbf{V}}_{\mathbf{0}}\mathbf{,}& \mathbf{if}\mathbf{}\mathbf{x}\mathbf{>}\mathbf{0}\mathbf{,}\end{array}$

a.Calculate the reflection coefficient, for the case E < V₀, and comment on the answer.

b. Calculate the reflection coefficient, for the case E >V₀.

c. For potential such as this, which does not go back to zero to the right of the barrier, the transmission coefficient is not simply $\raisebox{1ex}{$\left|F^2\right|$}\!\left/ \!\raisebox{-1ex}{$\left|A^2\right|$}\right.$(with A the incident amplitude and F the transmitted amplitude), because the transmitted wave travels at a different speed . Show that$T=\sqrt{\frac{E-V_0}{V_0}}\frac{\left|F^2\right|}{\left|A^2\right|}$,for E >V₀. What is T for E < V₀?

d. For E > V_0, calculate the transmission coefficient for the step potential, and check that T + R = 1.

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer relates to a particle's probability density in space and time.

Step potential is given as:

$$\begin{gathered} V\left(x\right)=0\mathrm{if}x\le 0 \\ V\left(x\right)=V_0\mathrm{if}x>0 \end{gathered}$$

The Schrodinger equation can be calculated as:

$-\frac{{\sout{h}}^2}{2m}\frac{d^2\Psi }{d{x}^2}+V\left(x\right)\Psi =E\Psi$

For $x\le 0$

role="math" localid="1658127639629" $-\frac{{\sout{h}}^2}{2m}\frac{d^2{\Psi }_2}{d{x}^2}=E{\Psi }_1$

For

$-\frac{{\sout{h}}^2}{2m}\frac{d^2{\Psi }_2}{d{x}^2}+V_0{\Psi }_2=E{\Psi }_2$

Let,

$k=\sqrt{\frac{2mE}{{\sout{h}}^2}}and\alpha =\sqrt{\frac{2m(V_0-E)}{{\sout{h}}^2}}$

For $x\le 0$

role="math" localid="1658127959257" $\frac{d^2\Psi }{d{x}^2}+k^2{\Psi }_1=0$

For $x>0$

$\frac{d^2{\Psi }_2}{d{x}^2}-{\alpha }^2{\Psi }_2=0$

For $x\le 0$

${\Psi }_1\left(x\right)=A{e}^{ikx}+B{e}^{-ikx}$

For $x>0$

${\Psi }_2\left(x\right)=F{e}^{ax}+G{e}^{-ax}$

In the above equations, F will be zero for E < V₀ ( $x\to +\infty$) the wave function will not be,then only wave function will be continuous across the boundry. Thus, the solution will be:

role="math" localid="1658128607114" $$\begin{gathered} {\Psi }_1\left(x\right)=A{e}^{ikx}+B{e}^{-ikx}\left(1\right) \\ {\Psi }_2\left(x\right)=G{e}^{-\alpha x}\left(2\right) \end{gathered}$$

Apply boundary conditions as:

Wave function and the derivative will be continous at x = 0, then:

${\Psi }_1\left(0\right)={\Psi }_2\left(0\right)$ (3)

${\left(\frac{d{\Psi }_1}{dx}\right)}_{x=0}={\left(\frac{d{\Psi }_2}{dx}\right)}_{x=0}$ (4)

Using the equations above, we get,

$$\begin{gathered} A+B=G \\ ik(A-B)=-\alpha G \end{gathered}$$

Solving the above equations and we get,

$$\begin{gathered} B=\frac{k-i\alpha }{k+i\alpha }A \\ G=\frac{2k}{k+i\alpha }A \end{gathered}$$

Substituting the values in equations 1 and 2, we get,

$$\begin{gathered} {\Psi }_1\left(x\right)=A{e}^{ikx}+A\left(\frac{k-i\alpha }{k+i\alpha }\right)e^{-ikx} \\ {\Psi }_2\left(x\right)=A\left(\frac{2k}{k+i\alpha }\right)e^{-ikx} \end{gathered}$$

The reflection coefficient can be calculated as:

$R={\left|\frac{B}{A}\right|}^2$

Substituting the values, and we get,

$$\begin{gathered} R={\left|\frac{k-i\alpha }{k-i\alpha }\right|}^2 \\ ={\left(\frac{\sqrt{k^2+{\alpha }^2}}{\sqrt{k^2+{\alpha }^2}}\right)}^2 \\ =1 \end{gathered}$$

Thus, the reflection coefficient is 1 when E< V₀.

The Schrodinger equation can be calculated as:

$-\frac{{\sout{h}}^2}{2m}\frac{d^2\Psi }{d{x}^2}+V\left(x\right)\Psi =E\Psi$

For $x\le 0$

$-\frac{{\sout{h}}^2}{2m}\frac{d^2{\Psi }_1}{d{x}^2}=E{\Psi }_1$

For $x>0$

$\frac{{\sout{h}}^2}{2m}\frac{d^2{\Psi }_1}{d{x}^2}+V_0{\Psi }_2=E{\Psi }_2$-

We know:

$k=\sqrt{\frac{2mE}{{\sout{h}}^2}},and\beta =\sqrt{\frac{2m(-V_0+E)}{{\sout{h}}^2}}$

For $x\le 0$

$\frac{d^2{\Psi }_1}{d{x}^2}+k^2{\Psi }_1=0$

For $x>0$

$\frac{d^2{\Psi }_2}{d{x}^2}+{\beta }^2{\Psi }_2=0$

For $x\le 0$

localid="1658132108137" ${\Psi }_1\left(x\right)=A{e}^{ikx}+B{e}^{-ikx}$

For $x>0$

localid="1658132126293" ${\Psi }_2\left(x\right)=F{e}^{i\beta x}+G{e}^{-i\beta x}$

In the above equations, G will be zero for E > V₀ ( $x\to \infty$),then only wave function will be continuous across the boundry. Thus, the solution will be:

localid="1658132145945" $$\begin{gathered} {\Psi }_1\left(x\right)=A{e}^{ikx}+B{e}^{-ikx} \\ {\Psi }_2\left(x\right)=F{e}^{i\beta x} \end{gathered}$$

Apply boundary conditions as:

Wave function and the derivative will be continous at x = 0, then:

$$\begin{gathered} {\Psi }_1\left(0\right)={\Psi }_2\left(0\right)\left(7\right) \\ {\left(\frac{d{\Psi }_1}{dx}\right)}_{x=0}={\left(\frac{d{\Psi }_2}{dx}\right)}_{x=0}\left(8\right) \end{gathered}$$

$$\begin{gathered} {\Psi }_1\left(0\right)={\Psi }_2\left(0\right) \\ \left(\frac{d{\Psi }_{}}{}\right) \end{gathered}$$

Using the equations above, we get,

$$\begin{gathered} A+B=F \\ ik(A-B)=i\beta F\left(13\right) \end{gathered}$$

Solving the above equations and we get,

$-B\left(1+\frac{k}{\beta }\right)=A\left(1-\frac{k}{\beta }\right)\left(14\right)$

The reflection coefficient can be calculated as:

$R={\left|\frac{B}{A}\right|}^2$

Substituting the values, and we get,

$$\begin{gathered} R={\left|\frac{1-(k/\beta )}{1+(k/\beta )}\right|}^2 \\ ={\left(\frac{\beta -k}{\beta +k}\right)}^2{\left(\frac{\beta -k}{\beta -k}\right)}^2 \\ =\left(\frac{{(\beta -k)}^4}{({\beta }^2+k^2)}\right) \\ \left(9\right) \end{gathered}$$

We know:

$k=\sqrt{\frac{2mE}{{\sout{h}}^2}},and\beta =\sqrt{\frac{2m(-V_0+E)}{{\sout{h}}^2}}$

$$\begin{gathered} k^2=\frac{2mE}{{\sout{h}}^2} \\ {\beta }^2=\frac{2m(-V+E)}{{\sout{h}}^2} \\ {\beta }^2-k^2=\frac{2m(-V+E)}{{\sout{h}}^2}-\frac{2mE}{{\sout{h}}^2} \\ {({\beta }^2-k^2)}^2={\left(\frac{2m(-V}{}\right)}^{} \end{gathered}$$₀)h²2

And

$$\begin{gathered} \beta -k=\sqrt{\frac{2m(-V_0+E)}{{\sout{h}}^2}}-\sqrt{\frac{2mE}{{\sout{h}}^2}} \\ \beta -k=\sqrt{\frac{2mE}{{\sout{h}}^2}}(\sqrt{-V_0+E}-\sqrt{E}) \\ {\left(\beta -k\right)}^4={\left(\sqrt{\frac{2m}{{\sout{h}}^2}}(\sqrt{-V_0+E}-\sqrt{E})\right)}^4 \end{gathered}$$

Substituting the values in equation 9, we get,

$R=\frac{{(\sqrt{E}-\sqrt{E-V_0})}^4}{V_0^2}$

Thus, the reflection coefficient is $R=\frac{{(\sqrt{E}-\sqrt{E-V_0})}^4}{V_0^2}$when E > V₀.

From the definition of the transmission coefficient:

$T=\frac{\mathrm{probability}\mathrm{of}\mathrm{transmitted}\mathrm{particle}\mathrm{to}\mathrm{the}\mathrm{right}\mathrm{of}\mathrm{the}\mathrm{barries}}{\mathrm{probability}\mathrm{of}\mathrm{finding}\mathrm{incident}\mathrm{particle}\mathrm{in}\mathrm{the}\mathrm{box}}$

Let's take the time interval dt here. The probability of finding the transmitted particle to the right of the barrier is $P_t$, and the probability of finding the incident particle in the box to the left of the barrier is $P_i$.

The above expression for transmission coefficient can be written as:

$T=\frac{P_t}{P_i}$

From the definition of probabilities:

$\frac{P_t}{P_i}=\frac{{\left|F\right|}^2{v}_t}{\left|A^2\right|v_i}$ (11)

Here v_iand v_t are the speed of the incident particle and transmitted particle, respectively.

From the definition of velocity:

$v=\frac{\sout{h}k}{2m}$

Thus the velocities can be calculated as:

$$\begin{gathered} v_t=\frac{\sout{h}\beta }{2m} \\ {v}_i=\frac{\sout{h}k}{2m} \end{gathered}$$

Substitute the values in equation 11, and we get,

role="math" localid="1658204796438" $\frac{P_t}{P_i}=\frac{{\left|F\right|}^2\frac{\sout{h}\beta }{2m}}{{\left|A\right|}^2\frac{\sout{h}k}{2m}}$ (12)

Substitute values, and we get,

$$\begin{gathered} \frac{P_t}{P_i}=\frac{{\left|F\right|}^2{\displaystyle \frac{\sqrt[\sout{h}]{{\displaystyle \frac{2m(-V_0+E)}{\sout{h}}}}}{2m}}}{{\left|A\right|}^2{\displaystyle \frac{\sqrt[\sout{h}]{{\displaystyle \frac{2mE}{{\sout{h}}^2}}}}{2m}}} \\ \frac{P_t}{P_i}=\frac{{\left|F\right|}^2\sqrt{(-V_0+E)}}{{\left|A\right|}^2\sqrt{\left(E\right)}} \end{gathered}$$

Thus, the transmission coefficient will be:

role="math" localid="1658205459586" $T=\frac{{\left|F\right|}^2\sqrt{(-V_0+E)}}{{\left|A\right|}^2\sqrt{\left(E\right)}}$

For E < V₀

F = 0 thus,

T = 0

Thus, the transmission coefficient is $T=\frac{{\left|F\right|}^2\sqrt{(-V_0+E)}}{{\left|A\right|}^2\sqrt{\left(E\right)}}$when E > V₀and zero when

E < V₀.

From equation 12, the transmission coefficient can be calculated as:

$$\begin{gathered} T=\frac{P_t}{P_i} \\ =\frac{{\left|F\right|}^2{\displaystyle \frac{\sout{h}\beta }{2m}}}{{\left|A\right|}^2{\displaystyle \frac{\sout{h}k}{2m}}} \end{gathered}$$

Solving the equation further:

$$\begin{gathered} T=\frac{P_t}{P_i} \\ =\frac{{\left|F\right|}^2{\displaystyle \beta }}{{\left|A\right|}^2{\displaystyle k}} \end{gathered}$$

From equations 13 and 14:

$$\begin{gathered} -B\left(1+\frac{k}{\beta }\right)=A\left(1-\frac{k}{\beta }\right) \\ A+B=F \end{gathered}$$

Substituting the value of B in the above expression, we get,

$$\begin{gathered} A-A\frac{\left(1-{\displaystyle \frac{k}{\beta }}\right)}{\left(1+\frac{k}{\beta }\right)}=F \\ F=A\left(1-\frac{\left(1-{\displaystyle \frac{k}{\beta }}\right)}{\left(1+\frac{k}{\beta }\right)}\right) \\ \frac{F}{A}=\frac{\left(1+\frac{k}{\beta }-1+\frac{k}{\beta }\right)}{\left(1+\frac{k}{\beta }\right)} \\ \frac{F}{A}=\frac{\left({\displaystyle \frac{2k}{\beta }}\right)}{\left({\displaystyle \frac{\beta +k}{\beta }}\right)} \\ \frac{F}{A}=\frac{2k}{\beta +k} \end{gathered}$$

The transmission coefficient can be written as:

$$\begin{gathered} T=\frac{P_t}{P_i} \\ ={\left(\frac{2k}{\beta +k}\right)}^2\frac{\beta }{k} \\ T=\frac{4k\beta }{{(\beta +k)}^2} \end{gathered}$$

The transmission coefficient can be written as:

$$\begin{gathered} T=\frac{P_t}{P_i} \\ ={\left(\frac{2k}{\beta +k}\right)}^2\frac{\beta }{k} \\ T=\frac{4k\beta }{{\left(\beta +k\right)}^2} \end{gathered}$$

Substituting values in the above expression, we get,

role="math" localid="1658209317653" $$\begin{gathered} T=\frac{\sqrt[4]{{\displaystyle \frac{2mE}{{\sout{h}}^2}}}\sqrt{{\displaystyle \frac{2m(-V_0+E)}{{\sout{h}}^2}}}}{{\left(\sqrt{{\displaystyle \frac{2m(-V_0+E)}{{\sout{h}}^2}}}+\sqrt{{\displaystyle \frac{2mE}{{\sout{h}}^2}}}\right)}^2} \\ T=\frac{4\sqrt{E}\sqrt{-V_0+E}}{{\left(\sqrt{E}+\sqrt{(-V_0+E)}\right)}^2}\times \frac{{\left(\sqrt{E}\sqrt{(-V_0+E)}\right)}^2}{{\left(\sqrt{E}\sqrt{(-V_0+E)}\right)}^2} \\ =\frac{4\sqrt{E}\sqrt{-V_0+E}\times {\left(\sqrt{E}\sqrt{(-V_0+E)}\right)}^2}{{\left(E+(-V_0+E)\right)}^2} \\ T=\frac{4\sqrt{E}\sqrt{-V_0+E}\times \left(\sqrt{E}-\sqrt{-V_0+E}\right)}{{\left(V_0\right)}^2} \end{gathered}$$

From subpart b, the reflection coefficient is:

$$\begin{gathered} R=\left(\frac{{\left(\beta -k\right)}^4}{{\left({\beta }^2-k^2\right)}^2}\right) \\ R+T=\left(\frac{{\left(\beta -k\right)}^4}{{\left({\beta }^2-k^2\right)}^2}\right)+\frac{4k\beta }{{\left(\beta +k\right)}^2} \\ =\frac{({\beta }^4+k^4-4{\beta }^{3}k+6{\beta }^2{k}^2-4\beta k^3)({\beta }^2+k^2+2k\beta )+\left(4k\beta \right)\left({\beta }^4+k^4-2{\beta }^2{k}^2\right)}{({\beta }^2+k^2+2k\beta )\left({\beta }^4+k^4-2{\beta }^2{k}^2\right)} \\ =\frac{\begin{array}{c}{\beta }^2{\beta }^4+k^2{\beta }^4+2k\beta {\beta }^4+k^4{\beta }^2+k^4{k}^2+k^{4}2k\beta -4{\beta }^{3}k{k}^2-4{\beta }^{3}k2k\beta \\ +6{\beta }^2{k}^2{\beta }^2+6{\beta }^2{k}^2{k}^2+6{\beta }^2{k}^{2}2k\beta -4\beta k^3{k}^2-4\beta k^3{k}^2-4\beta k^{3}2k\beta +4k\beta {\beta }^4\\ +4k\beta k^4-4k\beta 2{\beta }^2{k}^2\end{array}}{{\beta }^2{\beta }^4+{\beta }^2{k}^4-{\beta }^{2}2{\beta }^2{k}^2+k^2{k}^4-k^{2}2{\beta }^2{k}^2+2k\beta {\beta }^2+2k\beta k^4-2k\beta 2{\beta }^2{k}^2} \end{gathered}$$

Solving further as:

$$\begin{gathered} \begin{array}{c}\begin{array}{c}{\beta }^6+k^2{\beta }^4+2k{\beta }^5+k^4{\beta }^2+k^6+k^{5}2\beta -4{\beta }^{5}k-4{\beta }^3{k}^3-8{\beta }^4{k}^2\\ +6k^2{\beta }^4+6{\beta }^2{k}^4+12{\beta }^3{k}^3-4k^3{\beta }^3-4\beta k^5-8{\beta }^2{k}^4+4k{\beta }^5\end{array}\\ R+T=\frac{+4\beta k^5-8{\beta }^3{k}^3}{{\beta }^6+{\beta }^2{k}^4-2{\beta }^4{k}^2+k^2{\beta }^4+k^6-2{\beta }^2{k}^4+2k{\beta }^5+2\beta k^{5}4{\beta }^3{k}^3}\\ \end{array} \\ =\frac{{\beta }^6+{\beta }^2{k}^4-2{\beta }^4{k}^2+k^2{\beta }^4+k^6-2{\beta }^2{k}^4+2k{\beta }^5+2\beta k^{5}4{\beta }^3{k}^3}{{\beta }^6+{\beta }^2{k}^4-2{\beta }^4{k}^2+k^2{\beta }^4+k^6-2{\beta }^2{k}^4+2k{\beta }^5+2\beta k^{5}4{\beta }^3{k}^3} \\ R+T=1 \end{gathered}$$

Thus, the transmission coefficient is calculated as

$T=\frac{4\sqrt{E}\sqrt{-V_0+E}\times {\left(4\sqrt{E}\sqrt{(-V_0+E)}\right)}^2}{{\left(V_0\right)}^2}\mathrm{and}R+T=1.$