91影视

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:

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

The Schrodinger equation can be calculated as:

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

For $x鈮�0$

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

For

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

Let,

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

For $x鈮�0$

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

For $x>0$

$\frac{d^2{唯}_2}{d{x}^2}-{伪}^2{唯}_2=0$

For $x鈮�0$

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

For $x>0$

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

In the above equations, F will be zero for E < V₀ ( $x鈫�+鈭�$) 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" $$\begin{gathered} {唯}_1\left(x\right)=A{e}^{ikx}+B{e}^{-ikx}\left(1\right) \\ {唯}_2\left(x\right)=G{e}^{-伪x}\left(2\right) \end{gathered}$$

Apply boundary conditions as:

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

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

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

Using the equations above, we get,

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

Solving the above equations and we get,

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

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

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

The reflection coefficient can be calculated as:

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

Substituting the values, and we get,

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

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

The Schrodinger equation can be calculated as:

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

For $x鈮�0$

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

For $x>0$

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

We know:

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

For $x鈮�0$

$\frac{d^2{唯}_1}{d{x}^2}+k^2{唯}_1=0$

For $x>0$

$\frac{d^2{唯}_2}{d{x}^2}+{尾}^2{唯}_2=0$

For $x鈮�0$

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

For $x>0$

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

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

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

Apply boundary conditions as:

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

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

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

Using the equations above, we get,

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

Solving the above equations and we get,

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

The reflection coefficient can be calculated as:

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

Substituting the values, and we get,

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

We know:

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

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

And

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

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:

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

Substitute the values in equation 11, and we get,

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

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

Solving the equation further:

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

From equations 13 and 14:

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

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

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

The transmission coefficient can be written as:

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

The transmission coefficient can be written as:

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

Substituting values in the above expression, we get,

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

From subpart b, the reflection coefficient is:

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

Solving further as:

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

Thus, the transmission coefficient is calculated as

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