Question

Consider the potential $\mathit{V}\left(x\right)\mathbf{=}\mathbf{-}\frac{\sout{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{m}}\mathit{s}\mathit{e}\mathit{c}{\mathit{h}}^{\mathbf{2}}\left(ax\right)$where a is a positive constant, and "sech" stands for the hyperbolic secant

(a) Graph this potential.

(b) Check that this potential has the ground state

${\mathbf{\psi }}_{\mathbf{0}}\left(x\right)$and find its energy. Normalize and sketch its graph.

(C)Show that the function ${\mathbf{\psi }}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{A}\left(\frac{\mathrm{ik}-a\mathrm{ta}\mathrm{nh}\left(\mathrm{ax}\right)}{\mathrm{ik}+a}\right){\mathbf{e}}^{\mathbf{kx}}$

(Where $\mathbf{k}\mathbf{=}\sqrt{\mathbf{2}\mathbf{mE}}\mathbf{}\mathbf{i}\sout{\mathbf{h}}$as usual) solves the Schrödinger equation for any (positive) energy E. Since$\mathbf{ta}\mathbf{}\mathbf{nh}\mathbf{}\mathbf{z}\mathbf{\to }\mathbf{-}\mathbf{1}\mathbf{as}$as This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e., no term . What is the asymptotic form${\mathbf{\psi }}_{\mathbf{k}}\left(x\right)$ of at large positive x? What are R and T, for this potential? Comment: This is a famous example of a reflectionless potential-every incident particle, regardless of its energy, passes right through.

Short answer

(a) The graph of the potential is,

(b)Hence the energy is, $-\frac{\sout{h^2}{a}^2}{2m}$.

(c) For any wave function${\psi }_k\left(x\right)$ the transmission coefficient A is equal to 1, therefore, every state gets completely transmitted.

Step-by-step solution

Define the Schrödinger equation

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

Step 2: Plot the graph

(a)

Sketch of the potential for any parameter a is plotted below,

 Step 3 : Determination of the energy and graph

(b)

The given equation:

${\psi }_0\left(x\right)=Asech\left(ax\right)$

Differentiate with respect to x.

localid="1658302767211" $\frac{{\mathrm{d\psi }}_0}{\mathrm{dx}}=-\mathrm{Aa}\mathrm{sech}\left(\mathrm{ax}\right)\tanh \left(\mathrm{ax}\right)$

$\frac{{\mathrm{d\psi }}_0}{\mathrm{dx}}=-{\mathrm{Aa}}^2\left\{-\mathrm{sech}\left(\mathrm{ax}\right)\tan h^2\left(\mathrm{ax}\right)+sech\left(ax\right)sec{h}^2\left(ax\right)\right\}$ ...(1)

The given equation:

${\mathrm{H\psi }}_{\mathrm{k}}\left(\mathrm{x}\right)={\mathrm{E\psi }}_{\mathrm{k}}\left(\mathrm{x}\right)$

Now rewrite the Hamiltonian equation:

localid="1658303461512" $H{\psi }_0=-\frac{\sout{h}}{2m}\frac{d^2{\psi }_0}{d{x}^2}-\frac{{\sout{h}}^2{a}^2}{m}sec{h}^2\left(ax\right){\psi }_0$ (2)

Substitute the value of $\frac{d^2{\psi }_0}{d{x}^2}$ and ${\psi }_0$ in equation (2).

$$\begin{gathered} H{\psi }_0=\frac{\sout{h}}{2m}\times A{a}^2\left\{-sech\left(ax\right)\tan h^2\left(ax\right)+sec{h}^2\left(ax\right)\right\}-\frac{{\sout{h}}^2{a}^2}{m}Asec{h}^3\left(ax\right){\psi }_0 \\ H{\psi }_0=\frac{\sout{h^2}A{a}^2}{2m}\times A{a}^2\left\{-sech\left(ax\right)\tan h^2\left(ax\right)+2sec{h}^2\left(ax\right)\right\} \\ =\frac{\sout{h^2}A{a}^2}{2m}\times A{a}^2\left\{-sech\left(ax\right)\tan h^2\left(ax\right)+2sec{h}^2\left(ax\right)\right\} \end{gathered}$$

But $\tan h^2\left(ax\right)+sec{h}^2\left(ax\right)=1$so,

localid="1658304817569" $$\begin{gathered} H{\psi }_0=-\frac{\sout{h^2}A{a}^2}{2m}sech\left(ax\right)\left\{1\right\} \\ H{\psi }_0=-\frac{\sout{h^2}A{a}^2}{2m}sech\left(ax\right) \\ =\frac{\sout{h^2}{a}^2}{2m}{\psi }_0 \end{gathered}$$

Hence the energy is, $\frac{\sout{h^2}{a}^2}{2m}$.

For normalize ${\psi }_0$-

$$\begin{gathered} 1=A^2{\int }_{-\infty }^{\infty }sec{h}^2\left(ax\right)dx \\ =A^2\times \frac{1}{a}{\left[\tan h\left(ax\right)\right]}_{-\infty }^{\infty } \\ {A}^2=\frac{a}{2} \\ A=\sqrt{\frac{a}{2}} \end{gathered}$$

Determination of the Schrödinger equation

(c)

The given equation:

$$\begin{gathered} {\mathrm{\psi }}_{\mathrm{k}}\left(\mathrm{x}\right)=\mathrm{A}\left(\frac{\mathrm{ik}-\mathrm{a}\tanh \left(\mathrm{ax}\right)}{\mathrm{ik}+\mathrm{a}}\right){\mathrm{e}}^{\mathrm{ikx}} \\ \mathrm{Differentiate}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x}. \end{gathered}$$$$\begin{gathered} \frac{d{\psi }_k}{dx}=\frac{A}{ik+a}\left\{\left(ik-a\tan hax\right)ik-a^{2}sec{h}^{2}ax\right\}{e}^{ikx} \\ \frac{d{\psi }_k}{dx}=\frac{A}{ik+a}\left\{\left(ik-a\tan hax\right)ik-a^{2}sec{h}^{2}ax\right\}-a^{2}iksec{h}^{2}ax+2a^{3}sec{h}^{2}ax\tan hax \end{gathered}$$

For given equation-

$H{\psi }_k\left(x\right)=\left(-\frac{\sout{h^2}}{2m}\frac{{\partial }^2{\psi }_k}{\partial x^2}-\frac{{\sout{h}}^2{a}^2}{m}sec{h}^2\left(ax\right){\psi }_k\right)$

Differentiate with respect to x.

localid="1658306528413" $H{\psi }_k\left(x\right)=\left(-\frac{\sout{h^2}}{2m}\frac{{\partial }^2{\psi }_k}{\partial x^2}-\frac{{\sout{h}}^2{a}^2}{m}sec{h}^2\left(ax\right){\psi }_k\right)$

Substitute the value of $\frac{d^2{\psi }_k}{dx}$and ${\psi }_k$in the above equation

$H{\psi }_k\left(x\right)=\left(-\frac{\sout{h^2}}{2m}\frac{{\partial }^2{\psi }_k}{\partial x^2}-\frac{{\sout{h}}^2{a}^2}{m}sec{h}^2\left(ax\right){\psi }_k\right)$

$$\begin{gathered} =\frac{A}{ik+a}\left[-\frac{{\sout{h}}^{2}ik}{2m}\left\{-k^2-iak\tan hax-a^{2}sec{h}^{2}ax\right\}+\frac{{\sout{h}}^2{a}^2}{2m}iksec{h}^{2}ax\right]e^{ikx} \\ =\frac{A}{ik+a}\left\{-\frac{{\sout{h}}^2{a}^2}{m}sec{h}^{2}ax\tan ax-\frac{{\sout{h}}^2{a}^2}{m}sec{h}^{2}ax\left(ik-a\tan hax\right)\right\}{e}^{ikx} \end{gathered}$$

$$\begin{gathered} =\frac{A{e}^{ikx}}{ik+a}\frac{\sout{h^2}}{2m}\left(\begin{array}{c}i{k}^3-a{k}^2\tan hax+i{a}^{2}ksec{h}^{2}ax+i{a}^{2}ksec{h}^{2}ax-2a^{3}sec{h}^{2}ax\tan hax\\ -2i{a}^{2}ksec{h}^{2}ax+2a^{3}sec{h}^{2}ax\tan hax\end{array}\right) \\ =\frac{A{e}^{ikx}}{ik+a}\frac{\sout{h^2}}{2m}{k}^2\left(ik-a\tan hax\right) \end{gathered}$$

$$\begin{gathered} =\frac{\sout{h^2}{k}^2}{2m}{\psi }_k\left(x\right) \\ =E{\psi }_k\left(x\right) \end{gathered}$$

Evaluate further and get,

$H{\psi }_k\left(x\right)=E{\psi }_k\left(x\right)$

Observe the behaviour of the wave function

When$x\to +\infty$

$$\begin{gathered} \tan h\left(ax\right)\to 1 \\ whenx\to +\infty , \\ {\psi }_k\left(x\right)\to A\left(\frac{ik-a}{ik+a}{e}^{ikx}\right) \end{gathered}$$

$T={\left|\frac{ik-a}{ik-a}\right|}^2=1$

Thus, the conclusion is, for any wave function ${\psi }_k\left(x\right)$the transmission coefficient A is equal to 1, therefore, every state gets completely transmitted.