A particle of mass m and kinetic energy E > 0 approaches an abrupt potential drop V₀ (Figure 2.19).

(a)What is the probability that it will “reflect” back, if E = V₀/3? Hint: This is just like problem 2.34, except that the step now goes down, instead of up.

(b) I drew the figure so as to make you think of a car approaching a cliff, but obviously the probability of “bouncing back” from the edge of a cliff is far smaller than what you got in (a)—unless you’re Bugs Bunny. Explain why this potential does not correctly represent a cliff. Hint: In Figure 2.20 the potential energy of the car drops discontinuously to −V0, as it passes x = 0; would this be true for a falling car?

(c) When a free neutron enters a nucleus, it experiences a sudden drop in potential energy, from V = 0 outside to around −12 MeV (million electron volts) inside. Suppose a neutron, emitted with kinetic energy 4 MeV by a fission event, strikes such a nucleus. What is the probability it will be absorbed, thereby initiating another fission? Hint: You calculated the probability of reflection in part (a); use T = 1 − R to get the probability of transmission through the surface.

Schrodinger equation is given by,

$\sout{\mathrm{h}}\frac{\partial \mathrm{\psi }}{\partial \mathrm{t}}=-\frac{\mathrm{i}\sout{\mathrm{h}}}{2\mathrm{m}}\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}+\mathrm{V}(\mathrm{x},\mathrm{t})\mathrm{\psi }(\mathrm{x},\mathrm{t})-\infty <\mathrm{x}<\infty ,\mathrm{t}>0$

Where, the drop in potential energy is,

$V(x,t)=V\left(x\right)=\left\{\begin{array}{ll}0,& x\le 0\\ -V_0,& x>0\end{array}\right.$

Assuming,

$\psi (x,t)=\psi \left(x\right)\varphi \left(t\right)$

We get two separate equations as:

$$\begin{gathered} \sout{h}\frac{{\varphi }^{\text{'}}\left(t\right)}{\varphi \left(t\right)}=E \\ -\frac{\sout{h^2}}{2m}\frac{{\psi }^{\text{'}\text{'}}\left(x\right)}{\psi \left(x\right)}+v\left(x\right)=E \end{gathered}$$

Therefore the Schrodinger equation will be :

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

$$\begin{gathered} \frac{d^2\psi }{d{x}^2}=\frac{2m}{{\sout{h}}^2}(-E)\psi ,\mathrm{for}x\le 0 \\ \frac{d^2\psi }{d{x}^2}=\frac{2m}{{\sout{h}}^2}(-V_0-E)\psi ,\mathrm{for}x>0 \\ For,E=\frac{V_0}{3} \\ \frac{d^2\psi }{d{x}^2}=\frac{2m{V}_0}{3{\sout{h}}^2}(-E)\psi ,\mathrm{for}x\le 0 \\ \frac{d^2\psi }{d{x}^2}=\frac{8m{V}_0}{3{\sout{h}}^2}\psi ,\mathrm{for}x>0 \end{gathered}$$

Hence, the general solution would be:

$\psi \left(x\right)=\left\{\begin{array}{ll}Aexp\left(i\sqrt{\frac{2m{V}_0}{3{\sout{h}}^3}}x\right)+Bexp\left(-i\sqrt{\frac{2m{V}_0}{3{\sout{h}}^3}}x\right),& x\le 0\\ Cexp\left(i\sqrt{\frac{8m{V}_0}{3{\sout{h}}^3}}x\right)+Dexp\left(-i\sqrt{\frac{8m{V}_0}{3{\sout{h}}^3}}x\right),& x>0\end{array}\right.$

Solving the ordinary differential equation in t, we get $\varphi \left(t\right)=e^{-iEt/h}$, Hence $\psi (x,t)=\psi \left(x\right)\varphi \left(t\right)$is a linear combination of waves traveling to the left and right.

Here, we assume the wave to only be incident from the left, which yields D = 0.

Hence,

$\psi \left(x\right)=\left\{\begin{array}{ll}A{e}^{ikx}+B{e}^{-ikx}& x\le 0\\ C{e}^{2ikx}& x>0\end{array}\right.$

Where,$k=\sqrt{\frac{2m{V}_0}{3{\sout{h}}^2}}$

The reflection coefficient is defined as the magnitude of the ratio of the reflected probability current to the incident probability current.

Where the probability current is,

$$\begin{gathered} J(x,t)=\frac{i\sout{h}}{2m}\left(\psi \frac{\partial {\psi }^{*}}{\partial x}-{\psi }^{*}\frac{\partial \psi }{\partial x}\right) \\ J(x,t)=\frac{i\sout{h}}{2m}\left\{\left[\psi \right(x\left)e^{-Et/h}\right]\frac{\partial }{\partial x}\left[{\psi }^{*}\right(x\left)e^{iEt/h}\right]-\left[{\psi }^{*}\right(x\left)e^{iEt/h}\right]\frac{\partial }{\partial x}\left[\psi \right(x\left)e^{-iEt/h}\right]\right\} \\ J(x,t)=\frac{i\sout{h}}{2m}\left\{\left[\psi \right(x\left)e^{-Et/h}\right]\frac{d{\psi }^{*}}{\partial x}{e}^{iEt/h}-\left[{\psi }^{*}\right(x\left)e^{iEt/h}\right]\frac{d\psi }{dx}{e}^{-iEt/h}\right\} \\ J(x,t)=\frac{i\sout{h}}{2m}\left(\psi \frac{d{\psi }^{*}}{dx}-{\psi }^{*}\frac{d\psi }{dx}\right) \end{gathered}$$

Hence, the reflection coefficient,

localid="1658228499814" $$\begin{gathered} R=\left|\frac{{\displaystyle \frac{i}{2m}\left(B{e}^{-ikx}\right)\frac{d}{dx}\left(B^{*}{e}^{ikx}\right)-\left(B^{*}{e}^{ikx}\right)\frac{d}{dx}\left(B{e}^{-ikx}\right)}}{\frac{i}{2m}\left(A{e}^{ikx}\right)\frac{d}{dx}\left(A^{*}{e}^{-ikx}\right)-\left(A^{*}{e}^{-ikx}\right)\frac{d}{dx}\left(A{e}^{ikx}\right)}\right| \\ R=\left|\frac{{\displaystyle \left(B{e}^{-ikx}\right)\left(B^{*}{e}^{ikx}\right)-\left(B^{*}{e}^{ikx}\right)\left(B{e}^{-ikx}\right)}}{\left(A{e}^{ikx}\right)\left(A^{*}{e}^{-ikx}\right)-\left(A^{*}{e}^{-ikx}\right)\left(A{e}^{ikx}\right)}\right| \\ R=\left|\frac{2ikB{B}^{*}}{-2ikA{A}^{*}}\right| \\ R=\frac{{\left|B\right|}^2}{{\left|A\right|}^2} \end{gathered}$$

To determine one of the constants, the wave function needs to x = 0 be continuous at

i.e.

$\underset{x\to 0^{-}}{lim}\psi \left(x\right)=\underset{x\to 0^{+}}{lim}\psi \left(x\right)$

A + B = C

Let $\mu$be a very small number, and integrate the time-independent Schrodinger equation with respect to x from$-\mu$to$\mu$

$$\begin{gathered} {\int }_{-\mu }^{\mu }\frac{d^2\psi }{d{x}^2}dx={\int }_{-\mu }^{\mu }\frac{2m}{{\sout{h}}^2}\left[V\left(x\right)-E\right]\psi \left(x\right)dx \\ {\overline{)\frac{d\psi }{dx}}}_{-\mu }^{\mu }={\int }_{-\mu }^0\frac{2m}{{\sout{h}}^2}(-E)\psi \left(x\right)dx+{\int }_0^{\mu }\frac{2m}{{\sout{h}}^2}(-V_0-E)\psi \left(x\right)dx \\ {\overline{)\frac{d\psi }{dx}}}_{-\mu }^{\mu }=\frac{2m}{{\sout{h}}^2}(-E)\psi \left(0\right){\int }_{-\mu }^{0}dx+\frac{2m}{{\sout{h}}^2}(-V_0-E)\psi \left(0\right){\int }_0^{\mu }dx \\ {\overline{)\frac{d\psi }{dx}}}_{-\mu }^{\mu }=\frac{2m}{{\sout{h}}^2}(-E)\psi \left(0\right)\mu +\frac{2m}{{\sout{h}}^2}(-V_0-E)\psi \left(0\right)\mu \end{gathered}$$

Taking the limit as$\mu \to 0$

${\overline{)\frac{d\psi }{dx}}}_{0^{-}}^{0^{+}}=0$

Hence, $\frac{\partial \psi }{\partial x}$is continuous at x = 0

So,

$$\begin{gathered} \underset{x\to 0^{-}}{lim}\frac{d\psi }{dx}=\underset{x\to 0^{+}}{lim}\frac{d\psi }{dx} \\ ik(A-B)=2ikC \\ ik(A-B)=2ik(A+B) \\ B=-\frac{1}{3}A \end{gathered}$$

Therefore, the reflection coefficient, and hence, the probability that the mass will turn back at x = 0 is,

$$\begin{gathered} R=\left(\frac{B}{A}\right){\left(\frac{B}{A}\right)}^{*} \\ R=\left(-\frac{1}{3}\right){\left(-\frac{1}{3}\right)}^{*} \\ R=\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right) \\ R=\frac{1}{9} \end{gathered}$$

Thus, 1/9 is the probability that it will “reflect” back.

Here, the potential energy (V(x)) drops discontinuously from 0 to $-V_0$at x = 0, but comparing this to a car that drops off a cliff, its potential energy drops continuously from 0 to $-V_0$linearly.

i.e. $V_{car}\left(x\right)=-mgx$, where x is the vertical height.

Since a neutron has energy $E=4MeV$, which is equal to a third of $V_0=12MeV$, so, the previous analysis holds true.

The transmission coefficient is defined by the magnitude of the ratio of the transmitted probability current to the incident probability current.

$j(x,t)=\frac{i\sout{h}}{2m}\left(\psi {\frac{d\psi }{dx}}^{*}-{\psi }^{*}\frac{d\psi }{dx}\right)$

For,

$\psi \left(x\right)=\left\{\begin{array}{ll}A{e}^{ikx}+B{e}^{-ikx},& x\le 0\\ C{e}^{2ikx},& x>0\end{array}\right.$

Hence, the transmission coefficient,

$$\begin{gathered} T=\left|\frac{{\displaystyle \frac{{\displaystyle i\sout{h}}}{{\displaystyle 2m}}\left[\left(C{e}^{-ikx}\right)\frac{{\displaystyle d}}{{\displaystyle dx}}\left(C^{*}{e}^{-2ikx}\right)-\left(C^{*}{e}^{-2ikx}\right)\frac{{\displaystyle d}}{{\displaystyle dx}}\left(C{e}^{2ikx}\right)\right]}}{\frac{i\sout{h}}{2m}\left[\left(A{e}^{ikx}\right)\frac{d}{dx}\left(A^{*}{e}^{-ikx}\right)-\left(A^{*}{e}^{-ikx}\right)\frac{d}{dx}\left(A{e}^{ikx}\right)\right]}\right| \\ T=\left|\frac{{\displaystyle \left(C{e}^{2ikx}\right)(-2ik{C}^{*}{e}^{-2ikx})-\left(C^{*}{e}^{-2ikx}\right)\left(2ikC{e}^{2ikx}\right)}}{\left(A{e}^{ikx}\right)(-ik{A}^{*}{e}^{-ikx})-\left(A^{*}{e}^{-ikx}\right)\left(ikA{e}^{ikx}\right)}\right| \\ T=\left|\frac{-4ikC{C}^{*}}{-2ikA{A}^{*}}\right| \\ T=2\frac{{\left|C\right|}^2}{{\left|A\right|}^2} \end{gathered}$$

At : x = 0

role="math" localid="1658229056899" $$\begin{gathered} \underset{x\to 0^{-}}{lim}\psi \left(x\right)=\underset{x\to 0^{+}}{lim}\psi \left(x\right) \\ A+B=C \end{gathered}$$

And

$$\begin{gathered} \underset{x\to 0^{-}}{lim}\frac{d\psi }{dx}=\underset{x\to 0^{+}}{lim}\frac{d\psi }{dx} \\ ik(A-B)=2ikC \end{gathered}$$

Adding both the equations,

$$\begin{gathered} 2A=3C \\ C=\frac{2}{3}A \end{gathered}$$

Therefore, the transmission coefficient, i.e., the probability that the neutron will enter the nucleus is,

$$\begin{gathered} T=2\left(\frac{C}{A}\right){\left(\frac{C}{A}\right)}^{*} \\ T=2\left(\frac{2}{3}\right)\left(\frac{2}{3}\right) \\ T=\frac{8}{9} \end{gathered}$$

Thus, the transmission coefficient, i.e., the probability that the neutron will enter the nucleus is 8/9. Hence, R + T = 1