11th March 2021

(a)

For particle in ground state of infinite square well potential (particle in a box), we have its wavefunction given by $u(x) = \sqrt{\frac2L}\sin\left(\frac{\pi x}{L}\right)$ for $$0 \leq x \leq L$$.

The probability that the particle will be found in the region $$0 < x < \frac{L}3$$ is given by \begin{align*} \int_0^{L/3} \left\lvert u(x) \right\rvert^2\ dx &= \frac2L \int_0^{L/3} \sin^2 \left( \frac{\pi x}{L} \right)\ dx \\ &= \frac2L \left[ \frac{x}{2} - \frac{L\sin \left( \frac{2\pi x}{L} \right)}{4\pi} \right]_0^{L/3} \\ &= \frac2L \left[ \frac{L}{6} - \frac{L\sin \left( \frac{2\pi}{3} \right)}{4\pi} \right] \\ &= \frac{1}{3} - \frac{\sqrt{3}}{4\pi}. \end{align*}

(b)

We have \begin{align*} u(x) &= Ae^{-\beta x^2} \\ u'(x) &= -2A\beta x e^{-\beta x^2} \\ u''(x) &= 2A\beta (2\beta x^2 - 1) e^{-\beta x^2} \\ &= 2\beta(2\beta x^2 - 1) u(x) \end{align*} solving time-independent Schrödinger equation we have \begin{align*} -\frac{\hbar^2}{2m} 2\beta(2\beta x^2 - 1) + V(x) &= E \\ V(x) - \frac{\hbar^2}{m}(2\beta^2 x^2 - \beta) &= E \end{align*} substituting $$V(0) = 0$$ we obtain $$E = \frac{\hbar^2\beta}{m}$$, then we have expression for $$V(x)$$ as $V(x) = \frac{2\hbar^2\beta^2}{m}x^2.$

(c)

In each region the solution to time-independent Schrödinger equation takes the form $\psi(x) = \begin{cases} Ae^{i k_1 x} + Be^{-i k_1 x} &\text{ in region 1} \\ Ce^{i k_2 x} + De^{-i k_2 x} &\text{ in region 2} \\ Fe^{i k_1 x} &\text{ in region 3} \end{cases}$

where the wave number in region 1 and 3 is $$k_1 = \frac{\sqrt{2mE}}{\hbar}$$, in region two we have $$k_2 = \frac{\sqrt{2m(E + V_0)}}{\hbar} = \frac{2\pi}{\lambda_2}$$. As the beam is incident in the positive $$x$$-direction, there is only right-moving wave in region 3.

As $$\phi(x)$$ is continuous and differentiable at $$x =0$$ and $$x=L$$ we have \begin{align*} A + B &= C + D \\ i k_1 A - i k_1 B &= i k_2 C - i k_2 D \\ Ce^{i k_2 L} + D e^{-i k_2 L} &= F e^{i k_1 L} \\ i k_2 Ce^{i k_2 L} - i k_2 D e^{-i k_2 L} &= i k_1 F e^{i k_1 L} \end{align*} transmission coefficient is given by $$T = \left\lvert F/A \right\rvert^2$$. Solving these equations then rewriting $$k_1,k_2$$ in terms of $$E, V_0, \lambda_2$$ we obtain $T = \frac{4E(E+V_0)}{4E(E+V_0) + V_0^2\sin^2\left( \frac{2\pi L}{\lambda_2} \right)}.$ as $$2L = \lambda_2$$ it follows that $$T = 1$$ and $$R = 0$$.