# (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 \(R = 0\).