Question

Course: PHYS354 - Quantum mechanics 1 Year: 1439 17/18 Problems Sheet 3 TISE in 1D Infinite square well 1. What are the eigenfunctions and eigenvalues for the 1D box problem described in lectures if the ends of the box are at -L/2 and +L/2? 2. For which values of the real angle θ will the constant C-(e"-1) have no effect in caleulations involving the modulus ICl? 3. For the ID box problem, show that P is maximu at the values z, given by 2j+1 -/ -L, rI j=0, 1, 2, . . . ,n-1 lj = Step potential Using the general definitions of the transmission and reflection coefficients in terms of current densities show that T + R = 1. 4. 5. For scattering off a step potential, show that T 0, using the definition of the transmission cocfficient in terms of current densitics; Teran/Jinel Potential barrier 6. Show that T+R-F/AP+B/AP-1 using a brute force calculation. The expressions of F/A and B/A are given in Eqs. (3.65a) and (3.65b), respectively, in the lecture notes. Harmonic oscillator 7. Using the definitions of the ladder operatorscompute the following commutators: a,r , ti, a,荫 and [a+, p. 8. Compute the following commutators là, (a+)2시(à只a"] and [(a), (à*)2]

Answer 1

1.

$\textrm{Eigenfunctions}\ \psi_n = \sqrt{\frac{2}{L}}\cos \frac{n\pi}{L}x \\ \textrm{Eigenvalues}\ E_n = \frac{n^2\hbar^2\pi^2}{2mL^2}$

2. Constant C doesn't affect the calculation if .As $e^{in\pi} = -1\ \forall \ n \in \textrm{integers}$ , so we have $|C| = \frac{1}{2}|e^{in\pi} - 1| = \frac{1}{2}|-1 - 1| = \frac{1}{2}|-2| = \frac{1}{2}(2) = 1$

So for all integer multiples of pi, the constant C would not affect the calculation.

3. $\psi_n = \sqrt{\frac{2}{L}}\cos \frac{n\pi}{L}x$

$|\psi_n|^2 = \frac{2}{L}\cos^2 \frac{n\pi x}{L}$

To calculate the points x_j

$\left. \frac{d|\psi_n|^2}{dx}\right|_{x=x_j} = 0$

$\frac{2}{L}\left (2\cos \frac{n\pi}{L}x_j \right ) \left ( -\frac{L}{n\pi}\sin \frac{n\pi}{L}x_j \right )= 0$

$\sin \frac{2n\pi}{L}x_j = \sin j\pi$

$\frac{2n\pi}{L}x_j = j\pi$

but for x_j to be maximum, substituting the above value in second derivative of $|\psi_n|^2$ we would get posetive values.

$\left.\frac{d^2|\psi_n|^2 }{d^2x} \right |_{x=x_j} \ \textrm{is posetive}$

$\frac{2}{L}\frac{L}{n\pi}\frac{L}{2n\pi}\cos \frac{2n\pi x_j}{L} = \frac{L}{n^2\pi^2}\cos j\pi$

It is posetive and maximum for odd values of j. So j = 2j + 1. So we have

$\frac{2n\pi}{L}x_j = (2j + 1)\pi \Rightarrow x_j = \frac{2j + 1}{2n}L$

4.

$R = \left |\frac{\sqrt{E} - \sqrt{E-V}}{\sqrt{E} + \sqrt{E-V}} \right |^2$

$T =\left |\frac{2\sqrt{E}\sqrt{E-V}}{(\sqrt{E} + \sqrt{E-V})^2} \right |$

$R+ T = \frac{(\sqrt{E}-\sqrt{E-V})^2}{(\sqrt{E} + \sqrt{E-V})^2} +\frac{4\sqrt{E}\sqrt{E-V}}{(\sqrt{E} + \sqrt{E-V})^2}$

$R + T =\frac{(\sqrt{E}-\sqrt{E-V})^2+4\sqrt{E}\sqrt{E-V}}{(\sqrt{E} + \sqrt{E-V})^2}$

$\\(\sqrt{E}-\sqrt{E-V})^2+4\sqrt{E}\sqrt{E-V} = E + (E-V) - 2\sqrt{E}\sqrt{E-V} + 4\sqrt{E}\sqrt{E-V}$

$\\(\sqrt{E}-\sqrt{E-V})^2+4\sqrt{E}\sqrt{E-V} = E + (E-V) + 2\sqrt{E}\sqrt{E-V}$

$\\(\sqrt{E}-\sqrt{E-V})^2+4\sqrt{E}\sqrt{E-V} = (\sqrt{E}+\sqrt{E-V})^2$

$R + T =\frac{(\sqrt{E}+\sqrt{E-V})^2}{(\sqrt{E} + \sqrt{E-V})^2} = 1$