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HW 6: Particle in a Box

Table of useful integrals

Indefinite integralDefinite integral over [0,L][0, L]
sin(nx)cos(mx)dx=0\int\sin (nx) cos(mx)dx=0
(for any n and m)
0Lsin(nπxL)cos(nπxL)dx=0\int^{L}_0\sin ( \frac{n\pi x}{L}) cos( \frac{n\pi x}{L})dx=0
sin2(kx)dx=x2sin(2kx)4k\int sin^2 (kx)dx=\frac{x}{2}-\frac{sin(2kx)}{4k}
0Lsin2(nπxL)dx=L2\int^L_0 sin^2 \big( \frac{n\pi x}{L}\big)dx=\frac{L}{2}
xsin2(kx)dx=a24\int x\cdot sin^2 (kx)dx=\frac{a^2}{4}
0Lxsin2(nπxL)dx=L24\int^L_0 x\cdot sin^2 \big( \frac{n\pi x}{L}\big)dx=\frac{L^2}{4}
x2sin2(kx)dx=x36(x24k18k3)sin(2kx)xcos(2kx)4k2\int x^2 \cdot sin^2 (kx)dx=\frac{x^3}{6}-\Big(\frac{x^2}{4k}-\frac{1}{8k^3} \Big)sin(2kx)-\frac{xcos(2kx)}{4k^2}
0Lx2sin2(nπxL)dx=(L2πn)3(4π3n332nπ)\int^L_0 x^2 \cdot sin^2 \big( \frac{n\pi x}{L}\big)dx= \big( \frac{L}{2\pi n}\big)^3 \big (\frac{4 \pi^3 n^3}{3}-2n\pi \big )
(Optional exercise for the fearless souls) Give it a try and see if you can obtain these expressions by carrying out integrations explicitly. One approach is to use Euler’s relation with integration by parts.

Q-1

Using the table of integrals above along with lecture notes/book calculate the following quantities for the particle in a box described by the following wave-function ψn(x)=(2L)1/2sin(nπxL)\psi_n(x)=\Big(\frac{2}{L} \Big)^{1/2} sin \Big(\frac{n\pi x}{L}\Big).

Q-2

Particle in a box (PIB) system is a useful toy model for learning about the behaviour of electrons bound in atoms and molecules. Using the average quantities computed above comment on the meaning of the following variations between two extreme limits

Q-3

To solve this problem read carefully the end of chapter 3.5.

Particle in a box can also be a useful toy model for estimating excitation energies conugated molecules like butediene

CH2=CHCH=CH2CH_2=CH-CH=CH_2

. By taking LL to be the length of the linear chain and filling up energy levels by each pair of π\pi electrons (e.g 2 in level E1E_1 and 2 in E2E_2 for butediene) one can compute transitions from highest occupied state to the next unoccupied state. For butadiene it will be

hν=E3E2h\nu = E_3-E_2

Q-4

Normalize the discrete and continuous functions thereby making them proper probability distributions.

Q-5

In 2D and 3D particle in a box it is possible to have degenerate energy levels becasue of the symmetry of box. What are the possible degeneracies of the first four energy levels of a partcile in 3D box with