Homework 10
PHZ 3113
Due Friday, April 22, 2011
Chapters 12-13
1. Consider a single particle which is described in quantum mechanics by the
Schrodinger equation
~2 ∂ 2 ψ
∂ψ
−
= i~
2
2m ∂x
∂t
Also, assume that ψ(x, t) has boundary conditions ψ(x = 0, t) = ψ(x = l, t) =
0.
a) Using separation of variables, as described in class, show that the general solutions
can be written
∞
X
nπx −iEn t/~
an sin
ψ(x, t) =
e
l
n=1
where En =
~2 n2 π 2
.
2ml2
b) Now assume that we have an initial condition ψ(x, t = 0) = δ(x − x0 ) where
0 < x0 < l. Show that in this case, that for t > 0 the function ψ(x, t) is given
by,
ψ(x, t) =
∞ X
2
n=1
l
sin
nπx0
nπx −iEn t/~
sin
e
l
l
Notice that the solution you found in part b) is actually the Green function G(x, x0 ; t, 0)
for the particle in a box problem.
2. Boas, Chapter 13, Section 3, Problem 2
3. Boas, Chapter 13, Section 3, Problem 4
4. Boas, Chapter 13, Section 4, Problem 6
5. Boas, Chapter 13, Section 5, Problem 2
6. Boas, Chapter 13, Section 7, Problem 14
7. Boas, Chapter 13, Section 7, Problem 18
1
8. In class, we found that in the case of diffusion, if at t=0 we have u(x, y, t = 0) =
δ(x − x0 )δ(y − y 0 ), the subsequent distribution is given by the Green function
G(x, x0 , y, y 0 ; t) =
(x−x0 )2 +(y−y 0 )2
1
−
4α2 t
e
4πα2 t
At t = 0, we place a point source in the first quadrant at x = a and y = b (where
= 0 for x = 0, and u = 0
a > 0 and b > 0). We also have boundary conditions ∂u
∂x
for y = 0. Using the image method and the Green function given above, determine
the distribution u(x, y, t) in the first quadrant (x > 0 and y > 0). Sketch the source
and the images, using plus sign for sources and minus signs for any sinks.
2