1 Problem set 3

Handout: 9/12Due date: 9/19The following problems are from Levine 6th edition.Required:3.7, 3.9, 3.13, 3.18, 3.23, 3.25, 3.29, 3.33, 3.35, 3.36, 3.41, 3.43,Optional:3.1, 3.10, 3.17, 3.44

3.7)We have

(A+ B)f = Cf (1)

Af + Bf = Cf (2)

Af = Cf − Bf (3)

Af = (C − B)f (4)

A = C − B (5)

3.9)The two operators are given by A = x3 and B d

dx. Therefore, we have

ABf = x3d

dxf = x4f ′ (6)

=> AB = x3d

dx(7)

and

BAf =d

dx(x3f) = 3x2f + x3f ′ (8)

=> BA = 3x2 + x3d

dx(9)

3.13)We have

[A, B]f = (AB − BA)f = ABf − BAf (10)

and

[B, A]f = (BA− AB)f = BAf − ABf = −[A, B]f => [B, A] = −[A, B](11)

1

3.18)We have

A(B + C)f = A(Bf + Cf) (12)

= A(Bf) + A(Cf) (13)

= ABf + ACf (14)

= (AB + AC)f (15)

=> A(B + C) = AB + AC (16)

3.23)We know that Thf(x) = f(x+ h).a)

Th[f(x) + g(x)] = f(x+ h) + g(x+ h) = Thf(x) + Thg(x) (17)

andTh[cf(x)] = cf(x+ h) = cThf(x) (18)

which show that Th is linear.b) Here h = 1 and f(x) = x2. Therefore,

(T1T1 − 3T1 + 2)x2 = (x+ 2)2 − 3(x+ 1)2 + 2x2 = −2x+ 1 (19)

3.25)a)

d2

dx2exp(x) =

d

dxexp(x) = exp(x) (20)

with eigenvalue λ = 1.b)

d2

dx2x2 =

d

dx2x = 2 (21)

and is therefore not an eigenfunction.c)

d2

dx2sin(x) =

d

dxcos(x) = − sin(x) (22)

with eigenvalue λ = −1.d)

d2

dx23 cos(x) = −3

d

dxsin(x) = −3 cos(x) (23)

2

with eigenvalue λ = −1.e)

d2

dx2(sin(x) + cos(x)) =

d

dx(cos(x) − sin(x)) = −(sin(x) + cos(x)) (24)

with eigenvalue λ = −1.

3.29)We have pα = −i~(∂/∂α).a)

p3

y = (−i~)3(∂/∂y)3 = i~3∂3/∂y3 (25)

b)xpy − ypx = −i~x∂/∂y + i~y∂/∂x (26)

c)

(−i~x∂/∂y)2f(x, y) = −~2(x∂/∂y)(x∂f/∂y) = −~

2(x2∂2f/∂y2) (27)

=> (xpy)2 = −~

2(x2∂2/∂y2) (28)

3.33)a)

∫

2

0

|Ψ(x, t)|2dx (29)

b)∫

∞

0

∫

∞

0

∫

2

0

|Ψ(x, y, z, t)|2dxdydz (30)

c)∫

∞

0

∫

∞

0

∫

∞

0

∫

∞

0

∫

∞

0

∫

2

0

|Ψ(x, y, z, t)|2dx1dy1dz1dx2dy2dz2 (31)

3.35)The ground state has quantum numbers of nanbnc = 111. The first excitedstate would correspond to one of the quantum number raised to two. Sincewe know that the energy decreases as the box-length increases. Thus, the firstexcited state will be the one with the quantum number two in the directionof the longest edge with is c. Therefore,

hν =h2

8m

(

1

a2+

1

b2+

4

c2

)

−h2

8m

(

1

a2+

1

b2+

1

c2

)

(32)

=> v =3h

8mc2=

3(6.626 × 10−34Js

8(9.109 × 10−31kg)(6.00× 10−10m)2= 7.58 × 1014s−1 (33)

3

3.36)a)We need to evaluate the following integral in units of nm

∫

3

2

∫

2

1.5

∫

0.4

0

|ψ|2dxdydz (34)

∫

0.4

0

(2/a) sin2(πx/a)dx

∫

2

1.5

(2/b) sin2(πy/b)dy

∫

3

2

(2/c) sin2(πz/c)dz (35)

[

x

a−

sin(2πx/a)

2π

]0.4

0

×

[

y

b−

sin(2πy/b)

2π

]2

1.5

×

[

z

c−

sin(2πz/c)

2π

]3

2

(36)

= (0.3065)(0.09085)(0.3871) = 0.0108 (37)

b) The integration of y and z is over the complete region and thus equalsone. Therefore, the results is equal to the x integral of part a = 0.3065.c) The same as b.

3.41)For a linear combination of eigenfunctions of H to be an eigenfunction ofH, the eigenfunctions most have degenerate eigenvalues. For the particle incubic box this mean that they most have the same n2

x + n2

y + n2

z. This is thecase for (a) and (c) but not for (b).

3.43)a) The only state the gives this is energy is for (222). Thus the state isnondegenerate.b) The following states gives the energy (123), (132), (213), (231), (312),(321). Thus the degeneracy is 6.c) The following state have the energy: (115), (151), (511), (333). Thus, thedegeneracy is 4.

4