8-1
2

2
2 
2  2 n1  n2  n 3  
E
       
2m Lx  Ly  L z  





2 2

2
2
2
Lx  L , Ly  Lz  2L . Let
2  E0 . Then E  E0 4n 1  n 2  n 3 . Choose the quantum
8mL
numbers as follows:

n1
n2
1
1
1
2
1
2
2
2
1
1
1
2
1
1
2
1
2
2
1
3
E
E0
n3
1
1
2
1
2
2
1
2
3
1

6
9
9
18
12
21
21
24
14
14
ground state
first two excited states
*
*
*
next excited state
*
*
next two excited states
Therefore the first 6 states are  111 ,  121 ,  112 ,  122 ,  113 , and  131 with relative energies
E
 6 , 9, 9, 12, 14, 14. First and third excited states are doubly degenerate.
E0
8-2
(a)
n1  1 , n2  1 , n3  1


2
3 6.626  10 34 Js
3 2  2
3h2
E0 


2mL2
8 mL2 8 9.11  10 31 kg 2  10 10 m


(b)
8-3
n1  2 , n2
n1  1 , n2
n1  1 , n2
2
6h
E1 
2
8mL
 1 , n3  1 or
 2 , n3  1 or
 1 , n3  2
 2E0  56.4 eV
n 2  11
(a)
(b)
(c)
2  2  2 11 2  2 
E  

2 n 
2 


 mL2 
2
mL

n1
1
1
3
n2
1
3
1
n3
3
1 3-fold degenerate
1
 x   y  3 z 
 113  A sin  sin  sin 

 L 
 L 
 L 


2
 4.52  10 18 J  28.2 eV
 x  3 y   z 
 131  A sin  sin 
sin  
 L 
 L 
 L 
3 x   y   z 
sin  sin  
 L   L   L 
 311  A sin 
8-4
(a)
 x, y    1 x 2 y . In the two-dimensional case,   Asin k1 xsin k 2 y  where
n1 
n 2
k1 
E

2
  n1  n 2
2
(b)
and k2 
L
2
2
L
.

2
2mL
 22
If we let E0 
2 , then the energy levels are:
mL

8-5
(a)
n1
n2
1
1
1
2
2
1
2
2
 11
 12



 21

 22
n1  n 2  n 3  1 and
E111
(b)
E
E0
1
5
2
5
2
4


doubly degenerate
2
3 6.63  10 34
3 h2


 2.47  10 13 J  1.54 MeV
8mL2 8 1.67  10 27 4  10 28



22  12  12 h2

States 211, 121, 112 have the same energy and E 
 2E
2
and states 221, 122, 212 have the energy E 
(c)
8-6
8mL
22  22  12 h2


2
8mL
111
 3.08 MeV
 3E111  4.63 MeV .
Both states are threefold degenerate.
There is no force on a free particle, so that U r is a constant which, for simplicity, we take to
be zero. Substituting r, t    1 x 2 y  3 z t  into Schrödinger’s equation with U r  0
2
2
2
2

 
  

gives 
 r , t. Upon dividing through by

2 
2 
2 r , t   i
t
2m x
y
z 

 2 1x  2 y  3 z  it
 1 x 2 y  3 z t  we obtain 
. Each term in this




2m 
  t 
 1 x   2 y   3 z 

equation is a function of one variable only. Since the variables x, y, z, t are all independent,
each term, by itself, must be constant, an observation leads to the four separate equations
124
CHAPTER 8
with k22 =
QUANTUM MECHANICS IN THREE DIMENSIONS
2mE2
2
and k32 =
φ ( t ) = γ e − iω t with ω =
E
2mE3
2
. The equation for φ can be integrated once to get
and γ another indeterminate coefficient. Since the energy operator
( )
∂
∂
and i
φ = Eφ energy is sharp at the value E in this state. Also, since
∂t
∂t
 ∂2 
 ∂2 
 px2  = − 2  2  and − 2  2 ψ 1 = ( k1 )2 ψ 1 the magnitude of momentum in the x
 ∂x 
 ∂x 
direction is sharp at the value k1 . Similarly, the magnitude of momentum in the y and z
is [ E] = i
directions are sharp at the values k2 and k3 , respectively. (The sign of momentum also
will be sharp here if the mixing coefficients are chosen in the ratios
8-7
α1
= i , and so on).
β1
The stationary states for a particle in a cubic box are, from Equation 8.10
Ψ ( x , y , z , t ) = A sin ( k1 x ) sin ( k2 y ) sin ( k3 z ) e − iEt
= 0 elsewhere
0 ≤ x, y, x ≤ L
n1π
, etc. Since Ψ is nonzero only for 0 < x < L , and so on, the normalization
L
condition reduces to an integral over the volume of a cube with one corner at the origin:
where k1 =
L
L
L

2
1 = ∫ dx ∫ dy ∫ dz Ψ ( r , t ) = A 2  ∫ sin 2 ( k1 x ) dx ∫ sin 2 ( k2 y ) dy ∫ sin 2 ( k3 z ) dz 
0

0
0
L
Using 2 sin 2 θ = 1 − cos 2θ gives ∫ sin 2 ( k1 x ) dx =
0
L
L
1
−
sin ( 2 k1 x ) . But k1 L = n1π , so the last
2 4 k1
0
term on the right is zero. The same result is obtained for the integrations over y and z. Thus,
L 3
2 32
or A =
for any of the stationary states. Allowing
normalization requires 1 = A 2
2
L
()
()
the edge lengths to be different at L1 , L2 , and L3 requires only that L3 be replaced by the
12
12
( )
8 12
 2  2   2 
 8 
box volume L1 L2 L3 in the final result: A =       = 
where
 =
V
 L1  L2   L3 
 L1 L2 L3 
V = L1 L2 L3 is the volume of the box. This follows because it is still true that the wave must
vanish at the walls of the box, so that k1 L1 = n1π , and so on.
8-8
Inside the box the electron is free, and so has momentum and energy given by the de Broglie
relations p =
(
k and E = ω with E = c 2 p 2 + m2 c 4
)
12
for this, the relativistic case. Here
k = ( k1 , k2 , k3 ) is the wave vector whose components k1 , k2 , and k3 are wavenumbers
along each of three mutually perpendicular axes. In order for the wave to vanish at the walls,
the box must contain an integral number of half-wavelengths in each direction. Since
2π
and so on, this gives
λ1 =
k1
MODERN PHYSICS
λ
nπ
L = n1  1  or k1 = 1
 2 
L
λ
nπ
L = n2  2  or k2 = 2
 2 
L
λ
π
n
L = n3  3  or k3 = 3
 2 
L
Thus, p 2 =
 π c
= 

 L 
2
k2 =
2
{k12 + k22 + k32 } = 
{n12 + n22 + n32 } + ( mc 2 )
2


π 2 2
2
2
 {n1 + n2 + n3 } and the allowed energies are
L 
12
. For the ground state n1 = n2 = n3 = 1 . For an electron
confined to L = 10 fm , we use m = 0.511 MeV c 2 and c = 197.3 MeV fm to get
12
 (π ) ( 197.3 MeV fm )  2

+ ( 0.511 MeV )2 
E = 3 



10 fm
 

8-9
= 107 MeV .
L = [ l ( l + 1)]1 2
 6.63 × 10 −34 Js 
4.714 × 10 −34 Js = [ l ( l + 1)]1 2 

2π


( 4.714 × 10−34 ) ( 2π )2
l ( l + 1) =
2
( 6.63 × 10−34 )
2
= 1.996 × 101 ≈ 20 = 4 ( 4 + 1)
so l = 4 .
8-10
8-11
n = 4 , l = 3 , and ml = 3 .
(a)
L = [ l ( l + 1)]1 2
(b)
Lz = ml = 3 = 3.16 × 10 −34 Js
(a)
L = [ l ( l + 1)]1 2 ;
= [ 3 ( 3 + 1)]1 2
= 2 3 = 3.65 × 10 −34 Js
4.83 × 10 31 Js = [ l ( l + 1)]1 2
l +l =
2
( 4.83 × 1031 Js )
(1.055 × 10
−34
2
Js )
, so
≈ ( 4.58 × 1065 ) ≈ l 2
2
2
l ≈ 4.58 × 1065
(b)
With L ≈ l we get ∆L ≈
and
∆L 1
≈ = 2.18 × 10 −66
L
l
125

 2 1 x 

  E1
2m  1 x 

 2 x 
  E2
2m 
 2 x 

 2 3 x 
  E3
2m 
 3 x 
2
t  
i 
 E

 t  
 
This is subject to the condition that E1  E2  E3  E . The equation for  1 can be rearranged
d2 1  2mE1 

 x  , whereupon it is evident the solutions are sinusoidal
as
2  
  2 
 1
dx
2mE
 1 x   1 sin k1 x   1 cos k1 x  with k12  2 1 . However, the mixing coefficients 1 and 1


are indeterminate from this analysis. Similarly, we find
 2 y    2 sin k 2 y    2 cos k 2 y 
 3 z   3 sin k3 z   3 cos k 3 z 
2mE 2
2mE 3
and k32 
with k22 
2
2 . The equation for  can be integrated once to get




E
 t    e i t with   and  another indeterminate coefficient. Since the energy operator
 
  

and i   E energy is sharp at the value E in this state. Also, since
is E   i
 
t

 t
2 

 2 

2
px2  2  2  and 2  2  1  k1   1 the magnitude of momentum in the x direction
x 

 x 
is sharp at the value 
k1 . Similarly, the magnitude of momentum in the y and z directions are
sharp at the values 
k2 and k3 , respectively. (The sign of momentum also will be sharp here
 
if the mixing coefficients are chosen in the ratios
8-10
1
 i , and so on).
1
n  4 , l  3 , and m l  3 .
(a)
L  ll  1   33  1   2 3  3.65  10
(b)
34
Js
Lz  ml   3  3.16  10
1 2
12
34
Js
32
1   1 
 r      e r a 0
   a0 
1 2
8-12
(a)
(r)
r
(b)
2
The probability of finding the electron in a volume element dV is given by  dV .
Since the wave function has spherical symmetry, the volume element dV is identified
here with the volume of a spherical shell of radius r, dV  4 r 2 dr . The probability of
finding the electron between r and r  dr (that is, within the spherical shell) is
2
2
P   dV  4 r  dr .
(c)
2
P
r = a0
(d)
r
1  1 
 0 0
 4 
2
2 r a 0 2
2 r a 0 2
r dr   3  e
r dr
e
  dV  4    r dr  4  
 
a3 
a
2
2

0
0
Integrating by parts, or using a table of integrals, gives
3
 4  a0 3  2  

     1 .
  dV   3 2 
 a0   2   a0  

2
(e)
r2
2
2
P  4   r dr where r1 
r1
a0
3a
and r2  0
2
2
 4 r2
P   3  r2 e 2r a 0 dr
a0 r1
let z 
2r
a0
1 3 2 z
z e dz
2 1
3
1
  z 2  2z  2 e z
integrating by parts
1
2
17
5
  e 3  e 1  0.496
2
2


8-13


Z  2 for He 
(a)
For n  3 , l can have the values of 0, 1, 2
l  0  ml  0
l  1  m l  1, 0,  1
l  2  m l  2, 1, 0,  1,  2
(b)
All states have energy E3 
Z
2
13.6 eV
2
3
E3  6.04 eV .
8-14
Z  3 for Li 2 
(a)
(b)
n  1  l  0  ml  0
n  2  l  0  ml  0
and l  1  m l  1, 0,  1
3 2 
For n  1 , E1   2 13.6   122.4 eV
1 
32 
For n  2 , E2   2 13.6  30.6 eV
2 
8-16
For a d state, l  2 . Thus, m l can take on values –2, –1, 0, 1, 2. Since L z  ml  , Lz can be
2 ,   , and zero.
8-17
(a)
For a d state, l  2
L  ll  1   6

1 2
(b)
12
For an f state, l  3
L  ll  1   12
 
1 2
8-18
The state is 6g
(a)
1.055 10 34 Js 2.58  10 34 Js
n6
12
1.055  10 34 Js 3.65  10 34 Js
13.6 eV
2
n
(b)
En  
(c)
For a g-state, l  4
13.6
eV  0.378 eV
2
6
E6 
L  ll  1   4  5   20  4.47
1 2
8-21
12
(d)
m l can be –4, –3, –2, –1, 0, 1, 2, 3, or 4
Lz
ml
ml
12  
Lz  ml  ; cos   L 
20
ll  1

ml
4
3
2
1
0
1
2
3
4
Lz
4
3
2

0

2
3
4
 153.4 132.1 116.6 102.9 90 77.1 63.4 47.9 26.6
(a)
 2 s r 
32
1
42 
12
 1  
r  r 2a 0
. At r  a0  0.529  10 10 m we find
  2  e
a0   a0 
 2 s a0  
32
1
4 2 
12
 1 
 a 
 0 
2 1e
32
1 2
 1 
 0.380 
 a0 
32

1

 0.380

10
0.529  10
m 
8-22


 9.88  10
14
m
3 2
(b)
 2s a0   9.88  1014 m 3 2
(c)
Using the result to part (b), we get P2 sa0   4 a0  2 sa0   3.43  10
2
2
 9.75  10 29 m 3
2
R2 p r   Are r 2a 0 where A 
2
1
26
12
52
a0
Pr  r2 R22 p r  A 2 r 4 e r a 0


0
0
r   rP rdr  A 2  r 5e r a 0 dr  A 2 a60 5! 5a0  2.645 Å
10
m
1
.