6
Quantum Mechanics in One
Dimension
6-2
(a)
Normalization requires
L
A 2  L 
4 x 

2
2
2 2 x 
1    dx  A  4 L cos 
dx    4 L 1  cos 
 dx
 L 
 L 
2
4
4
  

2
.
L
so A 
(b)
L
8
L
8
L
x 
4 1 8 
4 x  
P    dx  A  cos 
dx     1  cos 
dx
 L 
 L 2 0 
 L  

0
0
2
2
2 2
2  L  2  L  4  x 
       sin 

L 8   L 4   L 
L
8
1 1

 0.409
4 2

0
 x  A cos kx  Bsin kx

6-6
x
 kA sin kx  kB cos kx
2
 k 2 A cos kx  k 2 Bsin kx
 x2
2m 
2mE 
 2 E U    2 A cos kx  Bsin kx


  
 
The Schrödinger equation is satisfied if
2
 x
2
2m 
  2 E U  or
  
2mE 
2
k A cos kx  Bsin kx    A cos kx  Bsin kx .
  

2 k 2
Therefore E 
.
2m

6-9
En 
n 2 h2
2
8mL
, so E  E2  E1 
3h2
2
8mL
E  3
1240 eV

nm c
2

8 938.28  106 eV c2 10 5 nm

2
 6.14 MeV
hc
1240 eV nm

 2.02  10 4 nm
 E 6.14  106 eV
This is the gamma ray region of the electromagnetic spectrum.

2 2
6-10
En 
n h
2
8mL


2
6.63  10 34 Js
h2

8mL2 8 9.11  10 31 kg 10 10 m



2
 6.03  10 18 J  37.7 eV
E1  37.7 eV
(a)
E2  37.7  22  151 eV
E3  37.7  3  339 eV
2
E 4  37.7  4 2  603 eV
hf 
(b)

hc

 En i  En f
1 240 eV  nm
hc

En i  En f
E n i  En f
For n i  4 , nf  1 , E n i  E n f  603 eV  37.7 eV  565 eV ,   2.19 nm
ni
ni
ni
ni
ni
4,
4,
3,
3,
2,
nf
nf
nf
nf
nf
 2 ,   2.75 nm
 3 ,   4.70 nm
 1 ,   4.12 nm
 2 ,   6.59 nm
 1 ,   10.9 nm
6-12
3 8 h 
 h2  2 2


L

 
1
and
2
E 

 8mL2 

 mc 

6-13
(a)
hc


12
 7.93  10 10 m  7.93 Å.
Proton in a box of width L  0.200 nm  2  10 10 m
6.626  10 34 J  s

 8.22  10 22 J
E1 
8m p L2 81.67  10 27 kg 2  10 10 m 2
2
h2

(b)
8.22  10 22 J
1.60  10 19 J eV
 5.13  10
3
eV
Electron in the same box:
6.626  10 34 J s 
E1 

8me L2 89.11  10 31 kg 2  10 10 m 2
h2
2
 1.506  10 18 J  9.40 eV .
6-16
(c)
The electron has a much higher energy because it is much less massive.
(a)
 x 
 x  A sin   , L  3 Å. Normalization requires
 L 
L
L
 x 
LA 2
2
1    dx   A 2 sin 2  dx 
 L 
2
0
0
12
2 
so A   
L 
 3
12
L3
2
2  3 
2 
2
2  x 
2

  0.195 5 .
dx
sin

dx


sin

d








 L  0
 L 
 0
 
8 
0
6

L3
P
100 x 
2 

 , A  
L 
 L 
12
(b)
  A sin 
P

L3
100  3
100  x 
200 
2  L 
1 100 1
2
 sin 
  sin d 


dx  
 3 
 L 
L 100  0
50  6
4
2
 sin 
0
1  1  2  1
3

 0.331 9
  
sin 


3
3 

3
400

200


1
.
3
Since the wavefunction for a particle in a one-dimension box of width L is given by
 n x 
2
2
2 n  x 
 n  A sin 
 ,
 it follows that the probability density is Px   n  A sin 
 L 
 L 
which is sketched below:
(c)
6-18
2
L
Yes: For large quantum numbers the probability approaches

2
3
2
From this sketch we see that Px is a maximum when
when
5
2
n x
L
n x
L


2
,
3 5
1 

,
,    m   or

2 
2
2
x
L 
1 
m  
2 
n 
Likewise, Px is a minimum when
x
6-24
m
  0, 1, 2, 3,  , n .
n x
 0 ,  , 2 , 3 ,   m or when
L
Lm
n
m
  0 , 1, 2, 3 ,  , n
After rearrangement, the Schrödinger equation is
U x  
d2 2m 
U x   E x  with
2  
  2 

dx
1
2 2
 x 2
gives
m  x for the quantum oscillator. Differentiating  x  Cxe
2
2
d
 2 x  x   C  x
dx
and
d2
dx
2

2 xd 
dx
 2 x  2 x Ce  x  2 x  x   6  x  .
2
2
2
2
m   2 2mE
2m
Therefore, for  x to be a solution requires 2 x   6  2 U x   E  
x  2 .
  




m
m
2 mE
Equating coefficients of like terms gives 2 
and 6  2 . Thus,  
and

2





3 2 3
2
2
2 2 x 2
dx where the
  . The normalization integral is 1    x  dx  2C  x e
E
2
m


second step follows from the symmetry of the integrand about x  0 . Identifying a with 2 in
1 4
32 3 
or C  
 .
  
At its limits of vibration x  A the classical oscillator has all its energy in potential form:
12
1   
the integral of Problem 6-32 gives 1  2C   
8 2 
2 
6-25
12
1
 2E 

1 
m  2 A 2 or A  
 . If the energy is quantized as E n  n   , then the
2
m  

2 
2

12
2n  1 
corresponding amplitudes are A n  
 .
 m  

E
6-29
(a)
Normalization requires




2



1    dx  C 2  e 2x 1  e x dx  C 2  e 2x  2e 3x  e 4x dx . The integrals are
2

0
0
1 1  1  C2
elementary and give 1  C2   2   
. The proper units for C are those of
2 3  4  12
length 1 2
thus, normalization requires C  12
12
nm
1 2
.
(b)
2
The most likely place for the electron is where the probability 
is largest. This is
d
also where  itself is largest, and is found by setting the derivative
equal zero:
dx
0
d
x
2x
x
x
 C e  2e
 Ce 2e  1 .
dx




The RHS vanishes when x   (a minimum), and when 2e x  1 , or x  ln 2 nm .
Thus, the most likely position is at x p  ln 2 nm  0.693 nm .
(c)
The average position is calculated from




2



x   x  dx  C 2  xe 2 x 1  e x dx  C2  x e 2 x  2e 3 x  e 4 x dx .
2

0
0

The integrals are readily evaluated with the help of the formula  xe ax dx 
1 1  1 
 13 
x  C 2   2     C2  . Substituting C2  12 nm


9
4
16
144 


x 
0
1
1
a2
to get
gives
13
nm  1.083 nm .
12
We see that x is somewhat greater than the most probable position, since the
probability density is skewed in such a way that values of x larger than x p are
weighted more heavily in the calculation of the average.
6-31
The symmetry of  x  about x  0 can be exploited effectively in the calculation of average
values. To find x
2

x   x  x  dx
2

We notice that the integrand is antisymmetric about x  0 due to the extra factor of x (an odd
function). Thus, the contribution from the two half-axes x  0 and x  0 cancel exactly,
leaving x  0 . For the calculation of x 2 , however, the integrand is symmetric and the halfaxes contribute equally to the value of the integral, giving


x   x 2  dx  2C2  x 2 e 2 x x 0 dx .
0
2
0
3
 x0 
Two integrations by parts show the value of the integral to be 2  . Upon substituting for
 2 

12

 x 2 
x
C , we get x
  0   0 . In calculating
2
 2 
the probability for the interval x to x we appeal to symmetry once again to write
2
2
 1  x 3 x 2
 2  2  0   0 and x  x 2  x
2
x0   2 
P
x
x
2 12
x
x 
2
2x x 0
dx  2C 2  0 e 2x x 0
  dx  2C  e
 2 
2
x
0
 1  e
2
 0.757
0
or about 75.7% independent of x0 .
1 4
6-32
2
a 
2 ax 2
The probability density for this case is  0 x   C0 e
with C 0   
 

m
and a 
. For


the calculation of the average position x   x  0 x  dx we note that the integrand is an
2

odd function, so that the integral over the negative half-axis x  0 exactly cancels that over
the positive half-axis ( x  0 ), leaving x  0 . For the calculation of x 2 , however, the
integrand x  0
2
2
is symmetric, and the two half-axes contribute equally, giving
x
2

2 ax 2
 2 C0  x e
2
0
12
2  1  
dx  2C 0   
4 a  a 

.
1

2
Substituting for C 0 and a gives x 2 

and x  x  x
2 a 2m 


6-33

2 12
12
  
 
 .
2 m 
(a)
Since there is no preference for motion in the leftward sense vs. the rightward sense,
a particle would spend equal time moving left as moving right, suggesting px  0 .
(b)
To find p2x we express the average energy as the sum of its kinetic and potential
px2
px2
 U 
 U . But energy is sharp in the
2m
2m
1
oscillator ground state, so that E  E0   . Furthermore, remembering that
2

1

2 2
2
from Problem 6U x   m  x for the quantum oscillator, and using x 
2
2m

  m 
1
1
2
32, gives U  m  2 x 2   . Then px  2mE0  U   2m  
.
 4 
2
2
4


energy contributions: E 
(c)

px  p2x  px

2 12
12
m 
 
 2 

12
6-34
  
From Problems 6-32 and 6-33, we have x  

2m 

12
12
12
m  
and px  
 . Thus,
 2 

   m 

xpx  
 for the oscillator ground state. This is the minimum



2m    2 
2
uncertainty product permitted by the uncertainty principle, and is realized only for the
ground state of the quantum oscillator.