14. Let |i = |i + |i, where  is an arbitrary complex constant.
(a) Express the positive quantity ||||2 = h|i in terms of  |i and |i, using the basic rules associated
with the inner product.
Since the inner product is antilinear in the bras, we can write, for |i = |i + |i
h|i = h|i + ∗ h|i = h|i + ∗ h|i + h|i + ∗ h|i
(b) Making the substitution  = −h|ih|i in this last expression yields
h|i = h|i −
h|i
h|i
h|i h|i
h|ih|i
h|i −
h|i +
h|i = h|i −
h|i
h|i
h|i h|i
h|i
But h|i ≥ 0. Thus, we conclude
h|i −
h|ih|i
≥0
h|i
or
h|i ≥
h|ih|i

h|i
Since h|i ≥ 0 this implies h|ih|i ≥ h|ih|i which is equivalent to
|h|i|2 ≤ kk2 kk2
or
|h|i| ≤ kk kk
which is Schwartz’s inequality. It becomes an equality only if |i and |i are parallel i.e., if there exists
a  such that |i = |i
R
(c) If  and  are two square integrable functions on  then their squared norms ||||2 =  ||2 =  2
R
2
and ||||2 =  || =  2 are finite. (Note here we use the abreviated notation kk = k|ik). The
squared norm of their sum is
|| + ||2 = ||||2 + ||||2 + h|i + h|i
p
but, if  = h|i =  +  then h|i + h|i = 2 ≤ 2 2 +  2 = 2 ||, i.e.,
h|i + h|i ≤ 2|h|i|
so
|| + ||2
≤  2 +  2 + 2|h|i| ≤  2 +  2 + 2 |||| · ||||
2
≤  +  + 2  = ( +  )
where in the last line we have used the Cauchy-Schwartz inequality. Thus, we find there exists a finite
constant  =  +  for which
|| + || ≤ 
Thus,  +  has a finite norm if  and  do. The set of square integrable functions is closed under
vector addition.
15. Consider the energy eigenvalue equation for a free particle of mass  moving throughout all space.
(a) Show how the energy eigenvalue equation reduces to the Helmholtz equation.
The energy eigenvalue equation takes the form ̂0  () =  ()  where with ̂0 = −~2 ∇2 2 this
becomes
−~2 2
∇  −  = 0
2
Multiplying through by −2~2 and introducing 2 = 2~2 this is the Helmholtz equation
∇2  + 2  = 0
(b) Working in spherical coordinates, and assuming a solution of separable form  () = () ( ) this
becomes
µ
¶¸
∙
()  2
1  ( )
 ( )  2
1  2  ( )
 () + 2
+
+ 2 () ( ) = 0
+

2


sin2 
2 tan 
2
Dividing both sides by  ()  ( ) this becomes
¶
µ 2
1
1  ( )

1  2  ( )
1 2
 () + 2
+
+ 2 = 0
+
 () 2
  ( ) 2 tan 

sin2 
2
This is almost separated except for the factor of 2 in the term involving  ( )  So we multiply
through by 2 and separate, to obtain
¸ ∙
¶¸
∙
µ 2
1  ( )
1
 2
  ( )
1  2  ( )
2
 () +  +
+
= 0
+
 () 2
 ( )
tan 

sin2 
2
2
The first term is a function of  alone. The 2nd term a function of the angular variables only. Thus,
each term must equal a constant, the sum of which add up to zero. Setting the first term equal to the
constant  ( + 1) and the second equal to the negative of this we obtain the two equations
 2
 () + 2 =  ( + 1)
 () 2
¶
µ 2
1
1  ( )
  ( )
1  2  ( )
+
= − ( + 1)
+
 ( )
tan 

sin2 
2
2
which are easily re-arranged to give
¸
∙
( + 1)
1 2
2
−
− 
 2
2
∙ 2
¸

1 
1 2
+

+
tan   sin2  2
2
= 0
= −( + 1)
(c) If we now assume that  ( ) =  () () the second equation can be written
∙ 2
¸
 
 
 2 
 2 +
= −( + 1) 
+
tan  
sin2  2

Dividing by   and multiplying by sin2  this becomes
∙
¸
1
1 2 
1 2 

2
+

=
−

+
(
+
1)
sin
 2
 tan  
 2
The right hand side is a function only of  the left hand side is a function only of . They both must
be equal to some separation constant, which we denote by 2  This gives two equations
¸
∙ 2
1 
2

 () = 0
+
+ ( + 1) −
sin2 
2 tan  
2 
+ 2  = 0
2
The equation for  () has the solutions  () =   To be single valued, we must have
 ( + 2) = (+2) = 
which requires that  ∈ {0 ±1 ±2   }. A standard normalization is to set  = (2)−12 so that
 () is square normalized for  ∈ (0 2).
16. Consider the energy eigenvalue problem () = () for a particle moving in a one-dimensional squarewell potential
 () = −0 ( − ||)
of width 2 where () denotes the step function: () = 0 for  ≤ 0 and () = 1 for   0
(a) The energy eigenvalue equation can be written in the form 00 − [() − ] = 0 where  = 2 ~2
and  = 2~2 . Integrating across the discontinuity in  at  =  we find
Z
+
−
 0
[ ()]  =

Z
+
−
[() − ]() 
for positive infinitesimal . Breaking the integral on the right into two parts, and setting ∆0 =
0 ( + ) − 0 ( − ), gives
0
∆ =
Z

−
[0 − ]()  −
Z
+
() 

where 0 = 20 ~2 . For infinitesimal , the function () can be replaced inside the integral by its
(continuous) value at  = . (Or we could invoke the mean value theorem which says that there exists
R
 ∈ ( ) such that   ()  = ( − )  ().) Thus, with either argugment as  → 0 we find
∆0 = {[0 − ] () −  ()} = (0 − 2 )() → 0
Since this vanishes as  → 0, we conclude that the derivative of  is continuous across the discontinuity
in  . Thus, we may take the boundary conditions at any point of finite discontinuity of  to be
continuity of () and its derivative. (Note that the argument breaks down if the potential jump is
infinite.)
(b) Bound state solutions correspond to motions in which the particle spends most of its time in the
neighborhood of the binding potential, and thus are square-normalizeable. Unbound solutions correspond to scattering solutions, in which the particle spends a relatively small amount of its time in the
neighborhood of the potential, and a large portion of its time moving towards or away from it. Such
solutions generally are not square normalizeable. The bound state energy eigenfunctions for the above
potential satisfy the equation
00 −  2  = 0 for   
00 + 2  = 0 for || ≤ 
00 −  2  = 0 for   −
where 2 = 2(0 −||)~2 , and where  2 = 2||~2 . Note that by assumption,   0, so || = −.
The solution in the interior region will be a linear combination of the two linearly independent solutions
() = ± , where here we assume that  is a positive constant. Thus the most general solution for
||   is of the form () =  + − . In the region to the right of the potential (  ), the
general solution will be a linear combination of the two linearly independent solutions () = ± ,
where  is a positive constant. Unfortunately, the positive exponential diverges as  → ∞, making
it physically unacceptable (a diverging exponential does not lie in the class of Fourier transformable
functions). Thus the most general physical solution in the region    is the decaying exponential
() = − . By a similar argument, we reject the negative exponential solution in the region   −,
which diverges as  → −∞. Thus, the most general physical solution for   − is also a “decaying”
exponential, () =  = −|| , that decays as  goes to increasingly negative values. Thus, the
form of a bound state solution is
⎧
−
for   
⎨
 + − for || ≤ 
 () =
⎩

for   −
The two relations associated with the requirement that  and its derivative be continuous at  = −
are
¡
¢
− = − + 
− =  − − 
while the two relations associated with the continuity of  and its derivative of 0 at  = + are
¡
¢
− =  + −
− = −  − − 
It is to be remembered that  and  are related. It is convenient to introduce a positive constant 0 ,
such that 02 = 20 ~2 and thus, we can write  2 = 02 − 2 . Solve the first set of equations above
for  and  in terms of :
 =  −(−)
 + 
2
 =  −(+)
 − 

2
The second set is equivalent to the first with the substitution  →  and  → −, hence
 = −−(+)
 − 
2
 = − −(−)
 + 
2
Equating the common ratio of  from each set we obtain the relation
2
 + 
 − 
= −2
 − 
 + 
from which follow two possible classes of solution
 − 
= ±2
 + 
depending on whether we take the positive or negative sign. The negative solution can be written
in the form  =  tan  Squaring this and expressing  in terms of  yields a condition on  alone,
namely
02 2 = 2 2 (1 + tan2 ) = 2 2 sec2 
The positive solution can be written  = − cot , which is equivalent to the relation
02 2 = 2 2 (1 + cot2 ) = 2 2 csc2 
(c) Solution: From the condition 02 2 = 2 2 sec2  set  0 = 0  and  =  so that we must show
that there is at least one solution to  20 =  2 sec2  or
 2 =  20 cos2 
As a function of  the left hand side of this equation starts at zero when  = 0 and rises montonically
to infinity as  → ∞. The right hand starts at the positive value  20 (i.e., above the left hand side)
and falls to zero (below the left-hand side) when  = 2. Thus between  = 0 and  = 2 there is a
point where the two curves must cross, showing that there is always at least one bound state solution
to this problem.
17. Consider a particle moving in 1-dimension, in a state described by the wave function.
() =  exp (− || )
where  is real and positive.) To determine the normalization, we write
Z ∞
Z ∞
Z ∞
2
2
2
2
−2||
1=
 | ()| = ||

 = 2 ||
−2  =  ||
−∞
−∞
0
which gives
1
= √

1
 () = √ exp (− || )

If the wavevector of the particle is measured the probability density  () to obtain the value  is given by
¯
¯2
¯
¯
 () = ¯̂ ()¯
where
̂ () =
=
=
=
Z ∞

 −|| −
−
√  () 
√
=


2
2
−∞
−∞
Z 0
Z ∞
  −
 − −
√
√
+

 

2
2
−∞
0
(∙
∙ − − ¸∞ )
¸0
1

 −

√
−
1 −  −∞
1 +  0
2
∙
¸ r

1
1
1
2
√
+
=
 1 + 2 2
2 1 −  1 + 
Z
∞
so
 () =
2
1
 (1 + 2 2 )2
which is peaked about  = 0 and falls to 1/4 of its maximum value at  = 1 As the width  of  ()
decreases, the width of  () increases and vice versa.