# Solutions to Problems 2.5 and 3.1        ```Solutions to Problems 2.5 and 3.1
2.5  A particle of mass m is trapped in a potential V  x     x  . Show that
there is only one bound state and find its energy.
Let the energy be E, which must be negative. Schr&ouml;dinger’s equation then gives
 2 d 2
  E  x     x   x  ,
2m dx 2
Which, away from x = 0, has solutions   x   e   x , where E    2  2 2m . We divide it
into two regions, region I with x  0 and region II with x  0 . To keep the wave
function from diverging, we must pick
 I  x   Ae  x and  II  x   Be   x .
If we integrate Schr&ouml;dinger’s equation across the boundary at x = 0, we find,



 2 d 2
dx   E    x  dx      x   x  dx ,
2m  dx 2


2
 II  0   I  0      0  .
2m 
Since the derivative of the wave function has a finite discontinuity, the wave function
must be continuous at the boundary, so we have  I  0    II  0  , which implies A  B .
Then the equation just derived for the discontinuity of the derivative become
m
2
  A  A    A , or   2 .
2m

Substituting this in for the equation for the energy, we find E    2  2 2m   m 2 2 2 .
3.1  Prove Schwartz’s inequality,  ,  ,     ,   ,  . You may prove it
however you want; however, here is one way to prove it. Expand out the inner
product of a  b with itself, which must be positive, where a and b are
arbitrary complex numbers. Then substitute in a   ,  and b    ,   .
Simplify, and you should have the desired result.
We take the suggestion given, hoping it will not lead us astray. We note that b is
*
real, so b*  b    ,   , while a is not, so a*   ,    ,   .
0   a  b , a  b 
 a*a  ,    a*b  ,   b*a  ,    b*b  , 
  ,   ,  ,     ,   ,   ,    ,   ,  ,     ,   ,   , 
   ,   ,  ,     ,   ,   ,  .
We now rearrange this and divide by  ,   to give  ,  ,     ,   ,  , the
desired relationship. The only detail that might be unclear is that in the ultimate step, we
divided by  ,   . This is valid, provided  ,    0 , which is guaranteed for   0 . Of
course, if   0 , then both sides of Schwartz’s inequality are zero, and the result is
trivially true.
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