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. Schrödinger’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 Schrödinger’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.