# Homework Help: Find the complete orthonormal set of eigenfunctions of the operators B-hat

1. Oct 16, 2011

### blueyellow

1. The problem statement, all variables and given/known data

A bound quantum system has a complete set of orthonormal, no-degenerate energy eigenfunctions u(subscript n) with difference energy eigenvalues E(subscript n). The operator B-hat corresponds to some other observable and is such that:

B u(subscript 1)=u(subscript 2)
B u(subscript 2)=u(subscript 1)
B u(subscript n)=0
n>3 or B=3

a) Find the complete orthonormal set of eigenfunctions of the operator B-hat (expand out the eigenvalues of B in terms of u, and do not neglect any solutions)
b) If B is measured and found to have the eigenvalue H, what is the expectation value of the energy in the resulting state?

3. The attempt at a solution

B u(subscript1)=u(subscript2)
B u(subscript 2)=u(subscript1)
(B^2) u*(subscript 2)=u(subscript2)
B^2 =1
B=1

2. Oct 16, 2011

### vela

Staff Emeritus
Remember B can be represented by a matrix relative to the {un} basis. Once you have the matrix, you can diagonalize it to find the eigenvalues of B.

3. Oct 17, 2011

### blueyellow

weirdly I can't edit the original post. I meant 'non-degenerate' rather than no-degenerate

4. Oct 17, 2011

### blueyellow

I don't know how to represent B as a matrix. I tried looking it up but the u(n) thing is confusing me. How does B relate to u(n)?

5. Oct 22, 2011

### blueyellow

B|t(subscript n)>=b(subscript n) |t(subscript n)>

|t(subscript n)>=sigma a(subscript i) |u(subscript i)>

|b(subscript i)>=sigma c(subscript n)|u(subscript n>

B-hat|b(i)>=sigma c(n) B-hat|u(n)>
B-hat|b(i)>=c(1) u(1)+c(2)u(1)=b|b(n)>

b c(1)=c(2)
b c(2)=c(1)

b=(c(1))/(c(2))
((c(1))^2)=(c(2))^2

n> or=3 b c(n)=0

c(n)=0, n>or=3

|b(1)>=c(|u(1)>+|u(2)>)
|b(2)>=c(|u(1)>-|u(2)>)

<b(1)|b(2)>=(c^2)(<u(1)|u(2)>+<u(2)|u(2)>)
=2(c^2)
<b(2)|b(2)>=c^2 (<u(1)|u(1)>+<u(2)|u(2)>)

c=1/(sqrt2)

|b(1)>=[1/(sqrt2)](|u(1)>+|u(2)>)
|b(2)>=[1/(sqrt2)](|u(1)>-|u(2)>)

I don't think I have answered the question completely though.