# Show that inner product is zero.

perishingtardi
Let A be a Hermitian operator with n eigenkets: $A|u_i\rangle = a_i |u_i\rangle$ for i=1,2,...,n.

Suppose B is an operator that commutes with A. How could I show that
$$\langle u_i | B | u_j \rangle = 0 \qquad (a_i \neq a_j)?$$

I have tried the following but not sure how to proceed:
$$AB - BA=0\\ \implies \langle u_i | AB | u_j \rangle - \langle u_i | BA | u_j \rangle = 0$$

Is ##A## Hermitian or is it self-adjoint? If it is the latter then the result is immediate from what you wrote down because for any self-adjoint operator ##A##, if ##|u_i \rangle## is an eigenket of ##A## with eigenvalue ##a_i## then ##\langle u_i|## is an eigenbra of ##A## with the same eigenvalue i.e. ##\langle u_i|A = a_i\langle u_i |## so you would be left with ##(a_i -a_j) \left \langle u_i|B|u_j \right \rangle = 0##.

perishingtardi
Actually I think I got it:

$$a_i \langle u_i |B|u_j \rangle - \langle u_i |B| u_j \rangle a_j = 0$$ using the Hermiticity of A for the first term, and then since $a_i \neq a_j$ we get
$$\langle u_i |B| u_j \rangle = 0$$

This is a technical point and I don't know if it matters for your class however it should be noted that while ##\langle u_i |A = a_i \langle u_i |## is certainly true if ##A## is self-adjoint, it won't necessarily be true if ##A## is only Hermitian.

perishingtardi
Is ##A## Hermitian or is it self-adjoint? If it is the latter then the result is immediate from what you wrote down because for any self-adjoint operator ##A##, if ##|u_i \rangle## is an eigenket of ##A## with eigenvalue ##a_i## then ##\langle u_i|## is an eigenbra of ##A## with the same eigenvalue i.e. ##\langle u_i|A = a_i\langle u_i |## so you would be left with ##(a_i -a_j) \left \langle u_i|B|u_j \right \rangle = 0##.

Thanks for that - I think we both posted at the same time! My question is related to quantum mechanics. I think physicists use the words Hermitian and self-adjoint interchangeably. I know that pure mathematicians will distinguish them but it's not important for my area. Thanks!