# Show that inner product is zero.

1. Oct 15, 2013

### 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:
$$AB - BA=0\\ \implies \langle u_i | AB | u_j \rangle - \langle u_i | BA | u_j \rangle = 0$$

2. Oct 15, 2013

### WannabeNewton

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 $(a_i -a_j) \left \langle u_i|B|u_j \right \rangle = 0$.

3. Oct 15, 2013

### 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
$$\langle u_i |B| u_j \rangle = 0$$

4. Oct 15, 2013

### WannabeNewton

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.

5. Oct 15, 2013

### perishingtardi

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!

6. Oct 15, 2013

### WannabeNewton

Alrighty then! Glad you got it worked out :)

7. Oct 15, 2013

### dextercioby

Just by using the Dirac bra-ket formalism (one of the most vicious inventions in the history of science) means that you're hiding a lot of mathematical beauty under the carpet. Let's proceed then and leave finesse aside and say that if A and B commute, then they have a common set of eigenvectors. Thus the ui's of A are the ui's of B, so that you need to prove that 2 eigenvectors pertaining to 2 different eigenvalues are orthogonal one on each other.