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## Main Question or Discussion Point

Let A be a Hermitian operator with n eigenkets: [itex]A|u_i\rangle = a_i |u_i\rangle[/itex] for i=1,2,...,n.

Suppose B is an operator that commutes with A. How could I show that

[tex]\langle u_i | B | u_j \rangle = 0 \qquad (a_i \neq a_j)?[/tex]

I have tried the following but not sure how to proceed:

[tex]AB - BA=0\\ \implies \langle u_i | AB | u_j \rangle - \langle u_i | BA | u_j \rangle = 0[/tex]

Suppose B is an operator that commutes with A. How could I show that

[tex]\langle u_i | B | u_j \rangle = 0 \qquad (a_i \neq a_j)?[/tex]

I have tried the following but not sure how to proceed:

[tex]AB - BA=0\\ \implies \langle u_i | AB | u_j \rangle - \langle u_i | BA | u_j \rangle = 0[/tex]