Verify that if $A$ and $B$ are diagonal in the same orthonormal basis, then $[A,B]=0$

Question

This is from Nielson and Chuang's textbook "Quantum Computation and Quantum Information".

They state the Simultaneous Diagonalisation Theorem: Suppose $A$ and $B$ are Hermitian operators. Then $[A,B]=AB - BA = 0$ iff there exists an orthonormal basis such that both $A$ and $B$ are diagonal with respect to that basis.

In the proof of this theorem, they say "you can easily verify that if $A$ and $B$ are diagonal in the same orthonormal basis, then $[A,B]=AB-BA=0$.

I think you could prove this by letting $U$ be a unitary operator performing the change of basis from orthonormal to canonical basis. Then $A = UDU^{*}$ and $B=UEU^{*}$ Then $$AB=UDU^{*}UEU^{*}=UDEU^{*}=UEDU^{*}=UEU^{*}UDU^{*}=BA$$ ($D$ and $E$ are diagonal)

$\newcommand\bra[1]{\left\langle#1\right|}\newcommand\ket[1]{\left|#1\right\rangle} $ However, the notation the book specifies is the diagonal representation $A=\sum_{i} a_{i} \ket{i} \bra{i}$ and $B = \sum_{i} b_{i} \ket{i} \bra{i}$ where $\ket{i}$ is some common orthonormal set of eigenvectors for $A$ and $B$. How do I prove that if $A$ and $B$ are diagonal, then $[A,B]=0$ using this notation?

Answer 1

$A B = (\sum_i a_i|i\rangle\langle i|)(\sum_j b_j|j\rangle\langle j|) = \sum_i \sum_j a_i b_j |i\rangle\langle i|j\rangle\langle j| = \sum_i a_i b_i |i\rangle\langle i|$ since $\langle i|j\rangle$ is 0 if $i \neq j$ and 1 if $i = j$ due to the basis being orthonormal. From there you can just reverse $a_i$ and $b_i$ since they are scalars and do the same algebra backwards to get $BA$.