Do mutual eigenkets imply commutation of two operators?

I have been working on this question. I have solved it, and I would like to check whether my line of reasoning is right or wrong

Question:

Prove that if there exists a mutual complete set of eigenkets of Hermitian operators $\hat{A}$ and $\hat{B}$ then $[\hat{A},\hat{B}]=0$.

Proof:

Let $\{|i\rangle\}$ be a complete set of mutual eigenkets for $\hat{A}$ and $\hat{B}$. Then $\hat{A}|i\rangle=a_i|i\rangle$ and $\hat{B}|i\rangle=b_i|i\rangle$, also since the set is complete that would mean any state $|\phi\rangle$ can be written as a linear combination of $|i\rangle$.

Also $[\hat{A},\hat{B}]|i\rangle= \hat{A}\hat{B}|i\rangle - \hat{B}\hat{A}|i\rangle$ an then using the above properties we can conclude that the commutator is 0.

Is this right? also what is the physical significance of this? Is is really necessary for the operators to be Hermitian?

Also, can one prove the converse?

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The Hermiticity of an operator is what gives its eigenbasis the additional property of orthonormality. Without it you would still know that arbitrary states can be written as a linear combination of the basis states, but you wouldn't know how to write it. –  David H Aug 22 '13 at 16:06
I still do not see the need for hermiticity of operators! No where in the proof I require hermiticity. So even if I have two operators and a mutual complete set of eigenkets. One can prove the above statement. Thanks for the the edit. –  Rahul CK Aug 22 '13 at 16:26
The reason they demand hermiticity (which isn't strictly necessary, see @joshphysics' answer) is simply because they want to use this property for observables in quantum mechanics, which are already required to be hermitian. They could have stated it "Prove that if there exists a mutual complete set of eigenkets of observables $\hat{A}$ and $\hat{B}$ then $[\hat{A},\hat{B}] = 0$." –  Wouter Aug 22 '13 at 17:44

1 Answer

You're right that there is a version of the result that does not require hermiticity.

Let's denote the Hilbert space in this question by $\mathcal H$ and assume it is finite-dimensional to avoid mathematical subtleties. When you write $$\hat A|i\rangle = a_i|i\rangle$$ you are using the fact that the basis staes $|i\rangle$ are egeinvectors of $\hat A$. It is not guaranteed that there exists a basis on $\mathcal H$ consisting of eigenvectors of just any old linear operator. There is, however, a theorem called the Spectral Theorem which guarantees that if an operator $\hat A$ is hermitian, then there is an orthornmal basis for $\mathcal H$ consisting of eigenvectors of $\hat A$.

However, note that the Spectral Theorem does not require the operator to be hermitian. There certainly are linear operators on $\mathcal H$ that are not Hermitian that nonetheless are diagonalizable in which case there is a basis for $\mathcal H$ consisting of eigenvectors of this operator. We could in fact prove the following:

Theorem. Let $\hat A$ and $\hat B$ be linear operators on a finite-dimensional Hilbert space $\mathcal H$, if there exists a basis for $\mathcal H$ consisting of simultaneous eigenvectors of linear operators $\hat A$ and $\hat B$, then $[\hat A, \hat B] = 0$.

In fact, the converse of this theorem is also true; if the operators commute and each is diagonalizable, then their exists a simultaneous eigenbasis for them.

In quantum mechanics, the physical significance of commuting, hermitian operators is that they can be measured simultaneously. When you make a measurement of each of the observables on the system, then after the measurement, the state of the system will be projected onto a simultaneous eigenspace of the two operators.

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