Uniqueness of eigenvector representation in a complete set of compatible observables

Question

Sakurai states that if we have a complete, maximal set of compatible observables, say $A,B,C...$ Then, an eigenvector represented by $|a,b,c....>$, where $a,b,c...$ are respective eigenvalues, is unique. Why is it so? Why can't there be two eigenvectors with same eigenvalues for each observable? Does maximality of the set has some role to play in it?

Answer 1

Assume that you have a maximal set $A,B,C,\ldots$ and two states $\phi_1$ and $\phi_2$ with the same set of eigenvalues in that set. Then construct the operator $Z = |\phi_1\rangle\langle\phi_1|$. Convince yourself that it would distinguish between $\phi_1$ and $\phi_2$, and that it would commute with all of $A,B,C,\ldots$ --- i.e. your original set was not maximal.

Answer 2

The answer relies on an algebraic property of uniqueness of eigenvectors and eigenvalues.

Non-degenerate case

Let us suppose that two such eigenvectors exist and the spectrum is non-degenerate. Let us call them $|a,b\rangle$ and $|a,b\rangle'$ (I consider just a couple of compatible observable for the sake of simplicity). If you claim that these two eigenvectors are independent, otherwise there exists a constant $c$ such that $|a,b\rangle=c|a,b\rangle'$, you must have

$$\langle a,b|A|a,b\rangle'=a\langle a,b|a, b\rangle'=0$$

and similarly

$$'\langle a,b|A|a,b\rangle=a'\langle a,b|a, b\rangle=0$$

and this will imply

$$'\langle a,b|A|a,b\rangle=\langle a,b|A|a,b\rangle'=(\langle a,b|A|a,b\rangle')^*=0$$

but the only operator having this property is $A=0$ and this is inconsistent with the starting hypothesis. This argument can be repeated also for $B$ proving the assertion.

Degenerate case

When there is a degeneracy, to a single eignevalue will correspond more than an eigenvector. So, let us suppose that these egievenctors for the eigenvalue $a$ are $|a,b,1\rangle$ and $|a,b,2\rangle$. We suppose we are able to write down from them a set of two orthogonal and normalized eigenvectors, e.g. through a Gram-Schmidt decomposition, that spans such a two-dimensional subspace.

Now, the above argument can be repeated for each eigenvector into the subspace as, if there is a third eigenvector, beyond those forming the two-dimensional subspace, this will be enough to prove that $A=0$.

Of course, all this is easy to extend to any number of operators.