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?

  • $\begingroup$ Cross-listed to TP.SE. $\endgroup$ – Qmechanic Aug 8 '12 at 17:49

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.

  • $\begingroup$ But Z need not be an Observable. Sakurai talks of maximal set of compatible observables. $\endgroup$ – user4235 Dec 5 '11 at 16:36
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    $\begingroup$ @Lakshya: any orthogonal projector is an observable; one-dimensional projectors are obviously orthogonal $\endgroup$ – Christoph Dec 5 '11 at 19:56
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    $\begingroup$ @Lakshya: What's your definition of observable? $\endgroup$ – Christoph Dec 6 '11 at 8:55
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    $\begingroup$ @Lakshya: An observable is just a self-adjoint linear operator, by definition. So as Christoph says, one-dimensional projectors are obviously observables. $\endgroup$ – mjqxxxx Dec 6 '11 at 22:10
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    $\begingroup$ Lakshya, A physical interpretation for you: The eigenvalues of a projection operator are 0 and 1, so they can be thought of as true/false values. Projection operators are as physically meaningful as position & momentum observables. In some sense, they're more physical, since lab experiments measure true/false more easily than precise location. $\endgroup$ – user1504 Dec 6 '11 at 23:24

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.

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    $\begingroup$ The spectrum may be degenerate. $\endgroup$ – user4235 Dec 5 '11 at 16:35
  • $\begingroup$ I have excluded this from the start but adding a degeneracy can only make more involved the proof without changing the essential. You just need a Gram-Schmidt decomposition on the different subspaces. $\endgroup$ – Jon Dec 5 '11 at 17:44
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    $\begingroup$ I still don't get you. In the non-degenerate case, it is absolutely trivial as we know that for a compatible observable the eigenvectors are same in this case and hence if you have |a,b> then any other eigenvector must be different because no other eigenvector has eigenvalue a w.r.t. A. I asked the question only for the non-trivial case and at the same type kept the question as general. $\endgroup$ – user4235 Dec 6 '11 at 4:29
  • $\begingroup$ I have fixed this. But my opinion is that this question is really trivial. $\endgroup$ – Jon Dec 6 '11 at 13:45
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    $\begingroup$ You have to prove that if the set is maximal, then each common eigenvector has a unique representation in terms of |a,b,c...>. For instance, in your special case, the maximal set boils down to [A,B]=0. Now you have to show that, in your example, |a,b,1> and |a,b,2> are indeed the same vector, i.e. any representation in terms of |a,b>, where a and b are fixed numbers, is unique. You don't have to prove that |a,b,1> is unique and |a,b,2> is unique. That's trivial! And that is where I believe that you have to use the maximality of the set of compatible observable. See the other solution. $\endgroup$ – user4235 Dec 6 '11 at 14:04

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