"Completeness" of eigenvectors in a complete, commuting set

Question

This question was originally the one below dashed line. Now after further discussions, it has boiled down to this question: Is the following construction possible? Suppose we have a 3 dimensional ket space in which the basic vectors are simultaneous eigenvectors of two observables $\xi_{1}$ and $\xi_{2}$ such that they can be represented as $|a,b\rangle , |a,d\rangle , |c,b\rangle$ where 'a' and 'c' are eigenvalues of $\xi_{1}$ and 'b' and 'd' are eigenvalues of $\xi_{2}$. Clearly, $\xi_{1}$ and $\xi_{2}$ form a complete, commuting set.

Or is it necessary that $|c,d\rangle$ eigenvector must exist? If this is the case then in 3D ket space, two observables can never form a complete, commuting set and the same thing can be generalized to any prime dimensional ket space. This leads to the conclusion that any prime dimensional ket space has either 1 observable having all distinct eigenvalues or it has p observables forming a complete commuting set each one having only one eigenvalue. And this thing is quite hard to believe unless there is a nice argument which can account for these weird conclusions coming from this.

If the above construction holds and $|c,d\rangle$ can be missing then I would like to ask as to why do we take n-fold integrals while writing the representation of an inner product of a ket and a bra when the set of eigenvectors forming complete, commuting set has continuous eigenvalues? The argument against it being that as some eigenvectors can be missing, the limits of each integral are dependent on other integrals and hence the notion of n-fold integrals breaks down. Another possibility is that we consider those missing eigenvectors as 0 vectors with the breaking of integral at discontinuity points being understood. Is it so or is it the former way if the above construction is correct and valid?

The text below the line and comments to the question may help in giving the context of the question more clearly.

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On Page 65 of his book, Dirac argues that we can generalize the results mentioned previously by taking a $u-v$ fold integral of a particular inner product mentioned there.

I have a problem that it can't be done as $\xi(v+1), \xi(v+2) ... \xi(u)$ may not be independent. And hence it's not necessary that for a fixed value of $\xi(v+1)$ there would exist eigenvectors as we vary other $\xi's$ and so the integral can't be formed as the limits of other integrals vary as we go on evaluating a particular integral.

As a counter-example in a discrete case, consider two observables which form a complete set with eigenvectors as $|a,b>, |a,d> and |c,b>$. Here, there is no eigenvector $|c,d>$.

I hope that I make myself clear that while passing on to continuous case, we might get same circumstance in which the notion of $u-v$ fold integral breaks down. May be I have largely wrong intuitive concepts as to when a multiple fold integral can be taken. If so, please rectify me. If the above counterexample is correct, then how would you justify the generalization that Dirac takes?

Answer 1

I don't have the book with me but I'm guessing that there are v observables $\xi_{1}$...$\xi_{v}$ which have discrete eigenvalues and u observables $\xi_{v+1}$...$\xi_{v+u}$ which have continuous eigenvalues. Assuming that they're talking about a complete set of commuting observables, the eigenkets are labelled $|\xi_{1}..\xi_{v},\xi_{v+1}..\xi_{v+u}\rangle$. Just restricting to the first 2 discrete observables for simplicity:

We have observables $\xi_{1}$ and $\xi_{2}$. Suppose $\xi_{1}$ has just 2 distinct eigenvalues $E_{10}$ and $E_{11}$ and $\xi_{2}$ has just 2 distinct eigenvalues $E_{20}$ and $E_{21}$. Then, since $\xi_{1}$ and $\xi_{2}$ are complete and commuting, the state space is spanned by the 4 vectors $|E_{10}E_{20}\rangle$, $|E_{11}E_{20}\rangle$, $|E_{10}E_{21}\rangle$ and $|E_{11}E_{21}\rangle$.

If I have understood it correctly, your worry is that there may be a case where you have just three independent eigenkets. Suppose this is the case, so we have $$|E_{11}E_{21}\rangle = \alpha|E_{10}E_{20}\rangle + \beta|E_{11}E_{20}\rangle + \gamma|E_{10}E_{21}\rangle$$ for some $\alpha, \beta, \gamma$.

Looking at the relationship of $|E_{11}E_{21}\rangle$ with the other 3 vectors in turn:

$$\xi_{1}|E_{11}E_{21}\rangle = E_{11}|E_{11}E_{21}\rangle; \xi_{1}|E_{10}E_{20}\rangle = E_{10}|E_{10}E_{20}\rangle$$ so $|E_{11}E_{21}\rangle$ and $|E_{10}E_{20}\rangle$ belong to different eigenvalues of $\xi_{1}$ so must be orthogonal

$$\xi_{2}|E_{11}E_{21}\rangle = E_{21}|E_{11}E_{21}\rangle; \xi_{2}|E_{11}E_{20}\rangle = E_{20}|E_{11}E_{20}\rangle$$ so $|E_{11}E_{21}\rangle$ and $|E_{11}E_{20}\rangle$ belong to different eigenvalues of $\xi_{2}$ so must be orthogonal

$$\xi_{1}|E_{11}E_{21}\rangle = E_{11}|E_{11}E_{21}\rangle; \xi_{1}|E_{10}E_{21}\rangle = E_{10}|E_{10}E_{21}\rangle$$ so $|E_{11}E_{21}\rangle$ and $|E_{10}E_{21}\rangle$ belong to different eigenvalues of $\xi_{1}$ so must be orthogonal

So $|E_{11}E_{21}\rangle$ can't be expressed as a linear combination of the other three vectors - it must be an independent vector.

Analogously in the continuous eigenvalue case, the eigenvalues of the different observables can be treated as independent, and hence integrated over independently.