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.

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?

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    $\begingroup$ I know it's annoying to tex, but for those of us who read the book a long time ago, but don't have it handy, a quick repeat of the context would help. $\endgroup$ – Ron Maimon Dec 13 '11 at 14:00
  • $\begingroup$ Have I understood your discrete example correctly: You have an observable $\xi_{1}$ - this has eigenvalues a and c, you have another commuting observable $\xi_{2}$ and that has eigenvalues b and d. Wouldn't the eigenkets of $\xi_{1}\otimes\xi_{2}$ be $|a,b\rangle$, $|c,b\rangle$, $|a,d\rangle$, $|c,d\rangle$ ? So when you pass to observables with continuous eigenvalues, a pair of commuting such observables can be integrated independently when forming inner products. $\endgroup$ – twistor59 Dec 13 '11 at 20:28
  • $\begingroup$ @twistor59 Suppose there are 3 independent eigenkets $\xi_{1} and \xi_{2}$ Now, if these eigenkets can be labelled as $|a,b\rangle , |a,d\rangle and |c,b\rangle$ These observables form a complete set but $|c,d\rangle$ eigenket is missing. Am I doing some big error? $\endgroup$ – user4235 Dec 14 '11 at 4:42
  • $\begingroup$ @Lakshya: I think what you're wanting there to be only 3 independent eigenkets, so $|c,d\rangle$ could be expanded in terms of $|a,b\rangle$, $|a,d\rangle$ and $|c,b\rangle$ ? But this is impossible since the eigenkets corresponding to different eigenvalues are orthogonal - so you can show $\langle c,d|a,b\rangle$=0, $\langle c,d|a,d\rangle$=0 and $\langle c,d|c,b\rangle$=0. If it's orthogonal to the other 3 vectors it's either zero or an independent vector. $\endgroup$ – twistor59 Dec 14 '11 at 9:42
  • $\begingroup$ @twistor59 Ok, so I got it. If there is no eigenvector corresponding to $|c,d\rangle$ then we take it to be zero vector. My main worry is this only. That is, if the whole ket space is 3 dimensional and those are the three orthonormal eigenkets that I mentioned, or in other words if there is no eigenvector that simultaneously corresponds to eigenvalue c of $\xi_{1}$ and eigenvalue d of $\xi_{2}$. So, in this case, do we assign $|c,d\rangle = 0$ ? But, if we do this in continuous case, we would have to split integral because of any discontinuities arising because of this assumption. $\endgroup$ – user4235 Dec 14 '11 at 12:16

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.

  • $\begingroup$ I still don't get it. $\xi_{1}$ and $\xi_{2}$ are both diagonalized with two of the diagonal entries for former being 'a' and for the latter being 'b'. These form a complete commuting set in 3 dimensional ket space whose eigenvectors I have mentioned above which are chosen as basic vectors in which the above mentioned diagonal representation of both the observables holds. Please clarify. $\endgroup$ – user4235 Dec 14 '11 at 12:57
  • $\begingroup$ Sorry, I meant to post this as a comment, not an answer - I'll flesh it out a bit.... $\endgroup$ – twistor59 Dec 14 '11 at 13:20
  • $\begingroup$ I have problem with the phrase "the state space is spanned by the 4 vectors" in the 2nd paragraph of your solution. Why are you assuming that there is an eigenvector which has eigenvalue $E_{11}$ when operated by $\xi_{1}$ and simultaneous eigenvalue $E_{21}$ when operated by $\xi_{2}$? Suppose that the original space is 3 dimensional with basic vectors $|E_{10},E_{20}\rangle , |E_{11},E_{20}\rangle and |E_{10},E_{21}\rangle$ and there exists no other eigenvector of either $\xi_{1} or \xi_{2}$. Why do you assume there are 4 eigenvectors? My construction gives you complete commuting set. $\endgroup$ – user4235 Dec 14 '11 at 15:05
  • $\begingroup$ @Lakshya, because I set it up with the statement "Suppose $\xi_{1}$ has just 2 distinct eigenvalues $E_{10}$ and $E_{11}$ and $\xi_{2}$ has just 2 distinct eigenvalues $E_{20}$ and $E_{21}$". The original space is 4 dimensional because I have 2 pairs of distinct eigenvalues. I need to check but I suspect this is the type of setup Dirac had in mind. $\endgroup$ – twistor59 Dec 14 '11 at 15:12
  • $\begingroup$ Ok, your statement is alright, but why are you again assuming that the original space is 4 dimensional? And how does this reason suffice that because you have two pairs of distinct eigenvalues, the original space is not 3 dimensional but 4 dimensional? I have constructed an example above in which the space is 3 dimensional but still the '2 pairs of distinct eigenvalues' requirement holds. Please expand on this. $\endgroup$ – user4235 Dec 14 '11 at 15:28

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