One thing that's always bothered me in Dirac's notation is that it assumes that the Hilbert space contains a "continuum basis" of vectors $|x\rangle$, which happen to be eigenvectors of an operator $X$ (which has no eigenvalues, only a continuous spectrum that spans the whole space). Their inner product is distribution-valued, with $\langle x'|x\rangle = \delta(x'-x)$. There's also the cryptic normalization property: $\int |x\rangle\langle x| dx = Id$. According to this question, some of these can't be made rigorous even with something like the concept of "Rigged Hilbert Spaces".

So, is there another approach to quantum mechanics in general that sidesteps this issue entirely, without loss of descriptive power?

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    $\begingroup$ The usage of that continuous basis isn't really a feature of the notation as such - writing vectors not as bras and kets would not prohibit any physicist from using the position basis. The "approach" to quantum mechanics that sidesteps this issue, or only uses its rigorous parts, is simply "rigorous" or "mathematical" quantum mechanics. I'm not exactly sure what kind of answer you're looking for for this question - the exact way to make a given step rigorous will depend on the step. $\endgroup$
    – ACuriousMind
    Commented Sep 30, 2016 at 22:24
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    $\begingroup$ I don't understand this questions because I use bra-key notation for vectors in a finite dimensional vector space all the time. $\endgroup$
    – DanielSank
    Commented Sep 30, 2016 at 22:44
  • $\begingroup$ @DanielSank The OP is not asking about finite dimensional vector spaces. With regards to ACuriousMind's note my impression is that the OP is asking if alternative formulations of QM are possible that avoid the use of such a continuum basis altogether. Not that I can give a meaningful answer to it. $\endgroup$
    – Jan Bos
    Commented Oct 1, 2016 at 1:57
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    $\begingroup$ @JanBos I thing Daniel is just objecting to an overly broad statement about Dirac's notation. There is nothing about it that intrinsically assumes a continuous basis, nor it is needed to do quantum mechanics with operators that have continuous spectra (not withstanding the difficulties with rigor that attend the work in any notation). $\endgroup$ Commented Oct 1, 2016 at 2:03
  • $\begingroup$ Dirac's notation is only a semi-rigorous convenient notation that helps in making some calculations faster; it is not necessary for the formulation of quantum mechanics. In fact, quantum mechanics can be formulated in a rigorous fashion without problems, using more sophisticated mathematical tools (such as C*-algebras of observables). $\endgroup$
    – yuggib
    Commented Oct 1, 2016 at 7:57

1 Answer 1


The problem, in my view does not lie with the notation as such, but rather with the physical scenario that is being studied. Space-time is continuous infinite. Therefore, one requires a basis that is also continuous and infinite. So, regardless of how one would represent such a basis in terms of a notation, its orthogonality condition would necessarily have to include a Dirac delta function. In the end, one can take one's hat off to people such as Dirac who came up with some mathematical system that makes it possible to do calculations that can lead to predictions, which in turn can be tested in experiments. The mere fact that such predictions often agree with these experimental results, seems to indicate that this way to calculate these quantities using this mathematical formalism must be correct to some extent. It then becomes a challenge to the mathematitians to try and come up with an axiomatic system that can lead to this formalism in a consistent manner. This often implies that one would need to stretch the notions of integrals, vector spaces and such so that the formalism can work in a strict mathematical sense. Whether or not that happens to be the case usually does not stop physicists from using the formalism.


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