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I'm currently reading Principles of Quantum Mechanics by Paul Dirac, specifically the 4th edition of 1958. There is a passage I'm having trouble to understand, I'll put the text here. For reference it's in the chapter of Dynamical variables and observables, section 10.

part 1

If I understood correctly, he wants to expand a ket vector with the eigenstates of an osservable with continuous spectrum as

$$ |X)= \int |\xi'x)d\xi' $$

the problems is that if we calculate the scalar product of two ket vector we have

$$ (X|Y)=\int \int (\xi'x|\xi''y) d\xi'd\xi'' $$

He than evaluates

$$ \int (\xi'x|\xi''y)d\xi' $$

and says that for the orthogonality theorem the integrand $(\xi'x|\xi''y)$ should be zero everywhere except when $\xi'=\xi''$, But this means that the integral is zero. The only way to have a non vanishing and finite integral is to have $(\xi'x|\xi'y)$ infintely great.

part 2

I get lost at this part

part 3

  1. First of all, I know that the Hilbert space can be a vector space of infinite dimensione, so why he says "the space of vectors when the vectors are of finite length and finite scalar product is called Hilbert Space"?

  2. Secondly, I've always read that one of the postulates of QM says that the bra and ket vectors belongs to an Hilbert space, so why he says "The vectors we now use form a more general space than a Hilbert Space"?

  3. and third, can someone explain to me what is equation (30)? What does that mean?

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2 Answers 2

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  1. There's a difference between the number of dimensions of a Hilbert space and the magnitude of the vectors within. A Hilbert space may be infinite-dimensional, but that does not mean the vectors can have an infinite norm. What Dirac is pointing at is the concept of a rigged Hilbert space. A rigged Hilbert space is necessary specifically for unbounded operators with a continuous spectrum (reference: "The role of the rigged Hilbert space in Quantum Mechanics", Rafael de la Madrid). I would check de la Madrid's work on this topic.

  2. This question is intrinsically related to the first, and the simplest answer is that the axiomatic approach everyone is taught is usually not complete. In general, you need a rigged Hilbert space. I suspect this has to do with the mathematical training of most physicists (including myself) not being very formal, so the distinction between a Hilbert space and a rigged Hilbert space may be lost. Also, it's only a necessary concept for some operators, and it's not necessary if all one wants to do is achieve correct results through calculation.

  3. It's simply stating that the norm of vectors is positive definite, but now for vectors within the more general space.

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  • $\begingroup$ So, if I understood correctly, since we need (29) to be finite, we need (ξ′|ξ′) to be infinitely great. But this would me that |ξ′ ) is of infinite length. So the space of ket (and bas) cannot be a Hilbert but a more general Hilbert Space called Rigged Hilber Space. Right? $\endgroup$
    – user424762
    Commented Aug 7 at 15:00
  • $\begingroup$ Equation (8) states that $(A|A)>0$ unless $|A)=0$. Using the sign $>0$ means including the possibility that it could be infinite ? And why is it related to equation (30)? $\endgroup$
    – user424762
    Commented Aug 7 at 15:08
  • $\begingroup$ No, $\langle A|A\rangle >0$ as stated in (8) is not including the possibility of an infinite norm, and it's also not stated in (30). Norms are always positive, finite numbers. What Dirac is doing is changing the definition of the norm (or length) to extend it to vectors that arise from continuous spectrums. $\endgroup$
    – agaminon
    Commented Aug 7 at 16:00
  • $\begingroup$ ok thank you so much $\endgroup$
    – user424762
    Commented Aug 7 at 17:10
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The Wikipedia article on Hilbert space mentions that

An element of a Hilbert space can be uniquely specified by its coordinates with respect to an orthonormal basis, in analogy with Cartesian coordinates in classical geometry. When this basis is countably infinite, it allows identifying the Hilbert space with the space of the infinite sequences that are square-summable. The latter space is often in the older literature referred to as the Hilbert space.

In the middle quote, Dirac seems to be alluding to the countably infinite nature of Hilbert space, while the continuum is non-countably infinite. Furthermore,

  1. Vectors being of finite length and finite inner product does not imply anything about finite dimensionality of the space.
  2. Quantum mechanics is a theory of quantized (discrete) systems. Treating continuous systems quantum mechanically is strictly speaking an abuse of that theory, but it turns out to work pretty well anyway. As an example, consider the plane wave, which is a good description of a free particle. This is not a normalizable state, and thus clearly does not live in Hilbert space. But it is useful nonetheless.
  3. Assuming that Eq. (30) holds, you have the grounds for introducing the delta function for continuous variable states $\langle x|y \rangle = \delta(x-y)$, which seems to be what Dirac is trying to do.
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    $\begingroup$ To clarify, $x$ and $y$ are not representing positions, they are just labels for two arbitrary kets $|X\rangle$ and $|Y\rangle$. $\endgroup$
    – agaminon
    Commented Aug 7 at 11:55
  • $\begingroup$ Indeed - or even more precisely, they do not necessarily represent position, but by allowing for continuous variables they can. $\endgroup$
    – Codename 47
    Commented Aug 7 at 12:43

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