2
$\begingroup$

My question is about the standard axiom on Hilbert's space in orthodoxal QM. It seems that this axiom appeares actually as an external pure mathematical axiom in all textbooks. Say, Mackey introduces it in his books as the axiom 7 and remarks about its substantiation like "why ... it works well". I'm wonder is it possible, based on some principal physical argument, to derive it? I know that Ludwig's book on QM has a subtitle to its 1st volume like a "derivation of Hilbert space structure". But it is very difficult to force one's way through his arguments.

$\endgroup$
  • 1
    $\begingroup$ related: physics.stackexchange.com/q/162704 and physics.stackexchange.com/q/155893 $\endgroup$ – Phoenix87 Jun 10 '16 at 13:46
  • 1
    $\begingroup$ Also related: physics.stackexchange.com/q/48469/2451 , physics.stackexchange.com/q/116595 and links therein. $\endgroup$ – Qmechanic Jun 10 '16 at 16:04
  • $\begingroup$ @CuriousOne Can you comment on this answer physics.stackexchange.com/questions/95528/… ? It claims that Hilbert spaces are really the fundamental structure and Fock space can be used only perturbatively in the case of interacting fields. Tbh I usually separated the use of Fock spaces and a first quantised Hamiltonian in my head, but some Fock spaces are Hilbert spaces I believe? $\endgroup$ – snulty Jun 10 '16 at 20:36
  • $\begingroup$ @snulty: Is Euclidean space the fundamental structure in Newtonian mechanics? Of course not. It's just a tool and, unfortunately, it's a misunderstood tool, at that, because what we really have is just a trivial application of a vector bundle over a trivial affine space. This becomes painfully obvious once we do relativity. In the same sense every attempt to understand physics by a simple Hilbert space approach is just as naive. Neither Hilbert nor Fock spaces are "it", we simply don't know, yet, what "it" is. $\endgroup$ – CuriousOne Jun 10 '16 at 20:50
6
$\begingroup$

It is actually possible to improve Mackey's approach completing a program started in the seventies by Jauch and Piron.

I remind you that a lattice is a partially ordered set $(\cal L, \leq)$ such that for every pair $a,b \in \cal L$ there exist $$ a\vee b := \sup\{a,b\}\in \cal L \quad \quad \mbox{and}\quad \quad a\wedge b := \inf\{a,b\}\in \cal L$$

Let us assume that the elementary propositions $p,q,\ldots$ of a quantum system, which can be experimentally tested producing the outcomes "Yes/No", give rise to the mathematical structure of a partially ordered set, where the order relation $p \leq q$ is the logical implication $p$ => $q$.

The family of all testable elementary propositions of a quantum system is actually an orthomodular, atomic, separable, irreducible lattice satisfying the so called covering property and all these features can be experimentally justified from the quantum phenomenology (some of them were already justified in Mackey's textbook where, essentially, only the structure of partially ordered set is exploited).

If the lattice includes at least four orthogonal atoms (*), it is possible to prove (theorem by Piron) that it coincides with the lattice of linear subspaces of a structure similar to the one of a Hermitian-scalar product vector space where the field of complex numbers is generalized to a division ring equipped with an involution and a sort of non-singular (Hermitian) scalar product is given. The order relation of the lattice is the theoretical set inclusion of subspaces.

The mentioned relevant subspaces $M$ forming the lattice are the ones closed (Maeda-Maeda theorem) with respect to the notion of orthogonal $\neg$ (corresponding to the orthocomplement of the lattice), $M = \neg (\neg M)$.

In this context $M \vee N$ is the closed subspace generated by $M$ and $N$ and $M \wedge N$ is the intersection of $M$ and $N$.

If, eventually, this vector space includes an infinite orthogonal system with constant norm, in view of the so-called Soler theorem (1995), the vector space can be proved to be a separable Hilbert space $H$ over $\mathbb R$, $\mathbb C$ or the division algebra of quaternions $\mathbb H$ and there is no further possibility.

The so constructed lattice of elementary propositions is nothing but the lattice ${\cal L}(H)$ of orthogonal projectors over $H$ since there is a one-to-one correspondence between closed subspaces and orthogonal projectors in a Hilbert space over $\mathbb R$, $\mathbb C$ or $\mathbb H$.

A quantum observable $A$ is nothing but a collections of such projectors $\{P_E\}$ parametrized by sets $E\subset \mathbb R$ denoting the set where the outcome of the measurement of $A$ stays. $A$ is realized as a self-adjoint operator by just integrating its corresponding collection of elementary propositions (the physical meaning of the spectral theorem is just this one). Here, a state is viewed as a probability measure over the lattice of orthogonal projectors. $\mu : {\cal L}(H) \ni P \mapsto \mu(P) \in [0,1]$.

The famous Gleason theorem, generalized by Varadarajan to the quaternionic case, establishes that any such measure $\mu$ is a statistical operator $\rho_\mu$ and $$\mu(P) = tr(\rho_\mu P)\quad \forall P \in {\cal L}(H)\:.$$

The extremal elements of this set of states are one-to-one with unit vectors up to phases (signs or quaternionic phases) as the standard QM assumes.

(See also this answer of mine)


(*) An atom is a proposition $a$ such that there is no further proposition $p$ with $p\leq a$, barring the always false proposition ${\bf 0}$. Atoms are one dimensional subspaces at the end of the reconstruction procedure.

$\endgroup$
  • $\begingroup$ This answer is nice, but I feel it hides much of the physics inside that "if" in "if the lattice includes at least four orthogonal atoms". The physical arguments for why you have that many independent logical propositions (i.e. as opposed to just classical logic) can surely also be used directly to justify the linearity of QM. Reformulating things is nice, as is having the cleanest and most waterproof routes possible to the known results, but this is still just a reformulation. $\endgroup$ – Emilio Pisanty Dec 28 '16 at 13:37
  • $\begingroup$ Sorry I do not uderstand well your comment. I think that the existence of four orthogonal atomis it is much less strong that assuming the existence of the Hilbert space from scratch. In this sense it is not a simple reformulation. Instead it points out the crucial requirements. Perhaps, I complety missed the sense of your comment... $\endgroup$ – Valter Moretti Dec 28 '16 at 13:46
  • $\begingroup$ I agree that it probably does point out a (minimal?) set of crucial requirements. However, it is nontrivially stronger than having a single orthogonal atom (which would give you classical logic), and you still need experiment to tell you which is the case; I'm curious to know which four atoms you think are a suitable starting point. My overall point was that postulating those four atoms as orthogonal will be as weird as postulating linearity to begin with; you can shift the weirdness around but you cannot (within the structures we know at the moment) explain it away. $\endgroup$ – Emilio Pisanty Dec 28 '16 at 13:54
  • $\begingroup$ I do not know if one atom guaratees classical logic. I agree that four atoms instead are crucial for linearity. This is an application of an important theorem of projective geometry. Varadarajan's book on geometry of QM, I think, deeply discusses these issues. I suggest you to have a look at it... $\endgroup$ – Valter Moretti Dec 28 '16 at 14:06
1
$\begingroup$

is it possible, based on some principal physical argument, to derive it?

No, not really. This has been one of the driving goals of the field of quantum foundations for multiple decades, but as yet we don't know of any substantially simpler, more intuitive, or even different principle from which to derive the linearity axiom of quantum mechanics.

We do know of a number of simpler and equivalent formulations, but they all produce equivalent theories and they have the weirdness embedded in some other part of the theory. The heart of the issue is that reality itself is weird, i.e. that matter exhibits wavelike behaviour including interference effects, and we know of no better way to phrase this than as the linear structure of Hilbert space.

If we do find a way to derive quantum mechanics from a simpler physical principle (in a way that is consistent with all of current experiments and known constraints, and which is validated by experiment itself) then trust me, you'll hear about it.

$\endgroup$
  • $\begingroup$ Well, but, for example, I'm not in trouble to substantiate an axiom of linearity. It looks quite naturaly because we have a lot of observations which are identified with words "superposition principle". So I'm looking for something like to such a kind of substantion to the Hilbert space axiom. Many other axioms (Mackey's, say) look to be absolutely natural wrt physics. The only exception (for me at least) is the axiom under question. $\endgroup$ – maximav Jun 10 '16 at 13:49
  • $\begingroup$ The Hilbert space axiom is the axiom of linearity (modulo the different but equivalent renumberings of the axioms in different sources): it states that there exists a space where you can form superpositions in the first place. $\endgroup$ – Emilio Pisanty Jun 10 '16 at 13:51
  • $\begingroup$ But what about subtitle of the known Ludwig's books on foundation for QM? The subtitle is "Derivation of Hilbert space structures". Book is hardly read but it seems that derivation does really exist in some way. $\endgroup$ – maximav Dec 28 '16 at 12:52
0
$\begingroup$

Hilbert spaces are needed because there exists physical systems that possess infinitely many eigenstates.

$\endgroup$
  • $\begingroup$ Please edit your answer to make it more explicit and enriched with arguments. Otherwise it will get downvoted or deleted. $\endgroup$ – rmhleo Jun 11 '16 at 8:23

protected by Qmechanic Jun 11 '16 at 0:53

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.