I read some while ago that, currently, eleven different formulations of quantum mechanics exist. Is this correct / accurate? If yes, can someone provide a pointer(s) (i.e. link(s)) to the various formulations?

  • $\begingroup$ Do you mean different mathematical formulations or different conceptual formulations? (A "conceptual formulation of QM" would be a wall of words telling you how to think about the mathematics as physics.) Either way, there is no standard list of "eleven different formulations". Whether you end up with eleven will depend on what you include, and even on whether you count certain versions as different. Please tell us where you read this, it will help us to understand what you're talking about. $\endgroup$ Nov 10, 2011 at 4:28
  • $\begingroup$ The paper Ron Maimon found looks like the source of the "11 formulations" because it lists 9+2. $\endgroup$ Nov 10, 2011 at 7:30
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    $\begingroup$ It makes no sense to count "different formulations of QM". It's adding apples to oranges, as well as seeds from the oranges and apples, plus some boxes of oranges and trucks used to move the apples. It's not a well-defined scientific question. Only a person who isn't thinking about the world scientifically may ask questions of this sort. There are different interpretations but they don't really differ physically as long as they're correct: Copenhagen, Consistent Histories, ... There are wrong "interpretations" such as the Bohm interpretation and tons of others in uncountable crackpot papers. $\endgroup$ Nov 10, 2011 at 7:50
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    $\begingroup$ It makes no sense to attribute a number to these wrong attempts to "revolutionize" quantum mechanics - there are thousands of nonsensical papers and each of them is a bit different than another, so you could count each of them as a separate "formulation". Aside from (mostly nonsensical) "interpretations" of QM, there are also meaningful ways to organize the equations of QM. Schrodinger picture, Heisenberg picture, Dirac interaction picture, Feynman's path integral are the 4 most important ones, related in many ways. But to say there are 4 is still a stretch. They're not well-separated objects. $\endgroup$ Nov 10, 2011 at 7:52
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    $\begingroup$ If the formulations are formulations of the 'same' thing, then they must be equivalent (mathematically). So, up to equivalence there is only one formulation. $\endgroup$
    – MBN
    Nov 10, 2011 at 13:17

1 Answer 1


The mathematical formulation of QM, as defined by Dirac, is a closed thing--- there is always a Hilbert space of states, and operators which act on the Hilbert space to produce physical changes. This mathematical scheme, however, is very general, and when you write down the description of specific quantum systems, you have to make some assumptions about the Hilbert space structure and the form of the linear operators. There are general classes of systems which are best defined in different formal schemes, I suppose that these formalisms are the quantum mechanical formulations you are talking about.

The historical formulations of quantum mechanics were matrix mechanics and wave mechanics, which are two pictures of the same mathematical structure. The two are unified and contained in Dirac's transformation theory formalism, which is what people learn today as "Quantum mechanics." This formulation includes all the others in a certain sense, because it defines quantum mechanics.

The path-integral formulation came later, and it is also mostly equivalent to the ordinary formulation, but as it is generally presented, it makes the extra assumption that the Hamiltonian is quadratic in the momentum, so that the Lagrangian can be expressed simply in terms of the trajectory. This assumption is not absolutely necessary, but it is mathematically convenient, and it is correct for nearly all applications, and when it is not, like the string world-sheet action in Nambu-Goto form, it can often be made correct using auxiliary variables. The path integral makes unitarity nontrivial, it is trivial in Dirac's formulation.

The path integral links quantum systems to non-quantum systems by analytic continuation, and this is surprising when looking at the pure Dirac formalism, so the path integral should count as a second formulation, truly different from Dirac's. The path integral formulation produces a natural description of gauge theories, which is very inconvenient in Dirac form.

The third formulation is more recent, and this is the PT-symmetric quantum mechanics. In abstract terms, PT symmetric QM is again just ordinary Dirac QM, just as path-integral QM is also Dirac QM. But it should count as a new formulation, because the metric on Hilbert space is defined dynamically, from the Hamiltonian, and you would never find it starting with a naive metric. The naive metric on Hilbert space makes the Hamiltonian seem to be non-Hermitian, and to verify that there is a Hermitian Hamiltonian requires work.

This formulation is only about 10 years old, and is very actively studied, but I think it is probably the most fundamental formulation, considering.

  • Dirac QM
  • Path integral
  • PT-symmetric QM

You also could formulate PT symmetric QM as a path integral, maybe you would count this as a fourth. But this gives three options. In addition, you can choose to formulate each of these as either acting on the Hilbert space of states, or on the space of all density matrices

  • States
  • Density matrices

The density matrix formulation is arguably more fundamental. It subsumes the formulation in terms of Wigner functions. So this is a choice of two options. Multiplying gives 6.

Different theories

There are different theories which are closely related to QM, which are generally not viable. These are most of the objective collapse theories, or nonlinear state evolution theories. But one deformation of quantum mechanics at least, is not ruled out by anything except theoretical principles, and this is the superoperator formulation

  • Superoperator formulation

The superoperator formulation generalizes the notion of Hamiltonian to the most general operator on the density matrix, rather than the state-vector. This formulation allows for a description of instantaneous decoherence, and it is developed in the 1970s in an obscure book which I read a long time ago, and whose author escapes me (I am sorry, it doesn't google for me).

Hawking also advocated a direct density matrix formulation as part of his program of information loss in black holes, but I don't think he cited the earlier book I am talking about (probably because he wasn't aware of it--- I stumbled across it in a library years ago). The name "superoperator" is not standard--- it is the name given in quantum information theory to the operator which multiplies the density matrix to give its time evolution. This is the quantum analog of the Fokker-Planck equation.

Here is a refernce with "nine formulations", with somewhat different ideas of what constitutes a formulation: http://www-physique.u-strasbg.fr/cours/l3/divers/meca_q_hervieux/Articles/Nine_form.pdf

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    $\begingroup$ Sorry, this is just a collection of random thoughts and concepts whose degree of validity and whose purpose in science is so completely different that I have to give it -1 for "complete mess". $\endgroup$ Nov 10, 2011 at 7:53
  • $\begingroup$ +1 I like this answer, especially with the link to the paper. $\endgroup$ Nov 10, 2011 at 14:12
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    $\begingroup$ @lubos is the paper linked in the answer a "mess"? $\endgroup$ Nov 10, 2011 at 14:13
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    $\begingroup$ @lubos: This is a collection of all valid formulations of QM that I am aware of. The superoperator formalism is the only one that isn't standard, and I can describe it in detail--- it is used in quantum information theory, and Hawking proposed it as a precise version of information loss. It certainly isn't fundamental--- there is no information loss, but it is certainly interesting in itself and has applications as a replacement for Caldera-Leggett decoherence formalism which is nonlocal in time and inaccurate for dense electronic systems. $\endgroup$
    – Ron Maimon
    Nov 10, 2011 at 15:36
  • $\begingroup$ I think there is nothing in your answer that deserves a -1 in any way! A +1 from me. $\endgroup$
    – user1355
    Nov 10, 2011 at 15:50

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