I am currently reading the book Symmetries, Lie Algebras and Representations: A graduate course for physicists by Jürgen Fuchs and Christoph Schweigert. On pp.11, they say:

Let us note that in quantum physics there can be more general symmetries which are not described by groups anymore. The presence of such symmetries can be understood as follows. In quantum physics, the commutative algebra of functions on the configuration space is traded for a non-commutative algebra of operators. Now for a classical dynamical system the symmetries can be viewed as acting on the points of the configuration space, hence they form a group, with the group multiplication provided by the composition of maps. In contrast, in quantum physics the configuration space is no longer present, so that this argument does not apply anymore.

I am a bit confused by the authors' argument here. For quantum physics, we still have the Hilbert space on which we can view the operators act. What's the significance here for the existence of configuration space? And can somebody explain it with a particular example, say, a specific quantum symmetry which can not be described by group and can reflect the authors' argument here?

  • $\begingroup$ a colleague, an ancient from cambridge where he got this curse, says that you don't understand what the authors mean and that the answers of this page are weak. He will write a page on his blog $\endgroup$ – user46925 Aug 17 '16 at 8:42
  • $\begingroup$ @igael, please let me know the link if he has given an answer. It would be better if he can answer this post right here. $\endgroup$ – Wein Eld Aug 17 '16 at 20:33
  • $\begingroup$ I hope to bring him here ... for my own, as usual, i upvoted everyone :) $\endgroup$ – user46925 Aug 17 '16 at 20:52

I cannot understand the claim. As a matter of fact, every quantum symmetry (see below) is the element of a group or, better, the image of the representation of a group. Due to Wigner-Kadison theorem, the symmetry is represented by a unitary or an anti unitary operator $U$. (This fact holds true also in the presence of superselection rules). Next define $G$ as the subgroup generated by $I $, $U$, $U^{-1}$ in the group of isometric surjective linear and antilinear maps in the Hilbert space of the system. $U$ is therefore the image through a (trivial) representation of an element of $G $. Since the group is defined by $U $ itself, there are no problems with phases and multiplicators and the representation is unitary instead of projective. Clearly, the symmetry is not part of a continuous symmetry group in general. However, if $U $ is unitary, it is always possible, via spectral theory, to write $U= e^{-iA} $ for some self-adjoint operator $A$. Therefore $U=V (t) $ for $t=1$, and where $V (t) $ is the one-parameter group generated by $A $. In this case $G $ can be re-defined as a one-dimensional Lie group (using MGZ theorem to define a Lie group structure) and $\{V(t)\}_{t \in \mathbb R}$ is a strongly continuous representation of that Lie group. Obviously the physical meaning of $A $ is dubious as the construction is quite artificial.

ADDENDUM. Perhaps the problem is with the notion of quantum symmetry. It is not clear what the authors mean by quantum symmetry.

However, there are in the literature three notions of (quantum) symmetry of a quantum system described in a complex separable Hilbert space.

N.B.: I am referring here to the general notion of quantum symmetry and not of quantum dynamical symmetry (exploited for instance in the statement of the quantum version of Noether theorem).

Wigner symmetry: a surjective map transforming rays of the Hilbert space to rays of the Hilbert space preserving the probability amplitudes.

Kadison symmetry: an automorphism of the lattice of orthogonal projectors of the Hilbert space (representing the elementary YES-NO observables of the quantum system) or, equivalently, a bijecitve convex-linear map from the convex body of (generally mixed) states into itself.

Jordan symmetry: a bijective map from the Jordan non-associative algebra of bounded self-adjoint operators into itself.

In the absence of superselection rules, the three notions coincide and give rise to the same mathematical statement: symmetries are all of unitary or anti unitary operators and the correspondence is one-to-one up to an arbitrary phase.

In the presence of superselection rules described by central projectors of the von Neumann algebra of observables (assuming that the centre of the lattice is atomic) the picture is essentially identical, but the phases may depend on the superselection sector.

In the presence of a gauge group (the algebra of observables in each superselection sector is a non maximal factor), the operators $U$ are defined up to elements of the commutant of the von Neumann algebra, but I am not sure that every symmetry can be described this way and that the three notions of symmetry still coincide.

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  • $\begingroup$ The point, I think, based on F&S' academic histories, is that there are interesting algebras being represented in $U(\mathcal{H})$, such as $U_q(\mathcal{sl}_n)$. The nature of these algebras provides more specific and interesting constraints than Wigner's theorem. $\endgroup$ – user1504 Aug 11 '16 at 23:31
  • $\begingroup$ Could you be more explicit please about the "interesting contraints"? $\endgroup$ – Valter Moretti Aug 12 '16 at 10:52
  • $\begingroup$ I was thinking of the connections between quantum groups and integrable systems. $\endgroup$ – user1504 Aug 12 '16 at 10:54
  • $\begingroup$ I understand, thanks, but it seems to be a very obscure way to discuss about quantum symmetries the quoted text by Jürgen Fuchs and Christoph Schweigert... $\endgroup$ – Valter Moretti Aug 12 '16 at 10:56
  • $\begingroup$ I agree. All I can say is that this seems to be common in the subject.. $\endgroup$ – user1504 Aug 12 '16 at 11:08

I think the authors are referring to quantum groups, which are to Lie groups as non-commutative geometry is to manifolds, hence the vague claims about quantum noncommutativity.

You can sometimes find QFTs whose Hilbert spaces are representations of quantum groups (by realizing the quantum group within the algebra of operators). Knowing that a Hilbert space is a quantum group representation has many of the same kinds of practical consequences as knowing that the Hilbert space is a Lie group representation: You can decompose into irreps, constrain inner products, and so forth. (It's all just linear algebra in the end.) For this reason, people sometimes get poetic and say that a quantum group action is a symmetry that doesn't come from a group.

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The passage appears to be non-sensical. By Wigner's theorem, every quantum symmetry can be represented by (anti-)unitary operators on the Hilbert space of states, and unitary operators have inverses which necessarily also are symmetries, so all symmetries form subgroups of the unitary group. The Hilbert space of quantum mechanics doesn't really behave that differently from the configuration space of classical mechanics.

However, it may well be that the authors are aiming at something different here: Given an abstract group of symmetries, the representation as unitary operators actually need not be an ordinary linear representation of the abstract group, but is allowed to be projective, as I discuss at length in this Q&A of mine.

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    $\begingroup$ Your answer is clear. But your guess that the author aim at the representation rather than the group itself should be taken with cautious for me. So I should wait for a bit longer to see if someone else can give a different answer. Thanks! $\endgroup$ – Wein Eld Aug 11 '16 at 13:13

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