# Interpretation of "superqubits"

+ 7 like - 0 dislike
717 views

Two very intriguing papers recently appeared on the arXiv, claiming that one can use "superqubits" -- a supersymmetric generalization of qubits -- to violate the Bell inequality by more than standard quantum mechanics would allow. (That is, they claim one can violate the Tsirelson bound, which says that the CHSH game can be won quantum-mechanically with probability at most cos2(π/8) ~ 0.85.) The first paper is by Borsten, Bradler, and Duff and the second is by Bradler.

Alas, I remain deeply confused about the physical meaning of these results, if any. As the authors define them, "superqubits" seem to involve amplitudes that can be Grassmann numbers rather than just complex numbers. While I know next to nothing about the topic, that seems like a fundamental departure from "supersymmetry" in the sense that high-energy physicists use the term! I take it that supersymmetry is "just" a proposed new symmetry of nature, alongside the many other symmetries we know, and doesn't involve tampering with the basic rules of quantum mechanics (or with spatial locality). In particular, in supersymmetric theories one still has unit vectors in a complex Hilbert space, unitary transformations, etc.

If that's correct, though, then what on earth could superqubits have to do with supersymmetry in physics---besides perhaps just repurposing some of the same mathematical structures in a totally different context? Is there any possibility that, if nature were supersymmetric in such-and-such a way, then one could do an actual experiment that would violate Tsirelson's bound?

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user Scott Aaronson
I think they use the term supersymmtery exactly the way it is used in high energy physics.

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user MBN
Well then, if nature is supersymmetric in the relevant way, could you or could you not do an actual experiment that would violate the Tsirelson bound? If you could, that's incredible! But I was surprised by the papers' failure to address what seems like such a basic question.

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user Scott Aaronson
I don't know, I should have said it explicitly that I was only voicing my impression from a first glance of the papers.

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user MBN
Dear MBN, they may be trying to say that their Grassmannization of the amplitudes is what SUSY actually does in physics but it ain't the case. This interpretation is simply invalid and Scott's dissatisfaction and surprise are of course fully justified. See my answer for more details.

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user Luboš Motl

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user MBN

+ 6 like - 0 dislike

I completely agree with Scott that this particular "Grassmannization" isn't equivalent to what supersymmetry is doing in physics. Supersymmetry is a constraint that picks a subset of theories – ordinary theories with ordinary bosonic and fermionic fields that are just arranged (and whose interactions are arranged) so that there is an extra Grassmann-odd symmetry. Because supersymmetric theories are a subset of more general theories, of course that all the general inequalities that hold for the more general theories hold for supersymmetric theories, too. And there are many new inequalities and conditions that hold for supersymmetric theories – but not fewer constraints.

In supersymmetric theories, what becomes Grassmann numbers are never probability amplitudes. Only particular observables are fermionic operators – operator counterparts of Grassmann-number-valued quantities in classical physics. These fermionic operators only have nonzero matrix elements between Grassmann-odd states and Grassmann-even states; for the same reason why bosonic operators only have nonzero matrix elements between states of the same grading. One may introduce a grading on the Hilbert space but the amplitudes are still complex commuting $c$-numbers.

There's a simple reason why probability amplitudes can't be Grassmann numbers. To get physical commuting quantities out of Grassmann numbers, one always has to integrate. That's why the Grassmann variables may be integration variables integrated over in Feynman's path integral; but that's also why they have to be set to zero if we're doing classical physics. There aren't any particular nonzero values of Grassmann numbers. On the other hand, probability amplitudes don't have to be integrated; their absolute values should be just squared to obtain the probabilities (or their densities such as differential cross sections).

So if their construction is consistent at all, it's just a mathematical analogy of superspaces at a different level – amplitudes themselves are considered "superfields" even though in genuine quantum physics, amplitudes are always complex numbers. That's why the inequalities can't be considered analogous to Bell-like inequalities and can't be applied to real physics. In particular, once again, Tsirelson's bound can't be violated by theories just because they're supersymmetric (in the conventional sense, just like the MSSM or type IIB string theory) because it may be derived for any quantum theory, whether it is supersymmetric or not, and supersymmetric theories are just a submanifold of more general theories for which the inequality holds.

I would point out that it wouldn't be the first time when Michael Duff and collaborators would be giving wrong interpretations to various objects related to quantum computation. Some formulae for the entropy of black holes mathematically resemble formulae for entangled qubits etc. But the interpretation is completely different. In particular, the actual information carried by a black hole is $A/4G$ nats i.e. the black holes roughly parameterize an $\exp(A/4G)$-dimensional space of microstates. That's very different (by one exponentiation) from what is needed for the quantum-information interpretation of these formulae in which the charges themselves play the role of the number of microstates.

So I think that at least Michael Duff has been sloppy when it came to the interpretation of these objects which was the source of his misleading comments about the "black hole entropy formulae emulating tasks in quantum computation". There may be mathematical similarities – I am particularly referring to the Cayley hyperdeterminant appearing both in quantum computing and black hole entropy formulae – but the black holes aren't really models of those quantum algorithms because their actual Hilbert space dimension is the exponential of what it should be for that interpretation and they're manipulating pretty much all the qubits at the same moment. The objects in the hyperdeterminant have completely different interpretations on the string theory and quantum computing side; there isn't any physical duality here, either.

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user Luboš Motl
answered Aug 21, 2012 by (10,278 points)
+ 4 like - 0 dislike

In mathematics a super lie algebra is defined as a graded algebra with commutators and anti-commutators satisfying a generalised Jacobi identity. No Grassman variables are needed in the defintion. The osp used for super qubits is a basic example of such an algrebra. This particular supersymmetry algebra cannot be used for SUSY theories in physics because spin statistics requires that the anticommuting part is a half integer spin representation of the Lorentz group. Other superlie algebras are used in SUSY that do have spinors, but they are still the same class of mathematical structure satisfiying the extended Jacobi identities.

If you want to describe the supersymmetry supergroup rather than the super lie algebra, then you need Grassmann variables for the anticommuting part. This also applies to the field variables used for the fermionic parts of the SUSY theory. These variables appear in the classical (first-quantised) field theory and this is no different from the formulation of non-supersymmetric theories with fermions. In the second-quantised theory the Hilbert space for the fermions is the usual Fock space and uses oridinary commuting complex numbers for the amplitudes.

The context of the two papers cited in the question is about possible generalised theories that are more non-local than quantum mechanics, see e.g. the reference to Popescu and Rohrlich. This would not be the case for the supersymmetry theories normally used in physics. Such theories are hypothetical and not related to anything observed in physics so far. Notice how the word "Hypothetical" is used in the first paper. One point of this work is to try to provide a more concrete model of how Tsirelson’s bound could be violated but it would have to go beyond ordinary quantum mechanics.

They do point out some theoretical work in condensed matter physics where osp supersymmetry may be emergent. I am not sure how this would work out if it was realised in nature but we are getting used to seeing emergent properties in condensed matter physics that seem to be paradoxically at odds with the underlying physics, e.g. fractional charges. A point I would make are that emergent physics in condensed matter may not be constrained by spin statistics because it is non-relativistic so supersymmetry groups such as osp could be an emergent supersymmetry. Another similar point is that equations in condensed matter physics are sometimes written in a form that looks like a relativistic formula but with a value of c that is less than the speed of light. Perhaps this would allow the emergence of some non-locality that goes beyond Tsirelson’s bound if you pretend that this c is the speed of light. However, a real violation of Tsirelson’s bound would require physics more general than quantum theory at a more fundamental level and these papers do not claim to have a full theory for that as far as I can tell.

This post imported from StackExchange Physics at 2014-07-24 15:47 (UCT), posted by SE-user Philip Gibbs
answered Aug 24, 2012 by (650 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.