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Natural Science Forum / Physics / Research / December 2004Parity always commute with energy operator?
Hi all, Recently I'm reading Streater and Wightman's "PCT, spin and statistics, and all that". In Chapter 3, they proved that the assumption of a definite transformation law for all the fields of a field theory uniquely fixes the corresponding transformation on the states up to a phase. Then we can always get a unitary operator that satisfies U(P)U(\Lambda, a)U(P)^{-1} = U(P\Lambda P^{-1}, Pa) Now, we can check the case when U(\Lambda, a) is an infinitesimal transformation, and we can always find that PHP^{-1} = H My question is, is the relation above indicate that P is a symmetry of the theory, I think the answer definitely is "no", so where the problem is, My guess is that P have to commute with Hamiltonian density, but I cannot prove that. BTW, Any one checked this paper? http://lanl.arxiv.org/pdf/quant-ph/0211123 It says any hermitian hamiltonian operator have parity. I think that have some relation with my problem, but I cannot figner out.:( > Now, we can check the case when U(\Lambda, a) is an infinitesimal > transformation, and we can always find that [quoted text clipped - 5 lines] > problem is, My guess is that P have to commute with Hamiltonian > density, but I cannot prove that. The point is, that you can build theories, which violate parity conservation, e.g., the standard model of electroweak interactions. It's good to read the original papers by Yang and Lee first, because there it is a little bit simpler to understand with currents. T. D. Lee, C.-N. Yang, Question of parity conservation in weak interactions, Phys. Rev. 104 (1956) 254, URL http://link.aps.org/abstract/PR/v104/i1/p254 T. D. Lee, C.-N. Yang, Parity nonconservation and a two component theory of the neutrino, Phys. Rev. 105 (1957) 1671, URL http://link.aps.org/abstract/PR/v105/i5/p1671 The same is of course true for the modern standard model, but there the theory is a little bit more involved (with Higgs mechanism and all that). > BTW, Any one checked this paper? > http://lanl.arxiv.org/pdf/quant-ph/0211123 > It says any hermitian hamiltonian operator have parity. I think that > have some relation with my problem, but I cannot figner out.:( Since the Hamiltonian of the standard model is, of course, hermitian and there is no parity conservation, this statement cannot be true, but I have not looked on the mentioned paper.  SignatureHendrik van Hees Fakultät für Physik Phone: +49 521/106-6221 Universität Bielefeld Fax: +49 521/106-2961 Universitätsstraße 25 http://theory.gsi.de/~vanhees/ D-33615 Bielefeld Hi, I have post two letters in the group a week ago, but they've not appeared yet. I think they may be lost. So here is another try. sorry if it is duplicated I've received a reply from Lubos Motl about another thread started by me (What is a symmetry in quantum theory). After reading his mail I think I can and should now make the problem more clear: Lubos Motl point out the example of neutrino, which certainly killed the statement that the unitary operator of parity can always been defined. But I think it's the only exception, am I right? The proof in Streater and Wightman require a definite transformation law for all the fields of a field theory. The field of neotrino don't have a definite transformation law under parity, so the theorem failed in case of neotrino. I just guess, so please correct me if I was wrong :). But I think without neotrino we can still construct theory with parity violation, or we can just give neotrino a mass, but I know nothing about that, so let forget neotrino and suppose we already got a theory of this kind, then the problem rise again. Since all the field have a definite transformation under parity, a unitary operator U(P) can always be defined up to a phase. and satisfy U(P)^-1 U(\Lambda, a) U(P) = U(P \Lambda P^-1, a), U(\Lambda, a) is the unitary operator of Lorentz transformation. They proved this property for vacuum state first, then use the field operatre to extend it to the whole Hilbert space. After we got this statement, just take the case of an infinitesimal transformation, we can always get U(P)^-1 H U(P) = H, H is the energe operator. so I "proved" that the parity operator can always commute with Hamiltonian. If the proof is correct, we cannot say that parity is a symmetry of a theory iff the parity operator is commute with Hamiltonian. BTW, I tried to construct a parity violated theory and prove its Hamiltonian do commute with parity operator. But now I mess around with the transformation properties of fields. Can those minus sign can always be cancel by taking dx -> -dx, since the intergrade is from -infinity to infinity? No its not just neutrinos, its the entire weak interaction that violates parity. In fact, it violates parity maximally. I believe the first observation of this was in beta decay in Cobalt 60. I really can't give you a good reason why this is so, I almost consider it axiomatic from a theoretical point of view, and a fact from an experiment. The whole thing becomes very apparent when you go through the QFT, the weak vertex couples right and left handed fermions in a way that say EM does not. But then again, the theory is pretty much designed to fit the experimental facts. Incidentally, a larger symmetry is broken (CP) in kaon systems, and there are even claims that CPT will be broken at some level (one has to go to axiomatic field theory for those cases). > No its not just neutrinos, its the entire weak interaction that > violates parity. In fact, it violates parity maximally. > > I believe the first observation of this was in beta decay in Cobalt > 60. http://physics.nist.gov/GenInt/Parity/cover.html > I really can't give you a good reason why this is so, I almost > consider it axiomatic from a theoretical point of view, and a fact [quoted text clipped - 7 lines] > there are even claims that CPT will be broken at some level (one has > to go to axiomatic field theory for those cases). First, some definitions. If you are sloppy with the definitions it all goes to mush. Then we show that a non-empty spacetime is *not* expected to be symmetric to parity inversion on mathematical grounds. CHIRALITY: Not superposable upon its mirror image. This is not quite rigorously equivalent to "inversion of *one* coordinate axis," which is the proper math. Chirality emerges from a minimum set of four non-coplanar points in 3-space. Chirality as a concept has been rigorously defined in space and time, and its rules mathematically and experiemtnally evaluated, J. Mol. Phys. 43 139 (1981) Angew. Chem. Int. Ed. 38 3418 (1999) Chem. Rev. 98(7) 2391 (1998) J. Am. Chem. Soc. 108 5539 (1986) Nature 405(6789) 932 (2000) PARITY: Not superposable upon its image with *all* coordinate axes inverted. Parity is a more restrictive subset of chirality and has no directional bias. You can find *incorrect* literature statements that parity is chirality plus a 180-degree rotation, http://www.mazepath.com/uncleal/invert.gif The rotation definition variation appears to be valid - but not if system rotation is quantized, as with angular momentum. Quantitative parity divergence: Any object or array of points with finite inertia can have a normalized measure of parity divergence (CHI) between it and its inversion quantitatively calculated, giving a number between CHI=zero (achiral, superposable mirror images) and CHI=1 inclusive (perfect party divergence). The overall theory is remarkably complex, http://www.mdpi.net/entropy/papers/e5030271.pdf http://www.mazepath.com/uncleal/petit.htm The validated implimentation is straightforward and has been reduced to public domain software, including output of symmetry diagnostics COR and DSI that bear on the meaning of the results, J. Math. Phys. 40(9) 4587 (1999) http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM http://petitjeanmichel.free.fr/itoweb.download.qcm.readme Now then... Why should spacetime be symmetric to parity transformation? Cross-products are asymmetric to chirality (EM interactions). If one wished to add persistent defects to a smooth background of spacetime, one simple way would be to tie knots. Chiral knots cannot internally jiggle to untie themselves, nor can they be rearranged to go planar or flat (Kuratowski's theorem), "Kuratowski's theorem" 821 hits ALL ATOMS are homochiral, which has been measured, http://arXiv.org/abs/cond-mat/0207627 Mendeleev Commun. 13(3) 129 (2003) http://socrates.berkeley.edu/~budker/PubList.html Phys. Rev. Lett. 82(12) 2484 (1999) Phys. Rev. Lett. 80(17) 3719 (1998) Rep. Prog. Phys. 60(11) 1351 (1997) Phys. Rev. A 52(3) 1895 (1995) Am. J. Phys. 56 1086 (1988) Given that gravitation is a non-flatness of spacetime that cannot be shielded, one might suspect it is the expression of a chiral defect of only one handedness (at least in part). If so, then there would be no evidence of chirality unless it were made to interact with a contrasted pair of of extremal parity divergence test bodies. Like a left shoe testing a left and right foot, an energy difference (diasteromeric interaction) would be then measured. An achiral sock (classic achiral composition test masses) would not show any difference between feet. It is possible to inexpensively fabricate paired test masses with perfect quantitative parity divergence [1-CHI around 10^(-15)] that are chemically and macroscopically identical. The emergent length at which such parity appears is around 0.5 nm. On a test mass volume basis, left-and right-handed [crystallographic space groups P3(1)21 and P3(2)21] alpha-quartz maximally satsifies these conditions. On a per atom basis, left-and right-handed tellurium is insignficantly better. Symmetry of gravitation to test mass parity inversion can be tested in existing apparatus, http://www.mazepath.com/uncleal/qz.pdf Somebody should look. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) > But I think [the neutrino] is the only exception, am I right? Not really. The quark sector, which has no neutrinos in it, violates not only the parity, but even the CP symmetry - by the phase in the CKM mass matrix. Let me return to the parity (P): the whole electroweak theory is parity violating because the "weak force gauge fields" only interact with the left-handed components of the particles and the right-handed components of the antiparticles, so to say. This left-right asymmetry in the interactions of the W and Z bosons with the fermions is what we usually mean by the adjective "chiral" in the context of gauge theories. The word "chiral" comes from Greek: "cheir" means a "hand". This gauge interaction with the left components only is true not only for the neutrinos, but also for all remaining leptons and quarks. The only special role of the neutrino is that the parity violation is obvious by assigning the quantum numbers - such as the helicity (the angular momentum around the axis of momentum) to a *free* particle. In other words, the parity violation resulting from the neutrinos is obvious even in *electrodynamics* (the very low-energy approximation of physics) and we don't even need to consider the electroweak weak interactions. Nevertheless, the weak force *does* exist and it implies as strong P-violation for charged laptons and quarks as it does for the neutrinos. > But I think without neotrino we can still construct theory with parity > violation, or we can just give neotrino a mass, ... If you don't have the right-handed component of the neutrino, you can only add Majorana masses - which essentially means that you identify the neutrino with the antineutrino. This unified particle has both helicities - left-handed and right-handed - and it can have a nonzero mass. Such a choice implies the possibility of a lepton number violation. > U(P)^-1 U(\Lambda, a) U(P) = U(P \Lambda P^-1, a), > U(\Lambda, a) is the unitary operator of Lorentz transformation. Maybe they proved it for the *free* Hamiltonian without neutrinos? The statement that U(P) commutes with the evolution operators is certainly wrong for more general theories - such as the Standard Model - so they could not have proved it. The Hamiltonian of the Standard Model contains some terms that are even under your parity operator, and some terms that are odd, and the sum simply does not commute with the operator of parity. > U(P)^-1 H U(P) = H, H is the energe operator. > > so I "proved" that the parity operator can always commute with > Hamiltonian. You have not proved anything, you just repeated a wrong statement many times which does not make it proved. U(P)^{-1} H U(P) is simply NOT equal to H. > If the proof is correct, we cannot say that parity is a symmetry of a > theory iff the parity operator is commute with Hamiltonian. This might have been another hint for you that your "proof" was not correct, especially because I explained it to you already in the previous post. > properties of fields. Can those minus sign can always be cancel by > taking dx -> -dx, since the intergrade is from -infinity to infinity? Nope, the kinetic and potential terms in the action never change the sign if you redefine x to -x - imagine that the integral has |dx| in it instead of dx. ______________________________________________________________________________ E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/ eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Superstring/M-theory is the language in which God wrote the world. Hi, > > But I think [the neutrino] is the only exception, am I right? > [quoted text clipped - 4 lines] > the left-handed components of the particles and the right-handed > components of the antiparticles, so to say. hehe, actually I wanted to say that I guess neutrino is the only field that do not have a definite transformation under parity. Since the theorem in Streater and Wightman require a definite transformation, so it cannot apply to neutrino. > > U(P)^-1 U(\Lambda, a) U(P) = U(P \Lambda P^-1, a), > > U(\Lambda, a) is the unitary operator of Lorentz transformation. > > Maybe they proved it for the *free* Hamiltonian without neutrinos? I don't think so. The theorem in Streater and Wightman only proved the exsitence of unitary operator satisfying the relationship above with U(\Lambda, a), but since Hamiltonian is one of the generator of U(\Lambda, a), I came to the result in the title... I think here H should not be restrict to the free Hamiltonian. > could not have proved it. The Hamiltonian of the Standard Model contains > some terms that are even under your parity operator, and some terms that > are odd, and the sum simply does not commute with the operator of parity. You mean Hamiltonian or Hamiltonian density? I see simillar argument about Hamiltonian density. (actually Lagrange density, but imho we can safely ignore the difference here.) Whether this statement will change after we intergrade it to Hamiltonian is what I feel confused now, maybe it's just a simple math work though. > times which does not make it proved. U(P)^{-1} H U(P) is simply NOT equal > to H. Maybe you are right. Now the situation is: 1) Streater and Wightman proved a theorem: the requirement that the fields a field theory have definite transformation laws under U(P) uniquely fixes that operator. especially satisfy: U(P)^-1 U(\Lambda, a) U(P) = U(P \Lambda P^-1, a). 2) Hamiltonian is a generator of U(\Lambda, a), so we can easily check the relationship between parity and Hamiltonian. 3) Parity is not a symmetry, so U(P)^(-1) H U(p) can not equal to H. My first guess is that we should use Hamiltonian *density* instead of Hamiltonian to define symmetry, since we get the motion equation from desity, and A symmetry should keep the motion equation unchanged . Or we just say 1) was wrong, but the theorem seems have a firm base. (Theorem 3-8, if anyone interested) Is there any other possibilities? Somewhere must be wrong:( Best Regards, LR Dear L.R., I think that your reasoning has very almost nothing to do with the real issue of the parity symmetry and parity violation. When we talk about the symmetry - especially parity that transforms the space - of course that we must talk about the whole integrated Hamiltonian; even in parity-symmetric theories, the Hamiltonian density at point "x" is transformed into the density at point "-x" which is a different number. The Hamiltonian density is not preserved, although in the simplest cases, the Hamiltonian density at "x" must be equal to the Hamiltonian density of the parity-transformed fields at "-x", and therefore the rule is simple. > hehe, actually I wanted to say that I guess neutrino is the only field > that do not have a definite transformation under parity. This sentence has no physical meaning. If you consider the electroweak theory - even if you complete the neutrinos into the full Dirac, left-right symmetric spinors - parity is simply not preserved. There exists no operator that deserves to be called "parity" because no operator acting in this geometric fashion commutes with the Hamiltonian. You might define *some* Z2 operator "P" but it would not commute with the Hamiltonian, and actually the precise choice of "P" would be ambiguous (even for quarks!). For example, you would not know whether the opposite components (left-handed vs. right-handed) of the quark field should have the same parity or the opposite one. According to everything you have written so far, the whole book by Streater and Wightman seems to be simply wrong, but I would have to check the real book in order to make this statement with certainty. Aren't the authors members of the Axiomatic Quantum Field Theory community? That would explain everything because the current AQFT is a pile of wrong statements. All the best Lubos ______________________________________________________________________________ E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/ eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Only two things are infinite, the Universe and human stupidity, and I'm not sure about the former. - Albert Einstein Hi, > I think that your reasoning has very almost nothing to do with the real > issue of the parity symmetry and parity violation. When we talk about the [quoted text clipped - 5 lines] > at "x" must be equal to the Hamiltonian density of the parity-transformed > fields at "-x", and therefore the rule is simple. I think you are right, thanks for clarify this for me.:) > acting in this geometric fashion commutes with the Hamiltonian. You might > define *some* Z2 operator "P" but it would not commute with the The theorem in Streater and Wightman has not proved that operator "P" is a Z2 operator, it seems that they treat the Z2 requirement as the physical meaning of operator "P", seems they cannot prove this for their "P" operator. Is that's the flaw in the proof? > Hamiltonian, and actually the precise choice of "P" would be ambiguous > (even for quarks!). For example, you would not know whether the opposite > components (left-handed vs. right-handed) of the quark field should have > the same parity or the opposite one. Streater and Wightman have said something about this: since spin 1/2 field cannot be observed directly, there is some freedom to choose the phase of state after P transformation, and physically identical choice can be related by a unitary operator. > According to everything you have written so far, the whole book by > Streater and Wightman seems to be simply wrong, but I would have to check > the real book in order to make this statement with certainty. Aren't the > authors members of the Axiomatic Quantum Field Theory community? That > would explain everything because the current AQFT is a pile of wrong > statements. :)))). Best Regards, LR > Lubos Motl point out the example of neutrino, which certainly killed > the [quoted text clipped - 9 lines] > correct me > if I was wrong :). Of course it is right, that the neutrinos within the standard model are massless lefthanded particles (here I negelect the masses, the neutrinos must have, because neutrino mixing is observed by SuperKamiokande, SNO etc.). So they are chiral fermions (or Weyl fermions), on which space reflection is not defined. Indeed, space reflection makes a lefthanded particle a righthanded particle, which in the case of neutrinos do not exist in nature. On my opinion it is more physical to argue with the interactions, whether they are parity conserving or not, because that's what is observed in experiment. As we know from the old days in 1956, the weak interactions indeed violate space reflection invariance and thus violate parity conservation. This was predicted by T. D. Lee and C. N. Yang from solving the "theta-tau" puzzle. Theta and tau in those days where particle names for the particles we nowadays call charged kaons (I hope I'm right here). The theta and the tau looked exactly the same (same life time, same mass etc.), but the one of them decays to two pions (positive parity) and the other to three pions (negative parity, pions are pseudoscalar mesons!). The puzzle was solved by the assumption, that the weak interaction violates parity conservation. All this was described in terms of the Fermi's theory four-fermion coupling theory of the beta-decay (1933) which was extended by Gamov and Teller shortly afterwards (1936). In 1957 the experiment by Wu showed explicitly that parity is violated by weak interactions. It could also established experimentally, within a relative short period, that the weak interactions are best described by the exchange of vector- and pseudovector currents and that the violation was maximal in the sense that the leptonic reactions where of the type (V-A) (i.e. vector minus axial vector current). Finally all this evolved in the famous standard model of electro-weak interactions (a nonabelian gauge theory, with massive gauge bosons, getting their mass from the Higgs-Kibble mechanism, invented by Salam, Glashow and Weinberg in the sixties). SignatureHendrik van Hees Cyclotron Institute Phone: +1 979/845-1411 Texas A&M University Fax: +1 979/845-1899 Cyclotron Institute, MS-3366 http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366 Dear LR, The answer to your query is not as Motl says (that the whole of the book, "PCT Spin and Statistics..." is wrong), but is explained on page 16 of the book. One must distinguish between the requirement that there is a unitary operator U(g) for a group element g, and the merely formal `substitution law' obtained by replacing symbols on a page with linear combinations of the same symbols and/or their complex conjugates. The latter is the way symmetries were formulated before Wigner's formalism was properly understood. To get a symmetry out of a substutution law, it is not enough simply to DO the transformation: we must say what happens when we do it! In the Feynman functional integration method, (the Lagrangian formalism) the Lagrangian must be invariant under the substitution. The trouble with this is (1) that the Feynman integral has no mathematical definition (2) that even heuristically, this is not sufficient to ensure a symmetry: anomalies can arise because of divergences, and also, there can be a spontaneous symmetry breakdown. In neither case does the expected unitary operator exist. In the Lagrangian formalism, the fields are classical, and no relation is assumed between fields at different times. Then the substitution can make sense, even when the Lagrangian is not invariant. If the Lagrangian is not invariant under the transformation g, then no unitary operator which transforms the field correctly for all time and space can be expected to exist. In the formalism in which equations of motion can be given a meaning, (called the Hamiltonian formalism) we can do the substitution, but it is not a symmetry if the equations of motion are not left invariant under the transformation. This is not sufficient, however. When there is a spontaneous breakdown of symmetry, the equations of motion are invariant under the substitution, but no unitary operator implementing the transformation exists. In Wigner's formalism, the theory is said to be invariant under a substitution law of the fields if there exists a unitary operator (independent of space and time) which implements the law (by unitary conjugation). Please read theorem 3-8, page 127 more carefully. We do not prove that the unitary operator U(I_s) exists, just because the substitution law can be written down. We ASSUME that U(I_s) exists. So what hapens when we try to do the same in a theory violating parity, for which the substitution law can be written down? The answer is that the transformation law is inconsistently defined, in the Hamiltonian formalism (Wightman theory is a generalisation of the Hamiltonian formalism, not the Lagrangian). There are relations between fields at one time and fields at a later time. The field at a later time is determined by the fields at an earlier time-slice (in models obeying what is called primitive causality). So the transformation of the operators at the earlier time determine the form of the transformation at the later time, and if the equations of motion are not invariant under the substitution, then we get an inconsistency with the substitution law at the later time. I hope that this helps. Raymo > The answer to your query is not as Motl says (that the whole of the > book, "PCT Spin and Statistics..." is wrong), [quoted text clipped - 9 lines] > (the Lagrangian formalism) the Lagrangian must be invariant under the > substitution. And the measure in the integral. The Lagrangian can be made invariant by choice; the possible problems are with the invariance of the measure - and it is there where anomalies cause problems. > The trouble with this is > (1) that the Feynman integral has no mathematical definition This is not quite true; it only holds for the handwaving definition given in typical QFT books. The Feynman integral has in many field theories of dimension <4 a well-defined meaning via analytic continuation of a Wiener-type integral. In dimension 4 this may still be the case, except that nobody has been able to prove it. But there is no nonexistence theorem that would forbid it. Arnold Neumaier |
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