Friday, June 10, 2016

quantum field theory - No monopoles in the Weinberg-Salam model


I'm reading Chapter 10.4 on the 't Hooft-Polyakov monopoles in Ryder's Quantum Field Theory.


On page 412 he explains why magnetic monopoles cannot appear in the Weinberg-Salam model. I'm I right by saying that he shows that the electromagnetic gauge group U(1)em is not compactly embedded into the U(1)×U(1) subgroup of SU(2)×U(1)?


He then immediately concludes that the first fundamental group of the unbroken symmetry, which is H=U(1)em, π1(H) must be trivial or doesn't exists. Could someone refer me why?


Comment: I know that in the SU(2)×U(1) ones must consider the second homotopy group from S2 to the orbit G/H=SU(2)×U(1)/U(1), where H is the isotropy group of a vacuum state, after symmetry breaking. But the second homotopy group of a quotient can be related through a exact series to the kernel of the map from π1(H) into π1(G).


What I do not understand is by which theorem for H having a non-compact covering group π1(H) must be trivial or non-existing (???)?



Answer



I) Consider a Yang-Mills type theory with gauge group G. In principle we can consider the same theory with its covering group ˜G, with π:˜GG. The covering group is by definition simply connected: π1(˜G) = {1}. Any representation ρ of G can naturally be viewed as a representation ρπ of ˜G. (Warning: the opposite is not true.) Hence the fields of the theory also transform under ˜G, and the theory is invariant under ˜G. In other words, we could in principle from the very beginning view ˜G as the gauge group of the theory, cf. Section III below.


II) Next assume a spontaneous symmetry breaking of the gauge group GH to a subgroup H. Define subgroup



˜H := π1(H)  ˜G.


Concretely for the electroweak theory, the gauge group is


G := SU(2)I×U(1)Y  U(1)I×U(1)Y;


the covering group


˜G = SU(2)I×(R,+)Y


is non-compact; and the unbroken electromagnetic subgroup


H := U(1)Q  U(1)I×U(1)Y


is irregularly/non-compactly/incommensurably/non-topologically embedded, cf. e.g. my Phys.SE answer here. Here we have assume that tangent to the Weinberg angle tanθWRQ is irrational. Now since the non-compact group H is not a subgroup of SU(2)I, the subgroup


˜H = (R,+)Q


is also non-compact. It is also simply connected



π1(˜H)  {1}.


III) Let us now view the non-compact gauge group ˜G in eq. (4) as the gauge group of electroweak theory. The non-compact gauge group of electromagnetism (5) is then replaced by the non-compact subgroup (6). Standard monopole analysis then shows that there are no magnetic monopoles in the electroweak theory


π2(˜G/˜H)  Ker(π1(˜H)π1(˜G)) (1) π1(˜H) (7) {1},


cf. e.g. Ref. 2.


IV) Now let us return to OP's question. Ref. 1 considers the system from the perspective of G and H rather than ˜G and ˜H, and recast the standard monopole analysis in this language. The problem is that a closed loop γG that wraps a U(1)G has a non-compact lift ˜γ˜G which is not a closed loop. Similarly, a closed loop γH that wraps a U(1)H has a non-compact lift ˜γ˜H which is not a closed loop.


Intuitively/heuristically, the relevant notion ˜π1(˜H) (of such non-compact "closed" paths) has more paths than just (7), and for the electroweak theory the relevant quantity


π2(G/H)  π2(˜G/˜H)  Ker(˜π1(˜H)˜π1(˜G))  {1}


is trivial, in agreement with eq. (8). See Ref. 1 for more details.


References:





  1. L.H. Ryder, Quantum Field Theory, 2nd eds., 1996; p. 412.




  2. S. Coleman, Aspects of Symmetry, 1985; p. 217 & 221.




  3. F.A. Bais, To be or not to be? Magnetic monopoles in non-abelian gauge theories, arXiv:hep-th/0407197. (Hat tip: Hunter.)





  4. S. Weinberg, Quantum Theory of Fields, vol. 2, 1996; p. 443-445.




  5. J. Preskill, Magnetic Monopoles, Ann. Rev. Nucl. Part. Sci. 34 (1984) 461-530; Sections 4.2 & 4.3. The PDF file is available here.




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