Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is this invariant under U(1) X SU(2) ?

  1. Jul 17, 2013 #1
    Could anyone explain why these are invariant under U(1) X SU(2)?

    H[itex]^{dagger}[/itex]H

    (H[itex]^{dagger}[/itex]H)[itex]^{2}[/itex]

    What is the condition for invariance under U(1) and similarly, under SU(2)?
    Unfortunately, I am not familiar with tensor contraction or tensor products...
     
  2. jcsd
  3. Jul 17, 2013 #2
    First I don't think you really need to understand tensor algebra to do this. Especially if you're in the QM forum, why not assume H is an operator?

    Second, if we transformed H under the transformation H-->UH, where U is some unitary operator, then what would Hdagger be after the transformation? Invariance of your expressions under the symmetry group means that if we took a transformation from the symmetry group, and then we substituted the transformed H and Hdagger into your expressions, we'd get the original expressions back.
     
    Last edited: Jul 17, 2013
  4. Jul 17, 2013 #3
    Hi Jolb,

    Well, I know that UU[itex]^{dagger}[/itex] and u[itex]^{dagger}[/itex]U are both equal to 1...
    Could we say that invariance is possibly like making a full rotation? And we end up where we started from?
     
  5. Jul 17, 2013 #4
    If H-->UH, then do you see that Hdagger --> (UH)dagger? What is (UH)dagger? Try simplifying/plugging this in.
     
  6. Jul 17, 2013 #5
    H[itex]^{dagger}[/itex] ?
     
  7. Jul 17, 2013 #6
    No! Where did U go?

    A general fact from linear algebra is that (AB)dagger = BdaggerAdagger. [This isn't hard to prove if you remember the dagger (adjoint) is just complex conjugation together with the transpose operation. I.e., Adagger=(A*)T=(AT)*.] Use this fact and the property of unitary operators in your post #3.

    Make sure you do not accidentally make the mistake of transforming things twice. Once you transform H to UH, you only need linear algebra to derive the transformed expressions for (UH)dagger, etc--you do not need to apply another U to something that already incorporated the substitution H-->UH.
     
    Last edited: Jul 17, 2013
  8. Jul 17, 2013 #7
    Thanks, Jolb! You mentioned adjoint- is the dagger the adjoint representation? I'm also trying to figure out if the Hilbert space consists of a real vector space and a complex vector space... and if these two spaces are A Dagger and A
     
    Last edited: Jul 17, 2013
  9. Jul 17, 2013 #8
    I would call the dagger the adjoint operation; a "representation" typically has to do with infinitesimal Lie groups or other symmetry groups. On a basic matrix algebra level, to perform the adjoint operation, you just take the arbitrary matrix (e.g. a Pauli spin matrix), then simply transpose the matrix and take its complex conjugate, and you have the matrix's adjoint. Clearly there are no advanced Lie techniques involved.

    In other words (Adagger)ij = (Aji)*, where the * denotes complex conjugation.
     
    Last edited: Jul 17, 2013
  10. Jul 17, 2013 #9
    Thank you! When does the Levi-Civita symbol come in?
     
  11. Jul 17, 2013 #10
    The Levi-Civita symbol doesn't really come in for the purposes of your problem. But the Levi-Civita symbol has wide applications, and two basic examples are in the formation of cross products of vectors and in finding the determinants of matrices.

    By the way, did you manage to prove the invariance of those expressions?

    Also, a hilbert space is a complex vector space, and, loosely speaking, its dual space is just the adjoint of the original space. Elements of the hilbert space are denoted by a ket |x> and the dual of |x> is the bra <x| where <x| = |x>dagger. [I apologize in advance to any mathematically minded readers (Wannabe) if I'm too loose with this description.]

    And finally, I wouldn't say that it's like going a "full rotation." If we use the analogy between unitary and rotation groups (keeping in mind that they're not exactly the same exact thing but strongly related), then the invariance would be more like undoing any rotation back to 0°. (Not completing the 360°.)
     
    Last edited: Jul 17, 2013
  12. Jul 17, 2013 #11
    :smile:Well... I'm trying to find invariances with Lie representation theory, and since I don't understand it well, I thought I'd start with this example... ? I was thinking the hypercharge played a role... If this H is a particle, would the hypercharge have to be a certain value in this example (maybe zero)? Working on a summer project...

    Aaah, good explanations, thank you... appreciate that!!!
     
  13. Jul 17, 2013 #12
    Hmmm, my memory on Gell-Mann's Eightfold Way is a bit fuzzy but I seem to remember that hypercharge and isospin correspond to various irreducible representations of symmetry groups the particle families obey. This is some advanced stuff though, so I would refer you to a textbook like Georgi's "Lie Algebras in Particle Physics" (a classic).
     
  14. Jul 17, 2013 #13
    I have it and it is too advanced!!!!! :smile: I don't understand any of it! Thanks Jolb, I'm really glad you mentioned adjoint... something went click!
     
  15. Jul 17, 2013 #14
    Well Georgi's book came from his lectures for a graduate course at Harvard, so even though it is supposed to be a "self-contained" intro, it is a pretty hard book. Unfortunately it is probably the best option. Other options that I used when I took a class on the topic were Hamermesh's "Group Theory and its Applications to Physical Problems" and Barnes' "Group Theory for the Standard Model of Particle Physics and Beyond". But they're hard too. Maybe someone else knows a better undergrad-level book (probably a real math book instead of a physics book).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why is this invariant under U(1) X SU(2) ?
Loading...