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How to find Internal Symmetries?

  1. Dec 7, 2014 #1
    For instance, how to find what internal symmetries a free relativistic lagrangian density has for complex scalar field?
     
  2. jcsd
  3. Dec 8, 2014 #2

    dextercioby

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    Well, this is a circular question, since Lagrangian (densities) are built on purpose to encompass the desired (rigid) symmetries. Think about the ##\phi\phi^*## term. What transformation of the field would leave it invariant? Then you can move to differentiated fields.
     
  4. Dec 8, 2014 #3
    Unitary transformation?
     
  5. Dec 8, 2014 #4

    dextercioby

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    Fancy language. Spell it out using 'common knowledge'. :)
     
  6. Dec 8, 2014 #5
    ## \phi \longrightarrow e^{i\alpha}\phi ##
    ## \phi^* \longrightarrow \phi^* e^{-i\alpha} ##

    ## (\phi^* e^{-i\alpha})(e^{i\alpha}\phi) = \phi^* \phi ##

    Also I could do it with the orthogonal matrix representation:

    $$ \phi^* \Omega^+ \Omega \phi = \phi^* \phi $$

    And this is all? there are no more kinds of internal symmetries?
     
  7. Dec 8, 2014 #6

    dextercioby

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    Yes, indeed. For the sake of the Standard Model SO(n) is not needed, a simple phase factor (U(1)) will do.
     
  8. Dec 9, 2014 #7

    ChrisVer

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    You can find more symmetries. In case for the phi* phi, you can also have the parity transformation:

    [itex] \phi \rightarrow - \phi [/itex]

    This is one reason someone won't write [itex]\phi^3 [/itex] terms a priori in a Lagrangian. Or instead of a [itex]U(1)[/itex] you can have a broken [itex]U(1)[/itex] in the case that it would correspond to a [itex]Z_N[/itex]...

    In fact you can find a lot of symmetries in a Lagrangian density. The reason is already stated, that the given Lagrangians will be built in order to contain your desired symmetries.

    Take for example the quarks. The quarks are in a [itex]3[/itex]-dimensional representation of [itex]SU(3)_{color}[/itex]. How can you make neutral quantities from combinations of [itex]3[/itex]?
    Take for example the [itex] 3 \otimes 3 = 6 \oplus \bar{3} [/itex]. Obviously the combination [itex]3 \otimes 3[/itex] cannot be decomposed to a singlet representation, but it contains the complex conjugate of [itex]3[/itex], that is the reason sometimes [itex]3 \otimes 3[/itex] are called anti-quarks.
    On the other hand the combination [itex] 3 \otimes \bar{3}= 8 \oplus 1[/itex] contains the 1-dimensional object [which transforms trivially under [itex]SU(3)[/itex] transformations] and so can "work". The result is that terms which belong to [itex]3, \bar{3}[/itex] can be combined in the Lagrangian to give you allowed terms. Similarly for [itex] 3 \otimes 3 \otimes 3 [/itex] (so combinations of 3 fields belonging to [itex]3[/itex] representation).
     
  9. Dec 9, 2014 #8
    You know I've always wondered why they don't have the real scalar field squared in the yukawa interaction term, is parity not relevant in that situation?
     
  10. Dec 10, 2014 #9

    ChrisVer

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    What do you mean? a term like : [itex] \phi^2 \bar{\psi} \psi [/itex]?
    one reason I see is that this term is non-renormalizable.
     
  11. Dec 10, 2014 #10
    Does that mean there's something fundamentally wrong with the term, or just that perturbation theory can't handle it?
     
  12. Dec 10, 2014 #11
    What are the dimensions of the spinors in terms of powercounting?
     
  13. Dec 10, 2014 #12

    ChrisVer

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    take the mass term : [itex]m \bar{\psi} \psi[/itex]...
    If [itex][m]=1[/itex] what should be [itex][\psi][/itex] in order to have a dimensionless action?

    As for the term, In general you can have such terms into an effective Lagrangian [insert some cut-off]. Why would you do that in the fundamental level? Take eg the case for which you don't write a Fermi-term: [itex]\bar{\psi} \psi \bar{\psi} \psi [/itex]. Both these operators are irrelevant.
     
    Last edited: Dec 10, 2014
  14. Dec 10, 2014 #13
    3/2, but where it comes? It is not a should be, it is a "it is". A spinor is 4 component complex vector. The half integer comes from the idea of bispinor?
     
  15. Dec 10, 2014 #14

    ChrisVer

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    it simply comes from asking the Dirac Lagrangian to be of dimension 4 (in 4D space).
    You are not counting degrees of freedom, but mass dimension... so I don't think the components can help you. A better way to write this is [itex] [\psi] = m^{3/2} [/itex] for not getting confused with a simple shorthanded 3/2 (half-integer) which would make you think of spinors.
    Of course the fact that you have spin 1/2 particles leads you to Dirac equation, and so an equation having the mass term as [itex]m \bar{\psi}\psi[/itex]
     
    Last edited: Dec 10, 2014
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