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I Unbroken electromagnetic symmetry

  1. Jan 21, 2017 #1
    Hi all,

    I'm studying electroweak spontaneous symmetry breaking at that time, see for instance Chang and Li's book ch 11. Have any one an idea that if the charge operator is defined by:

    ## Q = \int (- e^\dagger e + \frac{2}{3} u^\dagger u - \frac{1}{3} d^\dagger d ) d^3 x ,##

    and the isospin operator defined by :

    ## T_3 = \frac{1}{2} \int (\nu^\dagger_L \nu_L - e^\dagger_L e_L + u^\dagger_L u_L - d^\dagger_L d_L ) d^3 x, ##

    why when the electric charge operator acting on the vacuum expectation value ##\phi_0 = <0|\phi|0> = (0~~~~~~ v)^T ## it gives zero, i.e., ## Q <\phi>_0 = 0 ## , while when the isospin operator or the hypercharge = ##Q-T_3## acting on the VEV it doesn't vanish ?

    So that we say the electric charge still conserved after EWSSB while the hypercharge or isospin has been broken
     
  2. jcsd
  3. Jan 21, 2017 #2

    jambaugh

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    I'm not sure what you mean by "why" you are stating the case which is that the vacuum has non-zero hypercharge and weak-isospin charge. It is exactly the reason the vacuum breaks the weak-isospin (and hypercharge) symmetry. The vacuum being one of the degenerate lowest energy modes as opposed to the zero charge mode. It is the breaking of this symmetry in a specific direction which singles out the T3 charge rather than some other isospin component as the one contributing to electric charge.

    That's what's happening. "why?" as in why it happens this way is a bit tougher to answer.
     
  4. Jan 22, 2017 #3
    Ok. I just read this statement in Chang & Li's book , speficaly see Equ. (11.77), where they mentioned that ## Q <\phi>_0 = 0 ## , i thought this can be proven by really acting by Q or ##T_i ## operators on ## <\phi>_0= v ##, but this seems to me as acting by an operator on a number or a constant, so in both cases there will be no difference, that's what i thought about !
     
  5. Jan 26, 2017 #4
    Hi, I can't understand this sentence .. thanks
     
  6. Jan 26, 2017 #5

    jambaugh

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    Before the breaking of isospin symmetry there is a full spin(3)=su(2) isospin symmetry group and the isospin of an elementary particle field could be in any direction in an abstract isospin space. There was no distinction between u quarks, d quarks or various complex superpositions of these and likewise with the other quark and lepton modes. Now since this symmetry restoration is pre-vacuum definition it is rather inappropriate to speak of particles per se at this point but rather the gauge fields which will manifest the particles upon the system condensation.

    As with spatial spin there is a projective 2-sphere of Hermitian operators ([itex] u\sigma_x + v\sigma_y + w\sigma_z[/itex] for spin and [itex]uT_1 + vT_2 + wT_3[/itex] for isospin). The breaking of the symmetry selected out a specific isospin direction, which is analogous to the Ising spin model where a magnetic domain establishes a preferred spin direction due to local coupling, say in the z-direction defined by [itex]\sigma_z[/itex]. With the magnetic analogy there is still the U(1) = Spin(2) rotation symmetry about this z-axis and a splitting of the energies for the spin of a particle within that domain along the z-direction. Our isospin does similarly, selecting out the abstract T3 direction and its the T3 isospin components that have definite and distinct energy states, hence the mass discrepancy between the electron and electron neutrino etc. But in the isospin case, the mechanism of symmetry breaking is such that T3 is also correlated with electric charge. It is rather the full U(2) symmetry that is breaking with a correlation between the U(1) central subgroup (hypercharge gauge) and the T3 component of SU(2) isospin. I am not intimate with the mechanism but would refer you to the literature.

    As to your reference in Chang and Li's book, I need more context to answer and I do not have a copy. But keep in mind that the field component is now an operator on the Fock space ("2nd quantization") and for [itex]\phi[/itex] in particular it is a spinor of operators. Thus the vacuum expectation value [itex]\phi_0[/itex] is still an (iso)spinor and we can still act upon it non-trivially with the charge-isospin operators. Without the text I am not sure how the author is resolving these isospinor components so I can't point to further details. Quite possibly there's a bit of notational hanky panky to keep the exposition from being too tediously detailed.
     
  7. Feb 1, 2017 #6
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