I have seen this confusion surface in multiple questions, but neither contained the answer I was seeking. I hope you can help me understand where I'm going wrong.

Assume the electroweak symmetry is not broken, so that the electroweak force is described by a $\mathrm{SU}(2)_\text{L} \times \mathrm{U}(1)_\text{Y}$ symmetry.

  1. Can the electroweak force at this point be meaningfully separated into an weak isospin force $\mathrm{SU}(2)_\text{L}$ and a weak hypercharge force $\mathrm{U}(1)_\text{Y}$? Can they be considered to be as separate from each other as they are from $\mathrm{SU}(3)_{\text{QCD}}$?
  2. Does the symmetry $\mathrm{SU}(2)_\text{L}$ lead to two quantum numbers that differentiate particle states?

In the following question, asking about the charges of the weak force, the answer seems to indicate that the 1st and 3rd components of the weak isospin together are two such quantum numbers. Is it that simple? https://physics.stackexchange.com/a/262323/131633

I've been trying without result to find an explanation for the weak isospin with all its components, why it is defined in such a way, and why its third component seems so much more interesting. Is this still the case without a broken symmetry?


1 Answer 1

  1. Yes, in the unbroken theory you could treat the $\mathrm{SU}(2)$ and the $\mathrm{U}(1)_\text{Y}$ completely independently. The reason we usually still write $\mathrm{SU}(2)\times\mathrm{U}(1)_\text{Y}$ as "one gauge group" is that the residual electromagnetic $\mathrm{U}(1)_\text{em}$ is not the factor of this group, but a different one-parameter subgroup, and no $\mathrm{SU}(2)$ remains after breaking.

  2. I don't know what "Is the symmetry of the weak isospin part sufficient for a fundamental representation to form a basis with two orthogonal vectors?" is supposed to mean, but the representation theory of the weak isospin is exactly the same as that of ordinary 3D spin, since both are the representation theory of $\mathrm{SU}(2)$. An irreducible $\mathrm{SU}(2)$ representation is defined by the value of the Casimir, i.e. the square of the "total (iso)spin" $s$ (or $T$), which takes half-integer values $s$. Inside such an irrep, you have $2s+1$ different basic vectors, which you can conveniently label by choosing one of the three generators of $\mathrm{SU}(2)$ and just taking its eigenbasis, which will have labels $s_i\in\{-s,-s+1,\dots,s\}$. It's traditional to call the generator we choose to label the states the "third" or "$z$"-generator, but there's nothing special about it. Hence, a state under the weak force has two labels, $\lvert s,s_z\rangle$, or rather $\lvert T,T_3\rangle$ in the standard notation for isospin.

  • $\begingroup$ That clarifies a lot, as I am more familiar with spin. So similar to how e.g. a spin-3/2 particle state allows spins (3/2, 1/2, -1/2 , -3/2), a particle with total weak isospin T can have integer-spaced values T3 = (T, T - 1, ..., -T). So we can imagine three "spatial" components of isospin, although this can of course not be measured. I have also seen a "vector" for the weak isospin contaning three components, which are the total, the direction and the third component, and this was a source of confusion to me. $\endgroup$ Oct 25, 2016 at 16:22

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