In the Standard Model we have $SU(2)_I\times U(1)_Y$, where $U(1)_Y$ is weak hypercharge and $SU(2)_I$ is the symmetry group of weak isospin. Why do we introduce $U(1)_Y$ of weak hypercharge rather than $U(1)_\text{em}$ of electromagnetism?

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    $\begingroup$ We have to introduce the hypercharge because we want to have the correct theory, one that agrees with Nature, and Nature introduced the hypercharge as a more fundamental symmetry than the electric charge some 14 billion years ago. $\endgroup$ – Luboš Motl May 25 '16 at 12:09

Short answer: to accurately model reality.

Long answer: The weak interaction has several peculiar properties:

  1. The $W$ bosons are vector bosons (so the weak theory is likely a gauge theory)
  2. The $W$ bosons have electric charge
  3. The $W$ bosons have mass. (The $Z$ boson hadn't been observed experimentally; it was a prediction of the SM)
  4. The $W$ bosons couple chirally, meaning left-handed and right-handed fermions belong to different representations of the gauge group. In particular, left-handed fermions transform in the doublet representation, right-handed in the singlet (trivial) rep.
  5. EM is a gauge interaction that couples vectorially.

This leads to two problems that weren't understood before the 60s. The first problem is how to have massive gauge bosons (items 1 and 3). Naively a vector boson mass term violates gauge symmetry and unitarity. The second problem is how to get massive fermions (item 4). A fermion mass term couples left- and right-handed fermions, but those belong to different weak representations so the fermion mass term also breaks gauge invariance.

The realization of Higgs et al. and then later Glashow and Weinberg was that both of these problems can be solved by spontaneous symmetry breaking. The Higgs mechanism says that spontaneous symmetry breaking of a gauge symmetry leads to massive gauge bosons. And if the operator that gets a vacuum expectation value has the right quantum numbers you it can couple to the fermions in just the right way to make an effective fermion mass term.

The questions are, what pattern of spontaneous symmetry breaking corresponds to reality, and what kind of operator should get a vev?

Because of item 5, we need an SSB that leaves a $U(1)_{em}$ unbroken. Since there is only one Lie group with doublet irreps we also know the original gauge group should include an $SU(2)$ factor. Moreover, since the $W$ bosons have charge (item 2), the generator of $U(1)_{em}$ must not commute with some $SU(2)$ generators!

To fix item 4 and make a gauge-invariant mass term we need an operator to get a vev which transforms in the doublet of $SU(2)$. This rules out the symmetry breaking pattern $SU(2)\rightarrow U(1)_{em}$ (there is no generator of $SU(2)$ that leaves a non-trivial doublet invariant, meaning a doublet always breaks $SU(2)\rightarrow {1}$ which doesn't leave room for a photon).

The next simplest pattern to try is $SU(2)\times U(1)_Y\rightarrow U(1)_{em}$. We could achieve this with an uncharged doublet and $U(1)_Y = U(1)_{em}$, but then $[U(1)_{em},SU(2)]=0$ and the $W$ bosons wouldn't be charged. The only other way is to have the Standard Model pattern where the $U(1)_{em}$ generator is a linear combination of $U(1)_Y$ and some generator of $SU(2)$, which is indeed a possible pattern.

So, the symmetry breaking pattern $SU(2)\times U(1)_Y \rightarrow U(1)_{em}$ with $Q = Y + T_3$ is the minimal SSB pattern that matches general weak phenomenology. It's not the only possible mechanism, so it's really very nice that nature chose this one. It's also nice because it predicts the $Z$ boson.


Detailed answer

The thing is that you cannot really force in an a priori interpretation of the gauge group with which you extend your existing theory. You can only decide on the symmetries of your theory.

So keeping things very general you start by trying to gauge the symmetry group $SU(2)$. This has three generators and so gives rise to three independent currents. The bunch of symmetries associated with such group is commonly known as an isospin. This is not restricted to the weak interaction. There are, in fact, other objects with similar symmetries.

When you try to "squeeze" electromagnetism in your theory, you proceed by extending the group of symmetry. This means that your fields will then correspond to representation of the extended group. The natural way to do so is to go from $SU(2)$ to $SU(2)\times U(1)$.

Now remember, at this stage, it does not make much sense to say anything about the nature of the added $U(1)$ symmetry. And we must in fact analyse the new $SU(2)\times U(1)$ group as one single entity.

You must then proceed to write the most general element of the group which allows you to identify its generators. Then, when you gauge your group, the covariant derivative gets written accordingly (see Peskin and Schroeder, Introduction to Quantum Field Theory, p. 702 of 1995 edition):

$$D_\mu=\partial_\mu-igW^i_\mu T^i-ig'YB_\mu$$

where $W^i_\mu$ and $B^\mu$ are respectively the "vector potentials" associated with the $SU(2)$ and $U(1)$ symmetries and the factors in front are the respective charges of any given particle under the corresponding group.

Up to this point, we haven't had the leisure to assign any definite meaning to the symmetry groups used and we had to proceed rather mechanically.

We have to introduce the hypercharge because we want to have the correct theory

This is true. And by correct theory, we mean one that matches the observations. For our unification of weak interaction and EM to be a success, we thus want to recover, among the features of our new model, a boson with all the feature of the photon from electromagnetism.

To track down this photon, we start by noting that it has to be neutral under the action of the complete group. By introducing the above derivative in the Lagrangian for our theory (see P&S for details) we end up with a part that describes neutral currents proportional to

$$\mathcal{L}_{n.c}=\bar{\psi}\gamma_\mu[gT_3 W_3^\mu+g'YB^\mu]\psi~,$$

where $\psi$ denotes a larger object (an element of the representation of the total group), a bunch of particle mixed together by the electroweak interaction, e.g $\nu_e^L, \nu_e^R, e_L, e_R$ for the interaction of the electron with its neutrino.

To make EM appear, one then has to "shuffle" this neutral sector to get a part that is formally identical to the interaction part of the EM Lagrangian. This can be achieved by operating the rotation

$$ B_\mu = A_\mu \cos\theta_W - Z_\mu \sin\theta_W $$ $$ W_\mu^3 = A_\mu \sin\theta_W + Z_\mu \cos\theta_W $$

This reproduces the desired Lagrangian $\mathcal{L}_{\gamma}=\bar{\psi}\gamma_\mu eQ \psi A^\mu$ if one chooses the weak angle $\theta_W$ wisely.

It is really not difficult but in fear that this answer would become unbearably technical, I merely quote the result (again, from P&S):



$$Q = T^3+Y~.$$


The bottom line I hope I can make clear, is that you do not have a choice on the interpretation of the physical nature of any part of your symmetry group until you have worked out the Lagrangian. Then, only, can you assign the desired values to your free parameters to recover your physics.

Hence, the $U(1)_{EM}$ that is contained within the $SU(2)_L\times U(1)_Y$ is really a combination of aspects from both $SU(2)_L$ and $U(1)_Y$.

The terms Weak Isospin and Weak hypercharge might make more sense now after realising the that these are both needed aspects of the larger group that contains both weak and electromagnetic interactions.


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