Silly question on electromagnetism and electroweak interaction

Question

Introduction

So, consider the Maxwell action:

$$S_{\text{Electromagnetism}} = -\frac{1}{4\pi}\int F_{\mu\nu}F^{\mu\nu}dx^4.$$

Then, consider the Electroweak action$[1]$:

$$S_{\text{Electroweak}} = S_{\text{Isospin}} + S_{\text{Hypercharge}}$$ where,

$S_{\text{Hypercharge}} = -\frac{1}{4\pi}\int B_{\mu\nu}B^{\mu\nu}dx^4$

$S_{\text{Isospin}} = -\frac{1}{4\pi}\int W^{a}_{\mu\nu}W^{\mu\nu}_{a}dx^4.$

Furthermore, for both $S_{\text{Hypercharge}}$ and $S_{\text{Electromagnetism}}$, the gauge group is the $U(1)$.

My Question

So, since the gauge group of electromagnetism is $U(1)$ and the theory which "uses" the gauge group $SU(2)$ is $\mathcal{L} = -\frac{1}{4\pi}W^{a}_{\mu\nu}W^{\mu\nu}_{a}$, my question is:

• why do we need the $\mathcal{L} = -\frac{1}{4\pi}B_{\mu\nu}B^{\mu\nu}$?

My question can be rewritten in two other manners:

• Why the standard model requires a theory with strength tensor given by $B_{\mu\nu}$ but not the usual $F_{\mu\nu}$'s?

• Why $-\frac{1}{4\pi}F_{\mu\nu}F^{\mu\nu} \to -\frac{1}{4\pi}B_{\mu\nu}B^{\mu\nu}$ in standard model?

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$[1]$ https://en.wikipedia.org/wiki/Electroweak_interaction

Answer 1

Ultimately this is an experimental fact. But I will try to provide here the connection between experimental observations and the mathematical description.

It is an experimental fact that there are 4 electroweak currents: two charged (W+ & W-) and two neutral (Z & A). I am referring to them using the modern notation. A is the electromagnetic current.

• If the electromagnetic current were the u(1) part of the su(2) x u(1) algebra, then no current would be charged, because the generator of u(1) commutes with all generators of su(2). So your proposal is incompatible with observation.
• You could, alternatively, use the Cartan subalgebra of su(2), which is neutral, as A and the u(1) as the Z. That would yield the correct charge for W+ and W-, but it would also mean that every particle interacting with the charged currents would necessarily also interact with the electromagnetic A current.

It is an experimental fact that this is not the case: neutrinos interact through W+ & W-, but not through A.

The original argument due to Sheldon Glashow in the famous paper:

Partial-symmetries of weak interactions https://www.sciencedirect.com/science/article/abs/pii/0029558261904692

is a bit different. The W+ and W- currents only couple to the chiral part of the electron, and therefore violate parity. If the electromagnetic current was the neutral current of the su(2), then electromagnetic interactions would also violate parity.

It is an experimental fact that electromagnetic interactions do not violate parity.

The solution proposed by Glashow, that eventually gave him a Nobel prize (making the "silly" in the title of this topic a bit inappropriate) was that there is another u(1) interaction that couples to the anti-chiral part in such a way that the combination with the neutral current of su(2) does not violate parity (or makes the particle neutral: check the answer by Frederic Thomas to see how this works in practice).

Answer 2

In short, because experiment requires the presence of hypercharge, which is different from electromagnetic charge.

Notice that the (Electroweak) Standard Model is not only a $SU(2) \otimes U(1)$ gauge theory: it also has symmetry breaking with pattern $SU(2)_L \otimes U(1)_Y \to U(1)_{\text{em}}$, where the subscripts play the role of specifying that only left-handed particles carry $SU(2)$ charge and distinguishing hypercharge from electromagnetic charge.

The tricky bit is that the broken symmetries do not correspond to the $SU(2)_L$ group. Instead, two generators of $SU(2)_L$ are broken (leading to the masses of the $W^{\pm}$ bosons) and the third one mixes the remaining generator of $SU(2)_L$ with the generator of $U(1)_Y$ (leading to the mass of the $Z$ boson). We are left with one unbroken generator, which is a linear combination of the $U(1)_Y$ generator with the last generator of $SU(2)_L$. This unbroken generator generates $U(1)_{\text{em}}$. Notice then that while both hypercharge and electromagnetism are related to $U(1)$ symmetries, they are not the same symmetry (the fields transform differently under each one of them). This is the reason the standard model writes $B_{\mu\nu}$ instead of $F_{\mu\nu}$: $B_{\mu\nu}$ is not the electromagnetic strength tensor.

Before neutral currents were confirmed (i.e., before we knew about the $Z$ boson), there were proposals that did not use the tensor product with $U(1)$. I've often heard of the Georgi–Glashow model for the electroweak interactions as being an $SU(2) \to U(1)$ model, for example, despite the original paper writing $SO(3)$ instead of $SU(2)$ (probably either the original paper of some of the references I heard talking about it later were thinking about how $SO(3)$ and $SU(2)$ are locally isomorphic). In any way, in the Georgi–Glashow model one has only three generators, two of which are broken and associated to the $W^{\pm}$ bosons, while the remaining one is kept unbroken and represents the photon. Of course, this model was ultimately discarded because neutral currents do exist.

Answer 3

The post follows closely the book of Otto Nachtmann -- Elementary particle physics (German edition). The question is related with the fact how electrons and neutrinos are combined into representation vectors under SU(2) and U(1).

Although the particles mentioned here have mass, for this consideration they will be considered as massless. Furthermore, the acquisition of mass by means of the Higgs mechanism is not presented, it is just briefly mentioned when it is necessary.

Here only leptons of the first generation of SM particles are considered.

Let's start with SU(2):

A left-handed electron neutrino $\nu_{eL}$ and left-handed electron $e_L$ are combined into a duplett which transforms under SU(2) group transformations U:

$$\left( \begin{array}{c} \nu_{eL} \\ e_{L} \end{array}\right) \rightarrow U\left( \begin{array}{c} \nu_{eL} \\ e_{L} \end{array}\right)$$

whereas the right-handed electron $e_R$ is not affected by the SU(2) group (it is a singlet):

$$e_R \rightarrow e_R$$

In order to make the Lagrangian invariant under these local transformation a gauge field $\mathbf{W}_\mu = \mathbf{W}^a_\mu\frac{\sigma^a}{2}$ has to be introduced ($\sigma^a$ are the Pauli-matrices). This gauge field will provide us with 3 fields: $W^\pm_\mu = (W^1_\mu \pm i W^2_\mu)/\sqrt{2}$ and $W^3_\mu$. One could now suspect that the fields $W^\pm_\mu$ correspond to the 2 charged vector bosons and $W^3_\mu$ to the photon. However, $W^3_\mu$ does not couple to the right-handed electron $e_R$ (a photon undiscriminately would couple to $e_L$ and $e_R$), so we don't get what we want.

However, another local gauge transformation can be introduced to act on the 3 "particles" which is actually U(1):

$$\left(\begin{array}{c} \nu_{eL} \\ e_L \\ e_R\end{array}\right) \rightarrow e^{i\chi \mathbf{Y}}\left(\begin{array}{c} \nu_{eL} \\ e_L \\ e_R\end{array}\right)$$

$\mathbf{Y}$ is a matrix:

$$\mathbf{Y} =\left( \begin{array}{ccc} y_L & 0 & 0 \\ 0 & y_L & 0\\0 & 0 & y_R\end{array}\right)$$

This transformation can simply be achieved as phase transformation, so it is of the group U(1). We know the group U(1) from electromagnetism, we can associate it with a gauge field as in EM-theory, i.e. a 4-vector field called $B_\mu$. However, $B_\mu$ can't be the photon field neither since it couples to a neutral particle, the neutrino.

We now construct the gauge exchange bosons associated to the gauge fields. $W^\pm_\mu$ are the fields which can be associated with the vector bosons $W^+$ and $W^-$. But $W^3_\mu$ and the $B_\mu$-field combine to create the $Z^\circ$ particle (necessary to explain the neutral currents found by the experiments, remember LEP was running on the $Z^\circ$ resonance for years and generating millions of it) and the photon $\gamma$($g$ and $g'$ are the coupling constants of the gauge fields $W^a_\mu$ and $B_\mu$):

$$Z_\mu = (g W^3_\mu - g' B_\mu)/\sqrt{g^2+ g'^2}$$

and

$$A_\mu = (g W^3_\mu + g' B_\mu)/\sqrt{g^2+ g'^2}$$

It is important to note that the $\mathbf{Y}$ represents the weak Hypercharge which is different from the electrical charge by:

$$Q = \mathbf{Y} + T_3$$

the last being the third component of the weak Isospin.

Still all introduced gauge bosons are massless. Via the Higgs-mechanism they adopt mass in a rather subtle way so that $W^\pm$ and $Z^\circ$ adopt mass, whereas the photon $\gamma$ does not. In order to at least vaguely understand this one has to note that the Higgs field $\phi$ is a priori also a dublett that transforms under SU(2):

$$\phi = \left( \begin{array}{c} \phi_1 \\ \phi_2 \end{array}\right)$$

The trick to be used is to select a vaccum state of this field that breaks the SU(2) symmetry in a way that $W^\pm$ and $Z^\circ$ adopt mass, but the $\gamma$ not. This is done by:

$$\langle 0 | \phi | 0\rangle = \left( \begin{array}{c} 0 \\ \rho_0/\sqrt{2} \end{array}\right)$$

I will not go further into details, this has to be looked up in the corresponding textbooks. I recommend Otto Nachtmann -- Elementary Particle Physics and Aitchison & Hey for this subjects. The books on QFT present the SU(2) x U(1) gauge field towards the end, and in order to understand well what they say, one has to read at least 500 pages before this topic is discussed.

Recap: Once the SU(2) duplett is combined/created it is very difficult to find an additional transformation which could represent the U(1) symmetry of EM-theory. Because the SU(2) duplett is a combination of a neutral particle, the neutrino and a charged one, the left-handed electron $e_L$.

Actually, the U(1) out of SU(2) X U(1) is not the U(1) symmetry of EM-theory, it is the symmetry described by the weak Hypercharge $\mathbf{Y}$ and not the electric charge. The electric charge is a combination of Hypercharge and weak Isospin. Once one has realized that it is quite clear that the B field cannot be $A_{EM}$, the $A_{EM}$ has to be constructed out of 2 different fields, the $W^3$ and the $B$.

The quantum number $T_3$ distinguishes between the left-handed neutrino $\nu_{eL}$ where it is $T_3=1/2$, whereas for the left-handed electron it is $T_3=-1/2$. However, for the right-handed electron $e_R$ $T_3=0$. For completeness I give also the values for the weak Hypercharge: $y_L=-\frac{1}{2}$ and $y_R = -1$. Using these numbers the EM-charge $Q$ of all 3 particles neutrino $\nu_{eL}$ and left-handed electron $e_L$ and the right-handed electron can be computed correctly.