# Hypercharge for $U(1)$ in $SU(2)\times U(1)$ model

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I understand that the fundamental representation of $U(1)$ amounts to a multiplication by a phase factor, e.g. EM. I thought that when it is extended to higher dimensional representations, it would just become a phase factor times the identity matrix.

Can someone explain where the hypercharge comes into the $U(1)$ generator matrices in $SU(2)\times U(1)$ model, e.g. $Y = -(1/2) I$ in $(2, -1/2)$ representation? I don't quite understand where the "$-1/2$" comes from.

Where do all these hypercharges come from? What is the logic behind choosing a particular value like $-1/2$?

Hope someone can clarify.

This post imported from StackExchange Physics at 2015-04-18 05:36 (UTC), posted by SE-user curious

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1) Firstly, hypercharge usually refers to the (strong) hypercharge $Y$ related to the strong interaction and (strong) isospin $(\vec{I}^2,I_3)$. In contrast, the question(v1) deals with the weak isospin $(\vec{T}^2,T_3)$ and the weak hypercharge $Y_W$ of the electroweak gauge Lie group $SU(2)\times U(1)$ of the Glashow-Salam-Weinberg (GSW) model.

2) Secondly, there is a normalization issue. Some authors (e.g., Peskin and Schroeder, OP) define

$$Y_W~:=~Q-T_3,$$

where

$$Q~=~\frac{q}{e}$$

is the electric charge in units of the elementary charge $e$, while other authors (e.g., Wikipedia) has

$$Y_W~:=~2(Q-T_3).$$

If OP uses the latter convention, his "$-1/2$" would turn into the more natural "$-1$", ha-ha. :-) Well, I'm joking: A choice of normalization doesn't of course explain anything.

3) An elementary particle in the GSW model transforms in a Lie algebra representation

$$\rho: L~\to~ gl(V)$$

of the electroweak gauge Lie algebra $su(2) \oplus u(1)$. (There exist also Grand Unified Theories (GUT) with bigger gauge Lie algebras, e.g., $su(5)$, cf. arivero's comment. Here we shall however mainly focus on OP's original question(v1) for simplicity.) Recall that a Lie algebra element

$$x~=~\sum_a x^at_a~\in~ L$$

in an abstract Lie algebra $L$ can be written as a linear combination of some basis elements $t_a$. The $t_a$'s are also known as the Lie algebra generators. Obviously, if one changes the basis, one obtains a new set of generator $t^{\prime}_a$.

4) A non-Abelian gauge theory has a Lie algebra-valued gauge field

$$A_{\mu}~=~\sum_a A_{\mu}^a t_a.$$

In the case of the electroweak theory, the Lie algebra generators $t_a$ are

$$(T_1,T_2,T_3,Y_W),$$

where $(T_1,T_2,T_3)$ are the generators of $su(2)$. In this sense, the weak hypercharge $Y_W$ plays a dual role in the GSW model:

1. Firstly, $Y_W$ is a Lie-algebra generator for the $u(1)$ Lie subalgebra. In an irreducible representation $\rho: u(1)\to {\rm Mat}_{1 \times 1}(\mathbb{C})$ of the Lie algebra $u(1)$, it becomes some $1 \times 1$ matrix, e.g.,
$$\rho(Y_W)~=~-1/2.$$ Often, authors don't bother to mention the representation map $\rho$ explicitly.

2. Secondly, $Y_W$ describes a special type of physical charge, called weak hypercharge, which depends on the type of elementary particle.

As long as one doesn't change basis and sticks with the same normalization (cf. item 2 above!), these two dual roles are fully consistent.

This post imported from StackExchange Physics at 2015-04-18 05:36 (UTC), posted by SE-user Qmechanic
answered Nov 4, 2011 by (3,120 points)
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The hypercharge in the electroweak model is completely determined by the electric charge of the observed particles. In one usual normalization, the Peskin-Schroder one, it is just the average electric charge of all the particles contained in a weak SU(2) representation. The weak SU(2) representation is defined by the particles that can change into each other by weak interactions, so that the (left-handed) electron and the neutrino are partnered.

Weak SU(2) representations are like spin--- they have one state of each value of "L_z", called I_z, running from -m to m. For the purpose of this discussion, let's assume you have a weak SU(2) spin-2.5 particle, with hypercharge "Y". Then the actual values of the electric charge of the particles will be

-2.5 + Y, -1.5+Y , -.5 + Y , .5 + Y , 1.5 + Y, 2.5 + Y

Y is the offset of the electric charge, and the SU(2) spin value defines the range. The steps are always by one unit. This is the meaning of the formula

$$Q = I_z + Y$$

For the observed weak partners, the electron-neutrino and the electron, the charges are 0,-1, so the hypercharge is the average value, or -1/2. For the weak partners up-quark,down-quark the charges are 2/3,-1/3, so the hypercharge is the average of the two: 1/6.

But only the left-handed parts of the electron and the quarks are SU(2) partners. The right handed parts have no partner. The right handed parts have a Y which is just their electric charge. The right handed electron has a Y of -1, the right handed up quark Y=2/3, and the right handed down quark Y=-1/3. That's just so that they have the same electric charge as their left-handed partner, so that they can form a massive charged particle together.

One natural normalization of hypercharge is not Wikipedia's nor Peskin Schroeder's. This is in terms of the greatest rational value for which gives all the standard model particles integer hypercharges. This value is 1/6. In terms of multiples of 1/6, all the standard model particles have a hypercharge of 1,2,3,4 and 6 units.

But this assumes that the U(1) of hypercharge, with its crazy values, is fundamental, which is extremely unlikely. The most natural normalization choice comes from embedding SU(2) and SU(3) into SU(5) (or a higher GUT of the same type, like SO(10) or E6). In this embedding, you think of SU(5) as a 5 by 5 matrix, the top 2 by 2 block is the SU(2), the lower 3 by 3 block is SU(3), and the U(1) consists of all diagonal phase-matrices (a,a,b,b,b) where a^2b^3=1, so that this phase is generated by

$\mathrm{diag}(1/2, 1/2, -1/3, -1/3, -1/3)$

Meaning that if you rotate by the diagonal matrix with $(e^{i\theta/2},e^{i\theta/2}, e^{-i\theta/3}, e^{-i\theta/3}, e^{-i\theta/3})$ on the diagonal, you are still in SU(5), but you are not in SU(2)xSU(3).

The matter is the defining representation of SU(5), plus the antisymmetric two-tensor representation. The decomposition of these two representations explains the hypercharge assignments of the standard model easily and naturally. See this answer: Is there a concise-but-thorough statement of the Standard Model?

answered Nov 8, 2011 by (7,720 points)
edited Apr 18, 2015
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The short answer about the factor 1/2 is that at the end $e_L$ must get charge $-e$ while $\nu_L$ must get a charge of zero.

$-e\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)\left( \begin{array}{c} \nu_L \\ e_L \end{array} \right) ~~=~~ \left[-\frac{g}{2}\left( \begin{array}{cc} 1 & ~~0 \\ 0 & -1 \end{array} \right)\sin\theta_w ~~+~~ y\frac{g'}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\cos\theta_w\right]\left( \begin{array}{c} \nu_L \\ e_L \end{array} \right)$

Where the first term on the righthand side is the contribution from the third component ($\sigma^z$) of the SU(2) field and the second term stems from the hypercharge associated with the U(1) field. The mix of the two is determined by the Weinberg angle $\theta_w$

On the other hand, for the electric charge of $e_R$ we have

$-e\left(e_R\right) ~~=~~ y\,g'\cos\theta_w(e_R)$

So $e_L$ has only halve the hypercharge as $e_R$ because it gets the other halve of its electric charge from its mix with the third component of the SU(2) field. The solution for both is of course given by:

$g=\frac{-e}{\sin\theta_w}, ~~~~~~~~ g'=\frac{-e}{cos\theta_w}$

(Using Weinberg's sign notation) which simplifies the first equation to:

$-e\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)\left( \begin{array}{c} \nu_L \\ e_L \end{array} \right) ~~=~~ \left[\frac{e}{2}\left( \begin{array}{cc} 1 & ~~0 \\ 0 & -1 \end{array} \right) ~~-~~ y\frac{e}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\right]\left( \begin{array}{c} \nu_L \\ e_L \end{array} \right)$

This shows clearly that it gets 1/2 of the electric charge from the hypercharge of the U(1) field and half of the electric charge from the third component of the SU(2) field.

Regards, Hans

This post imported from StackExchange Physics at 2015-04-18 05:36 (UTC), posted by SE-user Hans de Vries
answered Nov 5, 2011 by (90 points)

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