3
$\begingroup$
  1. In electroweak theory, for the first generation of leptons, in the doublet $$\psi_L=\begin{pmatrix} \nu_{eL}\\ e_L \end{pmatrix}$$ we assign a non-abelian charge $I=\frac{1}{2}$. Is this a Noether charge?

  2. A $SU(2)$ transformation on the Lagrangian: $$\mathcal{L}^{ferm}=i\bar\psi_R\gamma^\mu D_\mu\psi_R+i\bar\psi_L\gamma^\mu D_\mu\psi_L$$ leads to this conservation of this charge. Right?

  3. When that corresponding charge operator acts on the one-particle (or multi-particle states) it gives us the corresponding charge. Right?

  4. The weak hyper charge $Y$ appears in the $U(1)$ transformation: $e^{\beta(x) Y/2}$ and $I$ appears in the $SU(2)$ transformation $e^{i\vec\alpha(x)\cdot\vec I}$? One can define, Noether charge (operator) for both these transformations, for $U(1)$ case it will be related to $Y$ and for $SU(2)$ case it will be related to $I$. But do we have one-particle and many-particle states here?

  5. On what states will these Noether charges act on?

$\endgroup$
4
$\begingroup$

This is an interesting question. I think the real difficulty is understanding the difference between Noether charges and what we normally call the charge of a particle. Noether charges are operators that when acting on states give the values which we normally call "charges" of different particles. In other words the charges of different particles are the eigenvalues of the Noether charge when acting on single particle states.

Noether charges can act on single or multiparticle states its just that the eigenvalues are the charges of those states. For example with the isospin Noether charge, acting on a state with 3 neutrinos will have an eigenvalue of $3/2 $.

To full appreciate the relationship between Noether charges and what we call charges of particles I think its important to see a detailed example. We derive this relation for isospin below. A Noether current associated with a gauge symmetry is the global version of that symmetry. Consider the $ SU(2) _L $ invariant Lagrangian: \begin{equation} {\cal L} = i \nu ^\dagger _L \bar{\sigma} ^\mu \partial _\mu \nu _L + i e ^\dagger _L \bar{\sigma} ^\mu \partial _\mu e _L + h.c. \end{equation} where $ e _L $ and $ \nu _L $ are two component Weyl fermions.

The conserved current associated with the symmetry is: \begin{equation} j ^\mu = \alpha ^\alpha \left( \begin{array}{cc}\bar{\nu} _L & \bar{e} _L \end{array} \right) \bar{\sigma} ^\mu t ^a \left( \begin{array}{c} \nu _L \\ e _L \end{array} \right) + h.c. \end{equation} where $ t ^a $ are the genarators of $ SU(2) $. This gives a conserved charge (we denote it $T$ since we want to associate it with isospin), \begin{equation} T = \alpha ^a \int d ^3 x \left( \begin{array}{cc}\bar{\nu} _L & \bar{e} _L \end{array} \right) t _a \left( \begin{array}{c} \nu _L \\ e _L \end{array} \right) + h.c. \end{equation} where $ \sigma ^0 $ is the identity and we keep it as it keeps the spinor indices consistent. We want to simplify this operator in an analogous way to what's done in QED for the EM charge (see for example Peskin Eq (3.113)).

Since this should hold for any value of $ a $, we actually have 3 conserved charges, $ T _1 , T _2 , $ and $ T _3 $. This is an analogue to angular momentum, where we have 3 quantities that are conserved, $ J _x , J _y , $ and $ J _z $. However, even though each of the $ T $'s are conserved, we can't measure all 3 simultaneously. We conventionally just study $ T _3 $, \begin{equation} T _3 = \frac{1}{2} \int d ^3 x \left( \nu ^\dagger _L \nu _L - e ^\dagger _L e _L \right) + h.c. \end{equation} This charge takes the form of a number operators for the neutrinos and another for the electrons but with a relative negative sign.

We can write it more explicitly by moving to four component notation which lets us use the familiar four-component expressions for the fields from e.g. Peskin pg 54). I should note that this can just as easily be done using the less familiar 2 component expressions. Nevertheless we have, \begin{equation} \int \,d^3x \bar{\nu} \gamma _0 P _L \nu = \int \frac{ d ^3 p }{ (2\pi)^4 } \frac{1}{ 2E _p } \left( a _p ^{ s \dagger } a _p ^{ s ' } \bar{u} _p \gamma _0 P _L u _p ^{ s ' } + b _p ^{ s \dagger } b _{ p} ^{ s ' } \bar{v} _p ^s \gamma _0 P _L v _{ p } ^{s' } \right) \end{equation} We now use the explicit forms of the spinors to get, \begin{align} u ^{s \dagger} _p P _L u _{ p } ^{ s ' } & = \xi ^{ s \dagger } p \cdot \sigma \xi ^{ s ' } \end{align} This expression simplifies nicely in the case of massless particles (which is the case here as $ SU(2) _L$ is conserved), \begin{align} u ^{s \dagger} _p P _L u _{ p } ^{ s ' } & = E _p (1 - 2 s ) \delta _{ s s' } \end{align} and we also have, \begin{align} v ^{s \dagger} _p P _L v _{ p } ^{ s ' } & = E _p (1 - 2 s ) \delta _{ s s' } \end{align} Therefore, \begin{align} \int \,d^3x \nu _L ^\dagger \nu _L & = \int \frac{ \,d^3p }{ (2\pi)^3 } \sum _s \left( a _p ^\dagger a _p + b _p ^\dagger b _p \right) E _p ( 1 - 2 s ) \\ & = \frac{1}{2} \int \frac{ \,d^3p }{ (2\pi)^3 } \left( a _p ^\dagger a _p + b _p ^\dagger b _p \right) \end{align} Both the particle and antiparticle of the neutrino contribute positively to the charge $ T _3 $. Furthermore, only one spin for each of the particles contributes. This is because we are working in the massless limit and we don't have right handed particles (or left handed antiparticles).

In total we have (we trivially add the Hermitian conjugate), \begin{equation} T _3 = \frac{1}{2} \int \,d^3x \left( ( a _p ^\dagger a _p + b _p ^\dagger b _p ) - ( c _p ^\dagger c _p + d _p ^\dagger c _p ) \right) \end{equation} where $ c ^\dagger $ and $ d ^\dagger $ are the creation operators of the left electron. Therefore, the positive term counts the number of neutrinos and the negative term counts the number of electrons. Each increases or decreases the charge by $ 1/2 $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.