5
$\begingroup$

In the Standard Model, the gauge covariant kinetic term for the left-handed lepton doublets $\psi_{iL}=(\nu_{iL},e_{iL})^T$, for example, is given by $$i\overline{\psi_{iL}}\gamma^\mu D_\mu\psi_{iL}.$$ Why is there no off-diagonal terms in the kinetic energy such as $i\overline{\psi_{iL}}\gamma^\mu D_\mu\psi_{jL}$ between two different generations $i$ and $j$. Such a kinetic term would also be $SU(2)_L\times U(1)_Y$ invariant. Am I wrong?

$\endgroup$
  • 1
    $\begingroup$ ...what is your definition of "generations"? The basis of "generations" is presumably precisely the basis in which the mass matrix is diagonal, which makes the question pointless. $\endgroup$ – ACuriousMind Dec 19 '16 at 17:13
  • 1
    $\begingroup$ @ACuriousMind-I think, I can start with a basis in which neither the kinetic terms nor the mass terms are diagonal. $\endgroup$ – SRS Dec 19 '16 at 17:20
6
$\begingroup$

The most general Lagrangian consistent with the symmetries indeed has a non-diagonal kinetic term however the point is that you can usually redefine the fields such that the kinetic term is the identity. The way this plays out depends on the fields we are interested in, i.e., whether its matter fields or gauge fields. We begin with discussing spin 1/2 fermions (the situation for scalars is similar) and then discuss the more involved case of gauge bosons.

Fermions

Consider a gauge theory with a set of fields, $ \psi _i $, with the same charge under a gauge symmetry. Lets see how this plays out. The most general Lagrangian you can write down is: \begin{equation} {\cal L} = i \lambda _{ ij} \bar{\psi} _i \gamma ^\mu D _\mu \psi _j - m _{ ij} \bar{\psi} _i \psi _j \end{equation} The reality condition requires that $ \lambda $ and $ m $ are both Hermitian matrices. We can always perform a field redefinition, $ \psi _i \rightarrow T _{ ij}\psi _j $. \begin{equation} {\cal L} \rightarrow ( T ^\dagger \lambda T ) _{ ij}\bar{\psi} _i \gamma ^\mu D _\mu \psi _j - ( T ^\dagger m T ) _{ ij} \bar{\psi} _i \psi _j \end{equation} For a Hermitian matrix, $ \lambda $, it is always possible to find a (non-unitary!) transformation $ T $ such that $ T ^\dagger \lambda T $ is the identity. This can be seen by splitting $ T $ up to into a unitary matrix, $ U $, that diagonalizes $ \lambda $ and a matrix $ R $ which is equal to the square root of the inverse of the eigenvalues of $ \lambda $:

\begin{equation} T ^\dagger \lambda T = R ^\dagger U ^\dagger \lambda U R = \sqrt{ \lambda _D ^{-1} } \lambda _{ D}\sqrt{ \lambda _D ^{-1} } = {\bf 1} \end{equation} where $ \lambda _D $ is defined as the diagonalized $ \lambda $. Therefore, we can write, \begin{equation} {\cal L} \rightarrow i \bar{\psi} _i \gamma ^\mu D _\mu \psi _i - ( T ^\dagger m T ) _{ ij} \bar{\psi} _i \psi _j \end{equation} Since the matrix $ T ^\dagger m T $ is completely generic we may as well just relabel it $M $ giving the usual starting Lagrangian, \begin{equation} {\cal L} = i \bar{\psi} _i \gamma ^\mu D _\mu \psi _i - M _{ ij} \bar{\psi} _i \psi _j \end{equation}

From this analysis one may think that we have completely fixed the basis of $ \psi $ and hence can no longer make field redefinitions. However, this is not the case. The kinetic term is still invariant under unitary transformation and hence we still have the freedom to rotate the $ \psi $ basis by a unitary matrix (which is often done in the discussion of the SM yukawa interactions).

Gauge fields

Gauge fields are trickier as there are symmetries which prevent certain terms from being written down (the gauge and mass matrices are often not generic). This results in subtle features.

Consider a theory with multiple U(1)'s: \begin{equation} {\cal L} = - \frac{1}{4} \lambda _{ ij} F _{ \mu\nu } ^i F ^{ j\,\mu\nu } + i \bar{\psi} _f D _\mu \gamma ^\mu \psi _f + m _{ ij} ^2 A ^i A ^j \end{equation} where the covariant derivative, $ D ^\mu \psi ^f \equiv ( \partial ^\mu -i e _a ^f X _a ^\mu ) \psi ^f $, has a sum over the couplings to the gauge fields. Note that if a gauge symmetry is unbroken its mass terms vanish. Reality of the Lagrangian sets $ \lambda $ to be real and symmetric. Analogous to the fermion case, it can be rotated to unity by an orthogonal transformation ($ O $) followed by the square root of the inverse of the eigenvalues of $ \lambda $, $ T = O R $. This matrix now rotates the different couplings, \begin{equation} {\cal L} = - \frac{1}{4} F _{ \mu\nu } ^i F ^{ i\,\mu\nu } + i \bar{\psi} _f ( \partial ^\mu - i e _a ^f T _{ ab} X _b ^\mu ) \gamma ^\mu \psi _f + ( T ^T m ^2 T ) _{ ij } A ^i A ^j \end{equation} Now if $ m ^2 _{ ij} $ and $ e _a ^f $ were the most generic terms possible this would be the end of the story. We would redefine the coupling as did above and move on. However, for gauge field we often have that one or more of these symmetries is unbroken and we want to write it in a basis which makes this obvious. From here its not obvious this is even possible. Lets see how that plays out.

The only rotation we can do leaving the kinetic term invariant is an orthogonal one, $ A ^\mu _i \rightarrow O' _{ ij} A _j ^\mu $. This gives a mass term, \begin{equation} {\cal L} _m \rightarrow ( O ^{ \prime T} T ^T m ^2 T O ' ) _{ij} A ^i A ^j \end{equation} Since the number of zero eigenvalues doesn't change by multiplication by invertible matrices, diagonalization will still yield the same number of massless vectors as we started with (and hence the diagonalization above didn't break any additional U(1)'s as it shouldn't). Performing this transformation: \begin{equation} {\cal L} _m \rightarrow M _i ^2 A ^i A ^i \end{equation} where $ M _i $ are the eigenvalues of the transformed vectors (and is only non-zero for the broken symmetries).

Now lets consider the covariant derivative term: \begin{equation} {\cal L} \supset e _a ^f T _{ ab} O _{ b c}'X _c ^\mu \bar{\psi} _f \gamma _\mu \psi _f \end{equation} Due to the mass diagonalization, we have already completely fixed the basis of the vectors and hence can no longer rotate things around. The fermions inevitably get a convoluted gauge coupling, \begin{equation} g _c ^f \equiv e _a ^f T _{ ab} O _{ b c}' \end{equation}

Lets now get a better feel for these matrices with a relevant example. Consider the case of U(1)$ \times $ U(1) where only $ 1 $ U(1) is broken ($ m _{11} ^2 = m _X ^2 , m _{ 12} ^2 = m _{ 21} ^2 = m _{ 2 2} ^2 = 0 $) and we have small mixing, $ \lambda = \left( \begin{array}{cc} 1 & \epsilon \\ \epsilon & 1 \end{array} \right) $. The unbroken (broken) U(1) is analogous to the photon (dark photon). In this case its easy to calculate the relevant matrices, \begin{equation} T = \frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{cc} ( 1 + \epsilon ) ^{ - 1/2} &0 \\ 0 & ( 1 - \epsilon ) ^{ - 1/2} \end{array} \right) \,,\quad O ' = \frac{1}{\sqrt{2}} \left( \begin{array}{cc} \sqrt{ 1 - \epsilon } & \sqrt{ 1 + \epsilon } \\ - \sqrt{ 1 + \epsilon } & \sqrt{ 1 - \epsilon } \end{array} \right) \end{equation} which gives, \begin{equation} T ^T \lambda T = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\,,\quad O^ { \prime T} T ^T m ^2 T O ' = \left( \begin{array}{cc} m _{ X} ^2 / ( 1 - \epsilon ^2 ) & 0 \\ 0 & 0 \end{array} \right) \end{equation} Now if we consider a fermion initially charged only under the unbroken U(1) we see it gains a coupling to the massive vector state: \begin{equation} {\mathbf{g}} ^{ ( 1 )}= O ^{ \prime T} T ^T \left( \begin{array}{c} 0 \\ e \end{array} \right) \simeq e \left( \begin{array}{c} - \epsilon \\ 1 \end{array} \right) \end{equation} while particles charged initially only under the massive U(1) don't gain a charge to the unbroken U(1): \begin{equation} {\mathbf{g}} ^{ ( 2 )}= O ^{ \prime T} T ^T \left( \begin{array}{c} e ' \\ 0 \end{array} \right) \simeq e' \left( \begin{array}{c} 1 \\ 0 \end{array} \right) \end{equation} The first effect is practically what happens when you consider a kinetically-mixed "dark photon": you have a coupling of the Standard Model fermions to the dark photon suppressed by a mixing parameter, $\epsilon$. Furthermore, we see a somewhat non-intuitive effect that the dark sector particles do not gain any such charge under the photon.

The more detailed example for the real Standard Model is complicated by the fact that the SM vector state ($B$) is a linear combination of a massive ($ Z $) and massless state ($ \gamma $). This leads to the possibility that the dark sector particles do gain a charge under the electromagnetism.

$\endgroup$
4
$\begingroup$

The one you point out is not considered a kinetic term, but a two-point interaction between the two lepton families.

To understand the physics then you need to diagonalize the kinetic term, so that you can understand what are the independent components of the fields.

If you know about Higgs mechanism, it is analogous to the fact that you have to decouple the two fields by going to the unitary gauge. [if you do not know about Higgs mechanism ignore this line]

$\endgroup$
  • $\begingroup$ rather than a two point interaction, since you are using the covariant derivative it is a two point interaction (two leptons) + a three point interaction (two leptons + gauge boson) $\endgroup$ – yoric Dec 19 '16 at 13:52
  • 1
    $\begingroup$ @yoric- It may not be called a kinetic term but it is a possible term to allow in a $SU(2)\times U(1)_Y$ theory. Isn't it? $\endgroup$ – SRS Dec 19 '16 at 13:55
  • $\begingroup$ yes but you don't know how to interpret the fields interacting in that way, so you diagonalize the kinetic term and call leptons the field after the diagonalization. $\endgroup$ – yoric Dec 19 '16 at 13:59

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.