- #1
tomdodd4598
- 137
- 13
- TL;DR
- Why does the Higgs coupling to up-type fermions take the form it does?
Hey there,
I was looking at the Higgs sector of the standard model, particularly its coupling to the fermions:
##\mathscr{L}_{ yukawa }=-\sum _{ a,b=1 }^{ 3 }{ \left( { Y }_{ ab }^{ u }{ \bar { Q } }_{ a }{ \hat { \varepsilon } }_{ 2 }{ H }^{ \dagger }{ u }_{ b }+{ Y }_{ ab }^{ d }{ \bar { Q } }_{ a }H{ d }_{ b }+{ Y }_{ ab }^{ e }{ \bar { L } }_{ a }H{ e }_{ b } \right) } +h.c.##
Where the ##{ Y }^{ f }## are the Yukawa coupling matrices, ##Q## holds the quark doublets, ##u## holds the up-type quark singlets, ##d## holds the down-type quark singlets, ##L## holds the lepton doublets, ##e## holds the charged lepton singlets, ##{ \hat { \varepsilon } }_{ 2 }## is the two-dimensional anti-symmetric tensor and ##H## is the Higgs doublet.
The up-type piece of this expression contains ##{ \hat { \varepsilon } }_{ 2 }## to 'flip' the Higgs doublet so that, when the Higgs acquires its non-zero VEV, ##H=\frac { 1 }{ \sqrt { 2 } } \left[ \begin{matrix} 0 \\ v+h \end{matrix} \right] ##, the up-type quarks correctly get their mass terms.
This question is already rather long-winded, but it is a simple one: is there a good reason for the choice of the anti-symmetric ##{ \hat { \varepsilon } }_{ 2 }## to flip the doublet over the symmetric ##\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}##?
I was looking at the Higgs sector of the standard model, particularly its coupling to the fermions:
##\mathscr{L}_{ yukawa }=-\sum _{ a,b=1 }^{ 3 }{ \left( { Y }_{ ab }^{ u }{ \bar { Q } }_{ a }{ \hat { \varepsilon } }_{ 2 }{ H }^{ \dagger }{ u }_{ b }+{ Y }_{ ab }^{ d }{ \bar { Q } }_{ a }H{ d }_{ b }+{ Y }_{ ab }^{ e }{ \bar { L } }_{ a }H{ e }_{ b } \right) } +h.c.##
Where the ##{ Y }^{ f }## are the Yukawa coupling matrices, ##Q## holds the quark doublets, ##u## holds the up-type quark singlets, ##d## holds the down-type quark singlets, ##L## holds the lepton doublets, ##e## holds the charged lepton singlets, ##{ \hat { \varepsilon } }_{ 2 }## is the two-dimensional anti-symmetric tensor and ##H## is the Higgs doublet.
The up-type piece of this expression contains ##{ \hat { \varepsilon } }_{ 2 }## to 'flip' the Higgs doublet so that, when the Higgs acquires its non-zero VEV, ##H=\frac { 1 }{ \sqrt { 2 } } \left[ \begin{matrix} 0 \\ v+h \end{matrix} \right] ##, the up-type quarks correctly get their mass terms.
This question is already rather long-winded, but it is a simple one: is there a good reason for the choice of the anti-symmetric ##{ \hat { \varepsilon } }_{ 2 }## to flip the doublet over the symmetric ##\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}##?
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