Find the adjoint and check its commutation relations

Familiarize with nested tables

It starts picking indexes from outside!

Table[
{i,k}, {i, {i1,i2}},{k, {k1,k2}}
]

Define the generators

?τ

Length[{a,b,c}]

τ[1]={{0,1},{1,0}}
τ[2]={{0,-I},{I,0}}
τ[3]={{1,0},{0,-1}}

The norm of the matrices is not unity, beware.

https://en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm

Table[ τ[i] //TeXForm, {i, 1, 3}]

Table[ Tr[ τ[i].ConjugateTranspose[ τ[i] ] ],{i,1,3}]

The SU(2) commutators table

[(Subscript[τ, i]),(Subscript[τ, j])]∝(Subscript[ϵ, ijk])(Subscript[τ, k])

Table[Commutator[τ,i,j]//MatrixForm,{i,1,3},{j,1,3}]

Table[Tr[
Commutator[τ, i, j].ConjugateTranspose[τ[k]]
], {k,1,3},{i, 1, 3}, {j, 1, 3}]

Let us define the above as the function to print the algebra multiplication (commutation) table

Clear[AlgebraTable]
(*AlgebraTable[τ_]:=Table[Tr[
  Commutator[τ, i, j].ConjugateTranspose[τ[k]]
  ], {k, 1, 3}, {i, 1, 3}, {j, 1, 3}]*)

The adjoin representation

Ad((Subscript[τ, i])) . T  : 
takes a generic T = (∑_k)(Subscript[t, k])(Subscript[τ, k]) to a new matrix:

T->[(Subscript[τ, i]),T]=(∑_k)(Subscript[t, k]) (Subscript[f, ijk]) (Subscript[τ, k])

A representation of the adjoint can be constructed explicitly as a set of matrices 3x3 recalling that the commutator is a matrix itself and can be expressed as a linear combination of the τ

That is to say the values of f define 3 matrices

(Subscript[(A^((i))), jk])=(Subscript[f, ijk])

where k,i,j take values in 1,2,3. This can be done for any algebra. In the case of SU(2)

(Subscript[f, ijk ])  = (Subscript[ϵ, ijk])  the Levi-Civita symbol

Clear[Adjoint]

?Commutator

AdjointSU2[i_]:=Table[Tr[
Commutator[τ, i, j].ConjugateTranspose[τ[k]]
], {j, 1, 3}, {k, 1, 3}]

Display of the matrices defined in this way from the fundamental commutation relation

Head[Tr[AdjointSU2[3]]]

Table[AdjointSU2[j] // MatrixForm, {j, 1, 3}]

AlgebraTable[τ]//MatrixForm

AlgebraTable[τ, Treatment -> Total]

AlgebraTable[τ,Treatment->Identity]

AlgebraTable[AdjointSU2,Treatment->Identity]

Are the same up to -128/4

4/128*AlgebraTable[Ad]

The Adjoint that we found respects the same multiplication table as the fundamental generators. It’s a valid representation.

Weights of the Fundamental

h[a_] := a*τ[3]

h[a]

Eigenvectors[h[a]]

Table[h[a].(ϕ[2][[i]])//MatrixForm,{i,1,2}]

Table[(ϕ[2][[i]]).h[a].(ϕ[2][[i]])//MatrixForm,{i,1,2}]

Table[h[a].(Eigenvectors[h[a]][[i]]) // MatrixForm, {i, 1, 2}]

Table[(Eigenvectors[h[a]][[j]]).h[a].(Eigenvectors[h[a]][[i]]) // MatrixForm, {i, 1, 2},{j,1,2}]

Weights of the Adjoint

h[a_]:=a*Ad[3]

h[a]//MatrixForm

Eigenvectors[ h[a] ]

Table[ 
h[a].(Eigenvectors[ h[a] ][[i]]),{i,{1,2,3}}
]

Table[ 
 (Conjugate[Eigenvectors[ h[a] ][[j]]]).h[a].(Eigenvectors[ h[a] ][[i]]), {i, {1, 2, 3}},{j, {1, 2, 3}}
 ]

Table[h[a].(ϕ[3][[i]])//MatrixForm,{i,1,3}]
{(({{0}, {-4 I a}, {0}})),(({{4 I a}, {0}, {0}})),(({{0}, {0}, {0}}))}


{1,-I,0}.h[a].{1,I,0}
-8 a
{1, I, 0}.h[a].{1, -I, 0}
8 a

(ϕ[3][[1]]).h[a].(ϕ[3][[2]])
4 I a
Table[ 
h[a].(ϕ[3][[i]]),{i,{1,2,3}}
]
{{0,-4 I a,0},{4 I a,0,0},{0,0,0}}