Normalization of SU(N) Group Generators

[LayMuon]

I am reading my textbook of QFT (Maggiore, Modern Introduction in QFT), and there is this statement:

"If T^a_R is a representation of the algebra and V a unitary matrix of the same dimension as T^a_R , then V T^a_R V^\dagger is still a solution o the Lie algebra and therefore provides an equivalent representation. We can fix V requiring that it diagonalizes the matrix D^{ab}(R) ≡ Tr (T^a_R T^b_R), so that Tr (T^a_R T^b_R) = C(R) \delta^{ab}."

I can't understand the second sentence, matrix D has different dimensions than V, how can it be used to diagonalize? Putting V within the trace doesn't make any sense, it would give unit matrix.

Thanks.

 

[Bill_K]

No you're right, there's no connection between the first and second sentences. It can't be the same V.

The correct statement is that D^ab = Tr(T^aT^b) is a symmetric matrix, and so there exists a linear transformation on the group indices a which brings it to a multiple of the identity.

 

[LayMuon]

Just a side question: The similarity transformation on particular representation generates all possible representations of the same dimensionality? If it is true then no way to tinker with representation of a given dimension to render D diagonal, right? But in that case how do you normalize it?

 

[haushofer]

Georgi's book on group theory treats this normalization issues in great detail in, if I remember correctly, chapter 2.

 

[Bill_K]

The similarity transformation on particular representation generates all possible representations of the same dimensionality? If it is true then no way to tinker with representation of a given dimension to render D diagonal, right? But in that case how do you normalize it?

The similarity transformation generates equivalent representations of a given set of generators T^a. But you can also reparametrize SU(N), leading to a different set of generators T'^a with different structure constants (but they still generate the same group.) The relationship between T^a and T'^a is a linear transformation, T'^a = C_b^a T^b where C_b^a is a constant matrix. It's this change of basis, C_b^a which gives you the freedom to diagonalize D^ab.

 

[LayMuon]

Thank you, Bill, I understand now.

I got Georgi's book, it seems to be a great book for a physicist. Thanks, haushofer.

 

[LayMuon]

Just one more question that i could not find in Georgi's book (probably don't look close enough). How do we prove that for fundamental representation C(R) = 1/2 for SU(N)? what difference "fundamental representation" makes? Can't we just choose any factor with appropriate linear transformation?

Georgi's 2nd chapter deals with normalization of the adjoint representation mostly.

Thanks again.

 

[LayMuon]

anybody can give me a reference on how to get these formulas in wikipedia?

http://en.wikipedia.org/wiki/Special_unitary_group#Generators

 

[Bill_K]

I believe you can get them by playing around with traces. You need to use the fact that the fundamental representation is n x n, so the n²-1 generators along with I form a complete set, and any n x n matrix can be expressed as a linear combination of these. Applying this:

T_aT_b ≡ α_ab I + β_abc T_c

Choosing the generators to obey the orthonormality condition Tr(T_aT_b) = ½ δ_ab gives you α_ab = (1/2n) δ_ab. Next,

Tr(T_aT_bT_c) = Tr((α_ab I + β_abd T_d)T_c) = ½ β_abc.

This gives you the fact that β_abc is cyclically symmetric, which means it is of the form if_abc + d_abc where f is totally antisymmetric and d is totally symmetric.

Finally, look at Tr(T_aT_bT_cT_d). Reduce this two ways: (1) T_awith T_b, then T_c with T_d, and (2) T_bwith T_c, then T_awith T_d should give you the expression for d_aced_bce.

 

[LayMuon]

thanks, Bill!

One thing that really bothers me is how do we get the one half in the exponential of the representation: e^{\frac{i}{2} \alpha_c T^c}? It is very important to clearly understand why special unitary condition yealds this 1/2.

For SU(2) with straightforward algebraic derivation it is easy to see, but how about the general case?

 

[Bill_K]

It's a question of normalization of the generators, and what convention you use. For example one of the generators of SU(3) is λ₃:
\left(\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&0\end{array}\right)
for which Tr(λ₃λ₃) = 2. This is NOT the convention used in the Wikipedia page. Rather they use T₃ = λ₃/2, for which Tr(T₃T₃) = 1/2. So you can write the finite rotation as Wikipedia would: exp(iαT₃), or as exp(iαλ₃/2) with the 2 present.