# Thread: Fun in the SU(N)

1. ## Fun in the SU(N)

OK, still plowing through a book on group theory and running up against another problem.

SU(N) matrices are NxN matrices that are both unitary and have determinant = 1. Unitary means that if you multiply the matrix by its complex conjugate transposed, then you get the identity matrix. Fairly simple. I understand all that. Made a few examples myself and played with them for fun.

Then the book brings up the Gell-Mann matrices. And that is where I'm lost. Not a single one has a determinant equal to 1. Nor do you get I if you multiply them by their Hermitian conjugate.

https://en.wikipedia.org/wiki/Gell-Mann_matrices

Obviously I am missing something basic. The Gell-Mann matrices are not SU(3), but then do they just help generate the proper SU(3) matrices through linear combination? Or what? Just confused at this point in the book.

TIA

SLD

2. Gell-Mann matrices are generators of the SU(3) group.
generator is something you multiply by i (complex number) and take exponent of to get element of the group G=exp(i*H)

Similar to exp(i*phi)
In physics generators associated with measurable quantities.

3. Originally Posted by barbos
Gell-Mann matrices are generators of the SU(3) group.
generator is something you multiply by i (complex number) and take exponent of to get element of the group G=exp(i*H)

Similar to exp(i*phi)
In physics generators associated with measurable quantities.
So, what are the actual matrices of SU(3)?

I read some more of the book and after going over it several times, I figured out that equation. Difficult to actually work out the actual matrices since they require an infinite sum. But I assume that it’s been done to a fairly high accuracy. Maybe that’s later in the book.

SLD

4. Originally Posted by SLD
Originally Posted by barbos
Gell-Mann matrices are generators of the SU(3) group.
generator is something you multiply by i (complex number) and take exponent of to get element of the group G=exp(i*H)

Similar to exp(i*phi)
In physics generators associated with measurable quantities.
So, what are the actual matrices of SU(3)?
obviously, unitary matrices with determinant=1
Generators are just useful way to look at the group.

I read some more of the book and after going over it several times, I figured out that equation. Difficult to actually work out the actual matrices since they require an infinite sum. But I assume that it’s been done to a fairly high accuracy. Maybe that’s later in the book.
SLD
I suspect you are doing all this as a hobby and skipping a lot of prerequisites, right?
Computing exponent of the matrix is actually not that difficult and is done with absolute precision in fixed amount of steps but it's rarely done in practice.

5. Originally Posted by barbos
obviously, unitary matrices with determinant=1
Generators are just useful way to look at the group.

I read some more of the book and after going over it several times, I figured out that equation. Difficult to actually work out the actual matrices since they require an infinite sum. But I assume that it’s been done to a fairly high accuracy. Maybe that’s later in the book.
SLD
I suspect you are doing all this as a hobby and skipping a lot of prerequisites, right?
Computing exponent of the matrix is actually not that difficult and is done with absolute precision in fixed amount of steps but it's rarely done in practice.
Yes. Just a hobby. Keeping my brain sharp. Or trying to. My undergraduate degree is in Mathematics, and I took linear algebra but we never learned that. This book has a preface in linear algebra and does go over some stuff I’ve never seen but wasn’t too difficult. The linear algebra I learned never involved complex numbers. But Hermitean Matrices aren’t that particularly difficult to understand.

SLD

SLD

6. It's true that linear algebra in practice taught first and in real numbers, simply because complex number analysis comes much later.

Hermitian operator have real eigenvalues and are analogous to real numbers. Unitary operators are analogous to complex rotations exp(i*phi)
Eigenvalues of unitary matrices have module of 1 obviously.

7. Originally Posted by SLD
So, what are the actual matrices of SU(3)?
I've read several books on Lie groups and Lie algebras, and I've never seen any expression for those matrices anywhere. My guess about them is that they are horribly complicated to calculate.

A Lie-group matrix D is calculated from its corresponding Lie-algebra matrix L with D = exp(i*L)

One can do this calculation by finding the eigenvalues V and eigenvectors M of L: L = M.V.M+, making D = M.exp(i*V).M+

In turn, L = sum over k of ck*Lk where the c's are coefficients and the L's are the Lie algebra's generators.

The calculation has been done for all the representations of SU(2)/SO(3), however:

If you have ever worked with the ladder-operator formulation of quantum-mechanical angular momentum, you have worked with representations of SU(2)/SO(3). The operators for angular momentum are this Lie algebra's generators.

8. Originally Posted by lpetrich
Originally Posted by SLD
So, what are the actual matrices of SU(3)?
I've read several books on Lie groups and Lie algebras, and I've never seen any expression for those matrices anywhere. My guess about them is that they are horribly complicated to calculate.
What do you mean by that? Unitary matrix consists of rows which are ortho-normal to each other.
In other words, any ortho-normal basis corresponds to unitary matrix.

9. Originally Posted by barbos
Originally Posted by lpetrich
Originally Posted by SLD
So, what are the actual matrices of SU(3)?
I've read several books on Lie groups and Lie algebras, and I've never seen any expression for those matrices anywhere. My guess about them is that they are horribly complicated to calculate.
What do you mean by that? Unitary matrix consists of rows which are ortho-normal to each other.
In other words, any ortho-normal basis corresponds to unitary matrix.
That is a constraint on the matrices, not a specification of the matrices in terms of some parameters.

For the SU(2) matrices, one can find them in terms of the unit quaternions:

D = x0*I - i*(x1*s1 + x2*s2 + x3*s3)

where 4-vector x = (x0,x1,x2,x3) has unit length, x.x = 1, and obeys the quaternion multiplication law. The matrices s1, s2, s3 are the Pauli matrices: {{0,1},{1,0}}, {{0,-i},{i,0}}, {{1,0},{0,-1}}. The matrix I is the identity matrix: {{1,0},{0,1}}.

There is a similar expression for the SO(3) matrices, but it is more complicated, and it contains a sort of square of the quaternion vector.

Quaternions are often used in computer 3D graphics, because it is relatively easy to smoothly change them -- it is much easier to keep a vector normalized than a matrix orthonormalized.

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