Intro to Group Theory for QFT
I have posted this in my journal but i thought it might be nice to put this text in this thread too...
Many students have difficulties understanding what 'transforming like a vector or tensor' really means in physics. Here is the solution...Before you begin, be sure that you know really well what a tensor is...if you do not, check out my '
what is a tensor' entry...
A spinor is a special kind of vector. I mean, it has the property that if you rotate it 360° you get the exact opposite (A ---> -A) of what you originally rotated. Rotate another 360° and you get where you started off in the first of the two rotations (A---> -A--->A).
Now let us look at the rotationgroup SO(3) or even any other group, it don't matter :
An object v transforms as a vector if you can write v' = Uv where U is a representation for the group in question, U represents a rotation. Another way to say this is if you transform an object under a certain group, the 'image' of this transformation will be a linear combination of the object that you transformed. So transforming like a vector really means that the object you transform will be written out as a linear combination of it's components after the transformation.
An object transforms as a tensor if you can write v'=UU'U''v
So this means that v transforms 'as a product of vectors' because of the multiple U-matrices.
Now, transforming like a spinor really means that the object tranforms like a vector (you know what that means) but not just any vector. This is a special case, where the U-matrix does not represent just any transformation but a transformation that gives you the opposite of the initial object after a rotation of 360°.
Rotations are generated by the J-operator. J = 1 for example means that the quantity at hand transforms like a vector under three dimensional rotations. And the other way around, if an object transforms like a vector under these 3-D rotations, you know it will have spin J =1 and thus three degrees of freedom. YES, because the fundamental representations will be (3*1)-matrices which have three components...Spin 2 is a tensor and Spin 0 is a scalar...ODD SPIN IS A SPINOR
One can recognize a spinor by the way it transforms under a group. If the generator is a Pauli-matrix you are done...Just like in the case of SU(3), if you now the generator is a GellMann matrix, you know you are working with anobject in the adjoint representation and these objects are GLUONS. Let us look into gluons :
There are 3 colors. Why ? Well, becauseSU(3) is the group of 3 x 3 unitary matrices with determinant 1. The most easy matrix such an SU(3) matrix can work on is the 3*1-colum-matrix (ythis one has three components and is called the fundamental representation). SU(3) is the symmetry group of the strong force. What this means is that, as far as the strong force is concerned, the state of a particle is given by a vector in some vector space on which elements of SU(3) act as linear (in fact unitary) operators.
We say the particle "transforms under some representation of SU(3)".
For example, since elements of SU(3) are 3 x 3 matrices,like i already said before , they can act on column vectors by matrix multiplication. This gives a 3-dimensional representation of SU(3). The quarks are represented by this 3*1-matrix. The antiquarks can be represented by row vectors because we can multiply a 3*3-matrix with a row vector on the LEFT side of the matrix.
The gluons are represented by the socalled adjoint representation which consits out of traceless 3*3matrices. It can be seen that a row of such a matrix represents one quark colour and a colom of such a matrix represents a anti-colour. each gluon is therefore constructed out of a colour-anticolour combination. Given that there are 3 such colours and anticolours, you would expect 9 gluons. However there are only eight . Can you see why ?
ps : you know that the colours are red green and blue and it is the postulate of QCD that the sum of these three represents colour-neutrality ! This is the main law that needs to be respected : in interactions : the sum of all involved colours must be WHITE
regards
marlon
ps : maybe others can add or correct ?