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SU(N) Vector vs. Spinor Representations

  1. Feb 25, 2005 #1
    I'm a little bit confused about the difference between the spinor and vector representations of SU(N)--I guess I could start with asking how a spinor and a vector differ: is this only a matter of how they transform under Lorentz transformations?

    Following up, the covariant derivative for a spinor of SU(2) is (i.e. for a scalar field [tex]\phi[/tex] that transforms as a spinor of SU(2)):

    [tex] D_\mu \phi = (\partial_\mu - i g A^a_\mu \tau ^a )\phi [/tex]

    While the covariant derivative for the vector representation of a scalar [tex]\phi[/tex] is:

    [tex] D_\mu \phi = \partial_\mu_a + g \epsilon_{abc} A^a_\mu \phi_c [/tex]

    (these are from Peskin and Schroeder p. 694-5, eq. (20.22) and (20.27) resp.)

    My understanding is that this means we have a scalar field [tex]\phi[/tex] that has a nonabelian gauge symmetry in some abstract (internal) SU(2) space.

    The spinor covariant derivative seems to make sense from the general definition of the covariant derivative:

    [tex]D_\mu = \partial_\mu - igA^a_\mu t^a[/tex]

    where [tex]t^a[/tex] is a generator of the gauge group. does this mean that the generator of the vector representation is something like [tex]\epsilon_{abc}[/tex]? Where does this [tex]\epsilon_{abc}[/tex] come from?

  2. jcsd
  3. Feb 25, 2005 #2


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    There is something really puzzling here.On normal basis,the matter field (in this case a scalar) would transform under the fundamental representation of the gauge symmetry group,which in this case is SU(2).The generators of the fundamental reperesentation of the group SU(2) are indeed [itex] T_{a}=\frac{1}{2}\sigma_{a} [/itex]...So they should be there...I don't know which kind (what spin bears) this fundamental representation is.


    P.S.I'll promiss to think about it.Any more comments are welcome.
  4. Mar 17, 2005 #3
    I actually just had this problem myself (on an exam) and was stuck! Luckily I figured it out!

    First, the covariant derivative is representation dependent:

    [tex]D_\mu = \partial_\mu - igA^a_\mu t_R^a[/tex]

    Where [tex]t_R^a[/tex] are the generaters for a particular representation R. Now when you have a field that transforms like a vector, you can use the fact that SU(2) is really the same as SO(3) (the infinitesimal transformations are measured by a vector).

    The Fundamental representation of SO(3) are the angular momentum generators J, that satisfy the commutation relations:

    [tex][J_i ,J_j ] = i\epsilon_i_j^k J_k[/tex]

    So the [tex]\epsilon^i^j^k[/tex] are the structure constants of SO(3). The Adjoint representation for any group has generators given by:

    [tex](t_A^a)^b^c=-i f^a^b^c[/tex]

    Where [tex]f^a^b^c[/tex] are the structure coeffiecients. So in particular for SO(3):

    [tex](t_A^a)^b^c=-i \epsilon^a^b^c[/tex]

    So then for the adjoint reprensentation of the "spin-one" (SO(3)) version of SU(2) we have:

    [tex]D_\mu = \partial_\mu + g\epsilon^a^b^c A^a_\mu [/tex]

    I hope this helps!

    -Laura :smile:
  5. Mar 17, 2005 #4
    Laura already explained to you the second part of your question.

    Let me add that 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°.

    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


  6. Mar 20, 2005 #5
    Since a long time ago, I've been looking for a simple explanation of what a spinor is, without much luck.
    So far, I had simple patched up together many definitions and different points of view until I finally had an acceptable idea on what a spinor was.
    Your explanation, Marlon, is, so far, the simplest and clearest I've seen, thanks a lot for that.

    Now, my question is, does ANY Lie group have a spinor representation?
    If not, wich ones do?
  7. Mar 20, 2005 #6
    No, for example : under addition, the real line is a Lie group.

    It is important to realize the the spinor representation is indeed the representation of a Lie Group because it is a rotation and any rotation can be put into a Lie Group like SO(3) and so on...

    Perhaps we should exclude the rotations (in the complex plane) that do not put an object onto itself after 360° of turning...but i am not sure of that anymore, it has been too long for me... :wink:

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