Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

    Thanks,
    Flip
     
  2. jcsd
  3. Feb 25, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.

    Daniel.

    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

    regards

    marlon
     
  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:


    regards
    marlon
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: SU(N) Vector vs. Spinor Representations
  1. Spinor representation (Replies: 2)

  2. SU(2) Spinor Question (Replies: 1)

  3. Spinor representations (Replies: 1)

Loading...