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Infinite dimensional representation of su(2)

  1. Apr 3, 2009 #1
    I'm trying to understand this paper which the author claimed that he had constructed an infinite dimensional representation of the su(2) algebra. The hermitian generators are given by

    [tex]J_x=\frac{i}{2}(\sqrt{N+1}a-a^\dagger\sqrt{N+1} )[/tex]
    [tex]J_y=-\frac{i}{2}(\sqrt{N+1}a+a^\dagger\sqrt{N+1} )[/tex]
    [tex]J_z=N+\frac{1}{2}[/tex]

    where the creation and the annihilation operators [tex]a^\dagger[/tex] and a satisfy the commutator relation [tex][a,a^\dagger]=-1[/tex] and [tex]N\equiv-a^\dagger a[/tex].

    My question is, what is the meaning of the expression [tex]\sqrt{N+1} [/tex]? So that I can proceed with the manipulation.
    Does it means the binomial expression
    [tex]\sqrt{N+1}= 1 + \frac{1}{2}N + ... [/tex]?

    I have been thinking along that line. How do we show that the representation is infinite dimensional?


    ps How do you create the superscript operator dagger in tex?
     
    Last edited: Apr 3, 2009
  2. jcsd
  3. Apr 3, 2009 #2

    Ben Niehoff

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    The dagger symbol is just \dagger. I don't know the answer to any of your other questions. :D
     
  4. Apr 3, 2009 #3
    Thanks I have done the necessary changes. They don't have the template for dagger.
     
  5. Apr 5, 2009 #4
    Hi,

    I think you are a little confused. The N in J_z represents the Number Operator which is defined as a^dagger*a
    Where as the N in the square root is just a number and is equal to the eigen value of the Number operator.

    Conventionally, that's what I would think of, when I see the equations you have written
    But of course, I haven't seen the paper you are referring to and I might be wrong. :smile:
     
  6. Apr 5, 2009 #5
    If N is just a number why the author bother to write the expression to the right of [tex]a^\dagger[/tex] .

    Yes I agree usually the number operator [tex]N=a^\dagger a[/tex]. But that is in the case of finite dimensional representation.

    For your information, the papers that I trying to understand are:

    1. Andre van Tonder, Ghosts as Negative Spinors, Nuc. Phys. B 645(2002) pp 371-386.
    2. Andre van Tonder, On the representation theory of negative spin, Nuc. Phys. B 645(2002) pp 387-402.
     
  7. Apr 5, 2009 #6
    What is attempted in your equations is known in physics as Schwinger's Boson representation of angular momentum algebra ( SU(2) in your case. ) The[tex] J_i[/tex] are the generators of this algebra.
    Therefore, one would conclude that the 'N' in the first two eqns are actually numbers:eigenvalue of the operator [tex]\hat{N} = a ^\dagger a [/tex], whose spectrum is infact the set of all non-negative integers, denoted by [tex]N[/tex].
    In the third eqn its actually the operator [tex]\hat{N}[/tex]. As u can see it doesn't matter if it is [tex]N[/tex] or, [tex]\hat{N}[/tex]. Both yields the same behavior. If it were a simple number, that eqn denotes (N+1/2)Id, where Id-> identity operator.
     
  8. Apr 6, 2009 #7
    OK I make a mistake. But it is in the expression Jy. There shouldn't be the imaginary number i. Other than that I think I have copied correctly.

    [tex]J_y=-\frac{1}{2}(\sqrt{N+1}a+a^\dagger\sqrt{N+1} )[/tex]

    If it true that [tex]\sqrt{N+1}[/tex] is just a number then

    [tex][J_x , J_y]=-\frac{i(N+1)}{4}[a , a^\dagger] = \frac{i(N+1)}{4}.[/tex]

    which is not equivalent to the su(2) algebra [tex][J_x , J_y]=iJ_z [/tex].


    However if I assume [tex]\sqrt{N+1}[/tex] to be an operator I'm able to verify the relation [tex][J_x , J_y]=iJ_z [/tex].
     
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