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A spherical tensor operator

  1. Oct 17, 2009 #1
    I ran into a problem reading Sakurais book about advanced quantum mechanics. I understand what a spherical tensor operator is, it's just an odd number of operators that transform in a nice way under rotation (or equivalently has some nice commutation relations with angular momentum). Sakurai defines a vector operator in several ways: by it's commutation relation [tex][V_i,J_j]=i\epsilon_{ijk}\hbar V_k[/tex], and he also defines a vector operator simply as a spherical tensor operator of rank 1.

    In my mind this is quite confusing however. By the first definition [tex]\textbf{J}=(J_x,J_y,J_z)[/tex] is clearly a vector operator, but J is not a spherical tensor operator (how would you define the three components in the first place?). It is true that you can construct 3 new operators from it: [tex]T_0^1=V_z, T_{\pm1}^1=\mp1/\sqrt{2}(V_x\pm iV_y)[/tex], and these 3 operators constitue a vector operator by definition 1. However, I can't understand why Sakurai claims this new vector operator to be equal to J. It could be that it was possible to make 3 other operators from J, and that these 3 operators also would constitute a vector operator.

    I guess my basic question is: Is the total angular momentum operator J a spherical tensor operator?
  2. jcsd
  3. Oct 17, 2009 #2


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    It depends on what the meaning of "is" is. As you wrote, given a vector operator [itex]\textbf{V}=(V_x,V_y,V_z)[/tex], we can define a spherical tensor operator of rank 1 [itex]\textbf{T}^1=(T_{-1}^1,T_0^1,T_{+1}^1)[/itex] via

    T_0^1=V_z, T_{\pm1}^1={\textstyle{\mp1\over\sqrt{2}}(V_x\pm iV_y)

    If we take this equality as the definition of "is", then any vector operator "is" a spherical tensor operator of rank 1. Since [itex]\textbf{J}[/itex] is a vector operator, then it "is" a spherical tensor operator of rank 1.
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