# A spherical tensor operator

1. Oct 17, 2009

### NewGuy

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 $$[V_i,J_j]=i\epsilon_{ijk}\hbar V_k$$, 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 $$\textbf{J}=(J_x,J_y,J_z)$$ 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: $$T_0^1=V_z, T_{\pm1}^1=\mp1/\sqrt{2}(V_x\pm iV_y)$$, 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. Oct 17, 2009

### Avodyne

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