Is the total angular momentum operator J a spherical tensor operator?

[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?

 

[Avodyne]

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 $\textbf{T}^1=(T_{-1}^1,T_0^1,T_{+1}^1)$ 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.

 

[Entropia]

I would first commend your efforts in trying to understand the concept of a spherical tensor operator and the confusion you have encountered while reading Sakurai's book. Let me provide some clarification on the matter.

A spherical tensor operator is a mathematical object that transforms in a specific way under rotations. It can be represented as a tensor with an odd rank, and its components are related to each other through a set of transformation rules. These operators are particularly useful in quantum mechanics, where they are used to describe the behavior of particles with angular momentum.

Now, to address your question, the total angular momentum operator J is indeed a spherical tensor operator. This can be seen from its transformation properties under rotations. As you have correctly mentioned, the components of J (Jx, Jy, Jz) do not individually satisfy the definition of a spherical tensor operator. However, when taken together as a vector operator, they do form a spherical tensor operator of rank 1.

Sakurai's definition of a vector operator is a specific case of a spherical tensor operator of rank 1. This is because, as you have pointed out, the components of J do not satisfy the definition of a spherical tensor operator individually. Therefore, in order to define J as a spherical tensor operator, we need to consider it as a vector operator.

I hope this helps to clarify your confusion. It is important to note that different textbooks or sources may use different definitions and notations for the same concept. It is always a good practice to carefully read and understand the context in which a particular definition is being used.