What are vector/tensors operators in QM or in general any vector space

[Sagnik.]

I am interested in knowing that in QM what vector/tensor operators are? In fact how do they differ from scalar operators?

 

[Bill_K]

A vector or tensor operator is a set of operators, such that when you perform a rotation they go into linear combinations of each other, the same way that the components of a vector or tensor do. Thus angular momentum J is a set of three operators J_x, J_y, J_z. Under a rotation about the z axis,
J_z' = J_z, J_x' = J_x cos θ + J_y sin θ,

J_y' = J_y cos θ - J_x sin θ.

Note this says nothing about any commutation relations that the operators might have.

 

[vanhees71]

In fact, there are some commutation relations, following from the fact that $\hat{V}_j$ are vector operators (you can generalize the following for tensor operators of any rank), namely those with angular-momentum operators:

$$[\hat{J}_k,\hat{V}_l]=\mathrm{i} \epsilon_{klm} V_m.$$

This is, because of the transformation relations you've just given, because rotations are generated by the angular-momentum operators. A rotation is given by the unitary transformation

$$\hat{U}(\vec{\varphi})=\exp(\mathrm{i} \hat{\vec{J}} \cdot \vec{\varphi}).$$

The rotation is given by

$$\hat{\vec{V}}'=\hat{U}(\vec{\varphi}) \hat{\vec{V}} \hat{U}^{\dagger}(\vec{\varphi}).$$

 

[Bill_K]

What I meant was, commutation relations may or may not exist among the operators themselves. I used the J's as an example of a vector operator, and they have the SO(3) commutators typical of angular momentum, but three operators which mutually commute could have been used instead.