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

1. Jun 8, 2012

### Sagnik.

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

2. Jun 8, 2012

### 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 Jx, Jy, Jz. Under a rotation about the z axis,
Jz' = Jz, Jx' = Jx cos θ + Jy sin θ,

Jy' = Jy cos θ - Jx sin θ.

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

3. Jun 8, 2012

### 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

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

4. Jun 8, 2012

### 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.