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

  1. Jun 8, 2012 #1
    I am interested in knowing that in QM what vector/tensor operators are? In fact how do they differ from scalar operators?
  2. jcsd
  3. Jun 8, 2012 #2


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    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.
  4. Jun 8, 2012 #3


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

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

    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

    [tex]\hat{U}(\vec{\varphi})=\exp(\mathrm{i} \hat{\vec{J}} \cdot \vec{\varphi}).[/tex]

    The rotation is given by

    [tex]\hat{\vec{V}}'=\hat{U}(\vec{\varphi}) \hat{\vec{V}} \hat{U}^{\dagger}(\vec{\varphi}).
  5. Jun 8, 2012 #4


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    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.
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