Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Understanding tensor operators

  1. May 18, 2014 #1

    CAF123

    User Avatar
    Gold Member

    The definition of tensor operator that I have is the following: 'A tensor operator is an operator that transforms under an irreducible representation of a group ##G##. Let ##\rho(g)## be a representation on the vector space under consideration then ##T_{m_c}^{c}## is a tensor operator in the irreducible representation ##c## if it transforms as follows: $$\rho(g) T_{m_c}^c \rho(g)^{\dagger} = (\rho_c(g))_{m_c m'_c} T_{m_c'}^c,$$ with summation over ##m_c'## implied.

    Can someone give me an example of a tensor operator realized in physics and the motivation for such a definition?

    Also, in that definition, what does it mean to say '...an operator that transforms under an irreducible representation of a group.'

    Thanks.
     
  2. jcsd
  3. May 18, 2014 #2
    Pretty much everything in physics is a tensor. Motivation is background independence.

    A representation is a set of matrices, right? Tensor is a vector for that matrix. Tensor is a vector that the representation matrix operates on.

    If you have some tensor and a transformation belonging to the group and its associated representation matrix, then the tensor under that transformation will be multiplied by that matrix.
     
  4. May 18, 2014 #3

    Bill_K

    User Avatar
    Science Advisor

    The electric and magnetic multipole moments of a nucleus.

    Also, the tensor force between two nucleons, a noncentral potential S12 that depends on the angles between the spin vectors of the two nucleons and the position vector connecting them.
     
  5. May 19, 2014 #4

    CAF123

    User Avatar
    Gold Member

    Are ##\rho(g)## some arbritary representations? And is ##\rho_c(g)## the representation of the state ##j=c##, a matrix of dimension ##(2c+1) \times (2c+1)##?

    The definition ##\rho(g) T_m^j \rho(g)^{\dagger} = (\rho_j (g))_{mm'} T_{m'}^j## may be rewritten in an infinitesimal form as $$[J_a, T_b^1] = i\epsilon_{abc} T_c^1\,\,\,\,\,\,\,\,(1)$$ How is this derived? It looks similar to the generic Lie algebra but the elements of the commutator are not both generators, as far as I understand - one is a tensor operator, the other is a generator.

    Edit: I derived an earlier result, that is $$e^{i\alpha_a J_a} T_k^1 e^{i\alpha_bJ_b} = T_k^1 + \epsilon_{alk} \alpha_a T_l^1$$ using (1). $$\rho(g) T_{m_c}^c \rho(g)^{\dagger} \equiv e^{i\alpha_a J_a} T_{m_c}^c e^{-i\alpha_b J_b} = e^{i\alpha_a J_a} T_{m_1}^1 e^{-i\alpha_b J_b},$$ where ##J_a## are the generators of some Lie algebra and ##T_{m_c}^c## is the tensor operator. The last equality follows from considering ##j=1## representation.

    Linearise for infinitesimal rotations gives $$(1+ i\alpha_aJ_a) T_{m_1}^1 (1 - i \alpha_b J_b)$$ and then multiplying out $$T_{m_1}^1 + i(\alpha_aJ_aT_{m_1}^1 - \alpha_b T_{m_1}^1J_b)$$

    I can get the result from here, but I explicitly used (1). Could someone explain why (1) is true?

    Thanks.
     
    Last edited: May 19, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Understanding tensor operators
  1. Tensor operators (Replies: 6)

  2. Tensor Operators (Replies: 2)

  3. Tensor operators in QM (Replies: 3)

Loading...