# Understanding tensor operators

1. May 18, 2014

### CAF123

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. May 18, 2014

### haael

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.

3. May 18, 2014

### Bill_K

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.

4. May 19, 2014

### CAF123

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