Tensor operator

"Spherical tensor operator" redirects here. For the closely related concept, seespherical basis.

Inpureandapplied mathematics,quantum mechanicsandcomputer graphics,atensor operatorgeneralizes the notion ofoperatorswhich arescalarsandvectors.A special class of these arespherical tensor operatorswhich apply the notion of thespherical basisandspherical harmonics.The spherical basis closely relates to the description ofangular momentumin quantum mechanics and spherical harmonic functions. Thecoordinate-freegeneralization of a tensor operator is known as arepresentation operator.

The general notion of scalar, vector, and tensor operators

In quantum mechanics, physical observables that are scalars, vectors, and tensors, must be represented by scalar, vector, and tensor operators, respectively. Whether something is a scalar, vector, or tensor depends on how it is viewed by two observers whose coordinate frames are related to each other by a rotation. Alternatively, one may ask how, for a single observer, a physical quantity transforms if the state of the system is rotated. Consider, for example, a system consisting of a molecule of mass$M$,traveling with a definite center of mass momentum,$p{\mathbf {\hat {z}} }$,in the$z$direction. If we rotate the system by$90^{\circ }$about the$y$axis, the momentum will change to$p{\mathbf {\hat {x}} }$,which is in the$x$direction. The center-of-mass kinetic energy of the molecule will, however, be unchanged at$p^{2}/2M$.The kinetic energy is a scalar and the momentum is a vector, and these two quantities must be represented by a scalar and a vector operator, respectively. By the latter in particular, we mean an operator whose expected values in the initial and the rotated states are$p{\mathbf {\hat {z}} }$and$p{\mathbf {\hat {x}} }$.The kinetic energy on the other hand must be represented by a scalar operator, whose expected value must be the same in the initial and the rotated states.

In the same way, tensor quantities must be represented by tensor operators. An example of a tensor quantity (of rank two) is the electrical quadrupole moment of the above molecule. Likewise, the octupole and hexadecapole moments would be tensors of rank three and four, respectively.

Other examples of scalar operators are the total energy operator (more commonly called theHamiltonian), the potential energy, and the dipole-dipole interaction energy of two atoms. Examples of vector operators are the momentum, the position, the orbital angular momentum,${\mathbf {L} }$,and the spin angular momentum,${\mathbf {S} }$.(Fine print: Angular momentum is a vector as far as rotations are concerned, but unlike position or momentum it does not change sign under space inversion, and when one wishes to provide this information, it is said to be a pseudovector.)

Scalar, vector and tensor operators can also be formed by products of operators. For example, the scalar product${\mathbf {L} }\cdot {\mathbf {S} }$of the two vector operators,${\mathbf {L} }$and${\mathbf {S} }$,is a scalar operator, which figures prominently in discussions of thespin–orbit interaction.Similarly, the quadrupole moment tensor of our example molecule has the nine components

$$Q_{ij}=\sum _{\ Alpha }q_{\ Alpha }\left(3r_{\ Alpha,i}r_{\ Alpha,j}-r_{\ Alpha }^{2}\delta _{ij}\right).$$Here, the indices$i$and$j$can independently take on the values 1, 2, and 3 (or$x$,$y$,and$z$) corresponding to the three Cartesian axes, the index$\ Alpha$runs over all particles (electrons and nuclei) in the molecule,$q_{\ Alpha }$is the charge on particle$\ Alpha$,and$r_{\ Alpha,i}$is the$i$-th component of the position of this particle. Each term in the sum is a tensor operator. In particular, the nine products$r_{\ Alpha,i}r_{\ Alpha,j}$together form a second rank tensor, formed by taking the outer product of the vector operator${\mathbf {r} }_{\ Alpha }$with itself.

Rotations of quantum states

Quantum rotation operator

Therotation operatorabout theunit vectorn(defining the axis of rotation) through angleθis

$$U[R(\theta,{\hat {\mathbf {n} }})]=\exp \left(-{\frac {i\theta }{\hbar }}{\hat {\mathbf {n} }}\cdot \mathbf {J} \right)$$

whereJ= (J_x,J_y,J_z)are the rotation generators (also the angular momentum matrices):

$$J_{x}={\frac {\hbar }{\sqrt {2}}}{\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}}\,\quad J_{y}={\frac {\hbar }{\sqrt {2}}}{\begin{pmatrix}0&i&0\\-i&0&i\\0&-i&0\end{pmatrix}}\,\quad J_{z}=\hbar {\begin{pmatrix}-1&0&0\\0&0&0\\0&0&1\end{pmatrix}}$$

and let${\widehat {R}}={\widehat {R}}(\theta,{\hat {\mathbf {n} }})$be arotation matrix.According to theRodrigues' rotation formula,the rotation operator then amounts to $$U[R(\theta,{\hat {\mathbf {n} }})]=1\!\!1-{\frac {i\sin \theta }{\hbar }}{\hat {\mathbf {n} }}\cdot \mathbf {J} -{\frac {1-\cos \theta }{\hbar ^{2}}}({\hat {\mathbf {n} }}\cdot \mathbf {J} )^{2}.$$

An operator${\widehat {\Omega }}$is invariant under a unitary transformationUif $${\widehat {\Omega }}={U}^{\dagger }{\widehat {\Omega }}U;$$ in this case for the rotation${\widehat {U}}(R)$, $${\widehat {\Omega }}={U(R)}^{\dagger }{\widehat {\Omega }}U(R)=\exp \left({\frac {i\theta }{\hbar }}{\hat {\mathbf {n} }}\cdot \mathbf {J} \right){\widehat {\Omega }}\exp \left(-{\frac {i\theta }{\hbar }}{\hat {\mathbf {n} }}\cdot \mathbf {J} \right).$$

Angular momentum eigenkets

The orthonormal basis set for total angular momentum is$|j,m\rangle$,wherejis the total angular momentum quantum number andmis the magnetic angular momentum quantum number, which takes values −j,−j+ 1,...,j− 1,j.A general state within thejsubspace

$$|\psi \rangle =\sum _{m}c_{jm}|j,m\rangle$$

rotates to a new state by:

$$|{\bar {\psi }}\rangle =U(R)|\psi \rangle =\sum _{m}c_{jm}U(R)|j,m\rangle$$

Using thecompleteness condition:

$$I=\sum _{m'}|j,m'\rangle \langle j,m'|$$

we have

$$|{\bar {\psi }}\rangle =IU(R)|\psi \rangle =\sum _{mm'}c_{jm}|j,m'\rangle \langle j,m'|U(R)|j,m\rangle$$

Introducing theWigner D matrixelements:

$${D(R)}_{m'm}^{(j)}=\langle j,m'|U(R)|j,m\rangle$$

gives the matrix multiplication:

$$|{\bar {\psi }}\rangle =\sum _{mm'}c_{jm}D_{m'm}^{(j)}|j,m'\rangle \quad \Rightarrow \quad |{\bar {\psi }}\rangle =D^{(j)}|\psi \rangle$$

For one basis ket:

$$|{\overline {j,m}}\rangle =\sum _{m'}{D(R)}_{m'm}^{(j)}|j,m'\rangle$$

For the case of orbital angular momentum, the eigenstates$|\ell,m\rangle$of the orbitalangular momentum operatorLand solutions ofLaplace's equationon a 3d sphere arespherical harmonics:

$$Y_{\ell }^{m}(\theta,\phi )=\langle \theta,\phi |\ell,m\rangle ={\sqrt {{(2\ell +1) \over 4\pi }{(\ell -m)! \over (\ell +m)!}}}\,P_{\ell }^{m}(\cos {\theta })\,e^{im\phi }$$

whereP_ℓ^mis anassociated Legendre polynomial,ℓ is the orbital angular momentum quantum number, andmis the orbital magneticquantum numberwhich takes the values −ℓ, −ℓ + 1,... ℓ − 1, ℓ The formalism of spherical harmonics have wide applications in applied mathematics, and are closely related to the formalism of spherical tensors, as shown below.

Spherical harmonics are functions of the polar and azimuthal angles,ϕandθrespectively, which can be conveniently collected into a unit vectorn(θ,ϕ) pointing in the direction of those angles, in the Cartesian basis it is:

$${\hat {\mathbf {n} }}(\theta,\phi )=\cos \phi \sin \theta \mathbf {e} _{x}+\sin \phi \sin \theta \mathbf {e} _{y}+\cos \theta \mathbf {e} _{z}$$

So a spherical harmonic can also be written$Y_{\ell }^{m}=\langle \mathbf {n} |\ell m\rangle$.Spherical harmonic states$|m,\ell \rangle$rotate according to the inverse rotation matrix$U(R^{-1})$,while$|\ell,m\rangle$rotates by the initial rotation matrix${\widehat {U}}(R)$.

$$|{\overline {\ell,m}}\rangle =\sum _{m'}D_{m'm}^{(\ell )}[U(R^{-1})]|\ell,m'\rangle \,,\quad |{\overline {\hat {\mathbf {n} }}}\rangle =U(R)|{\hat {\mathbf {n} }}\rangle$$

Rotation of tensor operators

We define the Rotation of an operator by requiring that the expectation value of the original operator${\widehat {\mathbf {A} }}$with respect to the initial state be equal to the expectation value of the rotated operator with respect to the rotated state,

$$\langle \psi '|{\widehat {A'}}|\psi '\rangle =\langle \psi |{\widehat {A}}|\psi \rangle$$

Now as,

$$|\psi \rangle ~\rightarrow ~|\psi '\rangle =U(R)|\psi \rangle \,,\quad \langle \psi |~\rightarrow ~\langle \psi '|=\langle \psi |U^{\dagger }(R)$$

we have,

$$\langle \psi |U^{\dagger }(R){\widehat {A}}'U(R)|\psi \rangle =\langle \psi |{\widehat {A}}|\psi \rangle$$

since,$|\psi \rangle$is arbitrary,

$$U^{\dagger }(R){\widehat {A}}'U(R)={\widehat {A}}$$

Scalar operators

A scalar operator is invariant under rotations:

$$U(R)^{\dagger }{\widehat {S}}U(R)={\widehat {S}}$$

This is equivalent to saying a scalar operator commutes with the rotation generators:

$$\left[{\widehat {S}},{\widehat {\mathbf {J} }}\right]=0$$

Examples of scalar operators include

• theenergy operator:$${\widehat {E}}\psi =i\hbar {\frac {\partial }{\partial t}}\psi$$
• potential energyV(in the case of a central potential only)$${\widehat {V}}(r,t)\psi (\mathbf {r},t)=V(r,t)\psi (\mathbf {r},t)$$
• kinetic energyT:$${\widehat {T}}\psi (\mathbf {r},t)=-{\frac {\hbar ^{2}}{2m}}(\nabla ^{2}\psi )(\mathbf {r},t)$$
• thespin–orbit coupling:$${\widehat {\mathbf {L} }}\cdot {\widehat {\mathbf {S} }}={\widehat {L}}_{x}{\widehat {S}}_{x}+{\widehat {L}}_{y}{\widehat {S}}_{y}+{\widehat {L}}_{z}{\widehat {S}}_{z}\,.$$

Vector operators

Vector operators (as well aspseudovectoroperators) are a set of 3 operators that can be rotated according to:

$${U(R)}^{\dagger }{\widehat {V}}_{i}U(R)=\sum _{j}R_{ij}{\widehat {V}}_{j}$$Any observable vector quantity of a quantum mechanical system should be invariant of the choice of frame of reference. The transformation of expectation value vector which applies for any wavefunction, ensures the above equality. In Dirac notation:$$\langle {\bar {\psi }}|{\widehat {V}}_{a}|{\bar {\psi }}\rangle =\langle \psi |{U(R)}^{\dagger }{\widehat {V}}_{a}U(R)|\psi \rangle =\sum _{b}R_{ab}\langle \psi |{\widehat {V}}_{b}|\psi \rangle$$where the RHS is due to the rotation transformation acting on the vector formed by expectation values. Since|Ψ⟩is any quantum state, the same result follows:$${U(R)}^{\dagger }{\widehat {V}}_{a}U(R)=\sum _{b}R_{ab}{\widehat {V}}_{b}$$Note that here, the term "vector" is used two different ways: kets such as|ψ⟩are elements of abstract Hilbert spaces, while the vector operator is defined as a quantity whose components transform in a certain way under rotations.

From the above relation for infinitesimal rotations and theBaker Hausdorff lemma,by equating coefficients of order$\delta \theta$,one can derive the commutation relation with the rotation generator:

$${\left[{\widehat {V}}_{a},{\widehat {J}}_{b}\right]=\sum _{c}i\hbar \varepsilon _{abc}{\widehat {V}}_{c}}$$whereε_ijkis theLevi-Civita symbol,which all vector operators must satisfy, by construction. The above commutator rule can also be used as an alternative definition for vector operators which can be shown by using theBaker Hausdorff lemma.As the symbolε_ijkis apseudotensor,pseudovector operators are invariantup toa sign: +1 forproper rotationsand −1 forimproper rotations.

Since operators can be shown to form a vector operator by their commutation relation with angular momentum components (which are generators of rotation), its examples include:

• theposition operator:$${\widehat {\mathbf {r} }}\psi =\mathbf {r} \psi$$
• themomentum operator:$${\widehat {\mathbf {p} }}\psi =-i\hbar \nabla \psi$$

and peusodovector operators include

• the orbitalangular momentum operator:$${\widehat {\mathbf {L} }}\psi =-i\hbar \mathbf {r} \times \nabla \psi$$
• as well thespin operatorS,and hence the total angular momentum$${\widehat {\mathbf {J} }}={\widehat {\mathbf {L} }}+{\widehat {\mathbf {S} }}\,.$$

Scalar operators from vector operators

If${\vec {V}}$and${\vec {W}}$are two vector operators, the dot product between the two vector operators can be defined as:

${\vec {V}}\cdot {\vec {W}}=\sum _{i=1}^{3}{\hat {V_{i}}}{\hat {W_{i}}}$

Under rotation of coordinates, the newly defined operator transforms as:$${U(R)}^{\dagger }({\vec {V}}\cdot {\vec {W}})U(R)={U(R)}^{\dagger }\left(\sum _{i=1}^{3}{\hat {V_{i}}}{\hat {W_{i}}}\right)U(R)=\sum _{i=1}^{3}({U(R)}^{\dagger }{\hat {V}}_{i}U(R))({U(R)}^{\dagger }{\hat {W}}_{i}U(R))=\sum _{i=1}^{3}\left(\sum _{j=1}^{3}R_{ij}{\widehat {V}}_{j}\cdot \sum _{k=1}^{3}R_{ik}{\widehat {W}}_{k}\right)$$Rearranging terms and using transpose of rotation matrix as its inverse property:$${U(R)}^{\dagger }({\vec {V}}\cdot {\vec {W}})U(R)=\sum _{k=1}^{3}\sum _{j=1}^{3}\left(\sum _{i=1}^{3}R_{ji}^{T}R_{ik}\right){\widehat {V}}_{j}{\widehat {W}}_{k}=\sum _{k=1}^{3}\sum _{j=1}^{3}\delta _{j,k}{\widehat {V}}_{j}{\widehat {W}}_{k}=\sum _{i=1}^{3}{\widehat {V}}_{i}{\widehat {W}}_{i}$$Where the RHS is the${\vec {V}}\cdot {\vec {W}}$operator originally defined. Since the dot product defined is invariant under rotation transformation, it is said to be a scalar operator.

Spherical vector operators

A vector operator in thespherical basisisV= (V₊₁,V₀,V₋₁)where the components are:$$V_{+1}=-{\frac {1}{\sqrt {2}}}(V_{x}+iV_{y})\,\quad V_{-1}={\frac {1}{\sqrt {2}}}(V_{x}-iV_{y})\,,\quad V_{0}=V_{z}\,,$$
using${\textstyle J_{\pm }=J_{x}\pm iJ_{y}\,,}$the various commutators with the rotation generators and ladder operators are:$${\begin{aligned}\left[J_{z},V_{+1}\right]&=+\hbar V_{+1}\\[1ex]\left[J_{z},V_{0}\right]&=0V_{0}\\[1ex]\left[J_{z},V_{-1}\right]&=-\hbar V_{-1}\\[2ex]\left[J_{+},V_{+1}\right]&=0\\[1ex]\left[J_{+},V_{0}\right]&={\sqrt {2}}\hbar V_{+1}\\[1ex]\left[J_{+},V_{-1}\right]&={\sqrt {2}}\hbar V_{0}\\[2ex]\left[J_{-},V_{+1}\right]&={\sqrt {2}}\hbar V_{0}\\[1ex]\left[J_{-},V_{0}\right]&={\sqrt {2}}\hbar V_{-1}\\[1ex]\left[J_{-},V_{-1}\right]&=0\\[1ex]\end{aligned}}$$

which are of similar form of$${\begin{aligned}J_{z}|1,+1\rangle &=+\hbar |1,+1\rangle \\[1ex]J_{z}|1,0\rangle &=0|1,0\rangle \\[1ex]J_{z}|1,-1\rangle &=-\hbar |1,-1\rangle \\[2ex]J_{+}|1,+1\rangle &=0\\[1ex]J_{+}|1,0\rangle &={\sqrt {2}}\hbar |1,+1\rangle \\[1ex]J_{+}|1,-1\rangle &={\sqrt {2}}\hbar |1,0\rangle \\[2ex]J_{-}|1,+1\rangle &={\sqrt {2}}\hbar |1,0\rangle \\[1ex]J_{-}|1,0\rangle &={\sqrt {2}}\hbar |1,-1\rangle \\[1ex]J_{-}|1,-1\rangle &=0\\[1ex]\end{aligned}}$$

In the spherical basis, the generators of rotation are:$$J_{\pm 1}=\mp {\frac {1}{\sqrt {2}}}J_{\pm }\,,\quad J_{0}=J_{z}$$

From the transformation of operators andBaker Hausdorff lemma:

${U(R)}^{\dagger }{\widehat {V}}_{q}U(R)={\widehat {V}}_{q}+i{\frac {\theta }{\hbar }}\left[{\hat {n}}\cdot {\vec {J}},{\widehat {V}}_{q}\right]+\sum _{k=2}^{\infty }{\frac {\left(i{\frac {\theta }{\hbar }}[{\hat {n}}\cdot {\vec {J}},.]\right)^{k}}{k!}}{\widehat {V}}_{q}=exp\left({i{\frac {\theta }{\hbar }}{\hat {n}}\cdot Ad_{\vec {J}}}\right){\widehat {V}}_{q}$

compared to

$U(R)|j,k\rangle =|j,k\rangle -i{\frac {\theta }{\hbar }}{\hat {n}}\cdot {\vec {J}}|j,k\rangle +\sum _{k=2}^{\infty }{\frac {\left(-i{\frac {\theta }{\hbar }}{\hat {n}}\cdot {\vec {J}}\right)^{k}}{k!}}|j,k\rangle =exp\left({-i{\frac {\theta }{\hbar }}{\hat {n}}\cdot {\vec {J}}}\right)|j,k\rangle$

it can be argued that the commutator with operator replaces the action of operator on state for transformations of operators as compared with that of states:

$U(R)|j,k\rangle =\exp \left({-i{\frac {\theta }{\hbar }}{\hat {n}}\cdot {\vec {J}}}\right)|j,k\rangle =\sum _{j',k'}|j',k'\rangle \langle j',k'|\exp \left({-i{\frac {\theta }{\hbar }}{\hat {n}}\cdot {\vec {J}}}\right)|j,k\rangle =\sum _{k'}D_{k'k}^{(j)}(R)|j,k'\rangle$

The rotation transformation in the spherical basis (originally written in the Cartesian basis) is then, due to similarity of commutation and operator shown above:$${U(R)}^{\dagger }{\widehat {V}}_{q}U(R)=\sum _{q'}{{D_{q'q}^{(1)}}(R^{-1})}{\widehat {V}}_{q'}$$

One can generalize thevectoroperator concept easily totensorial operators,shown next.

Tensor operators

In general, a tensor operator is one that transforms according to a tensor:$$U(R)^{\dagger }{\widehat {T}}_{pqr\cdots }^{abc\cdots }U(R)=R_{p,\ Alpha }R_{q,\beta }R_{r,\gamma }\cdots {\widehat {T}}_{ijk\cdots }^{\ Alpha \beta \gamma \cdots }R_{i,a}^{-1}R_{j,b}^{-1}R_{k,c}^{-1}\cdots$$where the basis are transformed by$R^{-1}$or the vector components transform by$R$.

In the subsequent discussion surrounding tensor operators, the index notation regarding covariant/contravariant behavior is ignored entirely. Instead, contravariant components is implied by context. Hence for an n times contravariant tensor:

$$U(R)^{\dagger }{\widehat {T}}_{pqr\cdots }U(R)=R_{pi}R_{qj}R_{rk}\cdots {\widehat {T}}_{ijk\cdots }$$

Examples of tensor operators

• TheQuadrupolemoment operator,$$Q_{ij}=\sum _{\ Alpha }q_{\ Alpha }(3r_{\ Alpha i}r_{\ Alpha j}-r_{\ Alpha }^{2}\delta _{ij})$$
• Components of two tensor vector operators can be multiplied to give another Tensor operator.$$T_{ij}=V_{i}W_{j}$$In general, n number of tensor operators will also give another tensor operator$$T_{pqr\cdots k}=V_{p}^{(1)}V_{q}^{(2)}V_{r}^{3)}\cdots V_{k}^{(n)}$$or,$$T_{i_{1}i_{2}\cdots j_{1}j_{2}\cdots }=V_{i_{1}i_{2}\cdots }W_{j_{1}j_{2}\cdots }$$

Note:In general, a tensor operator cannot be written as the tensor product of other tensor operators as given in the above example.

Tensor operator from vector operators

If${\vec {V}}$and${\vec {W}}$are two three dimensional vector operators, then a rank 2 Cartesian dyadic tensors can be formed from nine operators of form${\hat {T}}_{ij}={\hat {V_{i}}}{\hat {W_{j}}}$,$${U(R)}^{\dagger }{\hat {T}}_{ij}U(R)={U(R)}^{\dagger }({\hat {V_{i}}}{\hat {W_{j}}})U(R)=({U(R)}^{\dagger }{\hat {V}}_{i}U(R))({U(R)}^{\dagger }{\hat {W}}_{j}U(R))=\left(\sum _{l=1}^{3}R_{il}{\hat {V}}_{l}\cdot \sum _{k=1}^{3}R_{jk}{\hat {W}}_{k}\right)$$Rearranging terms, we get:$${U(R)}^{\dagger }{\hat {T}}_{ij}U(R)=\sum _{k=1}^{3}\sum _{l=1}^{3}\left(R_{il}R_{jk}{\hat {T}}_{lk}\right)$$The RHS of the equation is change of basis equation for twice contravariant tensors where the basis are transformed by$R^{-1}$or the vector components transform by$R$which matches transformation of vector operator components. Hence the operator tensor described forms a rank 2 tensor, in tensor representation,$${\hat {\mathbf {T} }}={\vec {V}}\otimes {\vec {W}}=({\hat {V}}_{i}{\hat {W}}_{j})(\mathbf {e} _{i}\otimes \mathbf {e} _{j})$$Similarly, an n-times contravariant tensor operator can be formed similarly by n vector operators.

We observe that the subspace spanned by linear combinations of the rank two tensor components form an invariant subspace, ie. the subspace does not change under rotation since the transformed components itself is a linear combination of the tensor components. However, this subspace is not irreducible ie. it can be further divided into invariant subspaces under rotation. Otherwise, the subspace is called reducible. In other words, there exists specific sets of different linear combinations of the components such that they transforms into a linear combination of the same set under rotation.In the above example, we will show that the 9 independent tensor components can be divided into a set of 1, 3 and 5 combination of operators that each form irreducible invariant subspaces.

Irreducible tensor operators

The subspace spanned by$\{{\hat {T}}_{ij}\}$can be divided two subspaces; three independent antisymmetric components$\{{\hat {A}}_{ij}\}$and six independent symmetric component$\{{\hat {S}}_{ij}\}$,defined as${\hat {A}}_{ij}={\frac {1}{2}}({\hat {T}}_{ij}-{\hat {T}}_{ji})$and${\hat {S}}_{ij}={\frac {1}{2}}({\hat {T}}_{ij}+{\hat {T}}_{ji})$.Using the$\{{\hat {T}}_{ij}\}$transformation under rotation formula, it can be shown that both$\{{\hat {A}}_{ij}\}$and$\{{\hat {S}}_{ij}\}$are transformed into a linear combination of members of its own sets. Although$\{{\hat {A}}_{ij}\}$is irreducible, the same cannot be said about$\{{\hat {S}}_{ij}\}$.

The six independent symmetric component set can be divided into five independent traceless symmetric component and the invariant trace can be its own subspace.

Hence, the invariant subspaces of$\{{\hat {T}}_{ij}\}$are formed respectively by:

1. One invariant trace of the tensor,${\hat {t}}=\sum _{k=1}^{3}{\hat {T}}_{kk}$
2. Three linearly independent antisymmetric components from:${\hat {A}}_{ij}={\frac {1}{2}}({\hat {T}}_{ij}-{\hat {T}}_{ji})$
3. Five linearly independent traceless symmetric components from${\hat {S}}_{ij}={\frac {1}{2}}({\hat {T}}_{ij}+{\hat {T}}_{ji})-{\frac {1}{3}}{\hat {t}}\delta _{ij}$

If${\hat {T}}_{ij}={\hat {V_{i}}}{\hat {W_{j}}}$,the invariant subspaces of$\{{\hat {T}}_{ij}\}$formed are represented by:

1. One invariant scalar operator${\vec {V}}\cdot {\vec {W}}$
2. Three linearly independent components from${\frac {1}{2}}({\hat {V}}_{i}{\hat {W}}_{j}-{\hat {V}}_{j}{\hat {W}}_{i})$
3. Five linearly independent components from${\frac {1}{2}}({\hat {V}}_{i}{\hat {W}}_{j}+{\hat {V}}_{j}{\hat {W}}_{i})-{\frac {1}{3}}({\vec {V}}\cdot {\vec {W}})\delta _{ij}$

From the above examples, the nine component$\{{\hat {T}}_{ij}\}$are split into subspaces formed by one, three and five components. These numbers add up to the number of components of the original tensor in a manner similar to the dimension of vector subspaces adding to the dimension of the space that is a direct sum of these subspaces. Similarly, every element of$\{{\hat {T}}_{ij}\}$can be expressed in terms of a linear combination of components from its invariant subspaces:

${\hat {T}}_{ij}={\frac {1}{3}}{\hat {t}}\delta _{ij}+{\hat {A}}_{ij}+{\hat {S}}_{ij}$

or

${\hat {T}}_{ij}={\frac {1}{3}}({\vec {V}}\cdot {\vec {W}})\delta _{ij}+\left({\frac {1}{2}}({\hat {V}}_{i}{\hat {W}}_{j}-{\hat {V}}_{j}{\hat {W}}_{i})\right)+\left({\frac {1}{2}}({\hat {V}}_{i}{\hat {W}}_{j}+{\hat {V}}_{j}{\hat {W}}_{i})-{\frac {1}{3}}({\vec {V}}\cdot {\vec {W}})\delta _{ij}\right)=\mathbf {T} ^{(0)}+\mathbf {T} ^{(1)}+\mathbf {T} ^{(2)}$

where:$${\widehat {T}}_{ij}^{(0)}={\frac {{\widehat {V}}_{k}{\widehat {W}}_{k}}{3}}\delta _{ij}$$$${\widehat {T}}_{ij}^{(1)}={\frac {1}{2}}\left[{\widehat {V}}_{i}{\widehat {W}}_{j}-{\widehat {V}}_{j}{\widehat {W}}_{i}\right]={\widehat {V}}_{[i}{\widehat {W}}_{j]}$$$${\widehat {T}}_{ij}^{(2)}={\tfrac {1}{2}}\left({\widehat {V}}_{i}{\widehat {W}}_{j}+{\widehat {V}}_{j}{\widehat {W}}_{i}\right)-{\tfrac {1}{3}}{\widehat {V}}_{k}{\widehat {W}}_{k}\delta _{ij}={\widehat {V}}_{(i}{\widehat {W}}_{j)}-T_{ij}^{(0)}$$

In general cartesian tensors of rank greater than 1 are reducible. In quantum mechanics, this particular example bears resemblance to the addition of two spin one particles where both are 3 dimensional, hence the total space being 9 dimensional, can be formed by spin 0, spin 1 and spin 2 systems each having 1 dimensional, 3 dimensional and 5 dimensional space respectively.These three terms are irreducible, which means they cannot be decomposed further and still be tensors satisfying the defining transformation laws under which they must be invariant. Each of the irreducible representationsT⁽⁰⁾,T⁽¹⁾,T⁽²⁾... transform like angular momentum eigenstates according to the number of independent components.

It is possible that a given tensor may have one or more of these components vanish. For example, the quadrupole moment tensor is already symmetric and traceless, and hence has only 5 independent components to begin with.

Spherical tensor operators

See also:Wigner–Eckart theorem

Spherical tensor operators are generally defined as operators with the following transformation rule, under rotation of coordinate system:

${\widehat {T}}_{m}^{(j)}\rightarrow U(R)^{\dagger }{\widehat {T}}_{m}^{(j)}U(R)=\sum _{m'}D_{m'm}^{(j)}(R^{-1}){\widehat {T}}_{m'}^{(j)}$

The commutation relations can be found by expanding LHS and RHS as:

$U(R)^{\dagger }{\widehat {T}}_{m}^{(j)}U(R)=\left(1+{\frac {i\epsilon {\hat {n}}\cdot {\vec {J}}}{\hbar }}+{\mathcal {O}}(\epsilon ^{2})\right){\widehat {T}}_{m}^{(j)}\left(1-{\frac {i\epsilon {\hat {n}}\cdot {\vec {J}}}{\hbar }}+{\mathcal {O}}(\epsilon ^{2})\right)=\sum _{m'}\langle j,m'|\left(1+{\frac {i\epsilon {\hat {n}}\cdot {\vec {J}}}{\hbar }}+{\mathcal {O}}(\epsilon ^{2})\right)|j,m\rangle {\widehat {T}}_{m'}^{(j)}$

Simplifying and applying limits to select only first order terms, we get:

${[{\hat {n}}\cdot {\vec {J}}},{\widehat {T}}_{m}^{(j)}]=\sum _{m'}{\widehat {T}}_{m'}^{(j)}\langle j,m'|{\vec {J}}\cdot {\hat {n}}|j,m\rangle$

For choices of${\hat {n}}={\hat {x}}\pm i{\hat {y}}$or${\hat {n}}={\hat {z}}$,we get:$${\begin{aligned}\left[J_{\pm },{\widehat {T}}_{m}^{(j)}\right]&=\hbar {\sqrt {(j\mp m)(j\pm m+1)}}{\widehat {T}}_{m\pm 1}^{(j)}\\[1ex]\left[J_{z},{\widehat {T}}_{m}^{(j)}\right]&=\hbar m{\widehat {T}}_{m}^{(j)}\end{aligned}}$$Note the similarity of the above to:$${\begin{aligned}J_{\pm }|j,m\rangle &=\hbar {\sqrt {(j\mp m)(j\pm m+1)}}|j,m\pm 1\rangle \\[1ex]J_{z}|j,m\rangle &=\hbar m|j,m\rangle \end{aligned}}$$Since$J_{x}$and$J_{y}$are linear combinations of$J_{\pm }$,they share the same similarity due to linearity.

If, only the commutation relations hold, using the following relation,$|j,m\rangle \rightarrow U(R)|j,m\rangle =exp\left(-{i{\frac {\theta }{\hbar }}{\hat {n}}\cdot {\vec {J}}}\right)|j,m\rangle =\sum _{m'}D_{m'm}^{(j)}(R)|j,m'\rangle$

we find due to similarity of actions of$J$on wavefunction$|j,m\rangle$and the commutation relations on${\widehat {T}}_{m}^{(j)}$,that:

${\widehat {T}}_{m}^{(j)}\rightarrow U(R)^{\dagger }{\widehat {T}}_{m}^{(j)}U(R)=exp\left({i{\frac {\theta }{\hbar }}{\hat {n}}\cdot ad_{\vec {J}}}\right){\widehat {T}}_{m}^{(j)}=\sum _{m'}D_{m'm}^{(j)}(R^{-1}){\widehat {T}}_{m'}^{(j)}$

where the exponential form is given byBaker–Hausdorff lemma.Hence, the above commutation relations and the transformation property are equivalent definitions of spherical tensor operators. It can also be shown that$\{ad_{{\hat {J}}_{i}}\}$transform like a vector due to their commutation relation.

In the following section, construction of spherical tensors will be discussed. For example, since example of spherical vector operators is shown, it can be used to construct higher order spherical tensor operators. In general, spherical tensor operators can be constructed from two perspectives.One way is to specify how spherical tensors transform under a physical rotation - agroup theoreticaldefinition. A rotated angular momentum eigenstate can be decomposed into a linear combination of the initial eigenstates: the coefficients in the linear combination consist of Wigner rotation matrix entries. Or by continuing the previous example of the second order dyadic tensorT=a⊗b,casting each ofaandbinto the spherical basis and substituting intoTgives the spherical tensor operators of the second order.

Construction using Clebsch–Gordan coefficients

Combination of two spherical tensors$A_{q_{1}}^{(k_{1})}$and$B_{q_{2}}^{(k_{2})}$in the following manner involving theClebsch–Gordan coefficientscan be proved to give another spherical tensor of the form:$$T_{q}^{(k)}=\sum _{q_{1},q_{2}}\langle k_{1},k_{2};q_{1},q_{2}|k_{1},k_{2};k,q\rangle A_{q_{1}}^{(k_{1})}B_{q_{2}}^{(k_{2})}$$

This equation can be used to construct higher order spherical tensor operators, for example, second order spherical tensor operators using two first order spherical tensor operators, sayAandB,discussed previously:

$${\begin{aligned}{\widehat {T}}_{\pm 2}^{(2)}&={\widehat {a}}_{\pm 1}{\widehat {b}}_{\pm 1}\\[1ex]{\widehat {T}}_{\pm 1}^{(2)}&={\tfrac {1}{\sqrt {2}}}\left({\widehat {a}}_{\pm 1}{\widehat {b}}_{0}+{\widehat {a}}_{0}{\widehat {b}}_{\pm 1}\right)\\[1ex]{\widehat {T}}_{0}^{(2)}&={\tfrac {1}{\sqrt {6}}}\left({\widehat {a}}_{+1}{\widehat {b}}_{-1}+{\widehat {a}}_{-1}{\widehat {b}}_{+1}+2{\widehat {a}}_{0}{\widehat {b}}_{0}\right)\end{aligned}}$$

Using the infinitesimal rotation operator and its Hermitian conjugate, one can derive the commutation relation in the spherical basis:$$\left[J_{a},{\widehat {T}}_{q}^{(2)}\right]=\sum _{q'}{D(J_{a})}_{qq'}^{(2)}{\widehat {T}}_{q'}^{(2)}=\sum _{q'}\langle j{=}2,m{=}q|J_{a}|j{=}2,m{=}q'\rangle {\widehat {T}}_{q'}^{(2)}$$and the finite rotation transformation in the spherical basis can be verified:$${U(R)}^{\dagger }{\widehat {T}}_{q}^{(2)}U(R)=\sum _{q'}{{D(R)}_{qq'}^{(2)}}^{*}{\widehat {T}}_{q'}^{(2)}$$

Using Spherical Harmonics

Define an operator by its spectrum:$$\Upsilon _{l}^{m}|r\rangle =r^{l}Y_{l}^{m}(\theta,\phi )|r\rangle =\Upsilon _{l}^{m}({\vec {r}})|r\rangle$$Since for spherical harmonics under rotation:$$Y_{\ell =k}^{m=q}(\mathbf {n} )=\langle \mathbf {n} |k,q\rangle \rightarrow U(R)^{\dagger }Y_{\ell =k}^{m=q}(\mathbf {n} )U(R)=Y_{\ell =k}^{m=q}(R\mathbf {n} )=\langle \mathbf {n} |D(R)^{\dagger }|k,q\rangle =\sum _{q'}D_{q',q}^{(k)}(R^{-1})Y_{\ell =k}^{m=q'}(\mathbf {n} )$$It can also been shown that:$$\Upsilon _{l}^{m}({\vec {r}})\rightarrow U(R)^{\dagger }\Upsilon _{l}^{m}({\vec {r}})U(R)=\sum _{m'}D_{m',m}^{(l)}(R^{-1})\Upsilon _{l}^{m'}({\vec {r}})$$Then$\Upsilon _{l}^{m}({\vec {V}})$,where${\vec {V}}$is a vector operator, also transforms in the same manner ie, is a spherical tensor operator. The process involves expressing$\Upsilon _{l}^{m}({\vec {r}})=r^{l}Y_{l}^{m}(\theta,\phi )=\Upsilon _{l}^{m}(x,y,z)$in terms of x, y and z and replacing x, y and z with operators V_xV_yand V_zwhich from vector operator. The resultant operator is hence a spherical tensor operator${\hat {T}}_{m}^{(l)}$.^This may include constant due to normalization from spherical harmonics which is meaningless in context of operators.

TheHermitian adjointof a spherical tensor may be defined as$$(T^{\dagger })_{q}^{(k)}=(-1)^{k-q}(T_{-q}^{(k)})^{\dagger }.$$There is some arbitrariness in the choice of the phase factor: any factor containing(−1)^±qwill satisfy the commutation relations.The above choice of phase has the advantages of being real and that the tensor product of two commutingHermitianoperators is still Hermitian.Some authors define it with a different sign onq,without thek,or use only thefloorofk.

Angular momentum and spherical harmonics

Orbital angular momentum and spherical harmonics

Orbital angular momentum operators have theladder operators:

$$L_{\pm }=L_{x}\pm iL_{y}$$

which raise or lower the orbital magnetic quantum numberm_ℓby one unit. This has almost exactly the same form as the spherical basis, aside from constant multiplicative factors.

Spherical tensor operators and quantum spin

See also:Spin-weighted spherical harmonicsandspinor spherical harmonics

Spherical tensors can also be formed from algebraic combinations of the spin operatorsS_x,S_y,S_z,as matrices, for a spin system with total quantum numberj= ℓ +s(and ℓ = 0). Spin operators have the ladder operators:

$$S_{\pm }=S_{x}\pm iS_{y}$$

which raise or lower the spin magnetic quantum numberm_sby one unit.

Applications

Spherical bases have broad applications in pure and applied mathematics and physical sciences where spherical geometries occur.

Dipole radiative transitions in a single-electron atom (alkali)

The transition amplitude is proportional to matrix elements of the dipole operator between the initial and final states. We use an electrostatic, spinless model for the atom and we consider the transition from the initial energy level E_nℓto final level E_n′ℓ′.These levels are degenerate, since the energy does not depend on the magnetic quantum number m or m′. The wave functions have the form,

$$\psi _{n\ell m}(r,\theta,\phi )=R_{n\ell }(r)Y_{\ell m}(\theta,\phi )$$

The dipole operator is proportional to the position operator of the electron, so we must evaluate matrix elements of the form,

$$\langle n'\ell 'm'|\mathbf {r} |n\ell m\rangle$$

where, the initial state is on the right and the final one on the left. The position operatorrhas three components, and the initial and final levels consist of 2ℓ + 1 and 2ℓ′ + 1 degenerate states, respectively. Therefore if we wish to evaluate the intensity of a spectral line as it would be observed, we really have to evaluate 3(2ℓ′+ 1)(2ℓ+ 1) matrix elements, for example, 3×3×5 = 45 in a 3d → 2p transition. This is actually an exaggeration, as we shall see, because many of the matrix elements vanish, but there are still many non-vanishing matrix elements to be calculated.

A great simplification can be achieved by expressing the components of r, not with respect to the Cartesian basis, but with respect to the spherical basis. First we define,

$$r_{q}={\hat {\mathbf {e} }}_{q}\cdot \mathbf {r}$$

Next, by inspecting a table of theY_ℓm′s, we find that for ℓ = 1 we have,

$${\begin{aligned}rY_{11}(\theta,\phi )&=&&-r{\sqrt {\frac {3}{8\pi }}}\sin(\theta )e^{i\phi }&=&{\sqrt {\frac {3}{4\pi }}}\left(-{\frac {x+iy}{\sqrt {2}}}\right)\\rY_{10}(\theta,\phi )&=&&r{\sqrt {\frac {3}{4\pi }}}\cos(\theta )&=&{\sqrt {\frac {3}{4\pi }}}z\\rY_{1-1}(\theta,\phi )&=&&r{\sqrt {\frac {3}{8\pi }}}\sin(\theta )e^{-i\phi }&=&{\sqrt {\frac {3}{4\pi }}}\left({\frac {x-iy}{\sqrt {2}}}\right)\end{aligned}}$$

where, we have multiplied eachY_1mby the radiusr.On the right hand side we see the spherical componentsr_qof the position vectorr.The results can be summarized by,

$$rY_{1q}(\theta,\phi )={\sqrt {\frac {3}{4\pi }}}r_{q}$$

for q = 1, 0, −1, where q appears explicitly as a magnetic quantum number. This equation reveals a relationship between vector operators and the angular momentum value ℓ = 1, something we will have more to say about presently. Now the matrix elements become a product of a radial integral times an angular integral, $$\langle n'\ell 'm'|r_{q}|n\ell m\rangle =\left(\int _{0}^{\infty }r^{2}drR_{n'\ell '}^{*}(r)rR_{n\ell }(r)\right)\left({\sqrt {\frac {4\pi }{3}}}\int \sin {(\theta )}d\Omega Y_{\ell 'm'}^{*}(\theta,\phi )Y_{1q}(\theta,\phi )Y_{\ell m}(\theta,\phi )\right)$$

We see that all the dependence on the three magnetic quantum numbers (m′,q,m) is contained in the angular part of the integral. Moreover, the angular integral can be evaluated by the three-Y_ℓmformula, whereupon it becomes proportional to the Clebsch-Gordan coefficient,

$$\langle \ell 'm'|\ell 1mq\rangle$$

The radial integral is independent of the three magnetic quantum numbers (m′,q,m), and the trick we have just used does not help us to evaluate it. But it is only one integral, and after it has been done, all the other integrals can be evaluated just by computing or looking up Clebsch–Gordan coefficients.

The selection rulem′ =q+min the Clebsch–Gordan coefficient means that many of the integrals vanish, so we have exaggerated the total number of integrals that need to be done. But had we worked with the Cartesian componentsr_iofr,this selection rule might not have been obvious. In any case, even with the selection rule, there may still be many nonzero integrals to be done (nine, in the case 3d → 2p). The example we have just given of simplifying the calculation of matrix elements for a dipole transition is really an application of the Wigner–Eckart theorem, which we take up later in these notes.

Magnetic resonance

The spherical tensor formalism provides a common platform for treating coherence and relaxation innuclear magnetic resonance.InNMRandEPR,spherical tensor operators are employed to express the quantum dynamics ofparticle spin,by means of an equation of motion for thedensity matrixentries, or to formulate dynamics in terms of an equation of motion inLiouville space.The Liouville space equation of motion governs the observable averages of spin variables. When relaxation is formulated using a spherical tensor basis in Liouville space, insight is gained because the relaxation matrix exhibits the cross-relaxation of spin observables directly.

See also

• Wigner–Eckart theorem
• Structure tensor
• Clebsch–Gordan coefficients for SU(3)

References

Sources

• P. T. Callaghan (2011).Translational Dynamics and Magnetic Resonance:Principles of Pulsed Gradient Spin Echo NMR.Oxford University Press.ISBN978-0-191-621-048.
• V. V. Balashov; A. N. Grum-Grzhimailo; N.M. Kabachnik (2000).Polarization and Correlation Phenomena in Atomic Collisions: A Practical Theory Course.Springer.ISBN9780306462665.
• J. A. Tuszynski (1990).Spherical Tensor Operators: Tables of Matrix Elements and Symmetries.World Scientific.ISBN978-981-0202-835.
• L. Castellani; J. Wess (1996).Quantum Groups and Their Applications in Physics: Varenna on Lake Como, Villa Monastero, 28 June-8 July 1994.Società Italiana di Fisica, IOS.ISBN978-905-199-24-72.
• Introduction to the Graphical Theory of Angular Momentum.Springer. 2009.ISBN978-364-203-11-99.
• A. R. Edmonds (1996).Angular Momentum in Quantum Mechanics(2nd ed.). Princeton University Press.ISBN978-0-691-025-896.
• L.J. Mueller (2011). "Tensors and rotations in NMR".Concepts in Magnetic Resonance Part A.38A(5): 221–235.doi:10.1002/cmr.a.20224.S2CID8889942.
• M.S. Anwar (2004)."Spherical Tensor Operators in NMR"(PDF).
• P. Callaghan (1993).Principles of nuclear magnetic resonance microscopy.Oxford University Press. pp. 56–57.ISBN978-0-198-539-971.

Further reading

Spherical harmonics

• G.W.F. Drake (2006).Springer Handbook of Atomic, Molecular, and Optical Physics(2nd ed.). Springer. p. 57.ISBN978-0-3872-6308-3.
• F.A. Dahlen; J. Tromp (1998).Theoretical global seismology(2nd ed.). Princeton University Press. p. appendix C.ISBN978-0-69100-1241.
• D.O. Thompson; D.E. Chimenti (1997).Review of Progress in Quantitative Nondestructive Evaluation.Vol. 16. Springer. p. 1708.ISBN978-0-3064-55971.
• H. Paetz; G. Schieck (2011).Nuclear Physics with Polarized Particles.Lecture Notes in Physics. Vol. 842. Springer. p. 31.ISBN978-364-224-225-0.
• V. Devanathan (1999).Angular Momentum Techniques in Quantum Mechanics.Fundamental Theories of Physics. Vol. 108. Springer. pp. 34, 61.ISBN978-0-7923-5866-4.
• V.D. Kleiman; R.N. Zare (1998)."5".A Companion to Angular Momentum.John Wiley & Sons. p. 112.ISBN978-0-4711-9249-7.

Angular momentum and spin

• Devanathan, V (2002). "Vectors and Tensors in Spherical Basis".Angular Momentum Techniques in Quantum Mechanics.Fundamental Theories of Physics. Vol. 108. pp. 24–33.doi:10.1007/0-306-47123-X_3.ISBN978-0-306-47123-0.
• K.T. Hecht (2000).Quantum mechanics.Graduate texts in contemporary physics. Springer.ISBN978-0-387-989-198.

Condensed matter physics

• J.A. Mettes; J.B. Keith; R.B. McClurg (2002)."Molecular Crystal Global Phase Diagrams:I Method of Construction"(PDF).
• B.Henderson, R.H. Bartram (2005).Crystal-Field Engineering of Solid-State Laser Materials.Cambridge Studies in Modern Optics. Vol. 25. Cambridge University Press. p. 49.ISBN978-0-52101-8012.
• Edward U. Condon,andHalis Odabaşı(1980).Atomic Structure.CUP Archive.ISBN978-0-5212-98933.
• Melinda J. Duer, ed. (2008)."3".Solid State NMR Spectroscopy: Principles and Applications.John Wiley & Sons. p. 113.ISBN978-0-4709-9938-7.
• K.D. Bonin; V.V. Kresin (1997)."2".Electric - Dipole Polarizabilities of Atoms, Molecules and Clusters.World Scientific. pp. 14–15.ISBN978-981-022-493-6.
• A.E. McDermott, T.Polenova (2012).Solid State NMR Studies of Biopolymers.EMR handbooks. John Wiley & Sons. p. 42.ISBN978-111-858-889-5.

Magnetic resonance

• L.J. Mueller (2011). "Tensors and rotations in NMR".Concepts in Magnetic Resonance Part A.38A(5): 221–235.doi:10.1002/cmr.a.20224.S2CID8889942.
• M.S. Anwar (2004)."Spherical Tensor Operators in NMR"(PDF).
• P. Callaghan (1993).Principles of nuclear magnetic resonance microscopy.Oxford University Press. pp. 56–57.ISBN978-0-198-539-971.

Image processing

• M. Reisert; H. Burkhardt (2009). S. Aja-Fernández (ed.).Tensors in Image Processing and Computer Vision.Springer.ISBN978-184-8822-993.
• D.H. Laidlaw; J. Weickert (2009).Visualization and Processing of Tensor Fields: Advances and Perspectives.Mathematics and Visualization. Springer.ISBN978-354-088-378-4.
• M. Felsberg; E. Jonsson (2005).Energy Tensors: Quadratic, Phase Invariant Image Operators.Lecture Notes in Computer Science. Vol. 3663. Springer. pp. 493–500.
• E. König; S. Kremer (1979). "Tensor Operator Algebra for Point Groups".Magnetism Diagrams for Transition Metal Ions.Lecture Notes in Computer Science. Vol. 3663. Springer. pp. 13–20.doi:10.1007/978-1-4613-3003-5_3.ISBN978-1-4613-3005-9.

External links

• (2012)Clebsch-Gordon (sic) coefficients and the tensor spherical harmonics
• The tensor spherical harmonics
• (2010)Irreducible Tensor Operators and the Wigner-Eckart TheoremArchived2014-07-20 at theWayback Machine
• Tensor operators
• M. Fowler (2008),Tensor Operators
• Tensor_Operators
• (2009)Tensor Operators and the Wigner Eckart Theorem
• The Wigner-Eckart theorem
• (2004)Rotational Transformations and Spherical Tensor Operators
• Tensor operators
• Evaluation of the matrix elements for radiative transitions
• D.K. Ghosh, (2013)Angular Momentum - III: Wigner- Eckart Theorem
• B. Baragiola (2002)Tensor Operators
• Spherical Tensors