Irreducible Spherical Tensors

An irreducible spherical tensor operator is a collection of operator components

$$T_q^{(k)}, \qquad q=-k,-k+1,\ldots,k,$$

that transform under rotations like a single spin-$k$ angular-momentum multiplet. The label $k$ is the rank. The label $q$ is the spherical component, analogous to a magnetic quantum number.

The broader Cartesian classification into scalars, vectors, and tensor operators is introduced in Scalar, Vector, and Tensor Operators. The commutator tests are collected in Commutators with Angular Momentum. This page starts after that classification and focuses on the irreducible spherical form needed for angular-momentum algebra.

The word irreducible means that the components do not contain several rotational types mixed together. A scalar is rank $0$, a vector is rank $1$ after changing from Cartesian to spherical components, and a quadrupole is a typical rank-$2$ tensor. These objects are the operator input for the Wigner–Eckart theorem and for angular-momentum selection rules.

Why Spherical Components

Cartesian components are often natural geometrically. A vector operator is written as

$$\mathbf V=(V_x,V_y,V_z).$$

But under a rotation about the $z$ axis, $V_x$ and $V_y$ mix. The combinations

$$V_0=V_z, \qquad V_{\pm1} = \mp\frac{1}{\sqrt2} \left( V_x\pm iV_y \right)$$

are better adapted to angular momentum because each component has a definite $q$ label. A $z$ rotation multiplies $V_q$ by a phase rather than mixing it with the other components.

This is the same simplification that motivates angular-momentum eigenstates $\lvert j,m\rangle$ instead of Cartesian basis vectors. Spherical tensor components are the operator analogue of angular-momentum basis states.

Rank and Component

In this chapter, the rank $k$ is a nonnegative integer:

$$k=0,1,2,\ldots.$$

For a fixed rank, there are $2k+1$ components:

$$q=-k,\ldots,k.$$

The first few cases are:

Rank	Components	Common operator type
$k=0$	$q=0$	rotational scalar
$k=1$	$q=-1,0,1$	vector operator
$k=2$	$q=-2,-1,0,1,2$	quadrupole tensor

The rank is not the matrix rank of an operator, and it is not merely the number of Cartesian indices. It is the angular momentum carried by the operator under rotations.

Finite Rotation Law

Let

$$U(R)$$

be the unitary operator implementing a rotation $R$ on the Hilbert space. With the active convention used here, an irreducible spherical tensor transforms as

$$U(R)\,T_q^{(k)}\,U(R)^\dagger = \sum_{q'=-k}^{k} D_{q'q}^{(k)}(R)\, T_{q'}^{(k)}.$$

Here $D^{(k)}(R)$ is the Wigner $D$ matrix in the spin-$k$ representation. For a rotation about the $z$ axis,

$$U_z(\phi) = \exp\!\left( -\frac{i\phi J_z}{\hbar} \right),$$

this reduces to

$$U_z(\phi)\,T_q^{(k)}\,U_z(\phi)^\dagger = e^{-iq\phi} T_q^{(k)}.$$

The inverse conjugation, used for expectation values in a rotated state, is

$$U(R)^\dagger T_q^{(k)}U(R) =\sum_{q'}D_{q'q}^{(k)}(R^{-1})T_{q'}^{(k)} =\sum_{q'}D_{qq'}^{(k)}(R)^*T_{q'}^{(k)}.$$

Both the conjugation and the exchange of the two indices matter. For a $z$ rotation this gives $e^{+iq\phi}T_q^{(k)}$, consistent with converting the Cartesian law $U^\dagger V_iU=R_{ij}V_j$ into spherical components.

Infinitesimal Rotation Law

The same transformation property can be encoded in commutators with angular momentum. The defining relations are

$$[J_z,T_q^{(k)}] = \hbar q\,T_q^{(k)},$$

and

$$[J_\pm,T_q^{(k)}] = \hbar \sqrt{k(k+1)-q(q\pm1)} \,T_{q\pm1}^{(k)}.$$

The second relation is understood to give zero when $q\pm1$ lies outside the range $-k,\ldots,k$. For example, the highest component satisfies

$$[J_+,T_k^{(k)}]=0,$$

and the lowest component satisfies

$$[J_-,T_{-k}^{(k)}]=0.$$

These equations say that the operator components themselves form an angular-momentum multiplet under the adjoint action of rotations.

Vector Operators

For a vector operator $\mathbf V$, the Cartesian commutator is

$$[J_i,V_j] = i\hbar \sum_{\ell} \epsilon_{ij\ell}V_\ell.$$

After changing to spherical components,

$$V_0=V_z, \qquad V_{+1} = -\frac{1}{\sqrt2}(V_x+iV_y), \qquad V_{-1} = \frac{1}{\sqrt2}(V_x-iV_y),$$

the rank-$1$ commutators become

$$[J_z,V_q] = \hbar q\,V_q, \qquad q=-1,0,1.$$

The ladder relations are

$$[J_+,V_{-1}] = \hbar\sqrt2\,V_0, \qquad [J_+,V_0] = \hbar\sqrt2\,V_{+1}, \qquad [J_+,V_{+1}] = 0,$$

with the corresponding lowering relations obtained by replacing $J_+$ with $J_-$.

This convention is the one used for electric-dipole components on Dipole Transitions.

Scalars and Hermitian Components

A rank-$0$ tensor has a single component $T_0^{(0)}$ and satisfies

$$[J_i,T_0^{(0)}]=0$$

for all $i=x,y,z$. Rotationally invariant Hamiltonians, central potentials, and scalar products such as $\mathbf A\cdot\mathbf B$ are scalar examples when the relevant vectors transform under the same rotations.

Hermiticity needs care in a spherical basis. Even if $\mathbf V$ is a Hermitian vector operator in Cartesian components, the spherical components are not individually Hermitian. Instead,

$$V_q^\dagger = (-1)^q V_{-q}.$$

For a Hermitian spherical tensor with the same phase convention, the analogous relation is

$$\left(T_q^{(k)}\right)^\dagger = (-1)^q T_{-q}^{(k)}.$$

This Hermitian-tensor condition is an additional property, not a consequence of irreducibility alone. In general the adjoints define a partner tensor $\widetilde T_q^{(k)}=(-1)^q(T_{-q}^{(k)})^\dagger$; the displayed relation holds when that partner equals $T^{(k)}$.

Building Tensors by Coupling

Irreducible tensors can be built by coupling lower-rank tensors with Clebsch–Gordan coefficients. If $A^{(k_1)}$ and $B^{(k_2)}$ are spherical tensors, define

$$\left[ A^{(k_1)} \otimes B^{(k_2)} \right]_Q^{(K)} = \sum_{q_1,q_2} \langle k_1q_1;k_2q_2|KQ\rangle\, A_{q_1}^{(k_1)} B_{q_2}^{(k_2)}.$$

The product order shown is part of the definition. The components of $A$ and $B$ need not commute, so exchanging the factors can change the operator. The allowed $K$ values obey the same angular-momentum addition rule:

$$|k_1-k_2| \le K \le k_1+k_2.$$

For two vectors, this gives

$$1\otimes1=0\oplus1\oplus2.$$

The three pieces correspond, in Cartesian language, to:

• a scalar trace-like part;
• an antisymmetric vector-like part;
• a symmetric traceless quadrupole-like part.

For example, the scalar part of two rank-$1$ tensors is proportional to the dot product:

$$\left[ A^{(1)} \otimes B^{(1)} \right]_0^{(0)} = -\frac{1}{\sqrt3} \mathbf A\cdot\mathbf B,$$

using the spherical convention above and the Condon-Shortley phase convention. Different phase conventions change intermediate signs but not the representation content.

Relation to Spherical Harmonics

Spherical harmonics are the coordinate-space model for irreducible rotational behavior. The functions $Y_{\ell m}(\theta,\phi)$ transform among themselves under rotations with rank $\ell$.

This is why multipole operators are often written as

$$Q_q^{(\ell)} \propto r^\ell Y_{\ell q}(\hat{\mathbf r}), \qquad q=-\ell,\ldots,\ell.$$

Up to normalization, $Q_q^{(\ell)}$ is a rank-$\ell$ spherical tensor. The common low ranks are:

• $\ell=0$: monopole or scalar part;
• $\ell=1$: dipole or vector part;
• $\ell=2$: quadrupole part.

A useful alternative to the unit-normalized $Y_{\ell q}$ is the Racah-normalized spherical harmonic

$$C_q^{(\ell)}(\hat{\mathbf r}) = \sqrt{\frac{4\pi}{2\ell+1}} Y_{\ell q}(\hat{\mathbf r}),$$

for which $C_0^{(1)}=\cos\theta$ and $\int|C_q^{(\ell)}|^2d\Omega=4\pi/(2\ell+1)$. The physical content is the same: the operator carries definite rotational rank $\ell$ and component $q$.

The Cartesian tensor decomposition separates trace, antisymmetric, and symmetric traceless parts. The coupled tensors above express those same ranks $0$, $1$, and $2$ in a basis adapted to $J_z$.

Relation to Wigner–Eckart

The Wigner–Eckart theorem applies when the operator has a definite tensor rank:

$$\langle\alpha',j',m'| T_q^{(k)} |\alpha,j,m\rangle.$$

The tensor labels imply the rotational selection rules

$$m'=m+q,$$

and

$$|j-k| \le j' \le j+k.$$

The theorem then states that the remaining $m$ dependence is fixed by Clebsch–Gordan algebra, while the dynamics is contained in a reduced matrix element. This page explains what it means for the operator to have the tensor labels $k,q$ in the first place.

Common Mistakes

• Confusing tensor rank $k$ with matrix rank or with the number of Cartesian indices.
• Forgetting the sign convention in $V_{\pm1}=\mp(V_x\pm iV_y)/\sqrt2$.
• Treating a reducible Cartesian tensor as if it were already one irreducible tensor.
• Applying the Wigner–Eckart theorem to an operator before identifying its rotational rank.
• Forgetting that $q$ is defined relative to a chosen quantization axis.
• Assuming Hermitian Cartesian components imply Hermitian spherical components one by one.
• Replacing $R$ by $R^{-1}$ without taking the full Hermitian adjoint of the Wigner $D$ matrix, including its index order.

Exercises

1. Verify the $q=+1$ vector component.

Use

$$[J_i,V_j] = i\hbar\epsilon_{ij\ell}V_\ell$$

and

$$V_{+1} = -\frac{1}{\sqrt2} (V_x+iV_y)$$

to show that $[J_z,V_{+1}]=\hbar V_{+1}$.

Solution

The Cartesian commutators give

$$[J_z,V_x]=i\hbar V_y, \qquad [J_z,V_y]=-i\hbar V_x.$$

Therefore

$$\begin{aligned} [J_z,V_x+iV_y] &= i\hbar V_y + i(-i\hbar V_x) \\ &= \hbar(V_x+iV_y). \end{aligned}$$

Multiplying by the constant $-1/\sqrt2$ gives

$$[J_z,V_{+1}] = \hbar V_{+1}.$$

2. Count rank-$2$ components.

List the allowed $q$ values for a rank-$2$ tensor and state what magnetic quantum-number changes those components can produce.

Solution

For $k=2$,

$$q=-2,-1,0,1,2.$$

In a Wigner–Eckart matrix element, the magnetic rule is

$$m'=m+q.$$

Thus the possible magnetic changes are

$$\Delta m=-2,-1,0,1,2.$$

3. Can coupling a vector to itself produce a rank-$1$ tensor? Compare position components with angular-momentum components.

Solution

The $Q=0$ component in the Condon–Shortley convention is

$$[V^{(1)}\otimes V^{(1)}]^{(1)}_0 =\frac{V_{+1}V_{-1}-V_{-1}V_{+1}}{\sqrt2}.$$

For position the components commute, so it vanishes. For $V=J$, the spherical components are $J_{+1}=-J_+/\sqrt2$ and $J_{-1}=J_-/\sqrt2$. Hence

$$[J^{(1)}\otimes J^{(1)}]^{(1)}_0 =-\frac{[J_+,J_-]}{2\sqrt2} =-\frac{\hbar J_z}{\sqrt2}.$$

It does not vanish. The difference comes from operator order, not from a different angular-momentum addition rule.

4. Use the triangle rule.

A rank-$2$ tensor acts on an initial state with $j=1/2$. Which final total angular momenta $j'$ are allowed by rotations?

Solution

The triangle rule is

$$|j-k| \le j' \le j+k.$$

With $j=1/2$ and $k=2$,

$$\frac32 \le j' \le \frac52.$$

The allowed half-integer values are

$$j'=\frac32,\frac52.$$

The magnetic label must also satisfy $m'=m+q$ for one of $q=-2,-1,0,1,2$.

References

• D. M. Brink and G. R. Satchler, Angular Momentum, 3rd ed., Oxford University Press, 1993.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• M. E. Rose, Elementary Theory of Angular Momentum, Wiley, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.