According to Ballentine, an irreducible tensor of degree k can be defined as a set of $2k + 1$ operators $\{T_q^{\;\;(k)}:(-k \le q \le k)\}$ satisfying the following commutation relations: $$ [J_\pm, T_q^{\;\;(k)}] = \hbar \left( (k \mp q)(k \pm q + 1) \right)^\frac{1}{2} T_{q \pm 1}^{\;\;(k)} $$ $$ [J_z, T_q^{\;\;(k)}] = \hbar q T_{q}^{\;\;(k)} $$ The usual examples employ scalars, vector operators and the irreducible subspaces of cartesian tensor operators. The irreducible tensors that arise from these constructs all have integral degree.

Is it possible to have irreducible tensors of half integral degree, for example $k=\frac{1}{2}$? If so is there a simple example?


1) Yes, I assume that OP is already familiar with the irreducible finite-dimensional representations $V_j$ of the Lie algebra $so(3)\cong su(2)$; that they are classified by a non-negative half-integer $j\in \frac{1}{2}\mathbb{N}_{0}$; and that the eigenstates $|j,m\rangle \in V_j$ for $\vec{J}^2$ and $J_z$ satisfy

$$ \vec{J}^2 |j,m\rangle ~=~ \hbar^2 j(j+1)|j,m\rangle , \qquad J_z |j,m\rangle ~=~ \hbar m|j,m\rangle ,$$ $$J_{\pm}|j,m\rangle ~=~ \hbar \sqrt{ (j \mp m)(j \pm m + 1)}|j,m\pm 1\rangle ,$$

where $-j\leq m\leq j$ such that $j-m\in\mathbb{Z}$.

2) Instead, it seems that OP is really asking about the tensor operators $T_q^{\;\;(k)}$ described in eq. (7.112) of the book Quantum mechanics: a modern development by Leslie E. Ballentine,

$$[J_{\pm}, T_q^{\;\;(k)}] ~=~ \hbar \sqrt{ (k \mp q)(k \pm q + 1)} T_{q \pm 1}^{\;\;(k)}, \qquad [J_z, T_q^{\;\;(k)}] ~=~ \hbar q T_{q}^{\;\;(k)}. \qquad (7.112) $$

3) A simple example of such operators

$$T_q^{\;\;(k)}:~V~\to~ V, \qquad V~=~ \oplus_{j\in \frac{1}{2}\mathbb{N}_{0}} V_j,$$

can be build as

$$T_q^{\;\;(k)}~:=~ | k,q\rangle \otimes \langle 0,0|,$$

where $k\in \frac{1}{2}\mathbb{N}_{0}$ is a non-negative half-integer, and $-k\leq q\leq k$ such that $k-q\in\mathbb{Z}$. It is straightforward to check that this example satisfies the wanted commutation relations (7.112).

4) Note that any set of $2k+1$ operators $T_q^{\;\;(k)}:V_0\to V_k$ with $-k\leq q\leq k$ that satisfy eq. (7.112) would have to be proportional to $| k,q\rangle \otimes \langle 0,0|$ with the same proportionality factor independent of $q$, due to Wigner–Eckart theorem.


Yes, these "tensors" are called "spinors" for $k=1/2$ or perhaps "spintensors" for $k\in{\mathbb Z}+1/2$ but they satisfy the very same algebra you wrote. It is somewhat misleading to ask about "a simple example". There's just one example – the example or the solution – for each value of $k$, up to the choice of a basis and up to the freedom to pick a greater degeneracy. And you have actually written it down.

The collection of $k=1/2$ operators is $2k+1=2$-dimensional and it's known as the spinor. The allowed values are $q=+1/2$ and $q=-1/2$ and your formulae actually describe the exact form of all the commutators for $k=1/2$ as well as any other allowed value of $k$ (integer or half-integer-valued).

Spinors are essential to describe elementary fermions, including electrons, neutrinos, and quarks (and even protons, neutrons, and all other composite spin-1/2 particles). These may be viewed as "many examples" but the maths how the $\vec J$ generators commute with the spinorial objects is exactly the same in all cases.

The reason why one can construct tensors with a "half-integer number of indices" is that the group $SO(3)$ of rotations, generated by the angular momentum, is isomorphic to $SU(2)/{\mathbb Z}_2$. The quotient by ${\mathbb Z}_2$ means that one only allows integer spins. However, if one allows representations of the whole $SU(2)\sim Spin(3)$, one also allows representations (and tensor operators) that change their sign after rotations by 360 degrees and they correspond to $k$ which differs by $1/2$ from an integer.

  • $\begingroup$ I am not sure if I understand correctly. You are saying that there is just one solution for each $k$? Let's take $k=1$ then there are quite some cases, for example the operators $\{J_+, J_-, J_z\}$ built from linear combinations of the angular momentum operators $\vec{J}$ and also the same combinations of the position operators $\vec{X}$ or the momentum operators $\vec{P}$. $\endgroup$ – Friedrich Nov 17 '11 at 17:43
  • $\begingroup$ I forgot some factors, it's actually $\{-J_+/\sqrt{2},J_-/\sqrt{2},J_z\}$, where $J_+$ and $J_-$ are the usual angular moment raising and lowering operators. $\endgroup$ – Friedrich Nov 17 '11 at 19:32
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    $\begingroup$ Dear @Friedrich, $J_\pm$ is the same thing as $J_x\pm i J_y$, up to the sign of $i$ and the overall normalization, so the difference between $\{J_x,J_y,J_z\}$ and $\{J_+,J_-, J_z\}$ is just a convention about the basis - which linear combinations of the three quantities are being written as the standardized ones. The 3D space of linear combinations of these 3 tensor operators is exactly the same in both cases. That's why we say that the $j=1$ irreducible representation of the group is unique. The same hold for any allowed $j$, too. $\endgroup$ – Luboš Motl Dec 12 '11 at 11:10

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