Basis definition

In the spherical bases denoted e₊, e₋, e₀, and associated coordinates with respect to this basis, denoted A₊, A₋, A₀, the vector A is:

${\displaystyle \mathbf {A} =A_{+}\mathbf {e} _{+}+A_{-}\mathbf {e} _{-}+A_{0}\mathbf {e} _{0}}$

(2)

where the spherical basis vectors can be defined in terms of the Cartesian basis using complex-valued coefficients in the xy plane:^[1]

${\displaystyle {\begin{aligned}\mathbf {e} _{+}&=-{\frac {1}{\sqrt {2}}}\mathbf {e} _{x}-{\frac {i}{\sqrt {2}}}\mathbf {e} _{y}\\\mathbf {e} _{-}&=+{\frac {1}{\sqrt {2}}}\mathbf {e} _{x}-{\frac {i}{\sqrt {2}}}\mathbf {e} _{y}\\\end{aligned}}\quad \rightleftharpoons \quad \mathbf {e} _{\pm }=\mp {\frac {1}{\sqrt {2}}}\left(\mathbf {e} _{x}\pm i\mathbf {e} _{y}\right)\,}$

(3A)

in which ${\displaystyle i}$ denotes the imaginary unit, and one normal to the plane in the z direction:

${\displaystyle \mathbf {e} _{0}=\mathbf {e} _{z}}$

The inverse relations are:

${\displaystyle {\begin{aligned}\mathbf {e} _{x}&=-{\frac {1}{\sqrt {2}}}\mathbf {e} _{+}+{\frac {1}{\sqrt {2}}}\mathbf {e_{-}} \\\mathbf {e} _{y}&=+{\frac {i}{\sqrt {2}}}\mathbf {e} _{+}+{\frac {i}{\sqrt {2}}}\mathbf {e_{-}} \\\mathbf {e} _{z}&=\mathbf {e} _{0}\end{aligned}}}$

(3B)

Commutator definition

While giving a basis in a 3-dimensional space is a valid definition for a spherical tensor, it only covers the case for when the rank ${\displaystyle k}$ is 1. For higher ranks, one may use either the commutator, or rotation definition of a spherical tensor. The commutator definition is given below, any operator ${\displaystyle T_{q}^{(k)}}$ that satisfies the following relations is a spherical tensor:

$${\displaystyle [J_{\pm },T_{q}^{(k)}]=\hbar {\sqrt {(k\mp q)(k\pm q+1)}}T_{q\pm 1}^{(k)}}$$

$${\displaystyle [J_{z},T_{q}^{(k)}]=\hbar qT_{q}^{(k)}}$$

Rotation definition

Analogously to how the spherical harmonics transform under a rotation, a general spherical tensor transforms as follows, when the states transform under the unitary Wigner D-matrix ${\displaystyle {\mathcal {D}}(R)}$, where R is a (3×3 rotation) group element in SO(3). That is, these matrices represent the rotation group elements. With the help of its Lie algebra, one can show these two definitions are equivalent.

${\displaystyle {\mathcal {D}}(R)T_{q}^{(k)}{\mathcal {D}}^{\dagger }(R)=\sum _{q'=-k}^{k}T_{q'}^{(k)}{\mathcal {D}}_{q'q}^{(k)}}$

Coordinate vectors

For the spherical basis, the coordinates are complex-valued numbers A₊, A₀, A₋, and can be found by substitution of (3B) into (1), or directly calculated from the inner product ⟨, ⟩ (5):

${\displaystyle {\begin{aligned}A_{+}&=\left\langle \mathbf {A} ,\mathbf {e} _{+}\right\rangle =-{\frac {A_{x}}{\sqrt {2}}}+{\frac {iA_{y}}{\sqrt {2}}}\\A_{-}&=\left\langle \mathbf {A} ,\mathbf {e} _{-}\right\rangle =+{\frac {A_{x}}{\sqrt {2}}}+{\frac {iA_{y}}{\sqrt {2}}}\\\end{aligned}}\quad \rightleftharpoons \quad A_{\pm }=\left\langle \mathbf {e} _{\pm },\mathbf {A} \right\rangle ={\frac {1}{\sqrt {2}}}\left(\mp A_{x}+iA_{y}\right)}$

(4A)

${\displaystyle A_{0}=\left\langle \mathbf {e} _{0},\mathbf {A} \right\rangle =\left\langle \mathbf {e} _{z},\mathbf {A} \right\rangle =A_{z}}$

with inverse relations:

${\displaystyle {\begin{aligned}A_{x}&=-{\frac {1}{\sqrt {2}}}A_{+}+{\frac {1}{\sqrt {2}}}A_{-}\\A_{y}&=-{\frac {i}{\sqrt {2}}}A_{+}-{\frac {i}{\sqrt {2}}}A_{-}\\A_{z}&=A_{0}\end{aligned}}}$

(4B)

In general, for two vectors with complex coefficients in the same real-valued orthonormal basis e_i, with the property e_i·e_j = δ_ij, the inner product is:

${\displaystyle \left\langle \mathbf {a} ,\mathbf {b} \right\rangle =\mathbf {a} \cdot \mathbf {b} ^{\star }=\sum _{j}a_{j}b_{j}^{\star }}$

(5)

where · is the usual dot product and the complex conjugate * must be used to keep the magnitude (or "norm") of the vector positive definite.

Orthonormality

The spherical basis is an orthonormal basis, since the inner product ⟨, ⟩ (5) of every pair vanishes meaning the basis vectors are all mutually orthogonal:

${\displaystyle \left\langle \mathbf {e} _{+},\mathbf {e} _{-}\right\rangle =\left\langle \mathbf {e} _{-},\mathbf {e} _{0}\right\rangle =\left\langle \mathbf {e} _{0},\mathbf {e} _{+}\right\rangle =0}$

and each basis vector is a unit vector:

${\displaystyle \left\langle \mathbf {e} _{+},\mathbf {e} _{+}\right\rangle =\left\langle \mathbf {e} _{-},\mathbf {e} _{-}\right\rangle =\left\langle \mathbf {e} _{0},\mathbf {e} _{0}\right\rangle =1}$

hence the need for the normalizing factors of ${\displaystyle 1/\!{\sqrt {2}}}$.

Change of basis matrix

The defining relations (3A) can be summarized by a transformation matrix U:

${\displaystyle {\begin{pmatrix}\mathbf {e} _{+}\\\mathbf {e} _{-}\\\mathbf {e} _{0}\end{pmatrix}}=\mathbf {U} {\begin{pmatrix}\mathbf {e} _{x}\\\mathbf {e} _{y}\\\mathbf {e} _{z}\end{pmatrix}}\,,\quad \mathbf {U} ={\begin{pmatrix}-{\frac {1}{\sqrt {2}}}&-{\frac {i}{\sqrt {2}}}&0\\+{\frac {1}{\sqrt {2}}}&-{\frac {i}{\sqrt {2}}}&0\\0&0&1\end{pmatrix}}\,,}$

with inverse:

${\displaystyle {\begin{pmatrix}\mathbf {e} _{x}\\\mathbf {e} _{y}\\\mathbf {e} _{z}\end{pmatrix}}=\mathbf {U} ^{-1}{\begin{pmatrix}\mathbf {e} _{+}\\\mathbf {e} _{-}\\\mathbf {e} _{0}\end{pmatrix}}\,,\quad \mathbf {U} ^{-1}={\begin{pmatrix}-{\frac {1}{\sqrt {2}}}&+{\frac {1}{\sqrt {2}}}&0\\+{\frac {i}{\sqrt {2}}}&+{\frac {i}{\sqrt {2}}}&0\\0&0&1\end{pmatrix}}\,.}$

It can be seen that U is a unitary matrix, in other words its Hermitian conjugate U^† (complex conjugate and matrix transpose) is also the inverse matrix U⁻¹.

For the coordinates:

${\displaystyle {\begin{pmatrix}A_{+}\\A_{-}\\A_{0}\end{pmatrix}}=\mathbf {U} ^{\mathrm {*} }{\begin{pmatrix}A_{x}\\A_{y}\\A_{z}\end{pmatrix}}\,,\quad \mathbf {U} ^{\mathrm {*} }={\begin{pmatrix}-{\frac {1}{\sqrt {2}}}&+{\frac {i}{\sqrt {2}}}&0\\+{\frac {1}{\sqrt {2}}}&+{\frac {i}{\sqrt {2}}}&0\\0&0&1\end{pmatrix}}\,,}$

and inverse:

${\displaystyle {\begin{pmatrix}A_{x}\\A_{y}\\A_{z}\end{pmatrix}}=(\mathbf {U} ^{\mathrm {*} })^{-1}{\begin{pmatrix}A_{+}\\A_{-}\\A_{0}\end{pmatrix}}\,,\quad (\mathbf {U} ^{\mathrm {*} })^{-1}={\begin{pmatrix}-{\frac {1}{\sqrt {2}}}&+{\frac {1}{\sqrt {2}}}&0\\-{\frac {i}{\sqrt {2}}}&-{\frac {i}{\sqrt {2}}}&0\\0&0&1\end{pmatrix}}\,.}$

Cross products

Taking cross products of the spherical basis vectors, we find an obvious relation:

${\displaystyle \mathbf {e} _{q}\times \mathbf {e} _{q}={\boldsymbol {0}}}$

where q is a placeholder for +, −, 0, and two less obvious relations:

${\displaystyle \mathbf {e} _{\pm }\times \mathbf {e} _{\mp }=\pm i\mathbf {e} _{0}}$

${\displaystyle \mathbf {e} _{\pm }\times \mathbf {e} _{0}=\pm i\mathbf {e} _{\pm }}$

Inner product in the spherical basis

The inner product between two vectors A and B in the spherical basis follows from the above definition of the inner product:

${\displaystyle \left\langle \mathbf {A} ,\mathbf {B} \right\rangle =A_{+}B_{+}^{\star }+A_{-}B_{-}^{\star }+A_{0}B_{0}^{\star }}$

General

• S. S. M. Wong (2008). Introductory Nuclear Physics (2nd ed.). John Wiley & Sons. ISBN 978-35-276-179-13.