A vector A in 3D Euclidean space R 3 Cartesian coordinate system in the standard basis e x e y e z coordinates Ax , Ay , Az :
A
=
A
x
e
x
+
A
y
e
y
+
A
z
e
z
{\displaystyle \mathbf {A} =A_{x}\mathbf {e} _{x}+A_{y}\mathbf {e} _{y}+A_{z}\mathbf {e} _{z}}
(1 )
or any other coordinate system with associated basis set of vectors. From this extend the scalars to allow multiplication by complex numbers, so that we are now working in
C
3
{\displaystyle \mathbb {C} ^{3}}
R
3
{\displaystyle \mathbb {R} ^{3}}
In the spherical bases denoted e + , e − , e 0 , and associated coordinates with respect to this basis, denoted A + , A − , A 0 , the vector A is:
A
=
A
+
e
+
+
A
−
e
−
+
A
0
e
0
{\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]
e
+
=
−
1
2
e
x
−
i
2
e
y
e
−
=
+
1
2
e
x
−
i
2
e
y
⇌
e
±
=
∓
1
2
(
e
x
±
i
e
y
)
{\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
i
{\displaystyle i}
imaginary unit , and one normal to the plane in the z direction:
e
0
=
e
z
{\displaystyle \mathbf {e} _{0}=\mathbf {e} _{z}}
The inverse relations are:
e
x
=
−
1
2
e
+
+
1
2
e
−
e
y
=
+
i
2
e
+
+
i
2
e
−
e
z
=
e
0
{\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
edit
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
k
{\displaystyle k}
T
q
(
k
)
{\displaystyle T_{q}^{(k)}}
[
J
±
,
T
q
(
k
)
]
=
ℏ
(
k
∓
q
)
(
k
±
q
+
1
)
T
q
±
1
(
k
)
{\displaystyle [J_{\pm },T_{q}^{(k)}]=\hbar {\sqrt {(k\mp q)(k\pm q+1)}}T_{q\pm 1}^{(k)}}
[
J
z
,
T
q
(
k
)
]
=
ℏ
q
T
q
(
k
)
{\displaystyle [J_{z},T_{q}^{(k)}]=\hbar qT_{q}^{(k)}}
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
D
(
R
)
{\displaystyle {\mathcal {D}}(R)}
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.
D
(
R
)
T
q
(
k
)
D
†
(
R
)
=
∑
q
′
=
−
k
k
T
q
′
(
k
)
D
q
′
q
(
k
)
{\displaystyle {\mathcal {D}}(R)T_{q}^{(k)}{\mathcal {D}}^{\dagger }(R)=\sum _{q'=-k}^{k}T_{q'}^{(k)}{\mathcal {D}}_{q'q}^{(k)}}
For the spherical basis, the coordinates are complex-valued numbers A + , A 0 , A − , and can be found by substitution of (3B 1 inner product ⟨, ⟩ (5
A
+
=
⟨
A
,
e
+
⟩
=
−
A
x
2
+
i
A
y
2
A
−
=
⟨
A
,
e
−
⟩
=
+
A
x
2
+
i
A
y
2
⇌
A
±
=
⟨
e
±
,
A
⟩
=
1
2
(
∓
A
x
+
i
A
y
)
{\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 )
A
0
=
⟨
e
0
,
A
⟩
=
⟨
e
z
,
A
⟩
=
A
z
{\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:
A
x
=
−
1
2
A
+
+
1
2
A
−
A
y
=
−
i
2
A
+
−
i
2
A
−
A
z
=
A
0
{\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 e i e j δij , the inner product is:
⟨
a
,
b
⟩
=
a
⋅
b
⋆
=
∑
j
a
j
b
j
⋆
{\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 .
edit
The spherical basis is an orthonormal basis , since the inner product ⟨, ⟩ (5 orthogonal :
⟨
e
+
,
e
−
⟩
=
⟨
e
−
,
e
0
⟩
=
⟨
e
0
,
e
+
⟩
=
0
{\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 :
⟨
e
+
,
e
+
⟩
=
⟨
e
−
,
e
−
⟩
=
⟨
e
0
,
e
0
⟩
=
1
{\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
1
/
2
{\displaystyle 1/\!{\sqrt {2}}}
Change of basis matrix
edit
The defining relations (3A transformation matrix U :
(
e
+
e
−
e
0
)
=
U
(
e
x
e
y
e
z
)
,
U
=
(
−
1
2
−
i
2
0
+
1
2
−
i
2
0
0
0
1
)
,
{\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:
(
e
x
e
y
e
z
)
=
U
−
1
(
e
+
e
−
e
0
)
,
U
−
1
=
(
−
1
2
+
1
2
0
+
i
2
+
i
2
0
0
0
1
)
.
{\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:
(
A
+
A
−
A
0
)
=
U
∗
(
A
x
A
y
A
z
)
,
U
∗
=
(
−
1
2
+
i
2
0
+
1
2
+
i
2
0
0
0
1
)
,
{\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:
(
A
x
A
y
A
z
)
=
(
U
∗
)
−
1
(
A
+
A
−
A
0
)
,
(
U
∗
)
−
1
=
(
−
1
2
+
1
2
0
−
i
2
−
i
2
0
0
0
1
)
.
{\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}}\,.}
Taking cross products of the spherical basis vectors, we find an obvious relation:
e
q
×
e
q
=
0
{\displaystyle \mathbf {e} _{q}\times \mathbf {e} _{q}={\boldsymbol {0}}}
where q is a placeholder for +, −, 0, and two less obvious relations:
e
±
×
e
∓
=
±
i
e
0
{\displaystyle \mathbf {e} _{\pm }\times \mathbf {e} _{\mp }=\pm i\mathbf {e} _{0}}
e
±
×
e
0
=
±
i
e
±
{\displaystyle \mathbf {e} _{\pm }\times \mathbf {e} _{0}=\pm i\mathbf {e} _{\pm }}
Inner product in the spherical basis
edit
The inner product between two vectors A and B in the spherical basis follows from the above definition of the inner product:
⟨
A
,
B
⟩
=
A
+
B
+
⋆
+
A
−
B
−
⋆
+
A
0
B
0
⋆
{\displaystyle \left\langle \mathbf {A} ,\mathbf {B} \right\rangle =A_{+}B_{+}^{\star }+A_{-}B_{-}^{\star }+A_{0}B_{0}^{\star }}
![{\displaystyle [J_{\pm },T_{q}^{(k)}]=\hbar {\sqrt {(k\mp q)(k\pm q+1)}}T_{q\pm 1}^{(k)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4ba041482aa1f9ed8b0cdda4af926ff6d25f55d)
![{\displaystyle [J_{z},T_{q}^{(k)}]=\hbar qT_{q}^{(k)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b39178911dcedd4c308338df03f1288096e6c81a)
