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© The scientific sentence. 2010

 Tensor operators
1. Definition of a rotated operator
Let's write the rotation operator as U(R). This operator
rotates a ket ψ> to the ket ψ'>.
If A is any operator, the rotated operator A' must be
unchanged with respect to its expectations before and
after rotation. That is:
<ψ'A'ψ'> = <ψAψ>.
We have then:
<ψAψ> = <ψU^{+}(R)A'U(R)ψ>. Therefore:
A = U^{+}(R)A'U(R) or A' = U(R)AU^{+}.
Hence:
The rotated operator of the operator A is A' = U(R)AU^{+},
where U(R) is the rotation operator.
2. Scalar Operators
A scalar operator is invariant under rotations. That is:
K = U(R)KU^{+} under any rotation operator U(R).
This definition implies that U(R) and K commute: [U(R),K] = 0.
The scalar operator commutes with any rotation operator U(R);
in particular with the rotation operator R(θ) = exp { i θ J/ħ}
or its infinitesimal rotation operator: R(dθ) = 1  idθJ/ħ;
hence, it commutes with the angular momentum J.
We have then a new definition:
A scalar operator K commutes with the
angular momentum J if
[J, K] = 0
3. Vector Operators
In threedimensional Euclidean space, a vector is
defined as a set of three numbers. Under rotation,
these numbers are transformed according certain rules.
Similarly, in quantum mechanics, we define a vector operator
as a vector of operators (that is, a collection of
three operators) with certain transformation
properties under rotations.
For its definition,a vector operator "V" must have
its expectation rotated according the definition of
the rotation of any ordinary vector. That is:
If ψ'> is the rotated state of
the original one ψ>, we have:
R <ψVψ> = <ψ'Vψ'>
In components form:
V'_{i} = <ψ'V_{i}ψ'> = Σ (j) R_{i}_{j} <ψV_{j}
Using ψ'> = U(R) ψ>,, the definition becomes:
R <ψVψ> = <ψU(R)^{+}VU(R)ψ>. So
R V = U(R)^{+}VU(R), or R^{1} V = U(R)VU^{+}(R)
(the adjoint of the rotation R is equal to its inverse R^{1})
The definition of a vector operator becomes:
R V = U^{+}(R)VU(R)
For an infinitesimal rotation about the zaxis
by an angle ε, we have:
V'_{i} = (V'_{x} , V'_{y} , V'_{z}) = R_{z}(ε) (V_{x} , V_{y} , V_{z})
= (V_{x}  εV_{y}, V_{y} + εV_{x}, V_{z})
Using the expression of the rotation operator
U(R) = 1  iεJ_{z}/ħ, to find:
U^{+}(R)V_{i})U(R) = (1 + iεJ_{z}/ħ)V_{i})(1  iεJ_{z}/ħ) =
V_{i} + (iε/ħ)[J_{z}, V_{i}]
Therefore:
i[J_{z}, V_{x}] =  ħ V_{y}
i[J_{z}, V_{y}] =  ħ V_{x}
The cyclic equivalents give the related commutation
relations of the components of any vector operator V:
[V_{i}, J_{j}] = iħε_{ijk}V_{k}
[V_{i}, J_{j}] = iħε_{ijk}V_{k}
ε_{ijk} is the LeviCivita element:
ε_{ijk} =
+1 for even permutations: 123,231,312
1 for odd permutations: 132,213,321
0 otherwise
4. Tensor Operators
A tensor operator is a matrix of operators. That is each element
of the matrix is an operator. For example, if we have:
A = (A_{1}, A_{2}, A_{3}), and
B = (B_{1}, B_{2}, B_{3}); we
can construct a 3x3 tensor T_{i}_{j} = A_{i}B_{j}.
This is the case of a rank2 tensor.
Before rotation, we have V_{i} and W_{j}, then
a tensor T_{i}_{j}. After rotation we have T'_{i}_{j}.
We are going to express the matrix elements of T', that is
T'_{i}_{j} function of T_{i}_{j}.
We have already V'_{i} = Σ(k) R_{i}_{k} V_{k}
W'_{j} = Σ(l) R_{j}_{l} W_{l}
Then:
T' = V' ⊗ W' and T'_{i}_{j} = V'_{i} ⊗ W'_{j} = Σ(k) R_{i}_{k} V_{k} Σ(l) R_{j}_{l} W_{l} =
Σ(k) Σ(l) R_{i}_{k} R_{j}_{l} V_{k}W_{l}
V_{i}W_{j} = T_{ij}, then:
T'_{i}_{j} = Σ(k)Σ(l) R_{i}_{k} R_{j}_{l} T_{kl}
Therefore:
T'_{i}_{j} = U^{+}(R)T_{i}_{j}U(R) = Σ(k)Σ(l) R_{i}_{k} R_{j}_{l} T_{kl}
Since the suffixes i, j, ... refer to Cartesian axes. Tensors
written this way are called Cartesian tensors. The
number of suffixes in "T" is the rank of the Cartesian
tensor (T_{i}_{j} has the rank 2). The rank n
tensor has 3^{n} components. A rank 3 Cartesian tensor
is transformed as:
T'_{i}_{j}_{k} = Σ(l) Σ(m) Σ(n) R_{i}_{l} R_{j}_{m} R_{k}_{n} T_{l}_{m}_{n},
having 27 components.
Now we will represent a Cartesian tensor in spherical coordinates
and talk about spherical tensor
4.1.Spherical Vector
The angular momentum eigenkets l,m> = Y^{l}_{m}(θ,φ),
called spherical harmonics can be expressed for l = 1:
1,0> = Y^{1}_{0}(θ,φ) = [3/4π]^{1} (z/r)
1,1> = Y^{1}_{1}(θ,φ) = +[3/4π]^{1}(x  i y)/2^{1/2} r
1,1> = Y^{1}_{1}(θ,φ) =  [3/4π]^{1}(x + i y)/2^{1/2} r
x, y , and z are considered position operators.
r = (x, y, z) is a vector operator that will be transformed
under rotation U(R) to the vector operator r' by rotating
the eigenstates l,m>
The eigenstates operators l,m> will be considered as
eigenstates operators j,m>.
j=1,m> → U(R(θ))j=1,m> = exp {iθJ/ħ} j=1,m> =
Σ(m') <j=1,m'exp {iθJ/ħ}j=1,m> j=1 m'>
with:
D^{(j)}_{m'}_{m}(Rθ)) = <j,m'exp {iθJ/ħ)}j,m>
We have:
j=1,m> → U(R(θ))j=1,m>Σ(m') D^{(j=1)}_{m'}_{m}(Rθ) j=1 m'>
Therefore:
If (V_{x},V_{y},V_{z}) are the operators components of
a vector operator V in Cartesian coordinates,
in spherical coordinates, they become:
V^{1}_{1}, V^{1}_{1}, V^{0}_{1}, with:
V^{1}_{1} = +(V^{x}  iV^{y})/[2]^{1/2}
V^{+1}_{1} = (V^{x} + iV^{y})/[2]^{1/2}
V^{0}_{1} = V^{z}
To work with the same notation T, we write:
T^{1}_{1} = +(V^{x}  iV^{y})/[2]^{1/2}
T^{+1}_{1} = (V^{x} + iV^{y})/[2]^{1/2}
T^{0}_{1} = V^{z}
These components are denoted T^{q}_{1}.
The definition :
V'_{i} = Σ (j) R_{i}_{j} V_{j}
becomes:
T'^{q} = Σ(q') R_{qq'} T_{q'}, or
T'^{q}_{1} = Σ(q') R_{qq'} T^{q'}_{1}
More generally, a rank k vector operator having 2k+1
(q varies from = k to +k) components is written as
T^{q}_{k}. If k = 2, we have a
spherical tensor.
4.2.Spherical Tensor
We ca generalize the latter result to define a spherical
tensor of rank k as a set of 2k + 1 operators:
T^{q}_{k}; q = k, ...+k, which
under rotation R, they are transformed with the
matrix of 2j+1 elements D^{(k)}_{mm'} = <j,mexp{iθJ/ħjm>} .
That is:
U(R)T^{q}_{k}U^{+}(R) = Σ(q) D^{(k)}_{q'q'}T^{q'}_{k}
Let's recall:
D^{(k)}_{mm'} = <j,mexp{iθJ/ħjm>}
U(R)T^{q}_{k}U^{+}(R) = Σ(q) D^{(k)}_{q'q'}T^{q'}_{k}
Using the expression of the infinitesimal rotation
operator U(R(dθ)) = 1  i εJ/ħ, we find the
commutation relations:
[J+, T^{q}_{k}] = +ħ [(kq)(k+q+1)]^{1/2} T^{q+1}_{k}
[J, T^{q}_{k}] = ħ [(k+q)(kq+1)]^{1/2} T^{q1}_{k}
[Jz, T^{q}_{k}] = ħ q T^{q}_{k}

