In this article spherical functions are replaced by polynomials that have been well known in electrostatics since the time of Maxwell and associated with multipole moments.[1] [2] [3] [4] [5] [6] [7] [8] In physics, dipole and quadrupole moments typically appear because fundamental concepts of physics are associated precisely with them.[9] [10] Dipole and quadrupole moments are:
\intxi\rho(x)dV
\int(3xixk-\deltaik)\rho(x)dV
\rho(x)
Octupole moment
\int3(5xixkxl-xl\deltaik-xk\deltail-xi\deltalk)\rho(x)dV
is used rather seldom. As a rule, high-rank moments are calculated with the help of spherical functions.Spherical functions are convenient in scattering problems. Polynomials are preferable in calculations with differential operators. Here, properties of tensors, including high-rank moments as well, are considered to repeat basically features of solid spherical functions but having their own specifics.
Using of invariant polynomial tensors in Cartesian coordinates, as shown in a number of recent studies, is preferable and simplifies the fundamental scheme of calculations[11] [12] [13] .[14] The spherical coordinates are not involved here. The rules for using harmonic symmetric tensors are demonstrated that directly follow from their properties. These rules arenaturally reflected in the theory of special functions, but are not always obvious, even though the group properties are general.[15] At any rate, let us recall the main property of harmonic tensors: the trace over any pair of indices vanishes.[16] Here, those properties of tensors are selected that not only make analytic calculations more compact and reduce 'the number of factorials' but also allow correctly formulating some fundamental questions of the theoretical physics.[14]
Four properties of symmetric tensor
Mi...k
A. Tensor is homogeneous polynomial:
Mi...k
lM | |
(kx)=k | |
i...k |
(x)
l
B. Tensor is symmetric with respect to indices;
C. Tensor is harmonic, i.e., it is a solution of the Laplace equation:
\DeltaMi...k(x)=0
D. Trace over any two indices vanishes:
Mii[...](x)=0
[...]
(l-2)
i=i
Components of tensor are solid spherical functions. Tensor can be divided by factor
rl
The multipole potentials arise when the potential of a point charge is expanded in powers of coordinates
xoi
ro
1 | |
\left|r-ro\right| |
1 | |
\left|r-ro\right| |
| ||||
= \sum | ||||
l(-1) |
1 | |
r |
= \sum | ||||
|
(l) | |
M | |
i...k |
(r)=
\sum | ||||||||||||||||||
|
l
x0i...x0k
⊗ l | |
=r | |
0 |
l
(l) | |
M | |
i...k |
(r)=
(l) | |
M | |
[i] |
(r)
\Delta
r2l+1
| = | ||||||||||
r2l+1 |
| |||||||||
r2l+1 |
-(l+1)
From the formula for potential follows that
| ||||||||||
r2l+3 |
=-\nablai
M(l)m...k(r) | |
r2l+1 |
There is an obvious property of contraction
(2l-1)!!(
⊗ l | |
r | |
0 |
(l) | |
,M | |
[i] |
(r))= (
(l) | |
M | |
[i] |
(r0),
(l) | |
M | |
[i] |
(r))
TheoremLet
\rho(x)
(\int\rho(r0)
⊗ l | |
r | |
0 |
dV | , | |
r0 |
| ||||||||||
l!r |
)= (\int\rho(r0
(l) | |
)M | |
[i] |
(r0)dV
, | ||
r0 |
| |||||||||
(2l-1)!!l!r2l+1 |
)
It is an advantage in comparing with using of spherical functions.
Example 1.
For the quadrupole moment, instead of the integral
\int\rho(r)(3xi
2 | |
x | |
k-r |
\deltaik)dV
3\int\rho(r)xixkdV
Formula for the tensor was considered in using a ladder operator.It can be derived using the Laplace operator. Similar approach is known in the theory of special functions. The first term in the formula, as is easyto see from expansion of a point charge potential, is equal to
(l) | |
M | |
[i] |
(r)=
(2l-1)!!x | |
i1 |
...x | |
il |
+...=(2l-1)!!r ⊗ +...
r
.
This form is useful for applying differential operators of quantum mechanics and electrostatics to it. The differentiation generates product of the Kronecker symbols.
Example 2
\Deltaxixk=2\deltaik
\Deltaxixkxm=2(\deltaikxm+\deltakmxi+\deltaimxk)
\Delta\Deltaxixkxmxn=8(\deltaik\deltamn+\deltakm\deltain+\deltaim\deltakn)
i=k
\Deltakr ⊗ | |
k!2k |
=\left\langle\left\langle
⊗ (l-2k) | |
\delta | |
[..] |
r ⊗ \right\rangle\right\rangle
,where the symbol
⊗ k
\deltaim
Following one can find the relation between the tensor and solid spherical functions. Two unit vectors are needed: vector
nz
z
nx\pminy=n\pm
(l) | |
(M | |
[i] |
⊗ (l-m) | |
(r),n | |
z |
⊗ m | |
n | |
\pm |
)=(l-m)!(x+iy)mr(l-m)
dm | |
dtm |
Pl(t)
\mid | ||||
|
where
Pl(t)
In perturbation theory, it is necessary to expand the source in terms of spherical functions. If the source is a polynomial, for example, when calculating the Stark effect, then the integrals are standard, but cumbersome. When calculating with the help of invariant tensors, the expansion coefficients are simplified, and there is then no need to integrals. It suffices, as shown in, to calculate contractions that lower the rank of the tensors under consideration. Instead of integrals, the operation of calculating the trace
\hat{T}r
\hat{T}r\left\langle\left\langle\deltaik
(l) | |
M | |
ik[m] |
\right\rangle\right\rangle=
(l-2) | |
(2l+3)M | |
[m] |
If the brackets contain several factors with the Kronecker delta, the following relation formula holds:
\hat{T}r\left\langle\left\langle
\delta | |
i1p1 |
...\delta | |
ikpk |
(l) | |
M | |
i1p1[m] |
\right\rangle\right\rangle=(2l+2k+1)\left\langle\left\langle
\delta | |
i2p2 |
...\delta | |
ikpk |
(l-2) | |
M | |
[m] |
\right\rangle\right\rangle
Calculating the trace reduces the number of the Kronecker symbols by one, and the rank of the harmonic tensor on the right-hand side of the equation decreases by two. Repeating the calculation of the trace k times eliminates all the Kronecker symbols:
\hat{T}r1...\hat{T}rk\left\langle\left\langle
\delta | |
i1p1 |
...\delta | |
ikpk |
(l) | |
M | |
[m] |
\right\rangle\right\rangle=(2l+2k+1)!!\left\langle\left\langle
(l-2k) | |
M | |
[m] |
\right\rangle\right\rangle
The Laplace equation in four-dimensional 4D space has its own specifics. The potential of a point charge in 4D space is equal to
1 | |
r2 |
1 | |
{(r-r0) |
2}
⊗ n | |
r | |
0 |
ak{M | |
(n) |
i...k}{r}={(-1)}n\nablai...\nabla
|
(1) | |
ak{M} | |
i |
(r)=2xi
(2) | |
ak{M} | |
ik |
(r)=2(4xixk-\delta)
(n) | |
ak{M} | |
[i] |
(r)
=(2n)!!x | |
i1 |
...x | |
in |
-...=(2n)!!r ⊗ -...
.Four-dimensional tensors are structurally simpler than 3D tensors.
Applying the contraction rules allows decomposing the tensor with respect to the harmonic ones.In the perturbation theory, even the third approximation often considered good. Here, the decomposition of the tensor power up to the rank l=6 is presented:
\bullet l=2
3xix
(2) | |
ik |
+
2\delta | |
r | |
ik |
\bullet l=3
5!!xixkx
(3) | |
ikm |
+
2\left\langle\left\langle\delta | |
3r | |
ik |
xm\right\rangle\right\rangle
\bullet l=4
7!!xixkxlx
(4) | |
iklm |
+5r2\left\langle\left\langleM
(2) | |
ik |
\deltalm\right\rangle\right\rangle
4\left\langle\left\langle\delta | |
+ 7r | |
ik |
\deltalm\right\rangle\right\rangle
\bullet l=5
9!!xixkxlxmx
(5) | |
iklmn |
+7r2\left\langle\left\langleM
(3) | |
ikl |
\deltamn\right\rangle\right\rangle+
4\left\langle\left\langle\delta | |
27r | |
ik |
\deltalmxn\right\rangle\right\rangle
\bullet l=6
11!!xixkxlxmxnxp=
(6) | |
=M | |
iklmnp |
+9r2\left\langle\left\langleM
(4) | |
iklm |
\deltanp\right\rangle\right\rangle+ 55r4\left\langle\left\langleM
(2) | |
ik |
\deltalm\deltanp\right\rangle\right\rangle+
6\left\langle\left\langle\delta | |
99r | |
ik |
\deltalm\deltanp\right\rangle\right\rangle
l=6
l=4
l
2l-1)!! | |
(l+1)!! |
(l) | |
(M | |
[i] |
(l) | |
,M | |
[i] |
)=
(l) | |
(M | |
i...k |
(x
(l) | |
),M | |
i...k |
(x))=
(2l)! | |
2l |
r
which arises when normalizing the states.
The decomposition of tensor powers of a vector is also compact in four dimensions:
\bullet n=2
4!!xix
(2) | |
ik |
+
2\delta | |
2r | |
ik |
\bullet n=3
6!!xixkxm=ak{M
(3) | |
} | |
ikm |
+
2\left\langle\left\langle\delta | |
8r | |
ik |
xm\right\rangle\right\rangle
\bullet n=4
8!!xixkxlx
(4) | |
iklm |
+6r2\left\langle\left\langleak{M}
(2) | |
ik |
\deltalm\right\rangle\right\rangle
4\left\langle\left\langle\delta | |
+ 16r | |
ik |
\deltalm\right\rangle\right\rangle
\bullet n=5
10!!xixkxlxmx
(5) | |
iklmn |
+8r2\left\langle\left\langleak{M}
(3) | |
ikl |
\deltamn\right\rangle\right\rangle+
4\left\langle\left\langle\delta | |
80r | |
ik |
\deltalmxn\right\rangle\right\rangle
\bullet n=6
12!!xixkxlxmxnxp=
(6) | |
=ak{M} | |
iklmnp |
+10r2\left\langle\left\langleak{M}
(4) | |
iklm |
\deltanp\right\rangle\right\rangle+ 72r4\left\langle\left\langleak{M}
(2) | |
ik |
\deltalm\deltanp\right\rangle\right\rangle+
6\left\langle\left\langle\delta | |
240r | |
ik |
\deltalm\deltanp\right\rangle\right\rangle
When using the tensor notation with indices suppressed, the last equality becomes
\bullet n=6
12!!x ⊗ (6
(6) | |
=ak{M} | |
[i] |
+10r2\left\langle\left\langleakak{M}
(4) | |
[i] |
\delta[..]\right\rangle\right\rangle+ 72r4\left\langle\left\langleak{M}
(2) | |
[i] |
⊗ 2 | |
\delta | |
[..] |
\right\rangle\right\rangle+
⊗ 3 | |
240r | |
[..] |
\right\rangle\right\rangle
Ladder operators are useful for representing eigen functions in a compact form.[18] [19] They are a basis for constructing coherent states [20] .[21] Operators considered here, in mani respects close to the 'creation' and 'annihilation' operators of an oscillator.
Efimov's operator
\hatD
\nabla | |||||||||||||
|
=-
| ||||||||||
r |
\hatDi
(l) | |
M | |
k...m |
(l+1) | |
(r)=M | |
ik...m |
(r)
\hatl=(r\nabla)
l
\hatDixk=3xix
2\delta{ik} | |
k-r |
\hatDi\hatDk\hatDm1=3[5xixkx
2(\delta | |
ik |
xm+\deltakmxi+\deltaimxk)]
As a result of an
l
\hatDi\hatDk...\hatDm1=M(l)ik...m
⊗ l | |
=D | |
[i] |
(l) | |
= M | |
[i] |
\hatL
(\hbar=1)
\hat{D}=\hat{l}r+i[r x \hat{L}]
\hatDi\hat
2 | |
D | |
i=r |
\Delta
\hatD
x
x\hatD=r2(\hatl+1)
\hatDx=r2\hatl
x
(l) | |
x | |
ik...m |
=2lr2
(l-1) | |
M | |
k...m |
l
The commutator in the scalar product on the sphere is equal to unity:
xD-Dx=r2
\nabla\hatD=(\hatl+1)(2\hatl+3)
\nablai\hat
(l) | |
M | |
ik...m |
=l(2l+1
(l-1) | |
)M | |
(k...m) |
l
The raising operator in 4D space
\hatak{D}i
(n) | |
ak{M} | |
k...m |
(r,\tau)=
(n+1) | |
ak{M} | |
ik...m |
(r,\tau)
(n+1) | |
=ak{M} | |
ik...m |
(y)
where
yi
i=1,2,3,4
y=(r,\tau), \left|y\right|2=\rho
2 | |
y |
=r2+\tau2
\hatn
\tau
\hatn=(r\nablar+\tau
\partial | |
\partial\tau |
)=y\nablay
In particular,
\hatak{D}i1=2yi, \hatak{D}iyk=2(4yiyk
2 | |
-\rho | |
y |
\delta)
\hatak{D}i\hatak{D}k\hatak{D}m=4!![6xixkx
2 | |
y |
(\deltaikxm+\deltakmxi+\deltaimxk)]
The scalar product of the ladder operator
\hatak{D}
y
y\hatak{D}=\hatny
2 | |
\rho | |
y |
, \hat
2 | |
ak{D}y=(\hatn-2)\rho | |
y |
\hatak{D}
\nabla
\nabla\hatak{D}=2(\hatn+2)2
\hatA
\hatA
\hatA=i(\tau\nablar-r
\partial | |
\partial\tau |
)
Separately for the 3D
r
\tau
\hatakDr=(\hatn+1)r+i[r x \hatL]+i\tau\hatA
\hatakD\tau=(\hatn+1)\tau+i(r\hatA)