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Legendre Polynomials

Legendre polynomials P_l(x) defined by the Rodrigues’ formula

P_l(x)={1\over2^l l!}{\d^l\over\d x^l}[(x^2-1)^l]

they also obey the completeness relation

(1)\sum_{l=0}^\infty {2l+1\over2}P_l(x')P_l(x)=\delta(x-x')

and orthogonality relation:

\int_{-1}^1 P_k(x) P_l(x) \d x = {2\delta_{kl} \over 2k+1}

Two Legendre polynomials can be expanded in a series:

P_k(x) P_l(x) = \sum_{m=|k-l|}^{k+l} \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1) P_m(x)

This was first proven by [Adams], where he shows:

P_k(x) P_l(x) = \sum_{m=|k-l|}^{k+l} {A(s-k) A(s-l) A(s-m)\over A(s)} {2m+1\over 2s+1} P_m(x)

where s={k+l+m\over 2} and

A(n) = {1\cdot3\cdot5 \cdot \dots \cdot (2n-1) \over 1\cdot 2\cdot 3\cdot \dots \cdot n} = {(2n)!\over 2^n (n!)^2} = {1\over 2^n}\binom{2n}{n}

The coefficient in the expansion can then be written using a 3j symbol as:

{A(s-k) A(s-l) A(s-m)\over A(s)} {1\over 2s+1} = = { {1\over2^{s-k}}\binom{2s-2k}{s-k} {1\over2^{s-l}}\binom{2s-2l}{s-l} {1\over2^{s-m}}\binom{2s-2m}{s-m} \over {1\over2^{s}}\binom{2s}{s} } {1\over 2s+1} = = {2^s\over2^{s-k+s-l+s-m}} { \binom{2s-2k}{s-k} \binom{2s-2l}{s-l} \binom{2s-2m}{s-m} \over \binom{2s}{s} } {1\over 2s+1} = = { \binom{2s-2k}{s-k} \binom{2s-2l}{s-l} \binom{2s-2m}{s-m} \over \binom{2s}{s} } {1\over 2s+1} = = { {(2s-2k)! \over ((s-k)!)^2} {(2s-2l)! \over ((s-l)!)^2} {(2s-2m)! \over ((s-m)!)^2} {(s!)^2 \over (2s)!} } {1\over 2s+1} = = {(2s-2k)! (2s-2l)! (2s-2m)! \over (2s+1)!} \left[{s! \over (s-k)! (s-l)! (s-m)!}\right]^2 = = \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2

So we will be just using the 3j symbol form from now on. We can now calculate the integral of three Legendre polynomials:

(2)\int_{-1}^1 P_k(x) P_l(x) P_m(x) \d x = = \int_{-1}^1 \sum_{n=|k-l|}^{k+l} \begin{pmatrix} k & l & n \\ 0 & 0 & 0 \end{pmatrix}^2 (2n+1) P_n(x) P_m(x) \d x = = \sum_{n=|k-l|}^{k+l} \begin{pmatrix} k & l & n \\ 0 & 0 & 0 \end{pmatrix}^2 (2n+1) {2\delta_{nm}\over 2n+1} = = 2 \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2

This is consistent with the series expansion:

P_k(x) P_l(x) = \sum_{m=|k-l|}^{k+l} {2m+1\over 2}\int_{-1}^1 P_k(x) P_l(x) P_m(x) \d x\,\, P_m(x) = = \sum_{m=|k-l|}^{k+l} \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1) P_m(x)

Any function f(x) (where -1\le x \le 1) can be expanded as:

f(x) = \sum_{l=0}^\infty f_l P_l(x) f_l = {(2l+1)\over 2} \int_{-1}^1 f(x) P_l(x) \d x

For the following choice of f(x) we get (for |t| \le 1):

f(x) = {1\over\sqrt{1-2xt+t^2}} f_l = {(2l+1)\over 2} \int_{-1}^1 {P_l(x)\over\sqrt{1-2xt+t^2}} \d x = {(2l+1)\over 2} \int_{|1+t|}^{|1-t|} {P_l\left(1-R^2+t^2\over 2 t\right)\over R} \left(-{R\over t}\right) \d R = = {(2l+1)\over 2 t} \int_{|1-t|}^{|1+t|} P_l\left(1-R^2+t^2\over 2 t\right) \d R = {(2l+1)\over 2 t} \int_{1-t}^{1+t} P_l\left(1-R^2+t^2\over 2 t\right) \d R = = t^l

Code:

>>> from sympy import var, legendre, integrate
>>> var("l R t")
(l, R, t)
>>> f = (2*l+1) / (2*t) * integrate(legendre(l, (1-R**2+t**2) / (2*t)),
...         (R, 1-t, 1+t))
>>> for _l in range(20): print _l, f.subs(l, _l).doit().simplify()
...
0 1
1 t
2 t**2
3 t**3
4 t**4
5 t**5
6 t**6
7 t**7
8 t**8
9 t**9
10 t**10
11 t**11
12 t**12
13 t**13
14 t**14
15 t**15
16 t**16
17 t**17
18 t**18
19 t**19

So the Legendre polynomials are the coefficients of the following expansion for |t| \le 1:

{1\over\sqrt{1-2xt+t^2}} = \sum_{l=0}^\infty P_l(x) t^l

Note that for |t| > 1 we get:

{1\over\sqrt{1-2xt+t^2}} = {1\over |t|}{1\over\sqrt{1-2x{1\over t}+\left({1\over t}\right)^2}} = {1\over |t|}\sum_{l=0}^\infty P_l(x) \left({1\over t}\right)^l = \sign t \sum_{l=0}^\infty P_l(x) t^{-l-1}

[Adams]Adams, J. C. (1878). On the Expression of the Product of Any Two Legendre’s Coefficients by Means of a Series of Legendre’s Coefficients. Proceedings of the Royal Society of London, 27, 63-71.

Example I

Very important is the following multipole expansion:

(3){1\over |{\bf r}-{\bf r'}|} ={1\over \sqrt{({\bf r}-{\bf r'})^2}} ={1\over \sqrt{r^2-2{\bf r}\cdot {\bf r'} + r'^2}} ={1\over r_>\sqrt{1-2\left(r_<\over r_>\right){\bf\hat r}\cdot {\bf\hat r'} + \left(r<\over r_>\right)^2}} = ={1\over r_>}\sum_{l=0}^\infty\left(r_<\over r_>\right)^l P_l({\bf\hat r}\cdot {\bf\hat r'}) =\sum_{l=0}^\infty {r_<^l\over r_>^{l+1}} P_l({\bf\hat r}\cdot {\bf\hat r'})

Where r_{>} = \max(r, r') and r_{<} = \min(r, r'). Assuming r > r', we get for the first few terms:

{1\over |{\bf r}-{\bf r'}|} ={1\over r}\left( P_0({\bf\hat r}\cdot {\bf\hat r'}) + P_1({\bf\hat r}\cdot {\bf\hat r'}){r'\over r} + P_2({\bf\hat r}\cdot {\bf\hat r'})\left(r'\over r\right)^2 + O\left(r'^3\over r^3\right) \right) = ={1\over r}\left( 1 + {\bf\hat r}\cdot {\bf\hat r'} {r'\over r} + \half\left(3({\bf\hat r}\cdot {\bf\hat r'})^2-1\right)\left(r'\over r\right)^2 + O\left(r'^3\over r^3\right) \right) = ={1\over r} +{{\bf r}\cdot {\bf r'}\over r^3} +{3({\bf r}\cdot {\bf r'})^2-r^2r'^2\over 2r^5} + O\left(r'^3\over r^4\right)

Example II

Let’s find the expansion of

f(x) = {e^{-\alpha \sqrt{1-2xt+t^2}}\over\sqrt{1-2xt+t^2}}

for |t| \le 1. We get:

f_l = {(2l+1)\over 2} \int_{-1}^1 {P_l(x)e^{-\alpha \sqrt{1-2xt+t^2}}\over\sqrt{1-2xt+t^2}} \d x = {(2l+1)\over 2} \int_{|1+t|}^{|1-t|} {P_l\left(1-R^2+t^2\over 2 t\right)e^{-\alpha R}\over R} \left(-{R\over t}\right) \d R = = {(2l+1)\over 2 t} \int_{|1-t|}^{|1+t|} P_l\left(1-R^2+t^2\over 2 t\right) e^{-\alpha R} \d R = {(2l+1)\over 2 t} \int_{1-t}^{1+t} P_l\left(1-R^2+t^2\over 2 t\right) e^{-\alpha R} \d R

Here is the result for the first few l:

f_0 & = \frac{\left(e^{2 \alpha t} -1\right) e^{- \alpha t - \alpha}}{2 \alpha t} \\ f_1 & = \frac{3}{2} \frac{\left(\alpha^{2} t e^{2 \alpha t} + \alpha^{2} t + \alpha t e^{2 \alpha t} + \alpha t - \alpha e^{2 \alpha t} + \alpha - e^{2 \alpha t} + 1\right) e^{- \alpha t - \alpha}}{\alpha^{3} t^{2}} \\ f_2 & = \frac{5}{2} \frac{\left(\alpha^{4} t^{2} e^{2 \alpha t} - \alpha^{4} t^{2} + 3 \alpha^{3} t^{2} e^{2 \alpha t} - 3 \alpha^{3} t^{2} - 3 \alpha^{3} t e^{2 \alpha t} - 3 \alpha^{3} t + 3 \alpha^{2} t^{2} e^{2 \alpha t} - 3 \alpha^{2} t^{2} - 9 \alpha^{2} t e^{2 \alpha t} - 9 \alpha^{2} t + X\right) e^{- \alpha t - \alpha}}{\alpha^{5} t^{3}} X = 3 \alpha^{2} e^{2 \alpha t} - 3 \alpha^{2} - 9 \alpha t e^{2 \alpha t} - 9 \alpha t + 9 \alpha e^{2 \alpha t} - 9 \alpha + 9 e^{2 \alpha t} -9

Expanding in t up to \operatorname{\mathcal{O}}\left(t^{7}\right) we get:

f_l & = e^{-\alpha} g_l \\ g_0 & = 1 + \frac{1}{6} \alpha^{2} t^{2} + \frac{1}{120} \alpha^{4} t^{4} + \frac{1}{5040} \alpha^{6} t^{6} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\ g_1 & = t + \alpha t + \frac{1}{10} \alpha^{2} t^{3} + \frac{1}{10} \alpha^{3} t^{3} + \frac{1}{280} \alpha^{4} t^{5} + \frac{1}{280} \alpha^{5} t^{5} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\ g_2 & = t^{2} + \alpha t^{2} + \frac{1}{3} \alpha^{2} t^{2} + \frac{1}{14} \alpha^{2} t^{4} + \frac{1}{14} \alpha^{3} t^{4} + \frac{1}{42} \alpha^{4} t^{4} + \frac{1}{504} \alpha^{4} t^{6} + \frac{1}{504} \alpha^{5} t^{6} + \frac{1}{1512} \alpha^{6} t^{6} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\ g_3 & = t^{3} + \alpha t^{3} + \frac{2}{5} \alpha^{2} t^{3} + \frac{1}{18} \alpha^{2} t^{5} + \frac{1}{15} \alpha^{3} t^{3} + \frac{1}{18} \alpha^{3} t^{5} + \frac{1}{45} \alpha^{4} t^{5} + \frac{1}{270} \alpha^{5} t^{5} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\ g_4 & = t^{4} + \alpha t^{4} + \frac{3}{7} \alpha^{2} t^{4} + \frac{1}{22} \alpha^{2} t^{6} + \frac{2}{21} \alpha^{3} t^{4} + \frac{1}{22} \alpha^{3} t^{6} + \frac{1}{105} \alpha^{4} t^{4} + \frac{3}{154} \alpha^{4} t^{6} + \frac{1}{231} \alpha^{5} t^{6} + \frac{1}{2310} \alpha^{6} t^{6} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\

Code:

>>> from sympy import var, legendre, integrate, exp, latex, cse
>>> var("l R t alpha")
(l, R, t, alpha)
>>>
>>> f = (2*l+1) / (2*t) * integrate(legendre(l, (1-R**2+t**2) / (2*t)) \
...         * exp(-alpha*R),
...         (R, 1-t, 1+t))
>>>
>>> for _l in range(3):
...     print "f_%d & =" %_l, latex(f.subs(l, _l).doit().simplify()), "\\\\"
...
f_0 & = \frac{\left(e^{2 \alpha t} -1\right) e^{- \alpha t - \alpha}}{2 \alpha t} \\
f_1 & = \frac{3}{2} \frac{\left(\alpha^{2} t e^{2 \alpha t} + \alpha^{2} t + \alpha t e^{2 \alpha t} + \alpha t - \alpha e^{2 \alpha t} + \alpha - e^{2 \alpha t} + 1\right) e^{- \alpha t - \alpha}}{\alpha^{3} t^{2}} \\
f_2 & = \frac{5}{2} \frac{\left(\alpha^{4} t^{2} e^{2 \alpha t} - \alpha^{4} t^{2} + 3 \alpha^{3} t^{2} e^{2 \alpha t} - 3 \alpha^{3} t^{2} - 3 \alpha^{3} t e^{2 \alpha t} - 3 \alpha^{3} t + 3 \alpha^{2} t^{2} e^{2 \alpha t} - 3 \alpha^{2} t^{2} - 9 \alpha^{2} t e^{2 \alpha t} - 9 \alpha^{2} t + 3 \alpha^{2} e^{2 \alpha t} - 3 \alpha^{2} - 9 \alpha t e^{2 \alpha t} - 9 \alpha t + 9 \alpha e^{2 \alpha t} - 9 \alpha + 9 e^{2 \alpha t} -9\right) e^{- \alpha t - \alpha}}{\alpha^{5} t^{3}} \\
>>> for _l in range(5):
...     result = f.subs(l, _l).doit().simplify() / exp(-alpha)
...     print "g_%d & =" %_l, latex(result.series(t, 0, 7)), "\\\\"
...
g_0 & = 1 + \frac{1}{6} \alpha^{2} t^{2} + \frac{1}{120} \alpha^{4} t^{4} + \frac{1}{5040} \alpha^{6} t^{6} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\
g_1 & = t + \alpha t + \frac{1}{10} \alpha^{2} t^{3} + \frac{1}{10} \alpha^{3} t^{3} + \frac{1}{280} \alpha^{4} t^{5} + \frac{1}{280} \alpha^{5} t^{5} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\
g_2 & = t^{2} + \alpha t^{2} + \frac{1}{3} \alpha^{2} t^{2} + \frac{1}{14} \alpha^{2} t^{4} + \frac{1}{14} \alpha^{3} t^{4} + \frac{1}{42} \alpha^{4} t^{4} + \frac{1}{504} \alpha^{4} t^{6} + \frac{1}{504} \alpha^{5} t^{6} + \frac{1}{1512} \alpha^{6} t^{6} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\
g_3 & = t^{3} + \alpha t^{3} + \frac{2}{5} \alpha^{2} t^{3} + \frac{1}{18} \alpha^{2} t^{5} + \frac{1}{15} \alpha^{3} t^{3} + \frac{1}{18} \alpha^{3} t^{5} + \frac{1}{45} \alpha^{4} t^{5} + \frac{1}{270} \alpha^{5} t^{5} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\
g_4 & = t^{4} + \alpha t^{4} + \frac{3}{7} \alpha^{2} t^{4} + \frac{1}{22} \alpha^{2} t^{6} + \frac{2}{21} \alpha^{3} t^{4} + \frac{1}{22} \alpha^{3} t^{6} + \frac{1}{105} \alpha^{4} t^{4} + \frac{3}{154} \alpha^{4} t^{6} + \frac{1}{231} \alpha^{5} t^{6} + \frac{1}{2310} \alpha^{6} t^{6} + \operatorname{\mathcal{O}}\left(t^{7}\right) \\

Example III

{e^{-{|{\bf r}-{\bf r'}|\over D}}\over |{\bf r}-{\bf r'}|} = {e^{-r_>\sqrt{1-2\left(r_<\over r_>\right) {\bf\hat r}\cdot {\bf\hat r'} +\left(r_<\over r_>\right)^2}\over D}\over r_>\sqrt{1-2\left(r_<\over r_>\right) {\bf\hat r}\cdot {\bf\hat r'} +\left(r_<\over r_>\right)^2}} = {1\over r_>} {e^{-\alpha \sqrt{1-2xt+t^2}}\over\sqrt{1-2xt+t^2}}

where:

\alpha & = {r_>\over D} \\ x & = {\bf\hat r}\cdot {\bf\hat r'} \\ t & = {r_<\over r_>}

Example IV

V(|{\bf r}_1-{\bf r}_2|) = {e^{-{|{\bf r}_1-{\bf r}_2|\over D}}\over |{\bf r}_1-{\bf r}_2|}

The potential V is a function of r_1, r_2 and \cos\theta only:

V(|{\bf r}_1-{\bf r}_2|) = V\left(\sqrt{r_1^2 - 2 {\bf r_1} \cdot {\bf r_2} + r_2^2}\right) = V\left(\sqrt{r_1^2 - 2 r_1 r_2\cos\theta + r_2^2}\right) = V(r_1, r_2, \cos\theta)

So we expand in the \cos\theta variable using the Legendre expansion:

V(|{\bf r}_1-{\bf r}_2|) = V(r_1, r_2, \cos\theta) = \sum_{l=0}^\infty V_l(r_1, r_2) P_l(\cos\theta)

where V_l(r_1, r_2) only depends on r_1 and r_2:

V_l(r_1, r_2) = {2l+1\over 2}\int_{-1}^1 V(|{\bf r}_1-{\bf r}_2|) P_l(\cos\theta) \d(\cos\theta) = = {2l+1\over 2}\int_{-1}^1 {e^{-{|{\bf r}_1-{\bf r}_2|\over D}}\over |{\bf r}_1-{\bf r}_2|} P_l(\cos\theta) \d(\cos\theta) = = {2l+1\over 2 r_1 r_2}\int_{|r_1 - r_2|}^{r_1+r_2} e^{-{r\over D}} P_l\left(r_1^2 - r^2 + r_2^2 \over 2 r_1 r_2 \right) \d r

In the limit D\to\infty we get:

V_l(r_1, r_2) \to {r_<^l\over r_>^{l+1}}

In general, the V_l(r_1, r_2) expressions are complicated. For the first few l we get:

V_0(r_1, r_2) = {D\over 2 r_1 r_2}\left(e^{-{|r_1 - r_2|\over D}} - e^{-{r_1 + r_2\over D}}\right) V_1(r_1, r_2) = \frac{3}{2} \frac{D \left(- D^{2} e^{2 \frac{r_{2}}{D}} + D^{2} - D r_{1} e^{2 \frac{r_{2}}{D}} + D r_{1} + D r_{2} e^{2 \frac{r_{2}}{D}} + D r_{2} + r_{1} r_{2} e^{2 \frac{r_{2}}{D}} + r_{1} r_{2}\right) e^{- \frac{r_{1}}{D} - \frac{r_{2}}{D}}}{r_{1}^{2} r_{2}^{2}}

In V_1(r_1, r_2) we assume r_1 \ge r_2.

Spherical Harmonics

Are defined for m \ge 0 by

Y_{lm}(\theta,\phi)=\sqrt{{2l+1\over4\pi}{(l-m)!\over(l+m)!}}\,P_l^m(\cos\theta)\,e^{im\phi}

where P_l^m are associated Legendre polynomials defined by

P_l^m(x)=(-1)^m (1-x^2)^{m/2}{\d^m\over\d x^m} P_l(x)

and P_l are Legendre polynomials. For m < 0 they are defined by:

Y_{lm}(\Omega) = (-1)^m Y_{l,-m}^*(\Omega)

Sometimes the spherical harmonics are written as:

Y_{lm}(\theta,\phi) = \Theta_{lm}(\theta) \Phi_m(\phi)

where:

\Phi_m(\phi) &= {1\over\sqrt{2\pi}} e^{im\phi} \\ \Theta_{lm}(\theta) &= \begin{cases} \sqrt{{2l+1\over2}{(l-m)!\over(l+m)!}}\,P_l^m(\cos\theta) & \mbox{for } m \ge 0 \\ (-1)^m \Theta_{l,-m}(\theta) & \mbox{for } m < 0 \\ \end{cases}

The spherical harmonics are ortonormal:

(4)\int Y_{lm}\,Y^*_{l'm'}\,\d\Omega = \int_0^{2\pi}\int_0^{\pi} Y_{lm}(\theta,\phi)\,Y^*_{l'm'}(\theta,\phi)\sin\theta\,\d\theta\,\d\phi = \delta_{mm'}\delta_{ll'}

and complete (both in the l-subspace and the whole space):

(5)\sum_{m=-l}^l|Y_{lm}(\theta,\phi)|^2={2l+1\over4\pi}

(6)\sum_{l=0}^\infty\sum_{m=-l}^lY_{lm}(\theta,\phi)Y_{lm}^*(\theta',\phi') ={1\over\sin\theta}\delta(\theta-\theta')\delta(\phi-\phi')= \delta({\bf\hat r}-{\bf\hat r'})

The relation (5) is a special case of an addition theorem for spherical harmonics

(7)\sum_{m=-l}^lY_{lm}(\theta,\phi)Y_{lm}^*(\theta',\phi')= {2l+1\over 4\pi}P_l(\cos\gamma)

where \gamma is the angle between the unit vectors given by {\bf\hat r}=(\theta,\phi) and {\bf\hat r'}=(\theta',\phi'):

\cos\gamma=\cos\theta\cos\theta'+\sin\theta\sin\theta'\cos(\phi-\phi') ={\bf\hat r}\cdot{\bf\hat r'}

Relations between complex conjugates is:

Y_{l m}^*(\Omega) = (-1)^m Y_{l,-m}(\Omega) (-1)^m Y_{l,-m}^*(\Omega) = Y_{lm}(\Omega)

Examples

\int_{-1}^1 P_k(x) \d x = \int_{-1}^1 P_k(x) P_0(x) \d x = 2\delta_{k0} \int Y_{k0}(\Omega) \d \Omega = \int Y_{k0}(\Omega) \sqrt{4\pi} Y_{00}(\Omega) \d \Omega = \sqrt{4\pi} \delta_{k0}

Gaunt Coefficients

We use the Wigner-Eckart theorem:

\braket{j m | T^k_q | j' m'} = (-1)^{j-m} \begin{pmatrix} j & k & j' \\ -m & q & m' \end{pmatrix} (j || T^k || j')

Where:

T^k_q = Y_{k q}

In order to calculate the reduced matrix element (j || T^k || j'), we evaluate the W-E theorem for m=q=m'=0:

\braket{j 0 | T^k_0 | j' 0} = (-1)^{j} \begin{pmatrix} j & k & j' \\ 0 & 0 & 0 \end{pmatrix} (j || T^k || j')

and also evaluate the left hand side explicitly:

\braket{j 0 | T^k_0 | j' 0} = \braket{j 0 | Y_{k 0} | j' 0} = \int Y_{j0}^*(\Omega) Y_{k0}(\Omega) Y_{j'0}(\Omega) \d \Omega = = \sqrt{(2j+1)(2k+1)(2j'+1)\over 4\pi} {1\over 4\pi} \int P_j(\cos\theta) P_k(\cos\theta) P_{j'}(\cos\theta) \sin\theta \d \theta \d \phi = = \sqrt{(2j+1)(2k+1)(2j'+1)\over 4\pi} {1\over 2} \int_{-1}^1 P_j(x) P_k(x) P_{j'}(x) \d x = = \sqrt{(2j+1)(2k+1)(2j'+1)\over 4\pi} \begin{pmatrix} j & k & j' \\ 0 & 0 & 0 \end{pmatrix}^2

where we used (2). Comparing these two results, we get:

(j || T^k || j') = (-1)^{-j} \sqrt{(2j+1)(2k+1)(2j'+1)\over 4\pi} \begin{pmatrix} j & k & j' \\ 0 & 0 & 0 \end{pmatrix}

and finally:

\int Y_{jm}^*(\Omega) Y_{kq}(\Omega) Y_{j'm'}(\Omega) \d \Omega = =\braket{j m | T^k_q | j' m'} = (-1)^{j-m} \begin{pmatrix} j & k & j' \\ -m & q & m' \end{pmatrix} (j || T^k || j') = = (-1)^{j-m} \begin{pmatrix} j & k & j' \\ -m & q & m' \end{pmatrix} (-1)^{-j} \sqrt{(2j+1)(2k+1)(2j'+1)\over 4\pi} \begin{pmatrix} j & k & j' \\ 0 & 0 & 0 \end{pmatrix} = = (-1)^{-m} \sqrt{(2j+1)(2k+1)(2j'+1)\over 4\pi} \begin{pmatrix} j & k & j' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} j & k & j' \\ -m & q & m' \end{pmatrix}

In order to evaluate other integrals of spherical harmonics, we just use the above result, for example:

\int Y_{l_1 m_1}(\Omega) Y_{l_2 m_2}(\Omega) Y_{l_3 m_3}(\Omega) \d\Omega = =(-1)^{m_1}\int Y_{l_1 -m_1}^*(\Omega) Y_{l_2 m_2}(\Omega) Y_{l_3 m_3}(\Omega) \d\Omega= =(-1)^{m_1} (-1)^{-(-m_1)} \sqrt{(2l_1+1)(2l_2+1)(2l_3+1)\over 4\pi} \begin{pmatrix} l_1 & l_2 & l_3 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l_2 & l_3 \\ -(-m_1) & m_2 & m_3 \end{pmatrix}= = \sqrt{(2l_1+1)(2l_2+1)(2l_3+1)\over 4\pi} \begin{pmatrix} l_1 & l_2 & l_3 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \end{pmatrix}

This is the most symmetric relation. It was first obtained by [Gaunt] (equation (9), p. 194, where he expanded the 3j symbols, so his formula is more complex but equivalent to the above).

It is useful to incorporate the selection rule m_1 + m_2 + m_3 = 0 of the 3j symbols into the formula and we get:

c^k(l, m, l', m') = \sqrt{4\pi \over 4k+1} \int Y_{lm}^*(\Omega) Y_{k, m-m'}(\Omega) Y_{l'm'}(\Omega) \d\Omega = = (-1)^{-m} \sqrt{4\pi \over 4k+1} \sqrt{(2l+1)(2k+1)(2l'+1)\over 4\pi} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix} = = (-1)^{-m} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix}

From the other selection rules of the 3j symbols it follows, that the c^k(l, m, l', m') coefficients are nonzero only when:

|l-l'| \le k \le l + l' l+l'+k = \mbox{even integer}

[Gaunt]Gaunt, J. A. (1929). The Triplets of Helium. Philosophical Transactions of the Royal Society of London, 228, 151-196.

Example I

c^0(l, m, l', m') =\sqrt{4\pi} \int Y_{lm}^*(\Omega) Y_{00}(\Omega) Y_{l'm'}(\Omega) \d\Omega =\delta_{l l'}\delta_{m m'}

Example II

\sum_{m=-l}^l c^k(l, m, l, m) = \sum_m \sqrt{4\pi \over 4k+1} \int Y_{lm}^*(\Omega) Y_{k0}(\Omega) Y_{lm}(\Omega) \d\Omega = = \sqrt{4\pi \over 4k+1} \int \sum_m |Y_{lm}(\Omega)|^2 Y_{k0}(\Omega) \d\Omega = = \sqrt{4\pi \over 4k+1} {2l+1\over 4\pi} \int Y_{k0}(\Omega) \d\Omega = = \sqrt{4\pi \over 4k+1} {2l+1\over 4\pi} \sqrt{4\pi} \delta_{k0} = = (2l+1) \delta_{k0}

Example III

c^k(l, m, l', m') = \sqrt{4\pi \over 4k+1} \int Y_{lm}^*(\Omega) Y_{k, m-m'}(\Omega) Y_{l'm'}(\Omega) \d\Omega = = \sqrt{4\pi \over 4k+1} \int \Theta_{lm}\Phi_m^* \Theta_{k, m-m'}\Phi_{m-m'} \Theta_{l'm'}\Phi_{m'} \sin\theta \d\theta \d\phi = = \sqrt{4\pi \over 4k+1} \int_0^\pi \Theta_{lm} \Theta_{k, m-m'} \Theta_{l'm'} \sin\theta \d\theta \int_0^{2\pi} \Phi_m^* \Phi_{m-m'} \Phi_{m'} \d\phi = = \sqrt{4\pi \over 4k+1} \int_0^\pi \Theta_{lm} \Theta_{k, m-m'} \Theta_{l'm'} \sin\theta \d\theta \left(1\over\sqrt{2\pi}\right)^3 \int_0^{2\pi} e^{-im\phi} e^{i(m-m')\phi} e^{im'\phi} \d\phi = = \sqrt{4\pi \over 4k+1} \int_0^\pi \Theta_{lm} \Theta_{k, m-m'} \Theta_{l'm'} \sin\theta \d\theta \left(1\over\sqrt{2\pi}\right)^3 \int_0^{2\pi} \!\!\!\d\phi = = \sqrt{2\over 4k+1} \int_0^\pi \Theta_{lm} \Theta_{k, m-m'} \Theta_{l'm'} \sin\theta \d\theta

Example IV

c^k(l, -m, l', -m') = = (-1)^{m} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ m & -m+m' & -m' \end{pmatrix} = = (-1)^{m}(-1)^{l+k+l'} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix} = = (-1)^{-m} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix} = c^k(l, m, l', m')

Where we used the fact, that l+k+l' is an even integer and (-1)^m=(-1)^{-m}. c^k is not symmetric in l m and l' m':

c^k(l', m', l, m) = (-1)^{-m'} \sqrt{(2l'+1)(2l+1)} \begin{pmatrix} l' & k & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l' & k & l \\ -m' & m'-m & m \end{pmatrix} = = (-1)^{-m'} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ m & m'-m & -m' \end{pmatrix} = = (-1)^{-m'} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix} = = (-1)^{m-m'} (-1)^{-m} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix} = = (-1)^{m-m'} c^k(l, m, l', m')

Few other identities:

c^k(l, 0, l', 0) = \sqrt{(2l+1)(2l'+1)} \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix}^2 \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix}^2 = {c^k(l, 0, l', 0) \over \sqrt{(2l+1)(2l'+1)}} = {c^{l'}(l, 0, k, 0) \over \sqrt{(2l+1)(2k+1)}} = {c^{l}(l', 0, k, 0) \over \sqrt{(2l'+1)(2k+1)}} c^k(l, 0, l', 0) = c^k(l', 0, l, 0)

Example V

\sum_{m'} \left(c^k(l, m, l', m')\right)^2 = = \sum_{m'} (2l+1)(2l'+1) \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix}^2 \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix}^2 = = (2l+1)(2l'+1) \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix}^2 \sum_{m'} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix}^2 = = (2l+1)(2l'+1) \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix}^2 {1\over 2l+1} = = (2l'+1) \begin{pmatrix} l & k & l' \\ 0 & 0 & 0 \end{pmatrix}^2 = =\sqrt{2l'+1\over 2l+1} c^k(l', 0, l, 0)

Example VI

(8)\sum_{m'}\sum_{q}\int Y_{l'm'}(\Omega) Y_{l'm'}^*(\Omega') Y_{kq}(\Omega) Y_{kq}^*(\Omega') Y_{lm}(\Omega') \d \Omega' = =\int {2l'+1\over 4\pi} P_{l'}({\bf \hat x}\cdot{\bf \hat x}') {2k+1\over 4\pi} P_k({\bf \hat x}\cdot{\bf \hat x}') Y_{lm}(\Omega') \d \Omega' = =\int {2l'+1\over 4\pi} {2k+1\over 4\pi} \sum_{\lambda=|l'-k|}^{\lambda=l'+k} \sqrt{2\lambda+1\over 2l'+1} c^k(l', 0, \lambda, 0) {4\pi \over 2\lambda+1} \sum_{\mu=-\lambda}^\lambda Y_{\lambda\mu}^*(\Omega') Y_{\lambda\mu}(\Omega) Y_{lm}(\Omega') \d \Omega' = = {2l'+1\over 4\pi} {2k+1\over 4\pi} \sum_{\lambda=|l'-k|}^{\lambda=l'+k} \sqrt{2\lambda+1\over 2l'+1} c^k(l', 0, \lambda, 0) {4\pi \over 2\lambda+1} \sum_{\mu=-\lambda}^\lambda Y_{\lambda\mu}(\Omega) \delta_{\lambda l} \delta_{\mu m} = = {2k+1\over 4\pi} \sqrt{2l'+1\over 2l+1} c^k(l', 0, l, 0) Y_{lm}(\Omega)

Where we used the following identities:

\sum_{m'} Y_{l'm'}(\Omega) Y_{l'm'}^*(\Omega') = {2l'+1\over 4\pi} P_{l'}({\bf \hat x}\cdot{\bf \hat x}') \sum_{q} Y_{kq}(\Omega) Y_{kq}^*(\Omega') = {2k+1\over 4\pi} P_k({\bf \hat x}\cdot{\bf \hat x}') P_k({\bf \hat x}\cdot{\bf \hat x}')P_{l'}({\bf \hat x}\cdot{\bf \hat x}') = \sum_{\lambda=|l'-k|}^{l'+k} \begin{pmatrix} k & l' & \lambda \\ 0 & 0 & 0 \end{pmatrix}^2 (2\lambda+1) P_\lambda({\bf \hat x}\cdot{\bf \hat x}') = = \sum_{\lambda=|l'-k|}^{\lambda=l'+k} \sqrt{2\lambda+1\over 2l'+1} c^k(l', 0, \lambda, 0) P_\lambda({\bf \hat x}\cdot{\bf \hat x}') = = \sum_{\lambda=|l'-k|}^{\lambda=l'+k} \sqrt{2\lambda+1\over 2l'+1} c^k(l', 0, \lambda, 0) {4\pi \over 2\lambda+1} \sum_{\mu=-\lambda}^\lambda Y_{\lambda\mu}^*(\Omega') Y_{\lambda\mu}(\Omega)

Note: using the integral of 3 spherical harmonics directly in (8):

\sum_{m'}\sum_{q}\int Y_{l'm'}(\Omega) Y_{l'm'}^*(\Omega') Y_{kq}(\Omega) Y_{kq}^*(\Omega') Y_{lm}(\Omega') \d \Omega' = =\sum_{m'} Y_{l'm'}(\Omega) Y_{k, m-m'}(\Omega) \sqrt{4\pi\over 2k+1} c^k(l, m, l', m')

doesn’t straightforwardly lead to the final result, as it is not obvious how to simplify things further.

Wigner 3j Symbols

Relation between the Wigner 3j symbols and Clebsch-Gordan coefficients:

\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} = {(-1)^{j_1-j_2-m_3}\over \sqrt{2j_3+1}} (j_1 m_1 j_2 m_2 | j_3 -m_3) (j_1 m_1 j_2 m_2 | j_3 m_3) = (-1)^{j_1-j_2+m_3}\sqrt{2j_3+1} \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & -m_3 \end{pmatrix}

They are nonzero only when:

m_1 + m_2 + m_3 = 0 j_1+j_2+j_3 = \mbox{integer (or even integer if $m_1=m_2=m_3=0$)} |m_i| \le j_i |j_1-j_2| \le j_3 \le j_1+j_2

They have lots of symmetries. The 3j symbol is invariant for an even permutation of columns:

\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} = = \begin{pmatrix} j_2 & j_3 & j_1 \\ m_2 & m_3 & m_1 \end{pmatrix} = = \begin{pmatrix} j_3 & j_1 & j_2 \\ m_3 & m_1 & m_2 \end{pmatrix}

For an odd permutation of columns it changes sign if j_1+j_2+j_3 is an odd integer:

\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} = = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_2 & j_1 & j_3 \\ m_2 & m_1 & m_3 \end{pmatrix} = = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_1 & j_3 & j_2 \\ m_1 & m_3 & m_2 \end{pmatrix} = = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_3 & j_2 & j_1 \\ m_3 & m_2 & m_1 \end{pmatrix}

and the same if you change the sign of the second row:

\begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix} = = (-1)^{j_1+j_2+j_3} \begin{pmatrix} j_1 & j_2 & j_3 \\ -m_1 & -m_2 & -m_3 \end{pmatrix}

Orthogonality relations:

\sum_{m_1 m_2} \begin{pmatrix} j_1 & j_2 & j \\ m_1 & m_2 & m \end{pmatrix} \begin{pmatrix} j_1 & j_2 & j' \\ m_1 & m_2 & m' \end{pmatrix} = {\delta_{jj'}\delta_{mm'} \over 2j+1}

As a special case, we get:

(9)\sum_{m'} \begin{pmatrix} l & k & l' \\ -m & m-m' & m' \end{pmatrix}^2 = {1 \over 2l+1}

Here is a script to check that the equation (9) works:

from sympy import S
from sympy.physics.wigner import wigner_3j

def doit(l, k, lp, m):
    s = 0
    for mp in range(-lp, lp+1):
        s += wigner_3j(l, k, lp, -m, m-mp, mp)**2
    print "%2d %2d %2d %2d  " % (l, k, lp, m), s, " ", S(1)/(2*l+1)

k = 4
lp = 3
print " l  k  lp m:  lhs   rhs"
for l in range(1, 6):
    for m in range(-l, l+1):
        doit(l, k, lp, m)

it prints:

l  k  lp m:  lhs   rhs
1  4  3 -1   1/3   1/3
1  4  3  0   1/3   1/3
1  4  3  1   1/3   1/3
2  4  3 -2   1/5   1/5
2  4  3 -1   1/5   1/5
2  4  3  0   1/5   1/5
2  4  3  1   1/5   1/5
2  4  3  2   1/5   1/5
3  4  3 -3   1/7   1/7
3  4  3 -2   1/7   1/7
3  4  3 -1   1/7   1/7
3  4  3  0   1/7   1/7
3  4  3  1   1/7   1/7
3  4  3  2   1/7   1/7
3  4  3  3   1/7   1/7
4  4  3 -4   1/9   1/9
4  4  3 -3   1/9   1/9
4  4  3 -2   1/9   1/9
4  4  3 -1   1/9   1/9
4  4  3  0   1/9   1/9
4  4  3  1   1/9   1/9
4  4  3  2   1/9   1/9
4  4  3  3   1/9   1/9
4  4  3  4   1/9   1/9
5  4  3 -5   1/11   1/11
5  4  3 -4   1/11   1/11
5  4  3 -3   1/11   1/11
5  4  3 -2   1/11   1/11
5  4  3 -1   1/11   1/11
5  4  3  0   1/11   1/11
5  4  3  1   1/11   1/11
5  4  3  2   1/11   1/11
5  4  3  3   1/11   1/11
5  4  3  4   1/11   1/11
5  4  3  5   1/11   1/11

Values of the 3j coefficients for a few special cases (use the symmetries above to obtain values for permuted symbols):

\begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix} &= (-1)^s \sqrt{(2s-2k)! (2s-2l)! (2s-2m)! \over (2s+1)!} {s! \over (s-k)! (s-l)! (s-m)!} \quad\quad\mbox{for $2s=k+l+m$ even} \\ \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix} &= 0 \quad\quad\mbox{for $2s=k+l+m$ odd} \\ \begin{pmatrix} j+\half & j & \half \\ m & -m-\half & \half \end{pmatrix} &= (-1)^{j-m-\half} \sqrt{j-m+\half \over (2j+1)(2j+2)} \\ \begin{pmatrix} j+1 & j & 1 \\ m & -m-1 & 1 \end{pmatrix} &= (-1)^{j-m-1} \sqrt{(j-m)(j-m+1) \over (2j+1)(2j+2)(2j+3)} \\ \begin{pmatrix} j+1 & j & 1 \\ m & -m & 0 \end{pmatrix} &= (-1)^{j-m-1} \sqrt{2(j+m+1)(j-m+1) \over (2j+1)(2j+2)(2j+3)}

Examples

\begin{pmatrix} j_3-\half & \half & j_3 \\ m_3-\half & \half & -m_3 \end{pmatrix} = \begin{pmatrix} j_3 & j_3-\half & \half \\ -m_3 & m_3-\half & \half \end{pmatrix} = \left. \begin{pmatrix} j+\half & j & \half \\ m & -m-\half & \half \end{pmatrix} \right|_{j=j_3-\half;m=-m_3} = = (-1)^{j_3-\half+m_3-\half}\sqrt{j_3-\half+m_3+\half\over (2 j_3-1+1) (2j_3-1+2)} = (-1)^{j_3+m_3-1}\sqrt{j_3+m_3\over 2 j_3 (2j_3+1)} \begin{pmatrix} j_3-\half & \half & j_3 \\ m_3+\half & -\half & -m_3 \end{pmatrix} = (-1)^{j_3-\half + \half + j_3} \begin{pmatrix} j_3 & j_3-\half & \half \\ m_3 & -m_3-\half & \half \end{pmatrix} = (-1)^{2j_3} \left. \begin{pmatrix} j+\half & j & \half \\ m & -m-\half & \half \end{pmatrix} \right|_{j=j_3-\half;m=m_3} = = (-1)^{2j_3} (-1)^{j_3-\half-m_3-\half}\sqrt{j_3-\half-m_3+\half\over (2 j_3-1+1) (2j_3-1+2)} = (-1)^{2j_3} (-1)^{j_3-m_3-1}\sqrt{j_3-m_3\over 2 j_3 (2j_3+1)} \begin{pmatrix} j_3+\half & \half & j_3 \\ m_3-\half & \half & -m_3 \end{pmatrix} = (-1)^{j_3+\half+\half+j_3} \begin{pmatrix} j_3+\half & j_3 & \half \\ m_3-\half & -m_3 & \half \end{pmatrix} = (-1)^{2j_3+1} \left. \begin{pmatrix} j+\half & j & \half \\ m & -m-\half & \half \end{pmatrix} \right|_{j=j_3;m=m_3-\half} = =(-1)^{2j_3+1}(-1)^{j_3-m_3+\half-\half}\sqrt{j_3-m_3+\half+\half \over (2j_3+1)(2j_3+2)} =(-1)^{2j_3+1}(-1)^{j_3-m_3}\sqrt{j_3-m_3+1 \over (2j_3+1)(2j_3+2)} \begin{pmatrix} j_3+\half & \half & j_3 \\ m_3+\half & -\half & -m_3 \end{pmatrix} = \begin{pmatrix} j_3+\half & j_3 & \half \\ -m_3-\half & m_3 & \half \end{pmatrix} = \left. \begin{pmatrix} j+\half & j & \half \\ m & -m-\half & \half \end{pmatrix} \right|_{j=j_3;m=-m_3-\half} = =(-1)^{j_3+m_3+\half-\half}\sqrt{j_3+m_3+\half+\half \over (2j_3+1)(2j_3+2)} =(-1)^{j_3+m_3}\sqrt{j_3+m_3+1 \over (2j_3+1)(2j_3+2)}

Multipole Expansion

Using (3) we get:

{1\over |{\bf r}-{\bf r'}|} =\sum_{l=0}^\infty{r_{<}^l\over r_{>}^{l+1}} P_l({\bf\hat r}\cdot {\bf\hat r'}) = \sum_{l,m}{r_{<}^l\over r_{>}^{l+1}} {4\pi\over 2l+1}Y_{lm}({\bf\hat r})Y_{lm}^*({\bf\hat r}')

where we used the formula:

\sum_m \braket{{\bf\hat r}|lm}\braket{lm|{\bf\hat r}'} ={2l+1 \over 4\pi} \braket{{\bf\hat r}\cdot{\bf\hat r'}|P_l}

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