Finding the equal of a expression
$begingroup$
We can write the equal of $psi_{11}psi_{11}$ as follows
$psi_{11}psi_{11}=sqrt {2}psi_{10}+dfrac{4}{sqrt{10}}psi_{12}$
where
$psi_{10}=sqrt {2}$
$psi_{11}=sqrt {6}(4t-1)$
$psi_{12}=dfrac{sqrt {10}left( 3, left( 4,t-1 right) ^{2}-1 right)}{ 2}$
$psi_{13}=dfrac{sqrt {14}}{2}(5( 4t-1 ) ^{3}-12t+3) $
$psi_{14}=dfrac{3sqrt {2}}{8}big( {35, ( 4t-1) ^{4}}-30( 4t-1) ^{2}+3big)$
Similary;
$psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$.
How can we find the equal of the followings in terms of $psi_{11}, psi_{12}$ etc. like above?
$psi_{13}psi_{11}=?$
$psi_{14}psi_{12}=?$
analysis
asked Dec 9 '18 at 16:51
HD239HD239
441314
$endgroup$
add a comment |
$begingroup$
We can write the equal of $psi_{11}psi_{11}$ as follows
$psi_{11}psi_{11}=sqrt {2}psi_{10}+dfrac{4}{sqrt{10}}psi_{12}$
where
$psi_{10}=sqrt {2}$
$psi_{11}=sqrt {6}(4t-1)$
$psi_{12}=dfrac{sqrt {10}left( 3, left( 4,t-1 right) ^{2}-1 right)}{ 2}$
$psi_{13}=dfrac{sqrt {14}}{2}(5( 4t-1 ) ^{3}-12t+3) $
$psi_{14}=dfrac{3sqrt {2}}{8}big( {35, ( 4t-1) ^{4}}-30( 4t-1) ^{2}+3big)$
Similary;
$psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$.
How can we find the equal of the followings in terms of $psi_{11}, psi_{12}$ etc. like above?
$psi_{13}psi_{11}=?$
$psi_{14}psi_{12}=?$
analysis
asked Dec 9 '18 at 16:51
HD239HD239
441314
$endgroup$
2
$begingroup$
What is "the equal" of $psi_{11}psi_{11}$?
$endgroup$
– Ben W
Dec 9 '18 at 16:53
$begingroup$
Your text is very unclear. Why do you need two indices while the first one is always 1 ? Moreover, in expression $psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$, one can simplify by $psi_{11}$... then...$psi_{12}= dfrac{4}{sqrt{10}}$ ?
$endgroup$
– Jean Marie
Dec 9 '18 at 19:06
add a comment |
$begingroup$
We can write the equal of $psi_{11}psi_{11}$ as follows
$psi_{11}psi_{11}=sqrt {2}psi_{10}+dfrac{4}{sqrt{10}}psi_{12}$
where
$psi_{10}=sqrt {2}$
$psi_{11}=sqrt {6}(4t-1)$
$psi_{12}=dfrac{sqrt {10}left( 3, left( 4,t-1 right) ^{2}-1 right)}{ 2}$
$psi_{13}=dfrac{sqrt {14}}{2}(5( 4t-1 ) ^{3}-12t+3) $
$psi_{14}=dfrac{3sqrt {2}}{8}big( {35, ( 4t-1) ^{4}}-30( 4t-1) ^{2}+3big)$
Similary;
$psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$.
How can we find the equal of the followings in terms of $psi_{11}, psi_{12}$ etc. like above?
$psi_{13}psi_{11}=?$
$psi_{14}psi_{12}=?$
analysis
asked Dec 9 '18 at 16:51
HD239HD239
441314
$endgroup$
We can write the equal of $psi_{11}psi_{11}$ as follows
$psi_{11}psi_{11}=sqrt {2}psi_{10}+dfrac{4}{sqrt{10}}psi_{12}$
where
$psi_{10}=sqrt {2}$
$psi_{11}=sqrt {6}(4t-1)$
$psi_{12}=dfrac{sqrt {10}left( 3, left( 4,t-1 right) ^{2}-1 right)}{ 2}$
$psi_{13}=dfrac{sqrt {14}}{2}(5( 4t-1 ) ^{3}-12t+3) $
$psi_{14}=dfrac{3sqrt {2}}{8}big( {35, ( 4t-1) ^{4}}-30( 4t-1) ^{2}+3big)$
Similary;
$psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$.
How can we find the equal of the followings in terms of $psi_{11}, psi_{12}$ etc. like above?
$psi_{13}psi_{11}=?$
$psi_{14}psi_{12}=?$
analysis
analysis
asked Dec 9 '18 at 16:51
HD239HD239
441314
asked Dec 9 '18 at 16:51
HD239HD239
441314
asked Dec 9 '18 at 16:51
HD239HD239
441314
asked Dec 9 '18 at 16:51
HD239HD239
441314
asked Dec 9 '18 at 16:51
HD239HD239
441314
441314
2
$begingroup$
What is "the equal" of $psi_{11}psi_{11}$?
$endgroup$
– Ben W
Dec 9 '18 at 16:53
$begingroup$
Your text is very unclear. Why do you need two indices while the first one is always 1 ? Moreover, in expression $psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$, one can simplify by $psi_{11}$... then...$psi_{12}= dfrac{4}{sqrt{10}}$ ?
$endgroup$
– Jean Marie
Dec 9 '18 at 19:06
add a comment |
2
$begingroup$
What is "the equal" of $psi_{11}psi_{11}$?
$endgroup$
– Ben W
Dec 9 '18 at 16:53
$begingroup$
Your text is very unclear. Why do you need two indices while the first one is always 1 ? Moreover, in expression $psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$, one can simplify by $psi_{11}$... then...$psi_{12}= dfrac{4}{sqrt{10}}$ ?
$endgroup$
– Jean Marie
Dec 9 '18 at 19:06
2
2
$begingroup$
What is "the equal" of $psi_{11}psi_{11}$?
$endgroup$
– Ben W
Dec 9 '18 at 16:53
$begingroup$
What is "the equal" of $psi_{11}psi_{11}$?
$endgroup$
– Ben W
Dec 9 '18 at 16:53
$begingroup$
Your text is very unclear. Why do you need two indices while the first one is always 1 ? Moreover, in expression $psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$, one can simplify by $psi_{11}$... then...$psi_{12}= dfrac{4}{sqrt{10}}$ ?
$endgroup$
– Jean Marie
Dec 9 '18 at 19:06
$begingroup$
Your text is very unclear. Why do you need two indices while the first one is always 1 ? Moreover, in expression $psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$, one can simplify by $psi_{11}$... then...$psi_{12}= dfrac{4}{sqrt{10}}$ ?
$endgroup$
– Jean Marie
Dec 9 '18 at 19:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your sequence of function $psi_{1a}(t)$ can be expressed in terms of Legendre polynomials
$$psi_{1a}(t) = sqrt{2(2a+1)} P_a(4t-1)$$
It is known that
$$int_{-1}^1 P_a(x) P_b(x) P_c(x) dx = 2 begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2$$
where
$begin{pmatrix}j_1 & j_2 & j_3\m_1 & m_2 & m_3end{pmatrix}$ is the Wigner 3j symbol.
Together with the orthogonality relation of Legendre polynomials
$$int_{-1}^1 P_a(x)P_b(x) dx =frac{2}{2a+1} delta_{ab}$$
One can deduce
$$psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b}
sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$$
For $(a,b) = (1,3)$ and $(a,b) = (2,4)$, this reduces to
$$begin{align}
psi_{11}psi_{13} &= sqrt{frac{54}{35}} psi_{12} + sqrt{frac{32}{21}} psi_{14}\
psi_{12}psi_{14} &= sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}
end{align}$$
I don't know offhand the formula for general Wigner 3j-symbols. I just ask a CAS to compute the coefficients for me. I remember the formula for $m_1 = m_2 = m_2 = 0$ is not that horrible, you need to dig that out yourself.
Update
I found the formula we need from L&L's Quantum Mechanics (Chapter XIV, addition of angular momentum).
After simplification, the formula isn't that bad.
When $2p = a+b+c$ is even, we have
$$begin{pmatrix}a & b & c\ 0 & 0 & 0end{pmatrix}
= (-1)^p left[frac{binom{2u}{u}binom{2v}{v}binom{2w}{w}}{(2p+1)binom{2p}{p}}right]^{1/2}
quadtext{ where }quad
begin{cases} 2u &= b + c - a \ 2v &= c + a - b\ 2w &= a+b - cend{cases}$$
and the 3j symbols vanish for odd $2p$.
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
$endgroup$
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
1
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan byload(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).
$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
Your sequence of function $psi_{1a}(t)$ can be expressed in terms of Legendre polynomials
$$psi_{1a}(t) = sqrt{2(2a+1)} P_a(4t-1)$$
It is known that
$$int_{-1}^1 P_a(x) P_b(x) P_c(x) dx = 2 begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2$$
where
$begin{pmatrix}j_1 & j_2 & j_3\m_1 & m_2 & m_3end{pmatrix}$ is the Wigner 3j symbol.
Together with the orthogonality relation of Legendre polynomials
$$int_{-1}^1 P_a(x)P_b(x) dx =frac{2}{2a+1} delta_{ab}$$
One can deduce
$$psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b}
sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$$
For $(a,b) = (1,3)$ and $(a,b) = (2,4)$, this reduces to
$$begin{align}
psi_{11}psi_{13} &= sqrt{frac{54}{35}} psi_{12} + sqrt{frac{32}{21}} psi_{14}\
psi_{12}psi_{14} &= sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}
end{align}$$
I don't know offhand the formula for general Wigner 3j-symbols. I just ask a CAS to compute the coefficients for me. I remember the formula for $m_1 = m_2 = m_2 = 0$ is not that horrible, you need to dig that out yourself.
Update
I found the formula we need from L&L's Quantum Mechanics (Chapter XIV, addition of angular momentum).
After simplification, the formula isn't that bad.
When $2p = a+b+c$ is even, we have
$$begin{pmatrix}a & b & c\ 0 & 0 & 0end{pmatrix}
= (-1)^p left[frac{binom{2u}{u}binom{2v}{v}binom{2w}{w}}{(2p+1)binom{2p}{p}}right]^{1/2}
quadtext{ where }quad
begin{cases} 2u &= b + c - a \ 2v &= c + a - b\ 2w &= a+b - cend{cases}$$
and the 3j symbols vanish for odd $2p$.
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
$endgroup$
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
1
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan byload(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).
$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
|
show 1 more comment
$begingroup$
Your sequence of function $psi_{1a}(t)$ can be expressed in terms of Legendre polynomials
$$psi_{1a}(t) = sqrt{2(2a+1)} P_a(4t-1)$$
It is known that
$$int_{-1}^1 P_a(x) P_b(x) P_c(x) dx = 2 begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2$$
where
$begin{pmatrix}j_1 & j_2 & j_3\m_1 & m_2 & m_3end{pmatrix}$ is the Wigner 3j symbol.
Together with the orthogonality relation of Legendre polynomials
$$int_{-1}^1 P_a(x)P_b(x) dx =frac{2}{2a+1} delta_{ab}$$
One can deduce
$$psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b}
sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$$
For $(a,b) = (1,3)$ and $(a,b) = (2,4)$, this reduces to
$$begin{align}
psi_{11}psi_{13} &= sqrt{frac{54}{35}} psi_{12} + sqrt{frac{32}{21}} psi_{14}\
psi_{12}psi_{14} &= sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}
end{align}$$
I don't know offhand the formula for general Wigner 3j-symbols. I just ask a CAS to compute the coefficients for me. I remember the formula for $m_1 = m_2 = m_2 = 0$ is not that horrible, you need to dig that out yourself.
Update
I found the formula we need from L&L's Quantum Mechanics (Chapter XIV, addition of angular momentum).
After simplification, the formula isn't that bad.
When $2p = a+b+c$ is even, we have
$$begin{pmatrix}a & b & c\ 0 & 0 & 0end{pmatrix}
= (-1)^p left[frac{binom{2u}{u}binom{2v}{v}binom{2w}{w}}{(2p+1)binom{2p}{p}}right]^{1/2}
quadtext{ where }quad
begin{cases} 2u &= b + c - a \ 2v &= c + a - b\ 2w &= a+b - cend{cases}$$
and the 3j symbols vanish for odd $2p$.
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
$endgroup$
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
1
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan byload(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).
$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
|
show 1 more comment
$begingroup$
Your sequence of function $psi_{1a}(t)$ can be expressed in terms of Legendre polynomials
$$psi_{1a}(t) = sqrt{2(2a+1)} P_a(4t-1)$$
It is known that
$$int_{-1}^1 P_a(x) P_b(x) P_c(x) dx = 2 begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2$$
where
$begin{pmatrix}j_1 & j_2 & j_3\m_1 & m_2 & m_3end{pmatrix}$ is the Wigner 3j symbol.
Together with the orthogonality relation of Legendre polynomials
$$int_{-1}^1 P_a(x)P_b(x) dx =frac{2}{2a+1} delta_{ab}$$
One can deduce
$$psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b}
sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$$
For $(a,b) = (1,3)$ and $(a,b) = (2,4)$, this reduces to
$$begin{align}
psi_{11}psi_{13} &= sqrt{frac{54}{35}} psi_{12} + sqrt{frac{32}{21}} psi_{14}\
psi_{12}psi_{14} &= sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}
end{align}$$
I don't know offhand the formula for general Wigner 3j-symbols. I just ask a CAS to compute the coefficients for me. I remember the formula for $m_1 = m_2 = m_2 = 0$ is not that horrible, you need to dig that out yourself.
Update
I found the formula we need from L&L's Quantum Mechanics (Chapter XIV, addition of angular momentum).
After simplification, the formula isn't that bad.
When $2p = a+b+c$ is even, we have
$$begin{pmatrix}a & b & c\ 0 & 0 & 0end{pmatrix}
= (-1)^p left[frac{binom{2u}{u}binom{2v}{v}binom{2w}{w}}{(2p+1)binom{2p}{p}}right]^{1/2}
quadtext{ where }quad
begin{cases} 2u &= b + c - a \ 2v &= c + a - b\ 2w &= a+b - cend{cases}$$
and the 3j symbols vanish for odd $2p$.
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
$endgroup$
Your sequence of function $psi_{1a}(t)$ can be expressed in terms of Legendre polynomials
$$psi_{1a}(t) = sqrt{2(2a+1)} P_a(4t-1)$$
It is known that
$$int_{-1}^1 P_a(x) P_b(x) P_c(x) dx = 2 begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2$$
where
$begin{pmatrix}j_1 & j_2 & j_3\m_1 & m_2 & m_3end{pmatrix}$ is the Wigner 3j symbol.
Together with the orthogonality relation of Legendre polynomials
$$int_{-1}^1 P_a(x)P_b(x) dx =frac{2}{2a+1} delta_{ab}$$
One can deduce
$$psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b}
sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$$
For $(a,b) = (1,3)$ and $(a,b) = (2,4)$, this reduces to
$$begin{align}
psi_{11}psi_{13} &= sqrt{frac{54}{35}} psi_{12} + sqrt{frac{32}{21}} psi_{14}\
psi_{12}psi_{14} &= sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}
end{align}$$
I don't know offhand the formula for general Wigner 3j-symbols. I just ask a CAS to compute the coefficients for me. I remember the formula for $m_1 = m_2 = m_2 = 0$ is not that horrible, you need to dig that out yourself.
Update
I found the formula we need from L&L's Quantum Mechanics (Chapter XIV, addition of angular momentum).
After simplification, the formula isn't that bad.
When $2p = a+b+c$ is even, we have
$$begin{pmatrix}a & b & c\ 0 & 0 & 0end{pmatrix}
= (-1)^p left[frac{binom{2u}{u}binom{2v}{v}binom{2w}{w}}{(2p+1)binom{2p}{p}}right]^{1/2}
quadtext{ where }quad
begin{cases} 2u &= b + c - a \ 2v &= c + a - b\ 2w &= a+b - cend{cases}$$
and the 3j symbols vanish for odd $2p$.
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
edited Dec 10 '18 at 7:41
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
answered Dec 9 '18 at 18:09
achille huiachille hui
96k5132258
96k5132258
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
1
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan byload(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).
$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
|
show 1 more comment
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
1
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan byload(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).
$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
$begingroup$
Thank you very much. Which CAS software? Could you share the code?
$endgroup$
– HD239
Dec 9 '18 at 19:19
1
1
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan by
load(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
@HD239 I use maxima, after loading the package clebsch_gordan by
load(clebsch_gordan), one can compute the 3j symbols using the function wigner_3j(j1,j2,j3,m1,m2,m3).$endgroup$
– achille hui
Dec 9 '18 at 19:32
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
I wrote a [Maple code][1] for calculating. But there is a wrong something. For example; In book; for $(a,b) = (2,2)$, we have begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} end{align} But in my code, for $(a,b) = (2,2)$, I find it as begin{align} psi_{12}psi_{12} &=sqrt{2}psi(1, 0)+{frac {20 psi left( 1,2 right) }{7sqrt{10}}} +{frac {6sqrt{2} psi left( 1,4 right) }{7}} end{align} [1]: mapleprimes.com/view.aspx?sf=255175_Answer/the_last_question.mw
$endgroup$
– HD239
Dec 11 '18 at 8:54
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Your book cannot be correct. $psi_{12}$ is a quadratic polynomial in $t$. So $psi_{12}psi_{12}$ is a quartic polynomial, the expression it gives $sqrt{2}psi_{10} + frac{20}{7sqrt{10}}psi_{12}$ is a quadratic polynomial...
$endgroup$
– achille hui
Dec 11 '18 at 9:55
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
$begingroup$
Dear @achille hui, is there something wrong in the following formula? $psi_{1a}psi_{1b} = sum_{c=|a-b|}^{a+b} sqrt{2(2a+1)(2b+1)(2c+1)}begin{pmatrix}a & b & c\0 & 0 & 0end{pmatrix}^2 psi_{1c}$ I am trying to check it for $ psi_{12}psi_{14}$. But it is not to equal $sqrt{frac{72}{49}}psi_{12} + sqrt{frac{4000}{5929}}psi_{14} + sqrt{frac{2250}{1573}}psi_{16}$. Could you more explain how to derive it the formula? I can't completely derive it.
$endgroup$
– HD239
Dec 17 '18 at 22:14
|
show 1 more comment
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$begingroup$
What is "the equal" of $psi_{11}psi_{11}$?
$endgroup$
– Ben W
Dec 9 '18 at 16:53
$begingroup$
Your text is very unclear. Why do you need two indices while the first one is always 1 ? Moreover, in expression $psi_{12}psi_{11}= dfrac{4}{sqrt{10}}psi_{11}$, one can simplify by $psi_{11}$... then...$psi_{12}= dfrac{4}{sqrt{10}}$ ?
$endgroup$
– Jean Marie
Dec 9 '18 at 19:06