Finding the equal of a expression












0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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


















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $2p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 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










  • $begingroup$
    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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032597%2ffinding-the-equal-of-a-expression%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $2p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 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










  • $begingroup$
    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


















2












$begingroup$

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 $2p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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 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










  • $begingroup$
    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
















2












2








2





$begingroup$

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 $2p$.






share|cite|improve this answer











$endgroup$



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 $2p$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 7:41

























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 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










  • $begingroup$
    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






  • 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












  • $begingroup$
    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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032597%2ffinding-the-equal-of-a-expression%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more