La transformación de una suma en un integrante de

Question

He publicado esto en la matemática de los foros. Tal vez usted me va a ayudar. He encontrado un duro artículo http://prola.aps.org/abstract/PR/v105/i3/p776_1 de huang yang y luttinger. Los autores comienza con la suma: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi ^2 a^2 \lambda ^2 }}{{V^2 }}\sum\limits_{} {'\frac{{\langle n_\alpha \rangle \langle n_\gamma \rangle \langle n_\lambda \rangle }}{{\frac{1}{2}\left( {k_\alpha ^2 + k_\beta ^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}\delta \left( {\vec k{}_\alpha + \vec k{}_\beta - \vec k{}_\gamma - \vec k{}_\lambda } \right)} $$ which represents the second order perturbation term of the energy of a bose gas. ${\langle n_\alpha \rangle }$ is: $$\langle n_\alpha \rangle = \sum\limits_{n = 0}^\infty {n\left( {ze^{ - \beta \varepsilon _\alpha } } \right)} ^n /\sum\limits_{n = 0}^\infty {\left( {ze^{ - \beta \varepsilon _\alpha } } \right)} ^n = \frac{{ze^{ - \beta \varepsilon _\alpha } }}{{1 - ze^{ - \beta \varepsilon _\alpha } }} $$ In the sum the terms with a vanishing denominator are omitted. There is also the restriction ${\vec k{}_\alpha \ne \vec k{}_\beta }\ \ $ and ${\vec k{}_\gamma \ne \vec k{}_\lambda }\ \ $. Now the passage that is unclear to me is the passage to the integral: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi a^2 \lambda ^2 }}{{V^2 }}\left( {\frac{{4V^3 }}{{\pi ^3 \lambda ^5 }}} \right)\sum\limits_{i,j,k}^\infty {\frac{{z^{j + k + l} }}{{\left( {j + k + l} \right)^{1/2} \left( {j - k} \right)l}}} \frac{{\partial J}}{{\partial u}} $$ where: $$J = \int\limits_0^\infty {\int\limits_0^\infty {dqdq\frac{{\cosh \left( {upq} \right)}}{{q^2 - p^2 }}} } e^{ - vq^2 - wp^2 } $$ and: $$u = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{2\left( {j - k} \right)l}}{{j + k + l}}} \right) $$ $$v = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l}}{{j + k + l}}} \right) $$ $$w = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l + 4jk}}{{j + k + l}}} \right) $$ $\frac{{\partial J}}{{\partial u}}$ se calcula aquí: http://math.stackexchange.com/questions/63534/difficult-integral Tal vez no tenemos que hacer térmica promedio de n y necesitamos : $$n = \frac{1}{{z^{ - 1} e^{\varepsilon /kT} - 1}} = \sum\limits_{n = 1}^\infty {\left( {ze^{ - \varepsilon /kT} } \right)^l } $$ so: $$\sum\limits_{} {'... = \sum\limits_{i,j,k = 1}^\infty {\sum\limits_{} ' } } z^{ijk} \frac{{e^{ - \frac{{\hbar ^2 k_a^2j }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\gamma ^2k }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\lambda ^2l }}{{2m}}} }}{{\frac{1}{2}\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}} $$ This explain the factor $z^{ijk}$ and the sum. How can we continue?Now we have to transform the sum' in a integral. The density of state depends of $kdk$ so i think we have to evaluate: $$\int\limits_0^\infty {\frac{{e^{ - \frac{{\hbar ^2 k_a^2j }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\gamma ^2k }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\lambda ^2l }}{{2m}}} }}{{\frac{1}{2}\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}k_a k_b k_\gamma k_\lambda } dk_a dk_b dk_\gamma dk_\lambda $$ Tenemos que evaluar las dos integrales, creo. ¿Tiene alguna sugerencia?

Answer 1

He tratado de resolver esta integral: $$F\left( {a,b,c,d} \right) = \int {\frac{{e^{ - \frac{{\manejadores ^2 }}{{2m}}\left( {ak_a^2 + bk_b^2 + ck_\gamma ^2 + dk_\lambda ^2 } \right)} }}{{\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}k_a k_b k_\gamma k_\lambda dk_a dk_b dk_\gamma dk_\lambda } $$ a partir de la cual podemos calcular nuestra integral de poner $b=0$. Tenemos: $$\frac{{\partial F}}{{\partial}} + \frac{{\partial F}}{{\partial b}} - \frac{{\partial F}}{{\partial c}} - \frac{{\partial F}}{{\partial d}} \propto \frac{1}{{abcd}} $$ Ahora ponemos: $$i=a+b+c+d$$ $$l=a+b-c-d$$ $$m=a-b$$ $$n=c-d$$ Así: $$\frac{\partial }{{\partial}} = \frac{\partial }{{\partial i}} + \frac{\partial }{{\partial l}} + \frac{\partial }{{\partial m}} $$ $$\frac{\partial }{{\partial b}} = \frac{\partial }{{\partial i}} + \frac{\partial }{{\partial l}} - \frac{\partial }{{\partial m}} $$ $$\frac{\partial }{{\partial c}} = \frac{\partial }{{\partial i}} - \frac{\partial }{{\partial l}} + \frac{\partial }{{\partial n}} $$ $$\frac{\partial }{{\partial d}} = \frac{\partial }{{\partial i}} - \frac{\partial }{{\partial l}} - \frac{\partial }{{\partial n}} $$ $$\frac{\partial }{{\partial}} + \frac{\partial }{{\partial b}} - \frac{\partial }{{\partial c}} - \frac{\partial }{{\partial d}} = 4\frac{\partial }{{\partial l}} $$ y: $$a=(i+l+2m)/4$$ $$b=(i+l-2m)/4$$ $$c=(i-l+2n)/4$$ $$d=(i-l-2n)/4$$ y la ecuación se convierte en: $$\frac{{\partial F}}{{\partial l}} \propto \frac{1}{{\left( {\frac{{i + l}}{2} + m} \right)\left( {\frac{{i + l}}{2} - m} \right)\left( {\frac{{i - l}}{2} + n} \right)\left( {\frac{{i - l}}{2} - n} \right)}} $$ La integración puede ser evaluado con derivan, y las sustituciones que se da una expresión divergente si ponemos $b=0$. Lo que está mal?

Answer 2

Creo que de $k_\alpha$$k_\beta$$x$$y$, lo $dk_\alpha\;dk_\beta = dx\;dy.$ Ahora convertir a coordenadas polares = $rdr\;d\theta$. La integral sobre la theta es trivial. Repita con los otros dos vectores de onda y se termina con las dos integrales.