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I posted this in the mathematical forums. Maybe you will help me. I found an hard article of yang huang and luttinger. The authors begins with the sum: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi ^2 a^2 \lambda ^2 }}{{V^2 }}\sum\limits_{} {'\frac{{\langle n_\alpha \rangle \langle n_\gamma \rangle \langle n_\lambda \rangle }}{{\frac{1}{2}\left( {k_\alpha ^2 + k_\beta ^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}\delta \left( {\vec k{}_\alpha + \vec k{}_\beta - \vec k{}_\gamma - \vec k{}_\lambda } \right)} $$ 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: $$u = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{2\left( {j - k} \right)l}}{{j + k + l}}} \right) $$ $$v = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l}}{{j + k + l}}} \right) $$ $$w = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l + 4jk}}{{j + k + l}}} \right) $$ $\frac{{\partial J}}{{\partial u}}$ is calculated here: Maybe we don't have to do thermal average of n and we need : $$n = \frac{1}{{z^{ - 1} e^{\varepsilon /kT} - 1}} = \sum\limits_{n = 1}^\infty {\left( {ze^{ - \varepsilon /kT} } \right)^l } $$ 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: $$\int\limits_0^\infty {\frac{{e^{ - \frac{{\hbar ^2 k_a^2j }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\gamma ^2k }}{{2m}}} e^{ - \frac{{\hbar ^2 k_\lambda ^2l }}{{2m}}} }}{{\frac{1}{2}\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}k_a k_b k_\gamma k_\lambda } dk_a dk_b dk_\gamma dk_\lambda $$ We have to evaluate two integrals, i think. Do you have any suggestion?

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Presumably the authors have started by addressing the problem in a limited case with discrete states and then taken a limit in which the discrete states become a continuum. The classic example is quantizing the electromagnetic field in a perfectly conductive box and then taking the limit as the box grows without bound to get the quantization of the EM field in free space. Indeed, this appears to be a very similar procedure. This was discussed a little as What's a good reference for this classical picture Feynman's talking about?. – dmckee Sep 13 '11 at 20:37

I have tried to solve this integral: $$F\left( {a,b,c,d} \right) = \int {\frac{{e^{ - \frac{{\hbar ^2 }}{{2m}}\left( {ak_a^2 + bk_b^2 + ck_\gamma ^2 + dk_\lambda ^2 } \right)} }}{{\left( {k_a^2 + k_b^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}k_a k_b k_\gamma k_\lambda dk_a dk_b dk_\gamma dk_\lambda } $$ from which we can calculate our integral putting $b=0$. We have: $$\frac{{\partial F}}{{\partial a}} + \frac{{\partial F}}{{\partial b}} - \frac{{\partial F}}{{\partial c}} - \frac{{\partial F}}{{\partial d}} \propto \frac{1}{{abcd}} $$ Now we put: $$i=a+b+c+d$$ $$l=a+b-c-d$$ $$m=a-b$$ $$n=c-d$$ So: $$\frac{\partial }{{\partial a}} = \frac{\partial }{{\partial i}} + \frac{\partial }{{\partial l}} + \frac{\partial }{{\partial m}} $$ $$\frac{\partial }{{\partial b}} = \frac{\partial }{{\partial i}} + \frac{\partial }{{\partial l}} - \frac{\partial }{{\partial m}} $$ $$\frac{\partial }{{\partial c}} = \frac{\partial }{{\partial i}} - \frac{\partial }{{\partial l}} + \frac{\partial }{{\partial n}} $$ $$\frac{\partial }{{\partial d}} = \frac{\partial }{{\partial i}} - \frac{\partial }{{\partial l}} - \frac{\partial }{{\partial n}} $$ $$\frac{\partial }{{\partial a}} + \frac{\partial }{{\partial b}} - \frac{\partial }{{\partial c}} - \frac{\partial }{{\partial d}} = 4\frac{\partial }{{\partial l}} $$ and: $$a=(i+l+2m)/4$$ $$b=(i+l-2m)/4$$ $$c=(i-l+2n)/4$$ $$d=(i-l-2n)/4$$ and the equation becomes: $$\frac{{\partial F'}}{{\partial l}} \propto \frac{1}{{\left( {\frac{{i + l}}{2} + m} \right)\left( {\frac{{i + l}}{2} - m} \right)\left( {\frac{{i - l}}{2} + n} \right)\left( {\frac{{i - l}}{2} - n} \right)}} $$ The integration can be evaluated with derive, and the substitutions gives an expression divergent if we put $b=0$. What is wrong?

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Think of $k_\alpha$ and $k_\beta$ as $x$ and $y$, so $dk_\alpha\;dk_\beta = dx\;dy.$ Now convert to polar coordinates = $rdr\;d\theta$. The integral over theta is trivial. Repeat with the other two wave vectors and you end up with two integrals.

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Hey, do you want me to actually have to type the trivial calculation in? Herp derp. – Carl Brannen Nov 20 '11 at 5:52
Sorry i respond you soo late.Can you give me the calculations??? – Pablo Dec 27 '11 at 21:58
There's no calculation to type in. Simply replace $dk_\alpha\;dk_\beta = r\;dr\;d\theta$ where $k_\alpha$ and $k_\beta$ run from between + and - infinity, while r runs from 0 to infinity and theta runs from 0 to 2 pi. This is the usual Cartesian to polar conversion. – Carl Brannen Jan 6 '12 at 15:48
The result is very difficul to write, i don't think it's soo easy to get it. I tried to use polar coordinates but i don't know how to integrate! – Pablo Jan 6 '12 at 23:27
Try using the substitution $\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} H(k_\alpha,k_\beta)\;dk_\alpha\;dk_\beta$ $= \int_{r=0}^\infty\int_{\theta=0}^{2\pi}H(r\cos(\theta),r\sin(\theta))r\;d\theta \; dr$ $= \int_{r=0}^\infty\int_{\theta=0}^{2\pi}H(r) r\;d\theta \; dr$ $= 2\pi \int_{r=0}^\infty H(r)\;r \;dr$. This is a method of simplifying an integral of H that depends on the requirement that H simplify. It reduces two integrals down to 1. Also note that if the original integral of H was from 0 to infinity instead of -infinity to +infinity, then you have to make some simple changes. – Carl Brannen Jan 8 '12 at 5:46

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