Why is $int d^3r_1 , d^3 r_2 , frac{F(vec{r}_1) F^*(vec{r}_2)}{|vec{r}_1 - vec{r}_2|}$ positive?












3












$begingroup$


I have the following quantity:



$$
I = int_{mathbb{R}^3 times mathbb{R}^3} d^3r_1 , d^3 r_2 , frac{F(vec{r}_1) F^*(vec{r}_2)}{|vec{r}_1 - vec{r}_2|}
$$



$I$ is obviously real by symmetry. How can I show that it is also positive semidefinite?










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  • 2




    $begingroup$
    The solution to Poisson's equation $nabla^2 J = F$ is $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-frac{1}{4pi}int J(r_2)nabla^2 J^*(r_2){rm d}r_2$. Integration by parts.
    $endgroup$
    – Winther
    Dec 22 '18 at 20:36










  • $begingroup$
    Very suave! If you want to write that up as an answer, I'll accept it.
    $endgroup$
    – Jolyon
    Dec 22 '18 at 20:40
















3












$begingroup$


I have the following quantity:



$$
I = int_{mathbb{R}^3 times mathbb{R}^3} d^3r_1 , d^3 r_2 , frac{F(vec{r}_1) F^*(vec{r}_2)}{|vec{r}_1 - vec{r}_2|}
$$



$I$ is obviously real by symmetry. How can I show that it is also positive semidefinite?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The solution to Poisson's equation $nabla^2 J = F$ is $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-frac{1}{4pi}int J(r_2)nabla^2 J^*(r_2){rm d}r_2$. Integration by parts.
    $endgroup$
    – Winther
    Dec 22 '18 at 20:36










  • $begingroup$
    Very suave! If you want to write that up as an answer, I'll accept it.
    $endgroup$
    – Jolyon
    Dec 22 '18 at 20:40














3












3








3





$begingroup$


I have the following quantity:



$$
I = int_{mathbb{R}^3 times mathbb{R}^3} d^3r_1 , d^3 r_2 , frac{F(vec{r}_1) F^*(vec{r}_2)}{|vec{r}_1 - vec{r}_2|}
$$



$I$ is obviously real by symmetry. How can I show that it is also positive semidefinite?










share|cite|improve this question









$endgroup$




I have the following quantity:



$$
I = int_{mathbb{R}^3 times mathbb{R}^3} d^3r_1 , d^3 r_2 , frac{F(vec{r}_1) F^*(vec{r}_2)}{|vec{r}_1 - vec{r}_2|}
$$



$I$ is obviously real by symmetry. How can I show that it is also positive semidefinite?







integration






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asked Dec 22 '18 at 19:57









JolyonJolyon

625




625








  • 2




    $begingroup$
    The solution to Poisson's equation $nabla^2 J = F$ is $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-frac{1}{4pi}int J(r_2)nabla^2 J^*(r_2){rm d}r_2$. Integration by parts.
    $endgroup$
    – Winther
    Dec 22 '18 at 20:36










  • $begingroup$
    Very suave! If you want to write that up as an answer, I'll accept it.
    $endgroup$
    – Jolyon
    Dec 22 '18 at 20:40














  • 2




    $begingroup$
    The solution to Poisson's equation $nabla^2 J = F$ is $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-frac{1}{4pi}int J(r_2)nabla^2 J^*(r_2){rm d}r_2$. Integration by parts.
    $endgroup$
    – Winther
    Dec 22 '18 at 20:36










  • $begingroup$
    Very suave! If you want to write that up as an answer, I'll accept it.
    $endgroup$
    – Jolyon
    Dec 22 '18 at 20:40








2




2




$begingroup$
The solution to Poisson's equation $nabla^2 J = F$ is $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-frac{1}{4pi}int J(r_2)nabla^2 J^*(r_2){rm d}r_2$. Integration by parts.
$endgroup$
– Winther
Dec 22 '18 at 20:36




$begingroup$
The solution to Poisson's equation $nabla^2 J = F$ is $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$ so the integral can be written $-frac{1}{4pi}int J(r_2)nabla^2 J^*(r_2){rm d}r_2$. Integration by parts.
$endgroup$
– Winther
Dec 22 '18 at 20:36












$begingroup$
Very suave! If you want to write that up as an answer, I'll accept it.
$endgroup$
– Jolyon
Dec 22 '18 at 20:40




$begingroup$
Very suave! If you want to write that up as an answer, I'll accept it.
$endgroup$
– Jolyon
Dec 22 '18 at 20:40










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$begingroup$

Recall that the solution to Poisson's equation $nabla^2 J = F$ is given by $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written
$$I = -4piint J(r_2)nabla^2 J^*(r_2){rm d}r_2 = 4piint |nabla J(r_2)|^2 {rm d}r_2 geq 0$$
where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.






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    1 Answer
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    3












    $begingroup$

    Recall that the solution to Poisson's equation $nabla^2 J = F$ is given by $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written
    $$I = -4piint J(r_2)nabla^2 J^*(r_2){rm d}r_2 = 4piint |nabla J(r_2)|^2 {rm d}r_2 geq 0$$
    where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Recall that the solution to Poisson's equation $nabla^2 J = F$ is given by $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written
      $$I = -4piint J(r_2)nabla^2 J^*(r_2){rm d}r_2 = 4piint |nabla J(r_2)|^2 {rm d}r_2 geq 0$$
      where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Recall that the solution to Poisson's equation $nabla^2 J = F$ is given by $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written
        $$I = -4piint J(r_2)nabla^2 J^*(r_2){rm d}r_2 = 4piint |nabla J(r_2)|^2 {rm d}r_2 geq 0$$
        where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.






        share|cite|improve this answer









        $endgroup$



        Recall that the solution to Poisson's equation $nabla^2 J = F$ is given by $J(r_2) = -frac{1}{4pi}int frac{F(r_1){rm d}r_1}{|r_1-r_2|}$. Using this the integral can be written
        $$I = -4piint J(r_2)nabla^2 J^*(r_2){rm d}r_2 = 4piint |nabla J(r_2)|^2 {rm d}r_2 geq 0$$
        where we used integration by parts (as usual assuming boundary terms vanish) to arrive at the last equality.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 20:43









        WintherWinther

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        20.7k33156






























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