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?
$begingroup$
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?
integration
asked Dec 22 '18 at 19:57
JolyonJolyon
625
$endgroup$
add a comment |
$begingroup$
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?
integration
asked Dec 22 '18 at 19:57
JolyonJolyon
625
$endgroup$
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
add a comment |
$begingroup$
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?
integration
asked Dec 22 '18 at 19:57
JolyonJolyon
625
$endgroup$
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
integration
asked Dec 22 '18 at 19:57
JolyonJolyon
625
asked Dec 22 '18 at 19:57
JolyonJolyon
625
asked Dec 22 '18 at 19:57
JolyonJolyon
625
asked Dec 22 '18 at 19:57
JolyonJolyon
625
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
$endgroup$
add a comment |
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1 Answer
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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.
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
$endgroup$
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.
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
answered Dec 22 '18 at 20:43
WintherWinther
20.7k33156
20.7k33156
add a comment |
add a comment |
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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