How to solve this Complex Integral using poles?
0
$begingroup$
I want to find the green's function of a free particle, which depends of the integral:
$$
I = frac{1}{4pi ²ir} int^{+infty}_{-infty} frac{ke^{ikr}}{E-frac{hbar²k²}{2m}+ieta} dk,.
$$
Then, using the cauchy principal value we remove the $eta$...The result is as follows:
$$
g=frac{e^{ifrac{sqrt{2mE}}{hbar}r}}{4pi r}
$$
complex-integration greens-function cauchy-principal-value
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
$endgroup$
add a comment |
0
$begingroup$
I want to find the green's function of a free particle, which depends of the integral:
$$
I = frac{1}{4pi ²ir} int^{+infty}_{-infty} frac{ke^{ikr}}{E-frac{hbar²k²}{2m}+ieta} dk,.
$$
Then, using the cauchy principal value we remove the $eta$...The result is as follows:
$$
g=frac{e^{ifrac{sqrt{2mE}}{hbar}r}}{4pi r}
$$
complex-integration greens-function cauchy-principal-value
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
$endgroup$
$begingroup$
Are you familiar with contour integration?
$endgroup$
– Mark Viola
Dec 9 '18 at 23:02
$begingroup$
I know that I need to use something like a contour integral + integral from -R to R with R-> Infinity.
$endgroup$
– Lucas Lopes
Dec 9 '18 at 23:34
add a comment |
0
0
0
$begingroup$
I want to find the green's function of a free particle, which depends of the integral:
$$
I = frac{1}{4pi ²ir} int^{+infty}_{-infty} frac{ke^{ikr}}{E-frac{hbar²k²}{2m}+ieta} dk,.
$$
Then, using the cauchy principal value we remove the $eta$...The result is as follows:
$$
g=frac{e^{ifrac{sqrt{2mE}}{hbar}r}}{4pi r}
$$
complex-integration greens-function cauchy-principal-value
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
$endgroup$
I want to find the green's function of a free particle, which depends of the integral:
$$
I = frac{1}{4pi ²ir} int^{+infty}_{-infty} frac{ke^{ikr}}{E-frac{hbar²k²}{2m}+ieta} dk,.
$$
Then, using the cauchy principal value we remove the $eta$...The result is as follows:
$$
g=frac{e^{ifrac{sqrt{2mE}}{hbar}r}}{4pi r}
$$
complex-integration greens-function cauchy-principal-value
complex-integration greens-function cauchy-principal-value
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
asked Dec 9 '18 at 22:50
Lucas LopesLucas Lopes
11
11
$begingroup$
Are you familiar with contour integration?
$endgroup$
– Mark Viola
Dec 9 '18 at 23:02
$begingroup$
I know that I need to use something like a contour integral + integral from -R to R with R-> Infinity.
$endgroup$
– Lucas Lopes
Dec 9 '18 at 23:34
add a comment |
$begingroup$
Are you familiar with contour integration?
$endgroup$
– Mark Viola
Dec 9 '18 at 23:02
$begingroup$
I know that I need to use something like a contour integral + integral from -R to R with R-> Infinity.
$endgroup$
– Lucas Lopes
Dec 9 '18 at 23:34
$begingroup$
Are you familiar with contour integration?
$endgroup$
– Mark Viola
Dec 9 '18 at 23:02
$begingroup$
Are you familiar with contour integration?
$endgroup$
– Mark Viola
Dec 9 '18 at 23:02
$begingroup$
I know that I need to use something like a contour integral + integral from -R to R with R-> Infinity.
$endgroup$
– Lucas Lopes
Dec 9 '18 at 23:34
$begingroup$
I know that I need to use something like a contour integral + integral from -R to R with R-> Infinity.
$endgroup$
– Lucas Lopes
Dec 9 '18 at 23:34
add a comment |
1 Answer
1
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oldest
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0
$begingroup$
I'll give you some hints:
- Get your integral in more combined form:
$$ I(r) = int frac{k e^{i k r}}{A - k^2} $$
here $A$ is constant. Forget about everything else till the end. - Think of the function $F(r)$, s. t.
$$ F'(r) = I(r) $$
- Draw some contours and see which one is better for $F(r)$. If nothing comes to mind, try literature, I would recommend: Russell L. Herman, An introduction to mathematical physics via oscillations
- Now return to your preintegral constant and substitute A.
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
I'll give you some hints:
- Get your integral in more combined form:
$$ I(r) = int frac{k e^{i k r}}{A - k^2} $$
here $A$ is constant. Forget about everything else till the end. - Think of the function $F(r)$, s. t.
$$ F'(r) = I(r) $$
- Draw some contours and see which one is better for $F(r)$. If nothing comes to mind, try literature, I would recommend: Russell L. Herman, An introduction to mathematical physics via oscillations
- Now return to your preintegral constant and substitute A.
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
$endgroup$
add a comment |
0
$begingroup$
I'll give you some hints:
- Get your integral in more combined form:
$$ I(r) = int frac{k e^{i k r}}{A - k^2} $$
here $A$ is constant. Forget about everything else till the end. - Think of the function $F(r)$, s. t.
$$ F'(r) = I(r) $$
- Draw some contours and see which one is better for $F(r)$. If nothing comes to mind, try literature, I would recommend: Russell L. Herman, An introduction to mathematical physics via oscillations
- Now return to your preintegral constant and substitute A.
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
$endgroup$
add a comment |
0
0
0
$begingroup$
I'll give you some hints:
- Get your integral in more combined form:
$$ I(r) = int frac{k e^{i k r}}{A - k^2} $$
here $A$ is constant. Forget about everything else till the end. - Think of the function $F(r)$, s. t.
$$ F'(r) = I(r) $$
- Draw some contours and see which one is better for $F(r)$. If nothing comes to mind, try literature, I would recommend: Russell L. Herman, An introduction to mathematical physics via oscillations
- Now return to your preintegral constant and substitute A.
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
$endgroup$
I'll give you some hints:
- Get your integral in more combined form:
$$ I(r) = int frac{k e^{i k r}}{A - k^2} $$
here $A$ is constant. Forget about everything else till the end. - Think of the function $F(r)$, s. t.
$$ F'(r) = I(r) $$
- Draw some contours and see which one is better for $F(r)$. If nothing comes to mind, try literature, I would recommend: Russell L. Herman, An introduction to mathematical physics via oscillations
- Now return to your preintegral constant and substitute A.
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
answered Dec 26 '18 at 8:50
Al PrihodkoAl Prihodko
436
436
add a comment |
add a comment |
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$begingroup$
Are you familiar with contour integration?
$endgroup$
– Mark Viola
Dec 9 '18 at 23:02
$begingroup$
I know that I need to use something like a contour integral + integral from -R to R with R-> Infinity.
$endgroup$
– Lucas Lopes
Dec 9 '18 at 23:34