Solving Quantum Mechanics Integrals: Prove This One

[maverick6664]

Hi all,
Reading a book on quantum mechanics, I cannot understand an integral.
Now I can understand this one.
http://www.geocities.jp/badtrans666/physics/int1.gif
however I don't understand this.
http://www.geocities.jp/badtrans666/physics/int2.gif
As a matter of course, x, t are real numbers.
Maybe I'm missing something easy...
Thanks in advance! (btw how can I display these as images? I'm using img tags)

 

[matt grime]

substitution x=t+i and not worrying too much seems to make it make more sense.

 

[arildno]

Consider the contour C in the complex plane consisting of the line segments (separately parametrized with "t"):
L_{1}: t+i, -a\leq{t}\leq{a}
L_{2}: a+(1-t)i, 0\leq{t}\leq{1}
L_{3}: -t, -a\leq{t}\leq{a}
L_{4}: a+ti 0\leq{t}\leq{1}

Consider now the (complex) integral \oint_{C}exp(-z*z)dz
where the complex variable z is to be evaluated along the contour C.
Since the integrand is an analytical function we have, by Cauchy's theorem:
\oint_{C}exp(-z*z)dz=0
This is analogous to the real (multi-)variable theorem that says that the integral of a gradient field along a closed contour is zero.

Furthermore, w the complex integral is additive, so we may split up the integral over C in 4 integrals over the 4 line segments:
\oint_{L_{1}}exp(-z*z)dz+\oint_{L_{2}}exp(-z*z)dz+\oint_{L_{3}}exp(-z*z)dz+\oint_{L_{4}}exp(-z*z)dz=0

Now, let us look at the limiting expression when we let "a" go towards infinity:
Every complex point on the vertical strips L_{2},L{4}[/tex] will get bigger and bigger modulus. But that means that the two integrals along these strips will decrease in value, reaching 0 in the limit.<br /> <br /> Thus, we are left with the expression:<br /> \int_{-\infty}exp(-(t+i)*(t+i))dt+\int_{\infty}exp(-t*t)dt=0<br /> Here, the upper limit in the first integral is infinity, whereas the upper limit in the second integral is negative infinity. (My keyboard is working against me!)<br /> <br /> Switching upper and lower limits in the last integral effects the identity you were after.

 

[maverick6664]

Thanks! It's so easy...

I noticed

exp(-(t+i)^2) is the conjugate of exp(-(-t+i)^2) !

arildno's explanation is very interesting.

 

[maverick6664]

Thanks! It's so easy...
I noticed
exp(-(t+i)^2) is the conjugate of exp(-(-t+i)^2) !
arildno's explanation is very interesting.

Let me practice tex on this forum:
exp(-(t+i)^2)
is the conjugate of
exp(-(-t+i)^2)

So

\int^{+\infty}_{-\infty} exp(-(x+i)^2)dx = \int^{+\infty}_{_\infty} exp(-x^2)dx

Oh! cool tex implementation!