How can we use the integration trick for normalisation of wave functions?

[bartieshaw]

Hi,

This integral seems to be coming up a fair bit in questions involving normalisation of wave functions

\int x^2\exp{(-ax^2)}dx

and my tutor and lecturer both say to just use the fact that

x^2\exp{(-ax^2)}=-\frac{d}{da}\exp{(-ax^2)}

my question simply is how do we use this. it may be obvious can you say

\int x^2\exp{(-ax^2)}dx=\int-\frac{d}{da}\exp{(-ax^2)}dx=\int-\frac{dx}{da}\exp{(-ax^2)}d=-\frac{dx}{da}\exp{(-ax^2)}

...?

cheers for any help,

Bart

 

[quasar987]

It's like this:

\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx=\int_{-\infty}^{+\infty}-\frac{d}{da}\exp{(-ax^2)}dx=-\frac{d}{da}\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx

And he assumes that you know what \int_{-\infty}^{+\infty}\exp{(-ax^2)}dx equals to.

 

[bartieshaw]

It's like this:

\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx=\int_{-\infty}^{+\infty}-\frac{d}{da}\exp{(-ax^2)}dx=-\frac{d}{da}\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx

And he assumes that you know what \int_{-\infty}^{+\infty}\exp{(-ax^2)}dx equals to.

well i do know that

\int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}

so then my next question is how do we work that out (rather than me just knowing it). i started off with integration by parts and then kept on going in circles gettin back to an integral similar to

\int_{-\infty}^{+\infty}x^2\exp{(-ax^2)}dx

this isn't an essential question now to what I am doing, just out of interest really...


cheers

bart

 

[quasar987]

You mean, how do we know that \int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}??

I did it there some time ago:

https://www.physicsforums.com/showthread.php?t=140779&highlight=integral

see post #5.

 

[quasar987]

Talking about the identity \int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}, Lord Kelvin once said "A mathematician is one to whom that is as obvious as two twice two makes four to you. Liouville was a mathematician." (quote taken from Spivak's calc. on manifolds)

Welcome to the club of the non-mathematicians.

 

[bartieshaw]

You mean, how do we know that \int_{-\infty}^{+\infty}\exp{(-ax^2)}dx=\sqrt{\frac{\pi}{a}}??

I did it there some time ago:

https://www.physicsforums.com/showthread.php?t=140779&highlight=integral

see post #5.

haha...

looks like i was maybe taking a little too simple approach trying to use integratoin by parts.

cheers for the help

bart

 

[HallsofIvy]

Lord Kelvin may have said "twice two is four" but he surely didn't say "two twice two"!