# Tricky Gauss Integral problem

• November 21st 2010, 02:37 AM
bugatti79
Tricky Gauss Integral problem
Hi, I am having difficulty getting to the end with this one...see attached

Thanks
• November 22nd 2010, 09:30 PM
Chris L T521
Quote:

Originally Posted by bugatti79
Hi, I am having difficulty getting to the end with this one...see attached

Thanks

Keep in mind that $\displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$. (*)

I take it you mean the original problem is $\displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk$, because the 2m term appears out of nowhere in the second step.

For the moment, let's focus on the exponent:

$-\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk$.

Rearranging, we get

$-\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right]$ (Verify)

Complete the square to get

$-\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$ (Verify)

With this, we can now evaluate this integral instead:

$\displaystyle\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2-\frac{amx^2}{m+2ai\hbar t}\right)\,dk$ $\displaystyle=\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2\right)\,dk$

Now apply the substitution $u=k-\dfrac{2amix}{m+2ai\hbar t}\implies \,du=\,dk$.

Thus, the integral becomes

$\displaystyle\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}u^2\right)\,du$.

Now apply (*) to get

$\exp\left(-\dfrac{amx^2}{m+2ai\hbar t}\right)\sqrt{\dfrac{4am\pi}{m+2ai\hbar t}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{m+2ai\hbar t}{4am\pi}}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{1}{4a\pi}+\dfrac{i\hbar t}{2m\pi}}}$

which appears to be different from the answer you have supplied....so maybe I made a mistake somewhere??

But this is the idea of how to approach this problem. Hope this helps!
• November 22nd 2010, 09:33 PM
bugatti79
Quote:

Originally Posted by Chris L T521
Keep in mind that $\displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$. (*)

I take it you mean the original problem is $\displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk$, because the 2m term appears out of nowhere in the second step.

For the moment, let's focus on the exponent:

$-\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk$.

Rearranging, we get

$-\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right]$ (Verify)

Complete the square to get

$-\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$ (Verify)

With this, we can now evaluate this integral instead:

$\displaystyle\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2-\frac{amx^2}{m+2ai\hbar t}\right)\,dk$ $\displaystyle=\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2\right)\,dk$

Now apply the substitution $u=k-\dfrac{2amix}{m+2ai\hbar t}\implies \,du=\,dk$.

Thus, the integral becomes

$\displaystyle\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}u^2\right)\,du$.

Now apply (*) to get

$\exp\left(-\dfrac{amx^2}{m+2ai\hbar t}\right)\sqrt{\dfrac{4am\pi}{m+2ai\hbar t}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{m+2ai\hbar t}{4am\pi}}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{1}{4a\pi}+\dfrac{i\hbar t}{2m\pi}}}$

which appears to be different from the answer you have supplied....so maybe I made a mistake somewhere??

But this is the idea of how to approach this problem. Hope this helps!

Hi Chris,

Thanks for your great help, I will have a look at it when I get the chance and will reply soon.

Cheers!
• November 25th 2010, 09:24 AM
bugatti79
Quote:

Originally Posted by Chris L T521
Keep in mind that $\displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$. (*)

I take it you mean the original problem is $\displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk$, because the 2m term appears out of nowhere in the second step.

For the moment, let's focus on the exponent:

$-\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk$.

Rearranging, we get

$-\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right]$ (Verify)

Complete the square to get

$-\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$ (Verify)

Yes your right, the 2m should be in there from the start.
I can see the algebra but i dont understand the idea of completing the square. Why do we have to re-arrange it when we already have it the form -(ak^2 +bk)?....
• November 25th 2010, 10:02 AM
Chris L T521
Quote:

Originally Posted by bugatti79
Yes your right, the 2m should be in there from the start.
I can see the algebra but i dont understand the idea of completing the square. Why do we have to re-arrange it when we already have it the form -(ak^2 +bk)?....

We have to rearrange it because completing the square only works when the coefficient of the $k^2$ term is 1.
• November 25th 2010, 12:32 PM
bugatti79
Quote:

Originally Posted by bugatti79
Yes your right, the 2m should be in there from the start.
I can see the algebra but i dont understand the idea of completing the square. Why do we have to re-arrange it when we already have it the form -(ak^2 +bk)?....

You mean the coefficient of k has to be 1 and not k^ as below in order to get du=dk?

$-\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$
• November 25th 2010, 12:43 PM
Chris L T521
Quote:

Originally Posted by bugatti79
You mean the coefficient of k has to be 1 and not k^ as below in order to get du=dk?

$-\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$

As I said in my previous response, you need to have the coefficient of the $k^2$ term to be 1 in order to complete the square. Thats why I did the following:

Quote:

Originally Posted by Chris L T521
For the moment, let's focus on the exponent:

$-\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk$.

Rearranging, we get

$-\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right]$ (Verify)

Now that its in the proper form, we complete the square -- i.e. $k^2+bk=k^2+bk+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 = \left(k+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$:

Quote:

Originally Posted by Chris L T521
Complete the square to get

$-\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$ (Verify)

So when you go back to the integral in question, the substitution becomes $u=k-\dfrac{2amix}{m+2ai\hbar t}\implies\,du=\,dk$

Then the rest of the problem becomes algebra once you apply the formula for the Gaussian.

Does this clarify things?
• November 25th 2010, 01:56 PM
bugatti79
Quote:

Originally Posted by Chris L T521
As I said in my previous response, you need to have the coefficient of the $k^2$ term to be 1 in order to complete the square. Thats why I did the following:

Now that its in the proper form, we complete the square -- i.e. $k^2+bk=k^2+bk+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 = \left(k+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$:

So when you go back to the integral in question, the substitution becomes $u=k-\dfrac{2amix}{m+2ai\hbar t}\implies\,du=\,dk$

Then the rest of the problem becomes algebra once you apply the formula for the Gaussian.

Does this clarify things?

aha, that is perfectly clear now. That is the first time I have seen the 'complete the square' used in an integral problem. Thanks alot Chris.