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

Thanks

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- Nov 21st 2010, 02:37 AMbugatti79Tricky Gauss Integral problem
Hi, I am having difficulty getting to the end with this one...see attached

Thanks - Nov 22nd 2010, 09:30 PMChris L T521
Keep in mind that $\displaystyle \displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$. (*)

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

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

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

Rearranging, we get

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

Complete the square to get

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

With this, we can now evaluate this integral instead:

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

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

Thus, the integral becomes

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

Now apply (*) to get

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

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! - Nov 22nd 2010, 09:33 PMbugatti79
- Nov 25th 2010, 09:24 AMbugatti79
- Nov 25th 2010, 10:02 AMChris L T521
- Nov 25th 2010, 12:32 PMbugatti79
- Nov 25th 2010, 12:43 PMChris L T521
As I said in my previous response, you need to have the coefficient of the $\displaystyle k^2$ 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. $\displaystyle k^2+bk=k^2+bk+\left(\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2 = \left(k+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$:

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

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

Does this clarify things? - Nov 25th 2010, 01:56 PMbugatti79