# Time development of a Gaussian integral help

1. Sep 5, 2012

### Don'tKnowMuch

Here is a link to a course which i am studying,

http://quantummechanics.ucsd.edu/ph130a/130_notes/node89.html#derive:timegauss

My problem comes from the k' term attached to Vsub(g) (group velocity). I used the substitution k' = k - k(0), factored out all exponentials with no k' dependence, as well grouped the remaining exponentials with a k' dependence with their respective orders. The trouble comes from the term in the second ψ(x,t) integral in the middle of the linked page; i cannot get the k' term on the group velocity*time term in the exponential to disappear. I thought i was being cautious. Any advice?

2. Sep 7, 2012

### Chopin

When performing a Gaussian integration, the linear term is always removed by completing the square, which allows you to shift the integration variable, and turn the integrand into a pure quadratic term.

Are you familiar with how to do this? If not, check the previous page in the link you provided--they go over how to do it. If so, though, then perhaps I've misunderstood your question, in which case it might be helpful if you can provide a little more detail on where you're getting hung up.

3. Sep 7, 2012

### Don'tKnowMuch

Thank you for you response!! Also sorry for the ambiguity of my query. I went over the integral from the page which is previous to the one that i linked and i understand it entirely. The integral on the page that i linked is essentially the same integral, only now t = 1 (instead of t = 0). So with the given substitutions from the linked page

1) K' = [κ - κ(0)] -----> where κ(knot) = κ(0))
2) A(κ) = e^ (-α*(κ')^2))
3) ω(κ) = ω(0) + dω(κ)/dκ * [κ - κ(0)] + (1/2!)*d^2ω(κ)/d^2 * [κ - κ(0)]^2

which is equal to...

ω(κ) = [ω(0) + V(group)*K' + β(K')^2] -------> where β = (1/2!)*d^2ω(κ)/d^2

I plug 1), 2), and 3) into the integral (∫ A(κ) * e^i(κx - ω(κ)t) dκ) to get...

∫ e^ (-α*(κ')^2)) * e^i(κx - [ω(0) + V(group)*K' + β(K')^2]t) dκ'

The next step is to multiply the integral by the exponential e^i(k(o)x)...

This method was used in the integral on the page that is previous to the one i linked. What this does is allows us to change the exponential e^iκx into e^iK'x because...

e^i[κx- κ(0)x] = e^iK'x -------> then e^iK'x * e^iκ(0)x = e^iκx

so using this idea the integral becomes...

∫ e^ (-α*(κ')^2)) * e^iK'x * e^-iω(0)t * e^-itV(group)*K' * e^-itβ(K')^2 * e^i(k(o)x) dκ'

Now it is time to factor out the terms which have no K' dependence and group the terms with a K' dependence according to their respective powers. I get...

e^i(k(o)x-iω(0)t) ∫ e^K'^2 (-α - itβ) * e^[iK'x -itV(group)*K'] dκ'

This is almost exactly what i should have. The only trouble is the V(group) term which has a pesky K' term attached to it. That whole exponential should read...

e^[iK'x -itV(group)]

Also, there is a sign error contained in the K'^2 term (which is not as worrisome as my first issue). In regards to both errors, have i made an algebra mistake somewhere? Not sure how to get rid of that damned K' term attached to the group velocity.

4. Sep 7, 2012

### Don'tKnowMuch

I do not think that completing the square is relevant at this point of the problem. It will be in the very next step. The author is trying to set up the integral in a way where completing the square is not necessary. By that i mean, he is cutting out the work of completing the square by setting up the integral in the same form as the integral from the page previous to the linked page.

5. Sep 7, 2012

### Chopin

Oh, ok, I see your question now. Actually I think you're right...the $v_g$ term should be multiplied by $k'$, since it's the linear term in the Taylor expansion of $\omega(k')$. I think the page just has a typo in it--the opening parenthesis in that second exponential ought to be shifted over, to give $e^{ik'(x-v_g t)}$.

That jives with the next line, where the substitutions are made--note that the linear term's substitution is written as $k'x \rightarrow k'(x-v_g t)$. That also makes sense intuitively, because it means that everything is exactly the same as before, except that the $x$ coordinate is getting shifted as time passes, at a rate of $v_g$, which is exactly what the group velocity is supposed to mean.

Also, I think your signs on the $k'^2$ term are correct as well--the page has another typo there. Notice that the final results are written correctly, though--it's turned into a $\alpha + i\beta t$ by the time the final equations are put together. So you were right on both counts!

Last edited: Sep 7, 2012
6. Sep 7, 2012

### Don'tKnowMuch

Awesome! Thanks for the extra set of eye's. It's like removing a splinter. Now i feel i can move on. Thanks Chopin! B.t.w., Chopin is a beast. Do you like Schumann? The piano/ string quintet's are this sh!+.

7. Sep 7, 2012

### Chopin

You're very welcome--nothing's more infuriating than getting tripped up by mistakes in the "correct" text that you're trying to follow.

I like what I've heard from Schumann, but actually, overall I'm embarrassingly unfamiliar with his work. I've tended to latch onto Chopin over the years, primarily because of how much I love the Nocturnes, but I really ought to force myself to branch out a little more. Also, as a pianist, I've been unfortunately biased towards works for solo piano (since those are the ones I can actually play), and haven't spent nearly enough time getting familiar with the ensemble literature. I'll have to look into those quintets--thanks for the suggestion!

8. Sep 7, 2012

### Don'tKnowMuch

*High five*. Thanks again for the help! Have a nice weekend.