Efficient Techniques for Solving Nasty Integrals in Quantum Mechanics

[dingo_d]

Homework Statement

I have a function in t=0 given by:

\psi(x,0)=\frac{1}{\pi^{1/4}\sqrt{\sigma}}e^{-\frac{(x-x_0)^2}{2\sigma^2}},

and I have to decompose it in eigenstates of harmonic oscillator given by:

u_n(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^nn!}}H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-\frac{1}{2}\frac{m\omega}{\hbar}x^2}

Homework Equations

\psi(x,0)=\sum_{n=0}^\infty c_n(0)u_n(x)

I use the fact that eigenfunctions of h.o. are ortonormal, and I can find:

c_n(0)=\int_{-\infty}^\infty\psi(x,0)u_n^*(x)dx

The Attempt at a Solution

And here comes the nasty part! Evaluating that integral. Now I can really move all the constants out, they play no vital role in the evaluation of that integral.

The integral is:

c_n(0)=\int_{-\infty}^\infty e^{-\frac{(x-x_0)^2}{2\sigma^2}}e^{-\frac{1}{2}\frac{m\omega}{\hbar}x^2}H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)dx

I can use the substitution \zeta=\sqrt{\underbrace{\frac{m\omega}{\hbar}}_{\lambda}}x, I can then sort things out a bit and get


\frac{1}{\sqrt{\lambda}}\int_{-\infty}^\infty e^{-\frac{(\zeta-\zeta_0)^2}{2\lambda\sigma^2}}e^{-\frac{\zeta^2}{2}}H_n(\zeta)d\zeta

Where I've used the substitution: \sqrt{\lambda}x_0=\zeta_0

Now I've solved similar integral by using generating function of http://en.wikipedia.org/wiki/Hermite_polynomials#Generating_function", but when I use that trick here I get:


\frac{1}{\sqrt{\lambda}}\int_{-\infty}^\infty e^{-\frac{(\zeta-\zeta_0)^2}{2\lambda\sigma^2}}e^{-\frac{\zeta^2}{2}}e^{-s^2+2s\zeta}d\zeta

The last part is the generating function. Now in some simpler exercises I would get some part with e^s and constants and Gauss integral, and then I would expand that part with Exp and use the generating function which has that similar part, compare it and voila! But here I get:

\int_{-\infty}^\infty e^{-\frac{(\zeta-\zeta_0)^2}{2\lambda\sigma^2}}e^{-\frac{\zeta^2}{2}}e^{-s^2+2s\zeta}d\zeta=\sqrt{2\pi}\sqrt{\frac{\lambda\sigma^2}{\lambda\sigma^2+1}}e^{s^2-\frac{(\zeta_0-2s)^2}{2+2\lambda\sigma^2}}


I have quadratic term in s and I would have different expansion for those parts with s and s^2, and I couldn't compare that to get the desired result :\

Can anyone point me in the right direction?

 

[fzero]

I think you want to write the formula for the generating function in a form like

e^{-a t^2 + b t} = \exp \left( -(\sqrt{a} t)^2 + 2 \frac{b}{2\sqrt{a}} (\sqrt{a} t) \right) = \sum_{n=0}^\infty H_n(b/(2\sqrt{a})) \frac{(\sqrt{a} t)^n}{n!}.

 

[dingo_d]

Where did you find that formula? I'll try with that :) Thank you very, very much! :virtual hug: :D

EDIT:

Hmmm ok so I tried with this and I got this:

\int_{-\infty}^\infty e^{-\frac{(x-x_0)^2}{2\sigma^2}-\frac{\lambda}{2}x^2}\underbrace{e^{-\frac{t^2}{4\lambda}+xt}}_{\textrm{gen. function from above}}dx=\frac{\sqrt{2\pi}\sigma}{\sqrt{1+\lambda\sigma^2}}e^{\frac{4tx_0\lambda-2x_0^2\lambda^2+t^2(\lambda\sigma^2-1)}{4\lambda(1+\lambda\sigma^2)}}

So when I put the expression with the sum inside instead of generating function I can get the original integral, but I still have terms with e^t^2 on the right side, I can't expand that into the sum and compare it with left hand side because the sums are not the same :\

 

[fzero]

Where did you find that formula? I'll try with that :) Thank you very, very much! :virtual hug: :D

It's just the standard generating function written with a change of variables.

So when I put the expression with the sum inside instead of generating function I can get the original integral, but I still have terms with e^t^2 on the right side, I can't expand that into the sum and compare it with left hand side because the sums are not the same :\

You have to expand the e^{t^2} terms into the series. I'm not sure how neat the general expression will look, but you can check the lowest powers to see how it works.

 

[dingo_d]

Yeah, but in the end I need to compare the two, and every quadratic term is messing it up :\

 

[fzero]

You can try splitting the original series into odd and even parts:

\sum_{n=0}^\infty a_n t^n = \sum_{k=0}^\infty ( b_k t^{2k} + c_k t^{2k+1} ).

 

[dingo_d]

I don't think that will help much. I have solved example scanned (ignore Croatian words :D).

I expand the solution of the integral, and compare it.

 

[fzero]

Is the expression in your post #3 incorrect? You seem to obtain a different result in your notes.

In any case, I'm not sure what you mean by "the sums are not the same." On one hand, the integrals that you want are the coefficients of a series in s. On the other hand you have integrated the generating function and can express that as a series in s with known coefficients.

 

[dingo_d]

Basically it's the initial Gaussian that's causing the problem. My assistant professor said I can neglect the st. deviation for the simplicity (put it to one). If I had \frac{m\omega}{\hbar}x instead of just x in it, I could use the substitution and I could add some parts... Basically I could have the same integral as in attachment.

Since adding a factor next to x in Gaussian will never change it's height, only width, and the original problem stated that the particle is described by a wave packet, do you think I could 'tweak' the initial Gaussian.

I'm loosing on generality, but I'm gaining on solvability :D

 

[dingo_d]

Is the expression in your post #3 incorrect? You seem to obtain a different result in your notes.

In any case, I'm not sure what you mean by "the sums are not the same." On one hand, the integrals that you want are the coefficients of a series in s. On the other hand you have integrated the generating function and can express that as a series in s with known coefficients.

Oh the notes are not the same problem :) Sorry if I hadn't stated that. In notes function at t=0 is described as oscillator in it's ground state.

By the sums are not the same, I meant that in the notes when I expand the solution which has e^{\zeta s} I can compare it with original integral with generating function in it. Since the generating function is given by:\exp (2\zeta s-s^2) = \sum_{n=0}^\infty H_n(\zeta) \frac {s^n}{n!}, putting it back in the original integral:
\int_{-\infty}^\infty e^{-s^2+2s\zeta}e^{-\zeta^2+\zeta\zeta_0-\frac{\zeta_0^2}{2}}d\zeta=\int_{-\infty}^\infty \sum_{n=0}^\infty\frac{1}{n!}H_n(\zeta)s^n e^{-\zeta^2+\zeta\zeta_0-\frac{\zeta_0^2}{2}}d\zeta=\sum_{n=0}^\infty\frac{1}{n!}s^n\int_{-\infty}^\infty e^{-\zeta^2+\zeta\zeta_0-\frac{\zeta_0^2}{2}} H_n(\zeta)d\zeta

Now expanding the solution (expanding the e^{\zeta s} part) and grouping you can compare the integral with the thing in the bracket on the paper.But in my original problem (the one I posted) I get solution from the integral which contains e^{~s^2} terms. If I would expand that I'd get \sum\frac{s^{2n}}{n!} and I cannot compare something that has different terms :\ Plus I mustn't have s parts in the final solution, since it comes from generating function.

I hope I cleared the misunderstanding :)

 

[fzero]

I don't want to get bogged down in the variable redefinitions, but I don't understand why you can't just derive an expression like

<br /> c_n(0) = \left. \frac{d^n}{ds^n} \sqrt{2\pi}\sqrt{\frac{\lambda\sigma^2}{\lambda\sigma^2+1}}e^{s^2-\frac{(\zeta_0-2s)^2}{2+2\lambda\sigma^2}} \right|_{s=0}.<br />

To do this, compare where the H_n(\lambda x) arises in the generating function sum and just make a Taylor expansion of the integral result.

 

[dingo_d]

But I need to get an exact solution :\ I'm not quite sure how you got that expression :|

 

[fzero]

You have

<br /> \exp (2\zeta s-s^2) = \sum_{n=0}^\infty H_n(\zeta) \frac {s^n}{n!}<br />

so if we define

<br /> I =\int_{-\infty}^\infty e^{-\frac{(x-x_0)^2}{2\sigma^2}}e^{-\frac{1}{2}\frac{m\omega}{\hbar}x^2}e^{-s^2+2s\zeta} dx<br />

then c_n(0) is proportional to the s^n term in the Taylor expansion around s=0. But that term is given, up to n!, by

\left. \frac{d^n I }{ds^n} \right|_{s=0}.

If you want an exact result, you can derive a closed-form expression for the nth term.

 

[dingo_d]

I'll try to derive it, if I get stuck I'll yell :D Thanks ^^

 

[dingo_d]

Ok, so I did what you said, and did the derivative for first few terms, and the expression just gets messier and messier :\ How can I deduce the closed-form expression from that?

For the first few terms (with \sigma=1 and \lambda=\frac{m\omega}{\hbar} and X=x_0)<br /> \begin{array}{l}<br /> n=0\quad \sqrt{2 \pi } \sqrt{\frac{\lambda }{\lambda +1}} e^{-\frac{\lambda <br /> X^2}{2 (\lambda +1)}} \\<br /> n=1\quad \frac{2 \sqrt{2 \pi } \lambda X e^{-\frac{\lambda X^2}{2 (\lambda<br /> +1)}}}{(\lambda +1)^{3/2}} \\<br /> n=2\quad \frac{2 \sqrt{2 \pi } \sqrt{\lambda } e^{-\frac{\lambda X^2}{2 (\lambda<br /> +1)}} \left(\lambda ^2+2 \lambda X^2-1\right)}{(\lambda +1)^{5/2}} \\<br /> n=3\quad \frac{4 \sqrt{2 \pi } \lambda X e^{-\frac{\lambda X^2}{2 (\lambda +1)}}<br /> \left(3 \lambda ^2+2 \lambda X^2-3\right)}{(\lambda +1)^{7/2}} \\<br /> n=4\quad \frac{4 \sqrt{2 \pi } \sqrt{\lambda } e^{-\frac{\lambda X^2}{2 (\lambda<br /> +1)}} \left(3 \left(\lambda ^2-1\right)^2+4 \lambda ^2 X^4+12 \lambda <br /> \left(\lambda ^2-1\right) X^2\right)}{(\lambda +1)^{9/2}}<br /> \end{array}<br />

I can see some reoccurring patterns, but it's just too messy :\

Help

 

[fzero]

I'm not sure how difficult it would be to turn those into closed form expressions. It would certainly be easier to try to work out the odd and even cases separately. It might also be illuminating to write down recurrence relations first.

In a certain sense, the differential expression for the coefficients is also closed form. Most of the time, that's how orthogonal polynomials are presented anyway, since closed-form series expansions aren't necessarily easier to work with.

 

[dingo_d]

And in the end my assistant professor said I can just as well put \psi(x,0)=\frac{1}{\pi^{1/4}\sqrt{\sigma}}e^{-\frac{(\sqrt{\lambda}x-x_0)^2}{2\sigma^2}}

With that my integral becomes the same as in the paper above...

Anyways thanks for all the help ^^