Quantum mechanics and an evil integral

[Beer-monster]

Not usre if this could have gone in the maths section as its a maths question based on a physics problem.
I'm trying to get to grips with the Rayleigh-Ritz variational method by trying a few examples from the books myself. The first was to try and recreate the solution of the infinite potential well.
I chose as directed the trial function
$$\gamma = (a^2-x^2)(1-cx^2)$$
where a is the width of the well from the origin (so the well is 2a wide in total) and c is a parameter which can be changed later.
The next step is to calculate the Energyfunction, unfortuantely that requires calculating:
$$\frac {\int (a^2-x^2)(1-cx^2) \frac{d^2}{dx^2}(a^2-x^2)(1-cx^2)dx}{\int(a^2-x^2)^2(1-cx^2)^2dx}$$
I can calculate the differential using the product rule to get
$$2c(a^2-x^2) - 8cx^2 - (1+cx^2)$$
uh ...I think (not sure about the middle term)
But after that I'm unsure how to procede in the numerator, trying by parts at the start usually leads to more chain rule derivations and more itegrals by parts and never seems to get me anywhere. I can't think of a suitable substitution to integrate with. Expanding all the factors leads to a huge string of numbers that is hard to keep track of.
Anyone have any clue they can lend me
Thanks

 

[inha]

I'd just use the brute force "calculate until your head explodes"-method. I mean all you've got to integrate are polynomials. Long and annoying polynomials, but still nothing difficult.

 

[Beer-monster]

Would that be the expand everything and do each polynomial seperately method of brain exploding or the keep everything in the brackets method. And can't I opt of the less painful shotgun in the mouth method

 

[inha]

I can't think of an easier way. I'm not advicing you to give up but this seems like an awfully uninstructive excercise.

 

[dextercioby]

Would it be easier if you part integrate...?

Daniel.

 

[Beer-monster]

That had occurred to me, but it seemed to require several steps of integration by parts and differentiation, and I wasn't sure it would simplify anything.

Does anyone have any tips as to how this method could work?

 

[SeReNiTy]

Just use your instincts and everything will turn out fine, it always does!

 

[TinTinn Tabulation]

Feeling somewhat masochistic, I be mainly trying to to solve Beer-Monster's Little Problemo !

$$\lambda (x)= \left( {a}^{2}-{x}^{2} \right) \left( 1-c{x}^{2} \right) -(1)$$

Taking derivative w.r.t.x of (1)

$$f(x)=\lambda'(x)=-2\,x \left( 1-c{x}^{2} \right) -2\, \left( {a}^{2}-{x}^{2} \right) cx -(2)$$

Taking 2nd derivative w.r.t.x of (1)

$$g(x)=\lambda''(x)=-2+10\,c{x}^{2}-2\, \left( {a}^{2}-{x}^{2} \right) c -(3)$$

Multiplying (1) by (3)

$$h(x)=\lambda(x) * \lambda''(x)=-2\,{a}^{2}+16\,{a}^{2}c{x}^{2}-2\,{a}^{4}c-14\,{a}^{2}{c}^{2}{x}^{4}+ 2\,{a}^{4}{c}^{2}{x}^{2}+2\,{x}^{2}-14\,{x}^{4}c+12\,{x}^{6}{c}^{2} -(4)$$

Integrating (4) w.r.t.x

$$i(x)={\frac {12}{7}}\,{c}^{2}{x}^{7}+ \left( {\frac {12}{5}}\, \left( -c{a} ^{2}-1 \right) c+1/5\,c \left( -2-2\,c{a}^{2} \right) \right) {x}^{5} + \left( 4\,c{a}^{2}+1/3\, \left( -c{a}^{2}-1 \right) \left( -2-2\,c{ a}^{2} \right) \right) {x}^{3}+{a}^{2} \left( -2-2\,c{a}^{2} \right) x -(5)$$

This looks a tad messy ! so substituting $\alpha=-2-2*a^2*c$

$$j(x)={\frac {12}{7}}\,{c}^{2}{x}^{7}+7/5\,c\alpha\,{x}^{5}+ \left( 1/6\,{ \alpha}^{2}-4-2\,\alpha \right) {x}^{3}+{a}^{2}\alpha\,x -(6)$$ :tongue2:

That concludes the Numerator(for the time being), turning now to Denominator;

Forming $\lambda^{2}(x)$

$$k(x)= \left( {a}^{2}-{x}^{2} \right) ^{2} \left( 1-c{x}^{2} \right) ^{2} -(7)$$

Expanding (7)

$$l(x)={x}^{8}{c}^{2}+ \left( -2\,{a}^{2}{c}^{2}-2\,c \right) {x}^{6}+ \left( {a}^{4}{c}^{2}+4\,c{a}^{2}+1 \right) {x}^{4}+ \left( -2\,{a}^{ 4}c-2\,{a}^{2} \right) {x}^{2}+{a}^{4} -(8)$$

Integrating (8) w.r.t.x

$$m(x)=1/9\,{c}^{2}{x}^{9}+ \left( -2/7\,{a}^{2}{c}^{2}-2/7\,c \right) {x}^{7 }+ \left( 1/5\,{a}^{4}{c}^{2}+4/5\,c{a}^{2}+1/5 \right) {x}^{5}+ \left( -2/3\,{a}^{4}c-2/3\,{a}^{2} \right) {x}^{3}+{a}^{4}x -(9)$$

Again substituting $\alpha=-2-2*a^2*c$ in (9)

$$n(x)=1/9\,{c}^{2}{x}^{9}+ \left( 1/7\,\alpha-2/7 \right) c{x}^{7}+ \left( - 2/5\,\alpha+1/20\,{\alpha}^{2}+1/5 \right) {x}^{5}+ \left( 1/3\,\alpha -2/3 \right) {a}^{2}{x}^{3}+{a}^{4}x -(10)$$

Forming our ratio of polynomials $o(x)=\frac{j(x)}{n(x)}$=(6)/(10)
$$o(x)=\frac{ \left( {\frac {12}{7}}\,{c}^{2}{x}^{7}+7/5\,c\alpha\,{x}^{5}+ \left( 1/6\,{\alpha}^{2}-4-2\,\alpha \right) {x}^{3}+{a}^{2}\alpha\,x \right)}{ \left( 1/9\,{c}^{2}{x}^{9}+ \left( 1/7\,\alpha-2/7 \right) c {x}^{7}+ \left( -2/5\,\alpha+1/20\,{\alpha}^{2}+1/5 \right) {x}^{5}+ \left( 1/3\,\alpha-2/3 \right) {a}^{2}{x}^{3}+{a}^{4}x \right)} -(11)$$

Tidying up (11)

$$p(x)=6\,{\frac {360\,{c}^{2}{x}^{6}+294\,c\alpha\,{x}^{4}+35\,{x}^{2}{ \alpha}^{2}-840\,{x}^{2}-420\,{x}^{2}\alpha+210\,{a}^{2}\alpha}{140\,{ c}^{2}{x}^{8}+180\,c{x}^{6}\alpha-360\,c{x}^{6}-504\,{x}^{4}\alpha+63 \,{x}^{4}{\alpha}^{2}+252\,{x}^{4}+420\,{a}^{2}{x}^{2}\alpha-840\,{a}^ {2}{x}^{2}+1260\,{a}^{4}}} -(12)$$

Taking the Limit as x tends to 0 of p(x)

$$\lim_{x\to 0}p(x)={\frac {1}{{a}}+O \left( {x}^{2} \right) - (13)$$

Taking the Limit as x tends to Infinity of p(x)

$$\lim_{x\to\infty}p(x)={\frac{108}{7x^2}+O \left( {x}^{-4} \right) -(14)$$

So familiar inverse-square law for large +- x .

To get a handle of the behaviour of p(x) within the potential well
Now need to find the poles and zeros of p(x)
For Zeros: Set j(x)=0, and solve for x.
For Poles: Set n(x)=0, and solve for x.

I'll leave that to you !

Did this make things any easier, or have I just wasted several hours ?
As in, say Dynamical Systems, it's sometimes better to just try and get a qualitative feel for the solution,
rather than strive for that ultimately defeating (and exhausting !) analytical solution !

Hmm. Not as easy as it first appears :yuck:

 

[Beer-monster]

I'm scared just looking at that.

However, I'm with you all the way up to 11/12. Although when I posted the problem I forgot to include the limits of integration are between -a and a (for the well). Perhaps that will make the integration neater?

I've moved on to work on a similar problem but for the harmonic oscillator, which has been set by the lecturer. Now that I see that I can do this by the mind-exploding expansion, I hope I can crack this new problem too.

Thanks

 

[Beer-monster]

Nope no luck

I've been trying this new problem for 4 days and still can't get it right

Its the same thing - the variational method however now the trial function is

$$\psi = (c^2-x^2)^2$$

With limits -c to c

Tried expanding it all out and proceding but I get large powers with ridiculous fractions (I think I had 11 and 762/4732ths at one point) which makes me think I'm going very wrong. By parts doesn't help as it doesn't seem to simplify anything. Can anyone think of a substitution?

 

[Physics Monkey]

Ugh! Why would anyone have you work such an awful problem! Whatever you do, don't expand it all out! It isn't actually that difficult to find the best value of c and the minimum value of $$\langle H \rangle$$. Here is a hint: try to extract the c dependence from the integrals leaving only definite integrals independent of c (just numbers). Write your answer in terms of these integrals but do not try to evaluate them yet. At the end of the day the answer cleans up rather nicely and you can calculate the relevant integrals numerically (general closed form solutions are also available for less trouble than it would be to actually calculate a single one by hand).

 

[Beer-monster]

Physics monkey: I'm not sure what you mean by extract the c dependence? Are you suggesting I divide c out of the function or try and factorise it out so that it appears outside the integral (which I'm not sure is possible in this case)?

 

[Physics Monkey]

Something like that, I'm suggesting you factor out the c dependence from the integrals (there are essentially three relevant integrals) by changing variables. This can be done because c is the only length scale in your trial wavefunction. Hint: try setting x = c u with u dimensionless.

 

[Beer-monster]

Similar to the substitution you make in solving a homoegeneous equation?

 

[Physics Monkey]

For instance, when calculating $$\int \psi^2 dx$$ you have to evaluate the integral
$$\int^c_{-c} (c^2 - x^2)^4 \,dx,$$
just try the suggested change of variable and see what happens. If you find something useful then try to repeat the procedure for the other two integrals you need to do.

 

[Beer-monster]

Right then

For the above integral I get $$c^9\int{(1-u^2)^2}du$$. Which is probably wrong but seems integratable.

For the Harmonic potential term I get $$c^6\int{u^2(1-u^2)}du$$
Which also seems do-able by parts.

However for the kinetic energy term I hit a bit of a problem as the second differential of the trial function has two different pwers of x making it hard to factorise.

I'm I in the ball park or at the swimming pool?

 

[Physics Monkey]

Don't be so quick to beat yourself up about it, you did the wavefunction term right. The integral you have left is, of course, completely doable since it is just a polynomial. However, you did get the potential energy term wrong. Think about your dimensions in the integral $$\int^c_{-c} x^2 (c^2 - x^2)^4 dx.$$
The point of all this is that you've gotten the c dependence out which is what you really care about (since you have to differentiate with respect to it). The leftover integral is just a number, call it $$I_1$$ for instance, and carry it through your calculation as a constant. At the end to the day, your answer for the best value of c and the expectation value of the Hamiltonian will involve these integrals just as it might involve pi or 3 or 7 or whatever. You can then calculate them numerically, or by directly expanding, or by playing some tricks with the integrals.

 

[Beer-monster]

Just to be sure. I do this substitution to take ouyt the c dependende. Refer to each integral as a constant and then differentiate with respect to c and fiddle to get the answer?

Sorry, this is just a far cry from what I've seen before, its taking a while to sink in.

 

[Physics Monkey]

Yes, that is exactly right. If you think about the system for a bit from a physical point of view, the characteristic length scale is c, so the expectation value of the Hamiltonian has got to look something like
$$n_1 \frac{\hbar^2}{2 m c^2} + n_2 \frac{m \omega^2 c^2}{2},$$
where n1 and n2 are just numbers. All you doing is leaving those numbers as simple symbols rather than immediately calculating them. At the end of the day, you can put all the numbers in at once.