Power Series Solution (SE)?

[nd]

Here's our equation:
[tex]\frac{d^2\psi}{du^2}+(\frac{\beta}{\alpha}-u^2)\psi=0[/tex]
This is the SE for the simple harmonic oscillator. My text goes through an elaborate solution to this DE and ends up resorting to a power series solution, not for psi, but for H, where [tex]\psi=H(u)e^{-u^2/2}[/tex]. The text also points out that no power series solution could be found by directly substituting in the SE for psi. However, Apostols THM6.13 (Volume II) states that any 2nd order ODE with analytic coefficients has a power series solution. Which is right?

 

[PBRMEASAP]

This is just a guess...i'm not a mathematician. Maybe they just meant that the recursion formula for the power series is very difficult or impossible to obtain if you stick it straight into the SE. By dividing out the asymptotic behavior as u->infinity, I think the resulting equation for H(u) is simpler. If Apostol says a power series solution exists, then it exists. He knows what's up.

 

[HallsofIvy]

Saying that the is a power series doesn't mean it is easy to find!

Certainly, if there exist a power series for H, then there mujst exist a power series for [tex]\psi=H(u)e^{-u^2/2}[/tex]. It's just much easier to find the power series for H.

 

[saltydog]

Hum. I don't know. This is what I got:

[tex]y(x)=\sum_{n=0}^\infty a_nx^n[/tex]

Letting [itex]k=\frac{\beta}{\alpha}[/itex]

[itex]a_0[/itex] is arbitrary
[itex]a_1[/itex] is arbitrary

[tex]a_2=-\frac{ka_0}{2}[/tex]

[tex]a_3=-\frac{ka_1}{6}[/tex]

And for [itex]n\geq4[/itex]

[tex]a_n=\frac{a_{n-4}-ka_{n-2}}{n(n-1)}[/tex]

This is just a quick check. I'd need to verify this by working through the substitutions again and then back-substituting the resulting series into the ODE to check it with real data with select choices for a0 and a1. Kinda sleepy. No lil' smilely faces for that?

 

[nd]

Yes, I got something similar to that. I was working out a problem where the point was to show that things get too messy when you try to get a solution directly. But that recursion formula seems just fine to me.

This is just a guess...i'm not a mathematician. Maybe they just meant that the recursion formula for the power series is very difficult or impossible to obtain if you stick it straight into the SE.

That's what I originally thought too.

 

[saltydog]

Letting:

[tex]y(x)=\sum_{n=0}^\infty a_nx^n[/tex]

Letting [itex]k=\frac{\beta}{\alpha}[/itex]

[itex]a_0[/itex] is arbitrary
[itex]a_1[/itex] is arbitrary

[tex]a_2=-\frac{ka_0}{2}[/tex]

[tex]a_3=-\frac{ka_1}{6}[/tex]

And for [itex]n\geq4[/itex]

[tex]a_n=\frac{a_{n-4}-ka_{n-2}}{n(n-1)}[/tex]

Just a follow up. Using the recursive formula above and setting k=2, a0=1, and a1=2, I calculated the first 50 terms of the series. A plot is attached. I then back-substituted the results into the ODE , range (-1,1), and obtained a discrepancy from 0 no larger than 10^-15. I believe this series is correct.

 

[arildno]

Now, the advantage of the "H(u)-method" espoused by the book and HallsofIvy, is that you peel of the exponential term.
What this means, is that you won't need as many terms for H(u) in order to gain a satisfactory degree of accuracy in the u-range your interested in, compared with the number of series terms you'll need for the power series for [tex]\psi[/tex]

 

[saltydog]

Now, the advantage of the "H(u)-method" espoused by the book and HallsofIvy, is that you peel of the exponential term.
What this means, is that you won't need as many terms for H(u) in order to gain a satisfactory degree of accuracy in the u-range your interested in, compared with the number of series terms you'll need for the power series for [tex]\psi[/tex]

Well I'm sorry Arildno but I'd like to see some proof of this please. A plot or two, a tabulated comparison of my series, the H(u) series and the discrepancy between the two as a function of n. You know something like that. I tell you what, I'll reduce the number of terms of my series down until the discrepancy just reaches 6 digits of accuracy in the range (-1, 1). I'll report it here. Do as you wish.

 

[saltydog]

Well that didn't take long. Thank God for Mathematica. I've attached a plot of the discrepancy of the back-substitution of 19 terms of the series. As you can see, the discrepancy is very low except at the end points where it reaches 10^-6.

 

[Data]

Well, the solutions to the Schrodinger equation that are physically acceptable (ie. they meet necessary boundary conditions) in this case actually have the form

[tex] \psi_n(u) = H_n(u)e^{-u^2/2}[/tex]

where n is a quantum number. In these cases, the functions [itex]H_n(u)[/itex] are actually just polynomials, not infinite power series (called Hermitian polynomials; [itex]H_0(u) = 1, \; H_1(u) = 2u, \; H_2(u) = 4u^2-2, \; . \ . \ .[/itex]). They satisfy the Hermite equation

[tex] \frac{d^2H_n}{du^2} - 2u\frac{dH_n}{du} + 2nH_n = 0[/tex]

Note that these are only the physically acceptable solutions (ie. boundary conditions have been applied implicitly to isolate the quantum number n), and there are more which undoubtedly do require a power series solution but that do not make sense in terms of physics.

Based on this the reason for isolating H is obvious, though: why resort to a power series solution to the DE when you can have a closed form that works in every case you're interested in?

If you tried to solve for [itex]\psi[/itex] using a power series solution, you'd still be left with a power series (as demonstrated above in this thread).