Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a differential equation of the form $$a y'' + b y/x = E y$$ (The origin is a 1D Schrödinger equation for a potential of the form $-1/x$). I am only interested in the ground state energy, i.e. the lowest order solution.

Is there a good, systematic way to tackle this? I used a lot of hand waving:

I said that for $x \rightarrow \infty$, the potential term is negligible and the equation is a simple homogeneous 2nd order ODE with constant coefficients, which has solution $e^{-kx}$ for some $k$. So as an overall ansatz I choose $$f(x)e^{-kx}$$, which yields $$a (f'' - 2k f' + k^2 f) + b f/x = E f$$.

I then argue -- that is where the hand-waving occurs -- that the ground state would have a polynomial of the lowest possible order for $f$. A constant (order $0$) is not possible, since then nothing cancels the $1/x$ in the equation, so I try the ansatz $f(x) = x$. With that, I can indeed solve the equation and obtain conditions for $k$ and $E$:

$$-2ka + b = 0$$ $$ak^2 = E$$

This allows me to solve for $k$ and $E$.

But is there a better, more rigorous way?

share|improve this question
    
There's always the Frobenius route... which you can use to derive the solutions in terms of confluent hypergeometric functions. –  J. M. Aug 7 '11 at 17:10
    
Since I'm a physicist and not a mathematician, would you briefly outline that route? –  Lagerbaer Aug 7 '11 at 17:12
    
This should be a quick review... it's also in Arfken and Weber. (FWIW, I ain't a mathematician either... :) ) –  J. M. Aug 7 '11 at 17:26
    
Ah, okay. So what that method does is writing $f(x)$ as a power series in $x$, which generates recursive equations for the coefficients. If I set a cut-off for the degree of the polynomial, this should then reproduce my result. –  Lagerbaer Aug 7 '11 at 19:59
add comment

1 Answer

up vote 1 down vote accepted

Let's assume that $a=1$ for simplicity. You can then, either use a CAS to solve this differential equation, or notice that it is a differential equation for a confluent hypergeometric functions $_1F_1(x)$ and $U(x)$.

Specifically the general solution to equation $y'' + \frac{b}{x} y = \mathcal{E}^2 y$ is

$$ y(x) = x e^{-x \mathcal{E}} \left( c_1 {}_1F_1(1 - \frac{b}{2\mathcal{E}}, 2, 2 x \mathcal{E}) + c_2 U( 1 - \frac{b}{2\mathcal{E}}, 2, 2 x \mathcal{E} ) \right) $$

Now, you could look up the asymptotic behavior of each independent solution (here and here) and choose indeterminates and the energy to satisfy needed boundary conditions.

You will find that $c_1$ must vanish due to decay at infinity, while $c_2$ is arbitrary. Behavior at the origin demands that $1 - \frac{b}{2\mathcal{E}}$ be a non-positive integer, giving you the spectrum. In that case the Tricomi function would degenerate into a polynomial.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.