Second order DE (hydrogen atom)

[fluidistic]

Homework Statement

The DE y''+\frac{2}{x}y'+ \left [ K+\frac{2}{x} - \frac{l(l+1)}{x^2} \right ]y=0, 0<x< + \infty. appears when working on the hydrogen atom. Find all the values of K (the eigenvalues) that generates solutions of the form \phi (x) such that \phi (x) remains finite when x tends to 0.

Homework Equations

I don't know.

The Attempt at a Solution

I don't know how to start. There's nothing said about the l's but I guess that they are natural numbers. Also why the solutions are phi's rather than y's, I don't know.

 

[fluidistic]

Can someone help me on this problem? It interests me a lot.
It seems I can't really use the method of undetermined coefficients because the terms in front of y' and y aren't constant. I can't use either the method of variation of parameters since I don't know any solution of the -homogeneous DE-, in fact I want to solve the homogeneous DE.
I see that I have an equation of the form y''+a(x)y'+b(x)y=0. Where b is a somehow special function in the sense that it also depends on K and l.

 

[fluidistic]

Ok I asked a friend about this problem, he told me to use power series solutions.
I've never done this before and I'm stuck near the end. I'd appreciate some guidance.
I proprose a solution of the form y(x)=\sum _{n=0}^{\infty} a_n x^n. I derivated this expression twice. Plugging back into the orignal DE and setting the 3 power series with the same starting "n" (in my case n=0), I reach \sum _{n=0}^{\infty} x^n \{ (n+2)(n+1)a_{n+2}+\frac{2}{x}(n+1)a_{n+1}+ \left [ K+\frac{2}{x}-\frac{l(l+1)}{x^2} \right ] a_n \} =0.
Since this is worth 0 for all x, I can conclude that (n+2)(n+1)a_{n+2}+\frac{2}{x}(n+1)a_{n+1}+ \left [ K+\frac{2}{x}-\frac{l(l+1)}{x^2} \right ] a_n must equal 0.

Doing this I reach the recurrence relation a_{n+2}= \frac{\left [\frac{l(l+1)}{x^2}-\frac{2}{x}-K \right ]}{(n+2)(n+1)} a_n-\frac{2}{x(n+2)} a_{n+1}.

I set n=0 to get a_2 in function of a_1 and a_0 and also n=1 to get a_3 in terms of a_1 and a_2.
But I don't have an idea on how to determine a_0 and a_1 so that I can really "run" my recurrence relation.
Any help is appreciated!

 

[jackmell]

I'm assuming the paramater "l" is an integer so the quantity l(l+1) is either zero or a positive integer. For now, just let l(l+1)=\nu. Let's first write yours as:

y''+\frac{2}{x}y'+\left(-\frac{\nu}{x^2}+\frac{2}{x}+K\right)y=0

The guiding principle for solving these is the "indicial" equation which I'll write as:

c^2+c-\nu=0

for which we solve for the roots:

\{c_1,c_2\}=\frac{-1\pm\sqrt{1+4\nu}}{2}

We then seek a solution of the form:

y(x)=\sum_{n=0}^{\infty}a_nx^{n+c},\quad c=\{c_1,c_2\}

Ok, now, can you substitute that sum with just "c" for now, into the DE, adjust all the sum indices so that the powers on x are the same, and then letting a_0 be arbitrary, come up with a recursive relation for a_n for n\geq 1?

 

[vela]

Look up the method of Frobenius if you need more background on solving the differential equation using a series.

 

[fluidistic]

Hey thanks for helping me jackmell. However I'm having troubles figuring out how you obtained your indicial or characteristic equation. Could you explain how you obtained it?
Edit: I just saw your post vela, I'm definitely heading up to Frobenius.

 

[jackmell]

I'm afraid I caused some confusion above and can't edit. I should have written the DE as:

x^2y''+2xy'+\left(Kx^2+2x-\nu\right)y=0

and now, substitute into that form, the expression:

y(x)=\sum_{n=0}^{\infty} a_n x^{n+c}

Now, the indicial equation is contrived so that we can set a_0 arbitrary. You can see how it's derived in any DE textbook. Tell you what, try getting these expressions:

a_0: \text{arbitrary}

a_1=-\frac{2a_0}{c^2+3c-\nu}

n\geq 2:\quad a_n=-\frac{Ka_{n-2}+2a_{n-1}}{(n+c)(n+c-1)+2(n+c)-\nu}

 

[fluidistic]

Ok guys thank you for your help, I'm finally back to this problem (I had to pass a final exam of another course, was too busy on it. Now that I'm done with it, I can concentrate on Mathematical methods and classical mechanics).
I've just downloaded a free online book on mathematical methods called "Mathematical Tools for Physics" by James Nearing. I've checked out Frobenius's method and indeed it looks like the way to go for my equation.
Using his notation (replace c by s from jackmell's notation), I get:
\sum _{l=0}^{\infty } a_l x^{l+s}[(l+s)(l+s+1)-v]+K\sum _{l=2}^{\infty } a_{l-2} x^{l+s}+2 \sum _{l=1}^{\infty } a_{l-1} x^{l+s}=0.
I assume that a_0 \neq 0 so that the coefficient in front of the lower power of x (it appear in the first series only, when l=0) must equal 0.
This gives me the inditial equation [s(s+1)-v]=0. Thus s=\frac{-1 \pm \sqrt {1+4v} }{2}.
Now I set the coefficients in front of x^{l+s} equal to 0.
I get: a_l [(l+s)(l+s+1)-v]+Ka_{l-2}+2a_{l-1}=0. And I'm stuck here. I can get a_{l+2} in terms of a_{l-1} and a_{l}.
So Jackmell, I don't know how to obtain your answer.

 

[vela]

Looks good so far. Separating out the first and second term of the series, you have
a_0[s(s+1)-v]x^s + [a_1((s+1)(s+2)-v)+2a_0]x^{s+1} +\sum_{l=2}^\infty [a_l((l+s)(l+s+1)-v)+Ka_{l-2}+2a_{l-1}]x^{s+l} = 0Setting the coefficient of the first term equal to 0 yields the indicial equation, which you both agree on, and the general recursion relation, which again you both agree on, comes from setting the coefficient in the summation equal to 0. Jackmell's second equation comes from setting the coefficient of x^s+1 equal to 0, though it looks like there's a term missing in the denominator.

 

[jackmell]

Ok, I made a typo in the equations for the coefficients. They should have been:

a_0: \text{arbitrary}

a_1=-\frac{2a_0}{c^2+3c+2-\nu}

n\geq 2:\quad a_n=-\frac{Ka_{n-2}+2a_{n-1}}{(n+c)(n+c-1)+2(n+c)-\nu}

and it's not hard to (visually) check by first solving it numerically for some values of the parameters, then comparing that to the series solution. However, using numerical methods, we cannot find a solution valid at the origin because of the singularity of the DE. That doesn't mean the solution there is necessarilly singular though. So I just solved it for y(1)=y1 and y'(1)=y2 and the parameter values below. And using the above expressions, the solutions agree. Here's the code I used.

k = 1; 
v = 1; 
y0 = 1; 
y1 = 1; 

mysol = NDSolve[{x^2*Derivative[2][y][x] + 2*x*Derivative[1][y][x] + (k*x^2 + 2*x - v)*y[x] == 0, y[1] == 1, 
    Derivative[1][y][1] == 1}, y, {x, 1, 10}]

ndsolveplot = Plot[y[x] /. mysol, {x, 1, 10}, PlotRange -> {{1, 10}, {-2, 2}}]

c1 = (-1 + Sqrt[1 + 4*v])/2.; 
Subscript[a, 0] = 1; 
Subscript[a, 1] = -((2*Subscript[a, 0])/(c1^2 + 3*c1 + 2 - v)); 
mya = Table[Subscript[a, n] = -((k*Subscript[a, n - 2] + 2*Subscript[a, n - 1])/((n + c1)*(n + c1 - 1) + 2*(n + c1) - 
        v)), {n, 2, 25}]; 

c2 = (-1 - Sqrt[1 + 4*v])/2.; 
Subscript[b, 0] = 1; 
Subscript[b, 1] = -((2*Subscript[b, 0])/(c2^2 + 3*c2 + 2 - v)); 
myb = Table[Subscript[b, n] = -((k*Subscript[b, n - 2] + 2*Subscript[b, n - 1])/((n + c2)*(n + c2 - 1) + 2*(n + c2) - 
        v)), {n, 2, 25}]; 

my1[x_] := Sum[Subscript[a, n]*x^(n + c1), {n, 0, 25}]
my2[x_] := Sum[Subscript[b, n]*x^(n + c2), {n, 0, 25}]

thek = Solve[{k1*my1[1] + k2*my2[1] == y0, k1*(D[my1[x], x] /. x -> 1) + k2*(D[my2[x], x] /. x -> 1) == y1}, {k1, k2}]

seriesplot = Plot[k1*my1[x] + k2*my2[x] /. thek, {x, 1, 10}, PlotRange -> {{1, 10}, {-2, 2}}]

Show[{ndsolveplot, seriesplot}]

Ok, so we got at least the series representation probably ok and to answer the OP's question about how to get the above expressions, well, you just have to substitute the series:

y(x)=\sum_{n=0}^{\infty} a_n x^{n+c}

into the DE and solve for them. Work a few simple ones first to get the hang of it. But I do not see how this series solution is going to help determine for what values of K the solution is bounded at the origin except to just run it for a range of k and the parameters and just try to determine this numerically. However, sometimes the series representation can be converted to a managagble function. Not sure about this one though.

 

[fluidistic]

Looks good so far. Separating out the first and second term of the series, you have
a_0[s(s+1)-v]x^s + [a_1((s+1)(s+2)-v)+2a_0]x^{s+1} +\sum_{l=2}^\infty [a_l((l+s)(l+s+1)-v)+Ka_{l-2}+2a_{l-1}]x^{s+l} = 0Setting the coefficient of the first term equal to 0 yields the indicial equation, which you both agree on, and the general recursion relation, which again you both agree on, comes from setting the coefficient in the summation equal to 0. Jackmell's second equation comes from setting the coefficient of x^s+1 equal to 0, though it looks like there's a term missing in the denominator.

Ah ok, that's a very clear explanation, thanks. Though if I set the coefficient of x^{s+1} equal to 0, I get a_1 [(s+1)(s+2)-v+2a_0]. Since I assume a_1 \neq 0, I can divide by it. This gives me a_0 = -\left ( \frac{1}{2} \right ) [(s+1)(s+2)+v]. I already have s in function of v so that I have a_0 in function of v. Now I don't see why would a_0 be arbitrary, am I missing something?

Edit: I'm reading your post jackmell, I didn't see it when I posted this one.

 

[jackmell]

The ability to set a_0 arbitrary comes from a consideration of the indical equation. If you take the first and second derivatives of that power series and substitute it into the DE you'll get after a change of index:

\sum_{n=0} a_n(n+c)(n+c-1)x^{n+c}+2\sum_{n=0}a_n(n+c)x^{n+c}+K\sum_{n=2}a_{n-2}x^{n+c}+2\sum_{n=1} a_{n+1}x^{n+c}-v\sum_{n=0} a_n x^{n+c}=0

Now consider just the a_0 term out of all of that. It's

a_0(c^2+c-v)=0

Well there you go, either a_0 has to be zero or the other one or both and we don't want a_0 to be zero so we let c^2+c-v (the indical equation) equal to zero. But if that's zero, then a_0 can be anything and the simplest, the one I used above, is to simply let it equal one.

 

[vela]

fluidistic, are you sure the differential equation you started with is correct? I checked my quantum textbook, and the equation for the radial function is slightly different.

 

[fluidistic]

fluidistic, are you sure the differential equation you started with is correct? I checked my quantum textbook, and the equation for the radial function is slightly different.

I don't know if it's correct (doesn't look like correct if you say it isn't) but I'm 100% sure it's the one of an exercise of my course.
You can see it in the attached file (exercise 1).

 

[vela]

Ah ok, that's a very clear explanation, thanks. Though if I set the coefficient of x^{s+1} equal to 0, I get a_1 [(s+1)(s+2)-v+2a_0]. Since I assume a_1 \neq 0, I can divide by it. This gives me a_0 = -\left ( \frac{1}{2} \right ) [(s+1)(s+2)+v]. I already have s in function of v so that I have a_0 in function of v. Now I don't see why would a_0 be arbitrary, am I missing something?

You can't get rid of a₁ altogether. You miswrote the coefficient. Only the first part is multiplied by a₁, so you can't factor it out of the entire expression.

Also, you can't assume a₁≠0. You can take a₀≠0 because by definition it's the coefficient of the lowest-order term of the series. If it were equal to 0, another term would be the lowest-order term of the series. You have no such requirement of a₁.

When you obtain the final answer, you'll find that all of the coefficients are proportional to a₀, so a₀ is essentially the arbitrary constant that appears when you solve a differential equation.

 

[vela]

The indicial equation is c^2+c - l(l+1)=0. You can solve that to find the two values of c, one of which you can exclude on physical grounds.

 

[fluidistic]

The indicial equation is c^2+c - l(l+1)=0. You can solve that to find the two values of c, one of which you can exclude on physical grounds.

Ok I finally reach a_1=-\frac{2 a_0}{(s+1)(s+2)-v} and a_l = \frac{-2 a_{l-1}-Ka_{l-2}}{(l+s)(l+s+1)-v} for l \geq 2.
The inditial equation gives me s=c= \frac{-1 \pm \sqrt {1+4v}}{2}. You are saying I can get rid of \frac{-1 - \sqrt {1+4v}}{2} and keep \frac{-1 + \sqrt {1+4v}}{2}? So I discard the negative value of s and take the positive... though I don't know why. I didn't relationate K and v to physical quantities yet. I'm guessing they are related to the energy of the H atom, but I'm not sure.

Edit: c or s must be positive indeed, I mean I'm not taking a Laurent series or something like that.

 

[vela]

Think about what happens as x approaches 0 if c is negative.

 

[fluidistic]

Think about what happens as x approaches 0 if c is negative.

Ok the whole function blows up to infinities and its first and second derivatives too.

Edit: Am I right if I say that |K| \leq 2, to answer the question of the problem?
This is my guess after examinating the behavior of a_l. But I'm not sure at all of this answer.

 

[vela]

No, that's not correct. The problem is essentially asking you to solve the radial equation and find the allowed energies.

 

[fluidistic]

No, that's not correct. The problem is essentially asking you to solve the radial equation and find the allowed energies.

Ah ok... so K is the energy.
I've checked out on the internet about the radial equation and in Cohen-Tanoudji's book too. The closest form I've seen compared to my equation is \left [ \frac{d^2}{d \rho ^2}-\frac{l(l+1)}{\rho ^2} - \lambda ^2 _{k,l} \right ] u_{k,l} (\rho )=0 in the latter book. So I have an extra y' with respect to this equation.
Hmm now that I look at other sources (like http://quantummechanics.ucsd.edu/ph130a/130_notes/node236.html) I can see the d/dr or y' term. In fact that source has exactly the same form as mine.
I'm so tired, I'm going to try to understand the given solution tomorrow.
Thank you guys for all your help so far.