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Solving Radial Schrodinger Equation

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data
    This is a (long) multi-part question working through the various stages of solving the radial Schrodinger equation and as such it would be impractical to type it all out here but I will upload the pdf (https://drive.google.com/open?id=0BwiADXXgAYUHOTNrZm16NHlibUU) of the problem set (its problem 2b.) and then refer to the various questions by recalling their latin numerals (i, ii, iii, iv, etc...)

    2. Relevant equations

    The Radial ODE: d^(2)u/dx^2 + (2/x - l(l+1)/x^2 - b^2)u = 0

    3. The attempt at a solution

    So I just started this problem and already am running into issues (will be updating this thread as I progress through the problems). Problem section i) tells us to show the asymptotic behavior of u(x) as x tends towards +inf is u ~ e^(+/-)bx. I imagine this means solve the Radial ODE for u in terms of x. Looking at the ODE one notices quickly its separable. Once separated it looks something like this: {d^(2)u}/u = -(-2/x - l(l+1)/x^2 - b^2)dx^2. In this form its clear we must integrate twice with respect to each variable in order to get something of the form u(x) = ... ... ...

    Now, integrating once I get: ln(u)du = {b^(2)*x - l(l+1)/x - 2ln(x)}dx

    This doesn't feel write to me though; It's hard to see how I'll get a clean exponential out of this. Furthermore the hint for this section states To do this omit the terms in 1/x and 1/x^2 which can be ignored relative to b^2 when x tends towards +inf. This leads me to believe I need to do some manipulation whilst the radial ODE is in its original form (as that's where the 1/x and 1/x^2 terms appear).

    Any help would make my day, I would be so grateful. I have been struggling with this problem for a little bit and can't seem to make any headway trying anything
     
  2. jcsd
  3. Dec 8, 2016 #2

    DrClaude

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    Staff: Mentor

    You need to modify the ODE taking into account that you are only looking at the behavior as x→∞.
     
  4. Dec 8, 2016 #3
    So, what you're saying, DrClaude, is that I need to cancel out the 1/x and 1/x^2 terms (since 1/x and 1/x^2 go to 0 very quickly when x--> infinity) which leaves me with the ODE d^(2)u/dx^2 = +b^2(u) and then assess how u(x) responds at infinity? If so, then I imagine I just follow through with integration which, after the first, leaves you with ln(u)du = +xb^(2)dx and a second integration yields u(ln(u)-1) = (x^2b^2)/2 which would not get me in the form u ~ e^(+/-)bx.

    Or perhaps did I misunderstand what you said (very real possibility)
     
  5. Dec 8, 2016 #4

    DrClaude

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    Staff: Mentor

    You're solving a differential equation, not integrating a derivative.
     
  6. Dec 8, 2016 #5
    I'm not following you I don't think.. doesn't solving a DE involve integrating derivatives? The problem asks you to show the form of u(x) as x →∞ is something like e±bx right, so I would imagine that means we need to solve the DE for u in terms of x. Looking at the DE now with the other terms cancelling I can see technically u = (d2u/dx)/b2 but that encodes no information about the x dependence. Sure, the x-dependence is implicit in that we're differentiating wrt x but it's not explicit and also not in the form I'm seeking. I'm looking for how to get u(x) in the form e±bx

    I apologize if I'm misunderstanding you further, I'm all sorts of confused right now.
     
  7. Dec 8, 2016 #6

    DrClaude

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    Staff: Mentor

  8. Dec 8, 2016 #7
    Ah yes! I was thinking that route but couldn't remember the name of the form of this equation-type. Thanks so much for your help DrClaude, I really appreciate it. I'll probably be working on this problem all night so feel free whenever to stop back again and offer up your sage advice, if you feel so inclined :)
     
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