• MxwllsPersuasns
In summary, DrClaude was struggling with a differential equation and needed help finding a solution. He found that he needed to modify the equation to take into account that he is only looking at the behavior as x→∞. Once he did this, he was able to find the solution in terms of x.
MxwllsPersuasns

## Homework Statement

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...)

## Homework Equations

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

## The Attempt at a Solution

[/B]
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

MxwllsPersuasns said:
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).
You need to modify the ODE taking into account that you are only looking at the behavior as x→∞.

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)

You're solving a differential equation, not integrating a derivative.

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.

MxwllsPersuasns said:
I can see technically u = (d2u/dx)/b2 but that encodes no information about the x dependence.
But all the dependence is there! You need a function u(x) that twice derived wrt x will give back itself times a constant.
Have a look at http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf

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 :)

## 1. What is the Schrodinger equation?

The Schrodinger equation is a mathematical equation that describes the behavior of quantum particles, such as electrons, in a given system. It was developed by Austrian physicist Erwin Schrodinger in 1926 and is a fundamental equation in quantum mechanics.

## 2. What is the radial Schrodinger equation?

The radial Schrodinger equation is a specific form of the Schrodinger equation that describes the behavior of particles in a spherically symmetric potential, such as the nucleus of an atom. It takes into account the radial distance from the nucleus and the potential energy of the particle.

## 3. How is the radial Schrodinger equation solved?

The radial Schrodinger equation can be solved using various mathematical techniques, such as separation of variables and the use of special functions. The solutions to the equation are known as wave functions and describe the probability of finding a particle in a particular location.

## 4. What is the significance of solving the radial Schrodinger equation?

Solving the radial Schrodinger equation allows us to understand the behavior and properties of quantum particles in a spherically symmetric potential. This is essential in many fields of science, such as chemistry, physics, and materials science, as it helps us to predict and control the behavior of particles and systems at a microscopic level.

## 5. Are there any limitations to the radial Schrodinger equation?

While the radial Schrodinger equation is a powerful tool in understanding quantum systems, it has its limitations. It only applies to systems with spherically symmetric potentials and does not take into account relativistic effects. To describe more complex systems, such as those involving multiple particles or strong interactions, other equations, such as the Dirac equation, must be used.

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