# Help needed with solutions to schrodinger equation

1. May 31, 2009

### vaibhav1803

1. The problem statement, all variables and given/known data
hey im vaibhav,16 an 12th grade..just for pastime i tried to solve schrodinger equation in the 1D 2D 3D spaces. i got the 1D solution(not quantum oscillator), i seperated in 2D by polar coordinates but there is a problem in the radial equation
as for 3D i know that the solutions are such that i wont be doing em till im a wee bit older
but im just curious as the polar and radial eqautions are simply horrifying to contemplate solving

2. Relevant equations
$$\hbar \frac{d}{dr} (\frac{rdR}{dr})+\frac{2m(Er+I)R}{\hbar}=$$$$\hbar$$a2
where;
R"=second Derivative of R wrt r..similar for R'..R being radial wavefunction
I=intensity factor for a central force field ie of circular/spherical symmetry..(basically a constant explaining the field for a particular body)
3. The attempt at a solution
after solving,normalizing(0,2\pi) i got the angular part of the 2D wavefunction as
Ya($$\theta$$)=$$\frac{exp(ia\theta)}{\sqrt{2\pi}}$$
where "a" is quantized as any integer
exp(x)=ex
also,
it would be helpful if anyone could suggest an easier way to get to the solutions of the radial, polar wavefunctions(im not that experienced..)..and also quantizing them..
also this link ive posted the same question..coz i wasnt able to move that from intro to advanced..(i need sum help wid site controls too..) so follow this link for the attachment..really sorry for the inconvenience guys..
thanks,
Vaibhav

Last edited by a moderator: Apr 24, 2017
2. May 31, 2009

### Hao

Solving the differential equations to obtain the wavefunctions in Quantum Mechanics is nontrivial - There is usually no 'easy' solution, although things can be made simpler if the differential equation in question can be converted to a standard form that matches another differential equation for which the solutions are known. In most cases, there is no closed form solution.

For the cases where there are closed form solutions, you need a knowledge of power series solutions.

There are many resources out there that can explain how to solve Schrodinger's equation in 3D, and a common example is for the particular case of spherical symmetry, and a 1/r potential (ie. Hydrogen), which is exactly solvable.

I think it would be very instructive if you work through this example for Hydrogen:
http://users.aber.ac.uk/ruw/teach/235/hatom.html [Broken]
and ask questions if you have any uncertainty about any of the steps taken.

As for quantization conditions, they usually fall out of the boundary conditions applied to the system. For example:

1) Function must be single valued - Quantization condition for $$\Theta(\phi) = e^{i m \phi}$$, where $$\Theta$$ is an azimuthal component.
2) Function must be normalizable (not infinite, and goes to 0 at infinity) - Quantization condition for Principal Quantum Shell.

And so on.

Last edited by a moderator: May 4, 2017
3. May 31, 2009

### vaibhav1803

thanks for the help..i am yet to learn about power series and differential equations i had tried solving just coz i saw this equation in the book..now since i just saw the no trivial nature of the solution..completely horrifying me..well its better that i wait for my undergraduate courses to start next year or so...
but i thot about reducing some terms in th radial part to zero...
but i dont get it how do i come back to the radial equation.. at any"r".??

Last edited: May 31, 2009
4. May 31, 2009

### Hao

It's good to learn beyond the syllabus if you have the time - keep it up!

Which terms in particular? Are you referring to the power series solution?

Last edited: May 31, 2009
5. May 31, 2009

### vaibhav1803

hehe thanks..but i had a doubt in the radial solution...how do i come back to R(r) from R(inf.)

6. May 31, 2009

### Hao

If you are referring to the asymptotic form of the equation as $$r \rightarrow \infty$$ in the link I provided, the purpose of doing so is to 'guess' what the solution will look like.

In this case, based on the easier differential equation that appears at large r, we 'guess' that the solution to the complete radial equation will have a $$e^{-a r}$$ dependence.

So we assume that $$R(r) = function(r) * e^{-a r}$$, and substitute this back into the complete radial solution to obtain a modified differential equation for $$function(r)$$.

The motivation for this step is technical, but is essentially there to simply the mathematics. Part of this motivation is that for a power series solution,
$$R(r) = \sum^{\infty}_{n=0} c_n r^n$$
We see that it is very difficult for R(r) to converge for large r due to the polynomials. However, we know that the wavefunction must go to zero at infinity, and we know that a negative exponential term goes to zero faster than any power of r. Hence, if we put in a negative exponential term, our hope is that the power series solution for the modified radial equation will have a finite number of terms, as convergence requires an infinite series.

Alternatively, from the asymptotic form of the radial equation, we know that the solution approaches a simple exponential term. So it makes sense to assume that R is something * exponential term.

7. Jun 1, 2009

### vaibhav1803

ohk..i get it all you want to say is that R(r) we assume to be monotonic in nature so it makes sense to assume an infinite power series..?? in product with another decreasing function..?

8. Jun 1, 2009

### vaibhav1803

i thot of a system where2m(Er+I)tends to 0 for very small masses practically ignorable or for a repulsive system..where E=-I/r..any ideas on existence of such systems..??

9. Jun 1, 2009

### Hao

R(r) is neither monotonically increasing nor decreasing - it is bounded by a negative exponential envelope, and within that envelope, R(r) can alternate between positive and negative.

The reason we assume an infinite power series is more because it is a standard method of solution for differential equations, and exactly why it is a valid method is, I believe, a very mathematical question.

For example, in the Simple Harmonic Equation,

$$\frac{d^2 y}{d x^2} = - \omega^2 y$$

we know the solution is of the form y = A sin(w x) + B cos(w x).

However, we could have just as easily assumed a power series solution for y, and putting that in, will will discover that the power series converges to both sin(w x) and cos(w x).

As long as the mass is non-zero, we cannot ignore the term in solving Schrodinger's equation. We can only examine the solutions and see what happens as mass approaches 0.

A system with zero mass exists - photons.

Unfortunately, when we are dealing with photons as the particles of interest, things get complicated.

Photons may be freely created and destroyed - there is no conservation law for number of photons. As a result, a treatment of photons (electromagnetism) is inherently multiparticle.

To deal with them, we therefore need to abandon Schrodinger's equation, and move on to Quantum Field Theory, which is much, much more complicated (Usually a graduate course).

On the other hand, introducing electromagnetic interactions for a massive particle is relatively straightforward - we introduce vector potential A for electromagnetism, and modify the momentum operator. However, the details of the how and why are perhaps beyond introductory quantum mechanics.

If you really want the wavefunction of a single photon, you can take a look at this article:
www.cft.edu.pl/~birula/publ/CQO7.pdf, but it is a fairly controversial topic:

Last edited: Jun 2, 2009
10. Jun 1, 2009

### vaibhav1803

yes i see as sin(wx) and cos(wx) have an altenate power series a very good one though
so i assume for the SHM equation
$$y= \sum^{\infty}_{i=0}c_{r} x^{r}$$
and solve accordindly..??
well thats good then...i think i can use power series fo a lot of other pending stuff then..
thanks a million for the help Hao..!!
xD
and what about repulsive systems..?..are there any such ones whose total energy is of the form
E=-I/r..? I being the intensity factor of the force field(central)..?

Last edited: Jun 1, 2009
11. Jun 1, 2009

### Hao

Such systems exist, but there aren't any bound states - one example is the scattering of protons off a heavy nucleus. We model the incoming proton as plane waves, and get a superposition of complicated solutions as the output.

12. Jun 1, 2009

### vaibhav1803

yes i could se that the plot of the wave function i obtained was an increasing one..
so bound states are out of question..the solution i got from my assumption was:
$$R(r)=a^2 r_{0}(\frac{r}{r_{0}} +log(\frac{r}{er_{0}}))$$

13. Jun 1, 2009

### Hao

What you are looking for is something called partial wave analysis for scattering:

http://www.physics.gla.ac.uk/~donnelly/files/xsection/Lecture7.htm [Broken]

Last edited by a moderator: May 4, 2017
14. Jun 1, 2009

### vaibhav1803

ohk thanks ill refer to that..thank you..xD