Solutions to schrodinger equation

Homework Statement

hey im vaibhav,16 an 12th grade..just as 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

Homework 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)

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..
thanks,
Vaibhav

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You're basically approaching the solution of the Schrodinger equation for the Hydrogen atom, which was very important work for quantum mechanics. You seem to be approaching it from a mathematical standpoint but you could get some help on this by considering the quantum mechanics themselves.

The derivation for this is presented well in my physical chemistry book. If you can get a copy of Physical Chemistry, by Peter Atkins & Julio de Paula, it should help. I believe the 2D and 3D solutions to the Schrodinger equation are in chapter 9, but it might be chapter 10.

Basically, the solution to the 3D requires spherical harmonics. If you want to try to derive this yourself you need to study them. But you are quite correct that the radial and angular parts can be separated. Once you have the spherical harmonics you can derive quantization from the angular momentum. Here it is best to work with the angular momentum operators, but I'm not sure if you know about that yet. You can derive them without operators but it's more intuitive and easier to understand the result if you use them. The surprising result is that angular momentum is quantized and so is the magnetic moment quantum number.

As for the 2D case, the boundary conditions are different than for the 1D case. In the 2D case, you seek functions that are 2*Pi periodic. This means that quantization follows the periodicity, but it also allows for the trivial solution. This makes sense if you consider a particle confined to a ring, it can have zero rotation on the ring.

Cheers.

sorryy for being such a newbie..but
question: can i find the three dimensional equation using the 2 D solution..??
thats the basis of solvin the 2D equation..

I need to review the material to answer for sure, but I have a meeting in a couple of minutes. In my recollection, you cannot apply the 2D solution to the 3D case without using the spherical harmonics. You can use similar cyclical boundary conditions for the angular momentum operator, but you can no longer know the direction of rotation, due to the commutators. (If this is confusing you, sorry. I'll try to explain it later).

Maybe the best way to work through this (admittedly difficult problem) is to start from what you have now. What do you have as a solution the 2D case? Let's figure that out first.

i am yet to figure how to extrapolate radial solution to the correct equation..
im gonna giv it a try soon enough but i hav an organic chem test to prepare for (alkyl halides..JEE probably...) so ill tell you about the solution later if that doesnt disturb ur schedule or...but this is one of the solutions i got from one of my assumptions as
2m(Er+I) tends to 0 i think its for a repulsive force field with total energy E=-I/r

i could se that the plot of this wave function i obtained was an increasing one..
so bound eigenstates are out of question..the solution i got from my assumption was:
$$R(r)=a^2 r_{0}(\frac{r}{r_{0}} +log(\frac{r}{er_{0}}))$$
and $$Y_a(\theta)=\frac{exp(ia\theta)}{\sqrt{2\pi}}$$
so, $$\Psi (r,\theta )=\frac{a^2 r_{0}}{\sqrt{2\pi}}(\frac{r}{r_{0}} +log(\frac{r}{er_{0}}))\{e^{(ia\theta)}}}$$

the only quantum number here is "a" azimuthal quantum no. this ones for a scattering system with repulsive forces..thats what i think

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