# Cant understand similar proof Quantum numbers from PDE (pdf attachment!)

1. Aug 27, 2011

### mohsin031211

I believe that this is similar to the proof of schrodinger equation to obtain quantum numbers, however i cannot seem to understand the relationship between n, l and m:

I have attached a pdf file on partial differential equations and on page 5, i cannot seem to understand why it is +n^2 and l,m are negative?

Also, on page 6 it states 'Hence the sumis also a solution. Note ℓ and m do not have to be integers and so
the above need not be a discrete sum. Also note that if ℓ → 0, cosine is replaced
by 1 and sine by x.' why is sine replaced by x, shouldn't it disappear as it equals to 0 rather than being replaced by x?

My final query is about equation 1.28, how the superposition principle is applied? Does it just form linear equations of the solutions and why is the solution for equation 1.27 cosh and sinh whereas for the rest it isn't?

Thank you so much in advance , whoever can clear this for me

#### Attached Files:

• ###### PDE.pdf
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2. Aug 27, 2011

### Fredrik

Staff Emeritus
He knows that $X''(x)/X(x)$ is equal to some real number for all x, so he needs to pick a notation for that number. He chooses $-l^2$ because it makes the solutions look nice: $\cos lx$ and $\sin lx$.

He knows that each term of (1.20) is a real number, and that they add up to 0. So they can't all have the same sign. He's showing you how to proceed if the first two happen to be non-positive and the third non-negative. That's why the Z equation has sinh and cosh solutions instead of sin and cos solutions.

I don't see what that limit has to do with anything, but 1 and x are the first-order approximations of cos x and sin x respectively. Recall that f(x)=f(0)+xf'(0)+... But maybe he's not talking about that at all. If you just set l=0 in (1.21), the X equation becomes X''(x)=0. It has x and 1 as solutions (actually, any first-degree polynomial). Maybe that's what he meant.

The superposition principle is the idea that if f and g are solutions, then so is af+bg where a and b are real numbers. I don't like calling it a "principle", because it follows from the fact that we're dealing with linear equations. Consider $X''(x)+l^2X(x)=0$. If we denote the operator that takes a function to its derivative by D, then the equation can be written as $(D^2+l^2)X=0$. The operator $D^2+l^2$ is clearly linear. So if $(D^2+l^2)f=0$ and $(D^2+l^2)g=0$, then $(D^2+l^2)(af+bg)=a(D^2+l^2)f+b(D^2+l^2)g=0$.