# Sloving the Hydrogen Atom

1. Dec 2, 2004

### emob2p

I have asked many people and searched many sources for the answer to the following question but have yet to reach a satisfactory conclusion.

When solving the hydrogen atom by separation of variables, for the angular equation, you come across the O.D.E.
d^2/(do)^2 (F) = -m^2 * F where o is the polar angle.

This is an easily solvable second order differential equation. The general solution is F(o) =A*e^imo + B*e^-imo where A and B are constants.

The problem is all the textbooks I've looked at only have the first term and just say m can positive or negative. This is fine, but why can't it be a linear combination of the two? Why must it only be one term? I've looked at the boundary conditions and have found no reason. If we allow for a combination of terms, it changes the wavefunction so that must be wrong. But where's the problem? Thanks.

2. Dec 2, 2004

### jujio77

That is a good question. Let's look at it this way. Assume you did use your value as the wavefunction. Then what would L_z operating on the wavefunction represent. If you choose "A=B=1", your wavefunction is no longer an eigenfunction of L_z, and you have to find a new set of operators (L_x,L_y,L_z).
When we do this we just choose the z-axis to be fixed.
I don't see any loss of generalization by choosing only e^{im\phi}.

3. Dec 2, 2004

### emob2p

I see that my wavefunction would no longer be an eigenfunction of L_z if we took the linear combination (the -mi(phi) messes things up). But why must the wavefunction be an eigenfunction of L_z? All I am saying is that a linear combination wavefunction solves the Schroedinger equation. That the wavefunction should be an eigenfunction of L-z seems to be an extra condition. From what I've learned this far, eigenfunctions of an operator only represent determinate states. But why should L_z yield a determinate state?

4. Dec 3, 2004

### emob2p

any thoughts?

5. Dec 3, 2004

### Staff: Mentor

Actually, the most general solution is a linear combination including all the possible values of m:

F(o) = A_0 + A_1*e^io + A_2*e^2io +...+ B_0 + B_1*e^-io + B_2*e^-2io +...

where A_0 is "A sub 0", etc.

Allow negative values of m, relabel the B's, and combine A_0 and B_0, and you get:

F(o) = A_1*e^io + A_2*e^2io +... + A_0 + A_(-1)*e^-io + A_(-2)*e^-2io +...

which is just the same thing, numerically, but you can describe it using a sum of only one kind of term instead of two.

6. Dec 3, 2004

### emob2p

I'm not quite sure how that solves the problem. True, the most general solution is a linear combination of all values of m. But for one value of m, F(o) still contains two terms: A_m*e^imo + A_-(m)*e^-imo.
Relabeling B does not remove the presence of two terms and it does not keep F from being an eigenfunction of L_z.

7. Dec 3, 2004

### emob2p

**Correction...with the two terms, F(o) is not an eigenfunction

8. Dec 3, 2004

### dextercioby

What is the problem,then?
1.Finding the eigenfunctions for Lz??
2.Finding the eigenfunctions for the Hydrogen atom Hamiltonian??

1.Piece of cake.The eigenvalue/eigenvector equation for Lz states
$$\hat{L}_{z}|\psi>=\hbar m |\psi>$$.Make the intelligent choise of representation in which Lz has both a familiar and a useful form
$$<r,\theta,\phi|\hat{L}_{z}|\psi>=\hbar m <r,\theta,\phi|\psi>$$.
$$\frac{\hbar}{i}\frac{\partial\psi (r,\theta,\phi)}{\partial\phi} =\hbar m \psi (r,\theta,\phi)$$.

2.The Hamiltonian of the Hydrogen atom has a particular form which allows us to write it as a sum from a term involving second derivatives wrt to "r" and the potential operator (a function of "r" as well) and a term proportional to the square of the angular momentum operator in the spherical coordinates of the coordinate representation of QM.

That's why i made that choise above.Now u know that $$\hat{H},\hat{L}^2,\hat{L}_{z}$$commute with each other,so they they have on orthonormal set of commune eigenvectors.So the Hamiltonian eigenvalue/eigenvector equation comprises the same functions $$\psi (r,\theta,\phi)$$ as does the diferential equation written above for Lz.The difference is that H is this particular representations depends on all 3 variables,as Lz depends only of the angle "phi".From this reason we are free to chose
$$\psi(r,\theta,\phi) =K(r,\theta)\chi(\phi)$$.

Substituting this particular form into the equation for the eigenfunctions of Lz,after symplifying the equation through the function K,gets u to the (made) "famous" (by this thread) function
$$\chi(\phi)=C\exp (im\phi)$$

The spectrum of Lz (the physically possible values of "m") is found to be the set of integer numbers.However,this spectrum is not infinite,as it is upper bounded by "l" and lower bounded by "-l" (as stated in the theory of angular momentum in QM).

9. Dec 3, 2004

### dextercioby

Conclusion:to the prior post and hopefully to the thread:

The eigenfunctions for the Lz operator are obviously not unique.In the case of the QM system called Hydrogen atom,these eigenfunctions are the normalized product between two particular functions:the Laguerre polynomials and the spherical harmonics.Being also eigenfunctions of the Hamilton operator,they describe physically stable (bound) quantum states of the Hydrogen atom.

Daniel.

10. Dec 3, 2004

### emob2p

Sorry to not let the issue rest, but I still think it is unresolved. I agree that H_ and L_z do not commute, but that does not necessarily mean they have a common set of eigenvectors. To my knowledge, only the inverse is true: if two operators have a common set of eigenvectors then they necessarily commute. All I am saying is that solving the Schrod eq. by separation of variables yields different angular functions than the eigenfunctions to the L_z operator. (The first includes the e^-imo term). If we knew the two operators necessarily shared the same eigenvectors, then we'd be home free since this would force the second term to vanish.

11. Dec 3, 2004

### dextercioby

Wow!!
1.Either u haven't read my post...It can be,it was long and maybe boring.
2.But if u have,u haven't done it properly.
What i'm saying is that in the case of the H atom (no spin for the electron,nor proton,no relativistic corrections or any other corrections whatsoever),the Hamiltonian commutes with both the Lsquared operator and Lz.Apparently u didn't read this line... Or wouldn't buy it...Perhaps i may have screwed up my credibility.Anyhow,there are smarter people on this forum to correct me when i'm wrong and if i'm wrong.
3.The 3 operators form what is called "a complete set of commuting selfadjoint linear operators".There is a powerful theorem in functional analysis which states that if a finite number of (self adjoint linear) operators commute with each other,hence form a CSCSALO (the acronym i wish to give for the expression above),then they admit a complete orthonormal system of commune eigenvectors (no acronym for this one).And viceversa.I wish to thank Pr.Saliu (my QM teacher) for pointing out and proving this theorem in his course as it is one the fundamental mathematical results which QM uses regulary...

So your knowledge should be a little "updated".Try the "refresh" button...
:tongue: Maybe it works...

Daniel.