Suppose V(x,y,z) = V(Sqrt[x^2+y^2+z^2])=V(r). This means that the potential depends only on the distance to the origin. An example might be the Coulomb potential, which is V(r)=-e/r. There are of course many other examples - nuclei have spherical potentials, but they are not Coulomb. A harmonic oscillator in 3-D looks like cr^2.
We will soon talk about a "spherical box", i.e. V(r) = 0 out to some radius r, and then it's infinite. Etc...
Now, for such a potential, rectangular (x,y,z) coordinates are a pain. But spherical coordinates are natural!
We need to go back to Physics 2140, and look up (or rederive),
the Laplacian in spherical coordinates. It's something of a pain, but here's what we got:
We plug this into (the first term of) the S.E.,
and once again we cross our fingers and try separation of variables,
this time
You take this, plug it into the S.E., divide the whole equation through by psi, and (as an extra trick this time) multiply through by r^2sin(theta) everywhere. This leaves us with
We don't get the beautiful simple separation like we got before in Cartesian coordinates for the square box, but it's not as bad as it looks. Consider the phi dependence. The only place phi appears is in the third term. Since it appears nowhere else, that term must just be equal to a constant. Let's call that constant -m^2, because it turns out that it must be negative. (That shouldn't be obvious right now, but as soon as we solve the differential equation for Phi, you'll see that it better be negative or we're in big trouble!) Once we've set that term to be a constant, we can take what we have left over, and then divide the whole equation through by sin^2(theta). This leaves:
The second two terms (on the left side) depend ONLY on theta, and not on r, while all the other terms depend on r, not on theta! Once again, the logic of separation says the two "theta only" terms must add up to a constant. We must give this constant a name also, and tradition is to call it -l(l+1). This is a really screwy name for a constant, but again, you can take my word for it that when we get to that differential equation, the constant must be negative, and even have that form (in fact, l will even have to be an integer!). Let me summarize what we have.
(The "m" on the r.h.s. is plain old mass, not the new "m quantum number" in the first equation. Oops! Bad choice of notation)
We have three linear ordinary differential equations, one for each of the three separated functions, R, Theta, and Phi. Only the R equation depends on the potential energy function! (That's because we're only considering the case V=V(r)) The first two functions are universal functions. In fact, the boundary conditions on Theta and Phi are also universal. E.g., the wave function must be single valued, which means Phi(phi)=Phi(phi+2Pi). If you look at the differential equation which Phi solves, it's our favorite one again, and has solutions
Now you see why the constant had to be negative. If it was positive, then Phi would be a real exponential, and then it couldn't satisfy Phi(phi)=Phi(phi+2Pi)! Not only that, but the solution we've got doesn't satisfy that unless m is an integer! Since single valuedness is always true, we've discovered Phi(phi) no matter what problem you're working on. If I let m be negative, that takes care of having both signs on the exponent. And this time, m=0 is o.k., because then Phi=constant, which is a perfectly acceptable wave function, and doesn't give us "zero particles".
The Theta equation is a bit more trouble. We encountered this very equation in 2140, (at least for the case m=0) It is the Legendre equation, if you just define x=cos(theta). (If m is not 0, it is called the associated Legendre equation) This differential equation, if you recall, has as solution two independent functions (which can be expressed as infinite power series in x). See Boas Ch. 12-2. But unless l is an integer, it turns out that those power series diverge at x=+/-1. Now, x=+/-1 is just the north or south pole in polar coordinates, and the wave function better not blow up anywhere. So this forces l to be an integer (as promised)
In that case Boas showed that ONE of the two possible functions still
blows up at
but the other solution is not only finite, it's a simple polynomial. (The
associated Legendre polynomials are a bit more complicated, but there's a
straightforward formula for them, see Boas 12.10) The bottom line is that
l=0,1,2,... is required absolutely generally, based only on keeping the
wave function finite, independent of V.
Thus the angular part of our wave function,
is given, for any spherically symmetric problem, in terms of 2 quantized
(integer) constants, m and l. It turns out that |m|<=l, because otherwise we
get a blowup of the associated Legendre equation. We sometimes call the
products the
's,
or "spherical harmonics". I can give you some specific examples, and this will
be good enough for our purposes here. (The "m" dependence is easy, we already
wrote it down above.) The Legendre polynomials we have to remember (or
rederive).
The spherical harmonics are normalized in a purely conventional way. The convention is to insist that
for reasons that will be (should be?) obvious?
Some results:
l=0 => m=0 only (since m<=l),
If l=1 => m=-1, 0, or +1. The corresponding solutions are
l=2 => m=-2,-1,0,+1, or +2:
etc...
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