[americanstrat4]

Can anyone provide a simplified explanation of the mathematics behind Schrödinger’s work?  

For example, how was he able to mathematically determine the areas where s clouds, p clouds, etc. would exist?  

How do the mathematics account for hybridizations and nodes within the electron clouds?

Just out of curiosity.

[gravenewworld]

In short, quantum mechanics in general is all about linear operators in Hilbert Space.  Hilbert space is at the heart of quantum.  I know more math than quantum, so I can not answer all of your questions-but I will tell you what I do know.  Anyway, eigenvalues that are associated with the wave equation correspond to energy levels of the atom.  Eigenvalues are sure to correspond with energy because the hamiltonian operator is a hermetian operator (it is it's own adjoint).  Hermetian operators in hilbert space have the nice property of having eigenvalues that are all real numbers.   The only problem with the operators that are used in quantum is the fact that some of them are not bounded operators, so a lot of the mathematics of hilbert spaces does not have to be true for the operators found in quantum. This is another reason why it is so hard to solve the s. equation. That is a little bit of what i do know.  

[americanstrat4]

Thanks.

[Donaldson Tan]

For example, how was he able to mathematically determine the areas where s clouds, p clouds, etc. would exist?  

Dr Schrodinger only determined those areas for the H atom - the simplest atomic structure

[americanstrat4]

So, the wave function cannot account for the behavoir of other elements?

[Borek]

Can, although using spdf orbitals that's only an approximation. But it was experimentally checked that the approximation is very good - something like four significant digits at least.

[pantone159]

To be a little more precise:

The Schroedinger equation is a differential equation that describes the wavefunctions that the electrons can have.  It can be written as:  

H * Psi = E * Psi

H is an 'operator' that contains the formula for kinetic and potential energy, and it has derivatives in it.  Operators act upon functions, resulting in new functions.  Often, H is written as a matrix, and Psi as a vector, and the action of the operator is multiplying the matrix times the vector.  H is the only thing you know when you start to solve the Schroedinger equation.  Psi is an unknown function (or vector), it is the wavefunction of the electron.  E is an unknown number (just a plain number), it is the energy of that wavefunction state.

Solving this equation, when you know H and nothing else, involves finding the small set of Psi functions that, when operated on by H, give back the SAME function, except scaled in size (by E).  (You calculate E at the same time.)  For a given H, there are only a limited set of Psi functions that come back the same, with only a change in size.  These functions that work are called 'eigenvectors' or 'eigenfunctions', and the amounts by which they are scaled by H are called 'eigenvalues', if you have heard of those names.

For example, you could imagine a Psi function that is a constant value for all points in space, that is, Psi(x,y,z) = c.  When you apply H to this function, you WON'T get something that is constant over all points in space, so you didn't just scale your function, so this Psi is no good as an answer.  Only certain, special, Psi's will work.  These are the allowed wavefunctions, your goal is to find them.  This limited set of answers is the 'reason' why the allowed states are quantized.

Now consider the hydrogen atom...  You can get the right H for this by considering the kinetic and potential energy of the electron, it turns out that H = -(hbar*hbar)/(2*m)*(d^2/dx2+d^/dy^2+d^2/dz^2) - e*e/r,
where d^2/dx^2 is the second derivative with respect to x, r = sqrt(x*x+y*y+z*z), e is the electron charge, m is an appropriate mass, and hbar is Plancks constant.

This is an easy enough H that is is possible to solve the equation in closed form (meaning you get an equation for the answer).  This is complicated, but the Psi functions you get are the familiar electron orbital shapes, the s,p,d,f orbitals.

If you want to look past hydrogen, say to helium, you have to adjust your H so that it accounts for the repulsion between the two electrons.  When you do this, H becomes complicated enough so that you can't get a formula for the wavefunctions anymore.  You can use a computer to calculate approximations of them, however.  The Schroedinger equation is still perfectly valid in this case, you just can't write the answers with a nice formula.  It turns out, happily, that the new wavefunctions look a lot like the hydrogen ones, which is why chemists can use them to talk about other atoms.

You can adjust this for as many electrons as you like.  Once you start getting into the heavier elements, Einstein's relativity becomes important, and the Schroedinger equation doesn't account for this, so it eventually breaks down.  Until then, though, it is still perfectly valid, just hard to solve.

This might be more complicated of an answer than you wanted, but I'm not sure how to make it any simpler.

[gravenewworld]

Just to make the idea of eigenfunctions and values clearer  take the derivative as an example

d/dx  is the derivative operator for a function f(x).

What f(x) solves this equation   d^2f(x)/dx^2=-f(x)? I bet you already know an answer if you took basic calculus.   A solution to this equation could be f(x)=sin(x). Notice that when you apply the operator to the function f(x)  you get f(x) back again with -1 in front of it.  sin(x) is an example of an eigenfunction for this operator and -1 would be the eigenvalue.  This is essentially what the schrodinger equation is all about.  The question to ask is what wave function satisfies HP=EP  (p=psi) for the hamiltonian operator H?

[FeLiXe]

I can tell you what hybridisation and nodes are.

Hybridisation is a linear combination of atomic orbitals. For Example you take 1 s and 1 p orbital and for each point in space you add the values that the s and p functions give. Then you get an sp-hybrid (not normalized). Square it to get the probability. If you take 1 s and Sqrt(2) times a p orbitals or Sqrt(3) times a p orbital than you get sp2 and sp3.

a nodal plane is an area where the wave function is 0. Take the function of the 2pz orbital (not normalized): r*Exp[-r/2]*Cos[th]. It will be 0 when th=90° which is the xy-plane. From this equation you can see that the 2pz orbital has a nodal plane, the xy-plane.

[FeLiXe]

really amazing is "deriving" the Schrödinger equation.

You start out with energy conservation:
(1): p^2/(2m) + V = E

p(x)...momentum
p^2/(2m)=mv^2/2...kinetic energy
V(x)...potential energy
E...total energy, must be constant because of energy conservation

We assume the wave function to be of the form:
Psi = Exp(i * 2 pi / lambda * x)

i ... imaginary unit
lambda ... wave length

according to DeBroglie we know:
lambda = h / p

Then we have:
Psi = Exp(i * 2 * pi * p / h * x)
the first and second derivatives are:
Psi' = Exp(i * 2 * pi * p / h * x) * i * 2 * pi * p / h = Psi * i * 2 * pi * p / h
Psi'' = Psi * (i * 2 * pi * p / h)^2 = - Psi * (2 * pi * p / h)^2

in other words:
p^2 * Psi = - Psi'' * (h / 2 pi)^2
with hbar = h / (2 pi)
(2): p * Psi = - Psi'' * hbar^2
(-d2/dx2 * hbar^2 is called the momentum operator)

now we multiply (1) by Psi
p^2*Psi/(2m) + V*Psi = E*Psi
insert (2) and you get the time independent 1 dimensional Schrödinger equation:
-hbar^2 / 2m * Psi(x)'' + V(x)*Psi(x) = E*Psi(x)

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this got kind of long, I am sorry

anyway the point is: the Schrödinger equation comes from the  law of energy conservation only that you insert properties of waves instead of particles. In my eyes it is amazing how this one thought produced the most productive equation in modern science.