You should not think of the Schrödinger equation as a true wave equation. In electricity and magnetism, the wave equation is typically written as
$$\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}$$
with two temporal and two spatial derivatives. In particular, it puts time and space on 'equal footing', in other words, the equation is invariant under the Lorentz transformations of special relativity. The one-dimensional time-dependent Schrödinger equation for a free particle is
$$ \mathrm{i} \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$
which has one temporal derivative but two spatial derivatives, and so it is not Lorentz invariant (but it is Galilean invariant). For a conservative potential, we usually add $V(x) \psi$ to the right hand side.
Now, you can solve the Schrödinger equation is various situations, with potentials and boundary conditions, just like any other differential equation. You in general will solve for a complex (analytic) solution $\psi(\vec r)$: quantum mechanics demands complex functions, whereas in the (classical, E&M) wave equation complex solutions are simply shorthand for real ones. Moreover, due to the probabilistic interpretation of $\psi(\vec r)$, we make the demand that all solutions must be normalized such that $\int |\psi(\vec r)|^2 dr = 1$. We're allowed to do that because it's linear (think 'linear' as in linear algebra), it just restricts the number of solutions you can have. This requirements, plus linearity, gives you the following properties:
You can put any $\psi(\vec r)$ into Schrödinger's equation (as long as it is normalized and 'nice'), and the time-dependence in the equation will predict how that state evolves.
If $\psi$ is a solution to a linear equation, $a \psi$ is also a solution for some (complex) $a$. However, we say all such states are 'the same', and anyway we only accept normalized solutions ($\int |a\psi(\vec r)|^2 dr = 1$). We say that solutions like $-\psi$, and more generally $e^{i\theta}\psi$, represent the same physical state.
Some special solutions $\psi_E$ are eigenstates of the right-hand-side of the time-dependent Schrödinger equation, and therefore they can be written as
$$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_E}{\partial x^2} = E \psi_E$$
and it can be shown that these solutions have the particular time dependence $\psi_E(\vec r, t) = \psi_E(\vec r) e^{-i E t/\hbar}$. As you may know from linear algebra, the eigenstates decomposition is very useful. Physically, these solutions are 'energy eigenstates' and represent states of constant energy.
- If $\psi$ and $\phi$ are solutions, so is $a \psi + b \phi$, as long as $|a|^2 + |b|^2 = 1$ to keep the solution normalized. This is what we call a 'superposition'. A very important component here is that there are many ways to 'add' two solutions with equal weights: $\frac{1}{\sqrt 2}(\psi + e^{i \theta} \phi)$ are solutions for all angles $\theta$, hence we can combine states with plus or minus signs. This turns out to be critical in many quantum phenomena, especially interference phenomena such as Rabi and Ramsey oscillations that you'll surely learn about in a quantum computing class.
Now, the connection to physics.
If $\psi(\vec r, t)$ is a solution to the Schrödinger's equation at position $\vec r$ and time $t$, then the probability of finding the particle in a specific region can be found by integrating $|\psi^2|$ around that region. For that reason, we identify $|\psi|^2$ as the probability solution for the particle.
- We expect the probability of finding a particle somewhere at any particular time $t$. The Schrödinger equation has the (essential) property that if $\int |\psi(\vec r, t)|^2 dr = 1$ at a given time, then the property holds at all times. In other words, the Schrödinger equation conserves probability. This implies that there exists a continuity equation.
If you want to know the mean value of an observable $A$ at a given time just integrate
$$ <A> = \int \psi(\vec r, t)^* \hat A \psi(\vec r, t) d\vec r$$
where $\hat A$ is the linear operator associated to the observable. In the position representation, the position operator is $\hat A = x$, and the momentum operator, $\hat p = - i\hbar \partial / \partial x$, which is a differential operator.
The connection to de Broglie is best thought of as historical. It's related to how Schrödinger figured out the equation, but don't look for a rigorous connection. As for the Hamiltonian, that's a very useful concept from classical mechanics. In this case, the Hamiltonian is a measure of the total energy of the system and is defined classically as $H = \frac{p^2}{2m} + V(\vec r)$. In many classical systems it's a conserved quantity. $H$ also lets you calculate classical equations of motion in terms of position and momentum. One big jump to quantum mechanics is that position and momentum are linked, so knowing 'everything' about the position (the wavefunction $\psi(\vec r))$ at one point in time tells you 'everything' about momentum and evolution. In classical mechanics, that's not enough information, you must know both a particle's position and momentum to predict its future motion.
The derivation is nice, because it shows the similarity between the time-independent Schrödinger equation and a standing-wave equation. I would like to point out some aspects of this derivation that may clarify its meaning and limitations.
First of all, this derivation has not derived the Schrödinger equation from classical mechanics. It is still introducing some new postulates, just they are different from Griffiths'. Namely, the authors are postulating:
- the de Broglie relation.
- that a time independent quantum system can be described by a wave equation.
Additionally, they will have to add some postulate to justify the time dependent Schrödinger equation.
Therefore I think that this is just a matter of taste. Griffiths needs fewer postulates, he abruptly gives the correct equation. Here they work slower towards the goal, but trying to give a better physical intuition of what is going on.
That said, let's talk about the derivation and what it means. It starts from the wave equation in one dimension. When talking about the wave equation one usually means that the velocity $v$ is a constant. It may be the speed of light in vacuum, the speed of sound in a particular medium, but it is a number fixed beforehand. Yet, it may be meaningful and useful to consider also cases where the speed varies with position $v=v(x)$. For example, if a sound wave moves from a region of colder air to a region of hotter air, its speed will increase. In this case, we consider an equation where the velocity is given by the energy relation of a point particle moving in a potential:
$$E = {1 \over 2}m v^2 + V(x) \implies v(x) = \sqrt{{2 \over m}(E-V(x))}$$
Imagine you have a point particle with energy $E$ in a potential $V(x)$. The velocity of the wave in every point will be the same as the velocity of the particle. Notice that this reasoning introduced a new parameter to the equation: the energy. Now, waves that have different energy will propagate at different speeds! This means that the waves will be dispersive. They will have a different phase velocity and group velocity.
We have now our wave equation and we would like to solve it
$${\partial^2 u \over \partial x^2} = {1 \over v(x)^2}{\partial ^2 u \over \partial t^2}$$
It will have many general solutions. It is possible that a subset of all solutions can be written as $u(x,t) = \psi(x)\cos(\omega t)$. It is also possible that no solution exists in this form. Or that it exists only form some value of $\omega$. We are free to plug this into the equation and verify whether any solution like this exists. Of course, the authors already know that this is the correct substitution, but this is in general a common method to solve differential equations. You just guess a solution and plug it in to see if you were right.
In this case, we plug the solution in the equation and find a new equation for $\psi$. $\psi$ doesn't depend on time, therefore our solution (if it exists) will be a standing wave.
Now you have an equation for $\psi$ that depends on two parameters: $\omega$ and $E$. Using the de Broglie relation you find out that $E=\hbar \omega$. Also, when trying to solve the equation for $\psi$ you find that not all values of $\omega$ (i.e. of $E$) allow a solution. You have found that a wave described by this equation cannot have any frequency (energy). The energy is quantized.
Best Answer
The issue is that the assumptions are fluid, so there aren't axioms that are agreed upon. Of course Schrödinger didn't just wake up with the Schrödinger equation in his head, he had a reasoning, but the assumptions in that reasoning were the old quantum theory and the de Broglie relation, along with the Hamiltonian idea that mechanics is the limit of wave-motion.
These ideas are now best thought of as derived from postulating quantum mechanics underneath, and taking the classical limit with leading semi-classical corrections. So while it is historically correct that the semi-classical knowledge essentially uniquely determined the Schrödinger equation, it is not strictly logically correct, since the thing that is derived is more fundamental than the things used to derive it.
This is a common thing in physics--- you use approximate laws to arrive at new laws that are more fundamental. It is also the reason that one must have a sketch of the historical development in mind to arrive at the most fundamental theory, otherwise you will have no clue how the fundamental theory was arrived at or why it is true.