What does the Schrodinger Equation really mean?

I like @Simon 's answer, but my personal favorite method to "derive" the Schrodinger equation is this.

Think of the quantum state as encoding some information about your system. That is to say some quantum version of a probability distribution defined on a vector space (Hilbert space).

What do we want of a meaningful probability distribution? First it must be always normalized so that mutually exclusive outcomes add up to probability 1. Second, we want all the probabilities corresponding to these outcomes to always be positive or at least 0. The most general form of a time evolution operator - that is to say an operator acting on your state at time $t_0$ takes it to $t_1$ is a so-called completely positive trace preserving map -http://en.wikipedia.org/wiki/Quantum_operation

This means essentially that the map meets all the requirements I stated (there are some subtleties, but it will take longer to explain).

Now, we may ask what sort of dynamical equation corresponds to this map? We want the equation to be Markovian, that is local in time so that the system does not depend on what happened a long time ago because this would violate locality in some sense.

Lindblad has shown that the most general form of such an equation is,

$$\dot\rho=-{i\over\hbar}[H,\rho]+\sum_{n,m = 1}^{N^2-1} h_{n,m}\left(L_n\rho L_m^\dagger-\frac{1}{2}\left(\rho L_m^\dagger L_n + L_m^\dagger L_n\rho\right)\right)$$

where $\rho$ is the state, $H$ is the Hamiltonian, $h_{m,n}$ are some rates and the $L_m$ are so called Lindblad operators which can be any operator.

However, as Banks, Susskind and Peskin have shown - http://adsabs.harvard.edu/abs/1984NuPhB.244..125B

this type of equation violates energy conservation or locality unless all $h_{m,n}$ are zero. If it violates energy conservation it can not describe a closed system which is invariant with respect to shifts in time. Therefore we set them to 0 and obtain just,

$$\dot\rho=-{i\over\hbar}[H,\rho],$$

which is the von Neumann equation, which reduces to the Schrodinger equation for pure states, $\rho=|\psi \rangle \langle \psi|$

This is a fairly basic approach suitable for students who have finished at least one semester of introductory Newtonian mechanics, are familiar with waves (including the complex exponential representation) and have heard of the Hamiltonian at a level where $H = T + V$. As far as I understand it has no relationship to Schrödinger's historical approach.

Let's take up Debye's challenge to find the wave equation that goes with de Broglie waves (restricting ourselves to one dimension merely for clarity).

Because we're looking for a wave equation we will suppose that the solutions have the form $$\Psi(x,t) = e^{i(kx - \omega t)} \;, \tag{1}$$ and because this is suppose to be for de Broglie waves we shall require that \begin{align} E &= hf = \hbar \omega \tag{2}\\ p &= h\lambda = \hbar k \;. \tag{3} \end{align}

Now it is a interesting observation that we can get the angular frequency $\omega$ from (1) with a time derivative and likewise wave number $k$ with a spacial derivative. If we simply define the operators1 \begin{align} \hat{E} = -\frac{\hbar}{i} \frac{\partial}{\partial t} \tag{4}\\ \hat{p} = \frac{\hbar}{i} \frac{\partial}{\partial x} \; \tag{5}\\ \end{align} so that $\hat{E} \Psi = E \Psi$ and $\hat{p} \Psi = p \Psi$.

Now, the Hamiltonian for a particle of mass $m$ moving in a fixed potential field $V(x)$ is $H = \frac{p^2}{2m} + V(x)$, and because this situation has no explicit dependence on time we can identify the Hamiltonian with the total energy of the system $H = E$. Expanding that identity in terms of the operators above (and applying it to the wave function, because operators have to act on something) we get \begin{align} \hat{H} \Psi(x,t) &= \hat{E} \Psi(x,t) \\ \left[ \frac{\hat{p}^2}{2m} + V(x) \right] \Psi(x,t) &= \hat{E} \Psi(x,t) \\ \left[ \frac{1}{2m} \left( \frac{\hbar}{i} \frac{\partial}{\partial x}\right)^2+ V(x) \right] \Psi(x,t) &= -\frac{\hbar}{i} \frac{\partial}{\partial t} \Psi(x,t) \\ \left[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+ V(x) \right] \Psi(x,t) &= i\hbar \frac{\partial}{\partial t} \Psi(x,t) \;. \tag{6}\\ \end{align} You will recognize (6) as the time-dependent Schrödinger equation in one dimension.

So the motivation here is

1. Write a wave equation.
2. Make the energy and momentum have the de Broglie forms, and
3. Require energy conservation

but this is not anything like a proof because the pass from variable to operators is pulled out of a hat.

As an added bonus if you use the square of the relativistic Hamiltonian for a free particle $(pc)^2 - (mc^2)^2 = E^2$ this method leads naturally to the Klein-Gordon equation as well.

1 In very rough language an operator is a function-like mathematical object that takes a function as an argument and returns another function. Partial derivatives obviously qualify on this front, but so do simple multiplicative factors: because multiplying a function by some factor returns another function.

We follow a common notational convention in denoting objects that need to be understood as operators with a hat, but leaving the hat off of explcit forms.

It stands for dynamics for strange quantum particle that can be expressed as a wave $\psi$. Since quantum theory is fundamentally probabilistic we must write down classical Hamiltonian in expectation values:$$\langle H\rangle=\langle T\rangle+\langle V(x,t)\rangle$$

In quantum mechanics we use different operators that act on state $\psi$ $$\langle H\rangle=\int i\hbar \frac{d}{dt}\psi \centerdot \overline{\psi}dx$$ $$\langle T\rangle=\int - \frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi \centerdot \overline{\psi}dx$$ $$\langle V(x,t)\rangle=\int V(x,t)|\psi|^2dx$$

[NOTE: if you dare, you can actually derive previous expressions with Fourier methods and lesser assumptions, but genereally we take them granted since Schrödinger's equation is a fundamental postulate in physics]

After using variational lemma you get: $$i\hbar \frac{d}{dt}\psi = - \frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi + V(x,t)\psi$$.