Fermi's golden rule still applies in the relativistic limit, and can be rewritten in a Lorentz invariant fashion. Starting with the transition probability
$$ W_{i\rightarrow f} = \frac{2\pi}{\hbar} |m_{if}|^2 \rho(E) \,,$$ to have $W$ Lorentz invariant we'd like both the matrix element $|m_{if}|^2$ and the density of final states $\rho(E)$ to be invariant.
This can be done by shifting a few terms around. A little bit of handwaving to motivate it: The wave function $\psi$ (which is in the matrix element) has to be normalized by $\int |\psi|^2 dV = 1$, which gives us a density (of probability to encounter a particle) of $1/V$. Now, a boosted observer experiences length contraction of $1/\gamma$, which changes the density to $\gamma/V$. To obtain the correct probability again, we should re-normalize the wave function to $\psi' = \sqrt{\gamma}\,\psi $ by pulling the Lorentz factor out.
So we intoduce a new matrix element $$|{\cal M}_{if}|^2 = |m_{if}|^2 \prod_{i=1}^n (2 \gamma_i m_i c^2) =|m_{if}|^2 \prod_{i=1}^n (2E_i)^2 $$ (this is for an $n$-body process). Now the transition probability (here in differential form) becomes:
$$
dW = \frac{2\pi}{\hbar} \frac{|{\cal M}_{if}|^2}{ (2E_1)^2 (2E_2)^2 \cdots}
\cdot \frac{1}{(2\pi\hbar)^{3n}}
\, d^3p_1 \, d^3p_2 \, \cdots
\delta({p_1}^\mu + {p_2}^\mu + \ldots - {p}^\mu )
$$
The delta function is there to ensure conservation of momentum and energy. Now we can regroup the terms:
$$
\Rightarrow \quad
dW = \frac{2\pi}{\hbar} \frac{|{\cal M}_{if}|^2}{ 2E_1 2E_2 \cdot \ldots}
\cdot d_\mathrm{LIPS}
$$
The density of states/"phase space" $d\rho$ is replaced by a relativistic version, sometimes called the Lorentz invariant phase space $d_\mathrm{LIPS}$, which is given by
$$
d_\mathrm{LIPS} = \frac{1}{(2\pi\hbar)^{3n}}
\prod_{i=1}^n \frac{d^3p_i}{ 2E_i }
\delta\left(\prod_{i=1}^n {p_i}^\mu - {p}^\mu \right) \,.
$$
The nice thing about the relativistic formula for $dW$ is that, in the case you are scattering particles off one another, it immediately shows us three important contributions: not only the matrix element and phase space, but also the flux factor $1/s$ (where $s = ({p_1}^\mu + {p_2}^\mu)^2$ is the Mandelstam variable, and in case the masses are negligible, $ s \approx 2 E $). This flux factor is responsible for the general $1/Q^2$ falling slope when you plot cross section over momentum transfer $Q = \sqrt{s}$, which comes entirely from relativistic kinematics.
Hope this answers your questions. Here is a presentation (PDF) that sums it up, with an explicit proof that it is Lorentz invariant.
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.
Best Answer
In non-relativistic Quantum Mechanics (NRQM), the dynamics of a particle is described by the time-evolution of its associated wave-function $\psi(t, \vec{x})$ with respect to the non-relativistic Schrödinger equation (SE) $$ \begin{equation} i \hbar \frac{\partial}{\partial t} \psi(t, \vec{x})=H \psi(t, \vec{x}) \end{equation} $$ with the Hamilitonian given by $H=\frac{\hat{p}^{2}}{2 m}+V(\hat{x}) .$ In order to achieve a Lorentz invariant framework (the SE is only Galilei NOT Lorentz invariant), a naive approach would start by replacing this non-relativistic form of the Hamiltonian by a relativistic expression such as $$ H=\sqrt{c^{2} \hat{p}^{2}+m^{2} c^{4}} $$ or, even better, by modifying the SE altogether such as to make it symmetric in $\frac{\partial}{\partial t}$ and the spatial derivative $\vec{\nabla} .$
However, the central insight underlying the formulation of Quantum Field Theory is that this is not sufficient. Rather, combining the principles of Lorentz invariance and Quantum Theory requires abandoning the single-particle approach of Quantum Mechanics.
Quantum Field Theory (QFT) solves both these problems by a radical change of perspective.
Remark 1: There are still some cases (however there are a lot subtleties), where one can use RQM in the single-particle approach. Then the SE is replaced by the e.g. Klein-Gordon equation. $$ (\Box+m^2)\;\psi(x)=0 $$ where $\psi(x)$ is still a wave-function.
Remark 2: The Schrödinger equation holds for SR. It's not the SE that fails, it's the non-relativistic Hamiltonian that fails. The Dirac equation is the SE, but with the Dirac Hamiltonian. The Schrodinger equation is valid. $$ i \hbar \frac{\partial \psi(x, t)}{\partial t}=\left(\beta m c^{2}+c \sum_{n=1}^{3} \alpha_{n} p_{n}\right) \psi(x, t)=H_\text{Dirac}\;\psi(x, t) $$