Causality is preserved, unless Tachyons exist.
Part 1:
STR doesn't assume causality. Causality is violated when you have a flow of information that goes back to the same place in space AND time, creating a contradiction. Both newtonian and STR guarantee causality. STR is more complex, but it still prevents anything from going back in time with respect to another's frame, as long as nothing starts out faster than light in the first place (i.e. Tachyons). The policeman work by creating a "light barrier" that prevents anything from being accelerated past the speed of light.
Part 2:
"C is the result of Simultaneous occurrence of A and B. In other words C occurs iff A and B are simultaneous."
You can't do this. Lets try: A and B are excited atoms that emit flashes of light. Suppose a detector halfway in-between explodes because both atoms flashed at once. The atoms and detector are stationary. Sounds like a simultaneity detector? Not in a frame moving with respect to the set up! In a moving frame, the atoms emit light at different times (light still goes the same speed in any frame), and the detector is still halfway in-between the atoms. But the set-up moves just the right amount in the time it takes for the flashes to converge that the detector will hit them at thier meeting point, and still explode.
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
- Write a wave equation.
- Make the energy and momentum have the de Broglie forms, and
- 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.
Best Answer
Correct, that does not constitute "causing" in this context. For the same reason I wouldn't call this sort of thing an "effect." It's really just a correlation.
In general, causality is just what you'd think it is: if event A causes event B, that means the physical state at (location,time) B has some logical/mathematical dependency on the physical state at (location,time) A, such that what happens at A influences what happens at B. Just how that dependency is expressed depends on which physical theory you're using. In classical mechanics, the dependency exists if a particle is able to travel from A to B. In quantum field theory, the dependency exists if quantum operators at A and B don't commute with each other. And so on. The exact mathematical expressions can get a bit technical if you're not familiar with the theories, but they're all expressing that same underlying idea.