The relative sign is not just a convention. Once you decide that $E$ is represented by $i\hbar \partial/\partial t$, there must be a minus sign in the formula for $p$, namely $p=-i\hbar \partial / \partial x$. Or vice versa.
First of all, there has to be $i$ or $-i$ in all the formulae because $\partial/\partial x$ is an anti-Hermitian operator (because of the minus sign in the integration by parts) and we need Hermitian operators (which has real, measured eigenvalues) for the energy, momentum, and others. What about the signs?
The only sign convention that was chosen in the early days of quantum mechanics was one for the energy; indeed, one could have replaced $i$ by $-i$ in that equation because $i$ and $-i$ play the same algebraic role: exchanging $i$ and $-i$ is an "outer automorphism" of complex numbers. But once this sign is fixed, all other signs are fixed, too. That includes minus signs in momentum, angular momentum, gauge transformations, Schrödinger's picture, Heisenberg's picture, Feynman's path integral, and any other formula of quantum mechanics. There is only one way to define quantum mechanics given a classical limit (with its sign conventions) we need to get.
The relative minus sign in $p,E$ may become invisible if you only act with second derivatives - squared momentum, squared energy - but it is visible if you act with first powers of the operators.
De Broglie's wave - or a wave associated with a particle - is proportional to
$$ \exp \left[ \frac{i}{\hbar} (\vec p\cdot \vec x-Et)\right] $$
Note that if you differentiate with respect to $x$ and $t$, and multiply the result by $i\hbar$, you get $-p$ and $E$, respectively. (Omit the vector signs if you want just one-dimensional space with one $x$ and one $p$.)
The relative sign between $\vec p\cdot \vec x$ and $Et$ in de Broglie's formula above is physically necessary because only $Et - \vec p \cdot \vec x$ is the correct Lorentzian inner product of the vectors $(E,\vec p)$ and $(t,\vec x)$: the relative minus sign comes from the opposite signs of space and time in the signature of spacetime. Note that the doubly relative sign in $(E,\vec p)$ and $(t,\vec x)$ can't be flipped because in relativity, $\vec pc^2/E$ is the velocity $\vec v$.
The argument above is for a relativistic interpretation of $E$. However, the non-relativistic kinetic energy is defined just as a shifted relativistic energy,
$$ E_{\rm nonrel} = E_{\rm rel} - mc^2 $$
so the coefficient (and sign) in front of $E$ remains unchanged and $\vec p$ is totally unchanged. One may also design related arguments analyzing where the de Broglie wave is moving. For it to be moving in the right direction, there has to be a minus sign in $Et-px$. It's related to the fact that the shape of objects moving by velocity $v$ only depends on $x-vt$ because $x=vt$ (without a minus sign, which becomes a minus sign if you move both terms to the same side) is the equation for their center of mass.
The hypersurfaces of constant phase of the de Broglie wave are orthogonal to the world lines of the particles which dictates the sign. After all, the waves' maxima and minima should be moving in the same direction as the particles themselves.
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
The important thing is the relative sign between the potential and the Laplacian.
Otherwise there are two square roots of $-1$ namely $\pm i.$ You could write $j=-i$ and then you have two perfectly good square roots of $-1$ you can use $i$ or you can use $j$.
For square roots of positive numbers you get things like $\pm\sqrt 2$ and there is a way to pick one consistently, pick the positive one. For $\pm\sqrt {-1}$ there is no way to prefer one over the other.
And in fact engineers like to use $j$ for $\sqrt{-1}$ and they like to write $e^{j\omega t}$ and physicists like to write $i$ for $\sqrt{-1}$ and use $e^{-i\omega t}$ but there isn't really a difference.
In real life you will meet people (usually engineers) that write oscillating waves like $E^{j\omega t}$ and you will meet other people that write oscillating waves like $E^{-i\omega t}.$ And they both are referring to a real wave that oscillates like $\cos (\omega t).$ The difference is one of convention, not of physics. It's like if you drew the x axis going to the right, drawing the y axis as going up (rather then down) is mere convention and means nothing. But you do need to stick to a convention and not mix and match.
In fact, selecting one of those two direction up versus down is all that is going on when you pick one of the two square roots. If you can't see this easily, it is because your favorite meaningless convention is so ingrained in your mind that you can't even see what you are doing. There are two square roots, and no one can tell them apart. You can draw them of the lien from 0 to 1 goes rightwards and then you pick the upwards direction and call it $+i$ and call downwards $-i$ but that is just a convention about how you like to draw things, in reality there are simply those two directions.
If you met an alien there would be literally no way to explain the difference between $i$ and $-i$ they would be different, like clockwise and counterclockwise, but the reason to have one of them have a $-$ or to have the word $counter$ is mere convention and history. It doesn't mean anything.
And I can't explain something that has no meaning. How would I explain clockwise other than people made sundials in the northern hemisphere during the northern summer and the sun rises in the east and people liked to keep doing that? Whoops, that used the words east and north, how is an alien going to no that other than from context and history, they are mere conventions. We called the direction the sun rises east and the north pole and the south pole are equally good poles, we put the north pole up on our globes by convention, not because that means anything.
What is correct is to have a sign difference between the Laplacian and the potential. What is correct is to make sure your momentum and Hamiltonian are related to your spatial and temporal derivatives correctly. If you like writing your waves like $e^{i(\vec k \cdot \vec x -\omega t)}$ then one choice will seem more natural to you. If you prefer to write your waves like $e^{j(\omega t -\vec k \cdot \vec x)}$ then other choices might seem more natural. But those aren't really different, they are both cosines as real parts and there is no way to distinguish between the two imaginary directions. Absolutely no way whatsoever. Some people even do quaternion quantum mechanics where the imaginary unit is a unit imaginary quaternion that can change from place to place, so it might point different ways in different places because then $+i$ and $-i$ are continuously connected to each other by rotating through the $i,k$ plane. That is different physics since it changes from place to place, but it shows that there is no meaning to picking one direction over the other.
You can write $i\hbar \frac{\partial}{\partial t} \Psi = -\frac{\hbar^2}{2m}\vec \nabla^2 \Psi + V \Psi$ and $\hat p = -i\hbar \nabla$ or else you can hit the first with a minus sign (no change in physics) and get $-i\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi - V \Psi$ and $\hat p = -i\hbar \nabla$ and then say hey, $\pm i$ are equally good and I dont' like writing so many minus signs, so I'll use $j=-i$ as my favorite square root and get $j\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi - V \Psi$ and $\hat p = j\hbar \nabla$ and all I did was write it in terms of the other square root which didn't change the physics.
No one is changing any physics. And both square roots are indeed equally good, and this is not deep. There is a minus sign difference between the Laplacian and the potential and there is no way around that. And you can call a square root of minus one anything you want because there are two and there is no way to know which is which. But like any convention you need to stick to it.
Now you don't have to worry about the relative sign between the two. Energy eigenstates will evolve like $e^{j\omega t}$ which is great if that is traditional in your field and annoying if $e^{-i\omega t}$ is traditional for your field, but deal with conventions or else get everyone to change (and make translations of old works as needed).
This is confusing, they aren't even really different equations. The equation $i\hbar \frac{\partial}{\partial t} \Psi = -\frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ and the equation $j\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ aren't saying different things, they both say that if you differentiate Psi and multiple by $\hbar$ then multiply by one (of the two equally good) square roots of minus one it is equal to $\frac{\hbar^2}{2m}$ times the Laplacian. They might disagree about where you want to draw that square root, if you should go clockwise or counterclockwise, but this is not physics, the equations are not saying different things.
A wave travels in the $+x$ direction if a point of constant phase travels in the $+x$ direction as time evolves. So a wave like $e^{j(\omega t-kx)}$ or a wave like $e^{i(kx-\omega t)}$ (they are the same wave) works fine to describe something with a fixed energy and momentum with positive momentum in the x direction. If you use the equation $i\hbar \frac{\partial}{\partial t} \Psi = -\frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ then you get $i\hbar(-i\omega) = -\frac{\hbar^2}{2m}(-k^2)$ and everything is fine, and you get $\hat p_x=\frac{\hbar}{i}\frac{partial}{\partial x}$ as usual for the reason that it gives $\hbar k =p >0$ for the state with positive $x$ momentum above.
If instead you use the equation $j\hbar \frac{\partial}{\partial t} \Psi = \frac{\hbar^2}{2m}\vec \nabla^2 \Psi $ then you get $j\hbar(j\omega) = \frac{\hbar^2}{2m}(-k^2)$ and everything is fine, and you get $\hat p_x=j\hbar\frac{\partial}{\partial x}$ for the reason that it gives $\hbar k =p >0$ for the state with positive $x$ momentum above.
Is this any different? No, you are just preferring to use the other root of $-1$.
That is a better question. It should be an operator that gives a positive number for a positive energy eigenstate if you want it to be an energy operator. So it should have the negative sign in front of the Laplacian.
I don't understand your concern. Your feelings (or your field's feelings) about which root of $-1$ you like best doesn't change whether something is Hermitian. And whether you slap a minus sign on it doesn't change whether it is Hermitian either. They are Hermitian.
But there is a similar convention that maybe you should address now. Why now? Because then you will less stuff to go recheck if your doubts are raised later in life. The question is about whether the inner product is $\int\overline\Psi\Psi$ or whether it is $\int\Psi\overline\Psi.$ Mathematicians prefer the latter when they are young mathematicians or are teaching young mathematicians. As far as I can tell it is because they like to say the word sesquilinear so they want it it be linear in the first argument. At first it doesn't make too much problem (well, it is nicer to write $\int\overline\Psi\frac{\hbar}{i}\frac{\partial}{\partial x}\Psi$ with each thing more clearly where it is because of what it is doing) but deep into operator algebra theory, the wrong choice turns out to create actual real headaches, so it would be better if everyone switched.
But there isn't really anything wrong, the two inner products are equally good (they are just complex conjugates of each other), and they are both there, making the same metric and everything. But the choice about which to be your favorite, that is similar to the $i$ versus $j$ issue, and I don't want it to create problems for you.