Paul,
This particular writing of the problem in the article I have always thought was sloppy as well. The most confusing part of the discussion is the statement "The continuity equation is as before". At first one writes the continuity equation as:
$$\nabla \cdot J + \dfrac{\partial\rho}{\partial t} = 0$$
Although the del operator can be defined to be infinite dimensional, it is frequently reserved for three dimensions and so the construction of the sentence does not provide a clear interpretation. If you look up conserved current you find the 4-vector version of the continuity equation:
$$\partial_\mu j^\mu = 0$$
What is important about the derivation in the wikipedia article is the conversion of the non time dependent density to a time dependent density, or rather:
$$\rho = \phi^*\phi$$
becomes
$$\rho = \dfrac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$
the intent is clear, the want to make the time component have the same form as the space components. The equation of the current is now:
$$J^\mu = \dfrac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)$$
which now contains the time component. So the continuity equation that should be used is:
$$\partial_\mu J^\mu = 0$$
where the capitalization of $J$ appears to be arbitrary choice in the derivation.
One can verify that this is the intent by referring to the article on probability current.
From the above I can see that the sudden insertion of the statement that one can arbitrarily pick $$\psi$$ and $$\dfrac{\partial \psi}{\partial t}$$ isn't well explained. This part the article was a source of confusion for me as well until one realized that the author was trying to get to a discussion about the Klein Gordon equation
A quick search of web for "probability current and klein gordan equation" finds good links, including a good one from the physics department at UC Davis. If you follow the discussion in the paper you can see it confirms that the argument is really trying to get to a discussion about the Klein Gordon equation and make the connection to probability density.
Now, if one does another quick search for "negative solutions to the klein gordan equation" one can find a nice paper from the physics department of the Ohio University. There we get some good discussion around equation 3.13 in the paper which reiterates that, when we redefined the density we introduced some additional variability. So the equation:
$$\rho = \dfrac{i\hbar}{2mc^2}(\psi^*\partial_t\psi - \psi\partial_t\psi^*)$$
(where in the orginal, c was set at 1)
really is at the root of the problem (confirming the intent in the original article). However, it probably still doesn't satisfy the question,
"can anyone show me why the expression for density not positive
definite?",
but if one goes on a little shopping spree you can find the book Quantum Field Theory Demystified by David McMahon (and there are some free downloads out there, but I won't link to them out of respect for the author), and if you go to pg 116 you will find the discussion:
Remembering the free particle solution $$\varphi(\vec{x},t) = e^{-ip\cdot x} = e^{-i(Et- px)}$$ the time derivatives are $$\dfrac{\partial\varphi}{\partial t} = -iEe^{-i(Et- px)}$$ $$\dfrac{\partial\varphi^*}{\partial t} = iEe^{i(Et- px)}$$ We have $$\varphi^*\dfrac{\partial\varphi}{\partial t} = e^{i(Et- px)}[-iEe^{-i(Et- px)}] = -iE$$ $$\varphi\dfrac{\partial\varphi^*}{\partial t} = e^{-i(Et- px)}[iEe^{i(Et- px)}] = iE$$ So the probability density is $$\rho = i(\varphi^*\dfrac{\partial\varphi}{\partial t} - \varphi\dfrac{\partial\varphi^*}{\partial t}) = i(-iE-iE) = 2E$$ Looks good so far-except for those pesky negative energy solutions. Remember that $$E = \pm\sqrt{p^2+m^2}$$ In the case of the negative energy solution $$\rho = 2E =-2\sqrt{p^2+m^2}<0$$ which is a negative probability density, something which simply does not make sense.
Hopefully that helps, the notion of a negative probability does not make sense because we define probability on the interval [0,1], so by definition negative probabilities have no meaning. This point is sometimes lost on people when they try to make sense of things, but logically any discussion of negative probabilities is non-sense. This is why QFT ended up reinterpreting the Klein Gordan equation and re purposing it for an equation that governs creation and annihilation operators.
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.
What about the case of free particle when V=0?
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).
What is the correct Schrödinger equation then?
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.
What will be the momentum operator?
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 $-1$.
What is the Hamiltonian operator?
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.
Are these operators Hermitian?
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.
Best Answer
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.
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.
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.