As you guessed, the probability density is not the same as transmission. In order to have non-zero transmission there has to be nonzero gradient: $$j=\frac{1}{2m}(\psi^\dagger \hat{p}\psi - \psi \hat{p}\psi^\dagger)$$
In the case of a constant probability, there is no gradient. You only have particle "buildup". The reason for this is that you are solving the time-independent Schrodinger equation, which gives the steady-state solutions. This means the system has had a long time to stabilize and reach a steady state. Since the time is so long (essentially infinite), the particle built up is really large (integral of probability from zero to +infinity). If E was lower than V, this build up would be limited to regions near x=0 and have a negative exponential form (also known as the evanescent part), but E=V is an edge case.
Had you solved the time-dependent Schrodinger equation (TDSE), you would observe the build up forms gradually in time. Specifically, assume a particle source at $-\infty$ (which makes this an open system), providing a constant particle flow at fixed energy $E=k^2/2m$ and a potential barrier $V(x)=V_0 u(x)$, where $u(x)$ is the step function. Start solving the TDSE with an initial condition at $t=0$ of $\psi(t=0,x)=\exp(ikx)u(-x)$, which is a crude approximation of the shape of the wavefunction right before it hits the potential barrier.
Mass is a number that relates quantitatively to how energy and momentum are balanced, e.g.
$$E=\sqrt{(c|\vec p|)^2+(mc^2)^2}$$
And the mass of a system is not the sum of the masses of the parts.
And particles don't have potential energy, systems do. Consider two massive particles of mass $m$ a distance d apart, there is some potential energy $U=-Gm^2/d^2$ but it only makes sense to associate it with the system not to randomly pick a particle and associate it with that. And potential energy is just for Newtonian physics anyway.
So why does a ground state hydrogen atom have less mass than the sum of the mass of an electron and of a proton? Because in its own frame it has no momentum but it has less energy. How do we know it has less energy? Because it gave off energy when it became bound. And it requires additional energy to become unbound.
So the mass of a system isn't the mass of the parts added up. Its just the parameter that makes $E=\sqrt{(c|\vec p|)^2+(mc^2)^2}$ work, you can even write it as $E^2-(c|\vec p|)^2=(mc^2)^2.$ And then for the mass of a system you use the energy of the system and the momentum of the system and it isn't the mass of the parts added up.
I understand what you mean by the mass of the system not being the mass of the components added up.
And keep in mind this is for special relativity.
But is there is always a change in mass of a system when its potential energy changes?
And in special relativity, potential energy doesn't make sense. When energy is exchanged it has to happen at the exact same when-where because otherwise if energy disappeared at one place and reappeared elsewhere then in some frame it wouldn't be conserved.
For example when an electron is pulled apart from its atom, or when the distance between two charges is changed?
So for instance the way work is done on electrically charged objects is that charged objects have kinetic energy, momentum, and rest energy and electromagnetic fields have their own field energy and field momentum. And what happens to charged objects is they move around. And what happens to fields is they evolve. And in regions where there is no charge the field energy just moves around in a conserved way. And in regions where there is neither charge nor current the field momentum just moves around in a conserved way. But in places with charge the field loses field energy when the charge gains kinetic energy and the field gains field energy when the charge loses kinetic energy. And in places with charge and/or current the field and the charges can exchanged momentum so that total momentum is conserved.
So there isn't potential energy in classical electromagnetism. If you want to see what is going on in the Schrödinger equation you need to learn the difference between a canonical momentum (which is a third type of momentum not really related to mechanical momentum of charges or field momentum of fields) and kinetic momentum, and how gauge choices affect the phase of the wavefunction of a charged object. All the physics is about the phase and the gauge together in the sense that it is their relationship that carries all the information.
But my description earlier is fine. The atom has less energy because it takes energy to ionize it and I gave up energy when it dropped down to the ground state.
In a Lagrangian, there is something that might look similar to a potential, but it is gauge dependent and the term in the Lagrangian have units of energy but are really just what they need to be to give Euler-Lagrange equations that match what we see.
Best Answer
The answer is ultimately no, but this is a reasonable idea, although old. This idea was floating around in the late 19th century, that the mass of the electron is due to the energy in the field around the electron.
The concept of potential energy is refined in field theories to field energy. The fields have energy, and this energy is identified with the potential energy of a mechanical system, so that if you lift a brick up, the potential energy of the brick is contained in the gravitational field of the brick and the Earth together.
This is important, because unlike kinetic energy, it is difficult to say where the potential energy is. If you lift a brick, is the potential energy in the brick? In the Earth? In Newton's mechanics, the question is meaningless both because things go instantaneously to different places, and also because energy is a global quantity with no way to measure the location. But in relativistic physics, the energy gravitates, and the gravitational field produced by energy requires that you know where this energy is located.
The upshot of all this is that potential energy is field energy, and you are asking if all the mass-energy of a particle is due to the fields around it.
This model has a problem if you think of it purely electromagnetically. Using a model where the electron is a ball of charge, and all the mass is electromagnetic field, you would derive, along with Poincare, Abraham and others that the total mass is equal to 4/3 of the E/c^2. The reason you don't get the right relativistic relation is because of the stresses you need to hold a ball of charge from exploding. The correct relation really needs relativity, and then you can't determine if the mass is all field.
The process of renormalization in quantum field theory tells you that part of the mass of the electron is due to the mass of the field it carries, but there are two regimes now. There is a long-distance regime, much longer than the Compton wavelength of the electron, where you get a contribution to the mass from the electric field which blows up as the reciprocal of the electron radius, and then there is the region inside the Compton wavelength, where you get the QED mass correction from electrons fluctuating into positrons, which softens the blowup to a log. The compton wavelength of the electron is 137 times bigger than the classical electron radius, so even with a Planck scale cutoff, not all the mass of the electron is field, because the blow-up in field energy is so slow at high energy.
So in quantum field theory, the answer is no--- the field energy is not the entire mass of the particle. But in another sense it is yes, because if you include the electron field too, then the total mass of the electron is the mass in the electron field plus the electromagnetic field.
Within string theory, you can formulate the question differently--- is there a measure of a field at infinity which will tell you the mass of the particle? In this case, it is the gravitational field, so that the far-away gravitational field tells you the mass.
But you probably want to know--- is the mass due to the combination of gravitational and electromagnetic field together? In this sense, since this is a classical question, it is best to think in classical GR.
If you have a charged black hole, there is a contribution to the mass of the black hole from the field outside, and a contribution from the black hole itself. As you increase the charge of the black hole, there comes a point where the charge is equal to the mass, where the entire energy of the system is due to the external fields (gravitational and electromagnetic together), and the black hole horizon becomes extremal. The extremal limit of black holes can be thought of as a realization of this idea, that all the mass is due to the fields.
Within string theory, the objects made out of strings and branes are extremal black holes in the classical limit. So within string theory, although it is highly quantum, you can say the idea that all the mass-energy is field energy is realized. This is not very great in giving you what the mass should be, because in the cases of interest, you are finding particles which are massless, so that all their energy is the energy in infinitely boosted fields. But you can take comfort in the fact that this is just a quantum regime of a system where the macroscopic classical limit of the particles are classical gravitational systems where your idea is correct.