I'm rusty on this stuff but it will go something like this (a real physicist may care to correct me if necessaryPlease Register or Log in to view the hidden image! ):- The wave function, ψ, is a solution to Schrödinger's equation, which has the general form -ih/2π . ∂/∂t ψ = Hψ. In this equation h is Planck's constant and H is the Hamiltonian operator for the system under consideration. This normally has the form -h²/8π²m. ∇² + V, in which V is the potential. For stable bound states H is not time-dependent and then one can use the time-independent version of Schrödinger's equation which is Hψ = Eψ, in which E is the energy of the state. So there are two issues to deal with before you can write down the wavefunction of a system. The first is to understand the dynamics of the system in question so that you can cast the Hamiltonian into the correct form, and the second is to solve the resulting partial differential equation. It is far from trivial to do this. However, almost all of QM avoids actually working these wave function out algebraically, since ψ has certain properties which follow from the nature of the wave equation that they are the solution to. So normally one deals with the equation and the known properties of its solutions, without ever needing to calculate these solutions explicitly. I suppose all this is a long way of answering "no" to your question. Please Register or Log in to view the hidden image!

I got myself mixed up thinking about this. If given a De Broglie wavelength in one frame, would a Lorentz transformation be used to find that wavelength in another frame? Please say yes, Or I will become more confused.

Does a mixed up foghorn sound off for drizzling rain? Please Register or Log in to view the hidden image!

I know nothing about applying relativity to QM, but as the de Broglie wavelength of an object is inversely proportion to its momentum, I presume that however the momentum transforms will tell you how the wavelength transforms. Again, we may need to wait for a real physicist - if there are any left. James seems not to be around.

take a spacetime as a length of x and the time taken to cover it is T. When the mass bends or stretch spacetime the x as x+dx. The time taken is still T. So when x proportional to T. Constant is velocity and when x+dx is proptional to T. The constant is acceleration. So what i am trying to imply which may be wrong is when the spacetime stretch there is generation of acceleration. So taking the spacetime bend around a mass it creates an acceleration field around mass. This also implies that the time flows at a constant rate.

Well you're the one with medical training. I wouldn't know.Please Register or Log in to view the hidden image!