Let’s consider why such a strange function appears similar to the wavefunction.
-If we square the wavefunction (cc also), then we get a density, i.e, where the ’particles’ in a system frequent. -It is something like the probability amplitude.
So, let’s break down the function: For a particle sitting at a location x, in a one dimensional, static potential \(V(x)\), he can consider his choices, move forward or backward. It takes a glance forward (fwd) at the potential ahead, this is the integral \[fwd(x)=\int_x^{x_{max}} V(x')dx'\]
If this integral is large, the potential ahead is large and the particle is likely to have a rough time traversing, forces in the opposing direction etc. So it checks behind (bwd), this is the intgral.
\[bwd(x)=\int_{x_{min}}^x V(x')dx'\]
If this measure is large he will have a rough time going backwards, if it is small it will be a nice downhill journey. He needs to make a decision, so take the ratio \[\frac{fwd}{bwd}\]
If this quantity is large, it would be better to go backward than forward, if the ratio \[\frac{bwd}{fwd}\]
is large it would be better to go forward. If we take the sum \[S=\frac{bwd}{fwd}+\frac{fwd}{bwd}\]
this quantity is now a measure of how much the particle wants to move. If the whole thing is large, one of the directions was preferrable. (Only one of the ratios can be large at once). Therefore, if the whole thing is large the particle will move and it’s density at that location will be small! So the wavefunction is like \(1/S\)! If S is small, then the particle was not inclined to move, the density at this location will be high!
If the ratio S was zero (I don’t know how) the particle would be glued to that position, infinitly inclinded to stay put. In fact it is important to piece in these possibilites...
We know that in the abscence of any potential the particle will not move, therefore, if( fwd=0 AND bwd=0) we must conclude the particle is stationary. In classical terms this would indicate a probability of 1 at that exact location, which would be a delta function infinatly high but 0 in width. In a quantum sense, this will spread out into an appropriate gaussian. If, say fwd only was zero, S will go to infinity and the wavefunction goes to zero. The particle just rolls off into the distance and won’t be found there again. (perhaps)
This then poses the question, how far ahead can a particle glance? If it knows it’s momentum as much as it can (not explicity taking time on just yet) then it would only be able to sample a space \(\frac{\hbar}{2\Delta p}\) in front or behind. What are the implications of this?
Appears to be extendable to calculus like questions, instead of infantesimal linear interpolation, use a juggling of areas proceeding in each direction.
Infinite Potential Well:
Step by step example. We are at the leftmost wall, backward is only infinity, forward is zero. This means any particles here must go forward, the density will be zero. Taking a few steps forward, ...tbc