I’d previously set out to show that the quantum mechanics of reality can be thought of as a reaction- diffusion process where some, unseen stuff (concentration ψ) is created at places where E>V, destroyed where E

The simple way to solve any quantum problem is to use discontinuous (discrete) calculus, or algebra. Most every quantum-relevant problem is time independent — that is the existence wave, ψ does not interact with itself. In this case, the time dependent parts (shown here and here) go away, and the problem can be set up to look like this:

(E-V) ψ =  -h²/8π²m d²ψ/dx²                      (1).

Here E is the energy of whatever you think is quantized, m is its mass, x is any possible location for the thing. V is the potential energy at the spot, h is plank’s constant, π = 3.14159, and ψ is a peculiar wave-stuff that is related to reality as follows: the probability of finding the thing at x equals ψ² at that spot.  Equation 1 is a one-dimensional form — a three-dimensional version looks similar.

Several books will show you how to solve Equation 1 using continuous calculus. Among the problems, is you end up with ψ=0 at most spots, and infinities because the smaller you locate space, the larger the energy. There are ways around this, but it’s a mess, and you can’t solve the general case of a general energy surface. Using discrete math is a lot easier to do and a lot easier to understand, since ψ has a more-real meaning. There is some loss of accuracy because you break up the space, but you can make this as small as you like, and you often don’t want to make it really small since you usually don’t know V all that well.

Lets do a very simple problem, the quantum particle in a 1-dimensional box where V=0 inside and ∞ elsewhere. Let the length of the box be L = 2 ∆x, and solve only in one spot, the middle, for the energy, and the value of ψ there. From discrete math for this case, d²ψ/x² = – 2ψ/∆x². Equation 1 becomes

(E) ψ =  -h²/8π²m (-2ψ/∆x²)     (2).

and we find E =  h²/4π²m∆x² = h²/π²mL². There is only one energy solution for equation 1 when we approximate with only one point in space and this energy is the ground state energy (nearly). A more accurate calculation, with 2 points will give a more accurate value, and will give a second energy level, the “first excited state.” For a one point solution, there are only two possible  ψ values, ±1. We see this because ψ2 = 1, and since there is only on spot where the thing can be in this approximation, at that spot ψ =√1 = ±1.

Now let’s do a better job by assuming there are two spots in the box.  ψ1 and ψ2, separated, as before by ∆x with ψ = 0 elsewhere. To make life easier, lets call the term,

-h²/8π²m = D,      (3)

since this is the form of a diffusion equation. The values of d²ψ/dx² are going to be (ψ₂ -2ψ1)/∆x² and (ψ1-2ψ2)/∆x² at point 1 and 2 respectively. Equation 1 becomes

(E-V)∆x² ψ1 =  D (ψ2 -2ψ1) and     (4)

(E-V) ∆x² ψ2 =  D (ψ1-2ψ2)         (5).

We also note that:     (ψ1)²+ (ψ2)²= 1               (6).

This is to say, for a 2 spot space, the thing is either going to be found at 1 or 2 with a probability related to the value of ψ at the two locations.  If V is not zero at some location, we solve for ψ and E simultaneously by creating a two-dimensional space of ψ1 vs ψ2, where you know these values are never going to be bigger than 1 nor smaller than -1. You can use the monte carlo method (and similar, relaxation versions)to find values of E,  ψ1, and ψ2 that fit equations 4-6.

Alternately, you can solve equation 4 for ψ1 in terms of ψ2

ψ1 =  D ψ2/((E-V1)∆x² +2D).

If you insert this into equation 6, it’s easy to solve the quadratic relation for ψ2 in terms of E. And once you have this, you can solve Equation 5 for E. As a quick version of this, if V is small, you can guess that the values for ψ are either √2/2 for both locations, or √2/2 for one and -√2/2 for the other. 

Solving the particle in the box quanta by linear approximation with 2 points. Would be exact if π=3.

If V= 0 you’ll find this matches all the equations, and that, in the first case, E = -D/∆x², or in the second case,  E = -3D/∆x². The more points you include the more possible values of E you find. Since ∆x here equals L/3, ∆x² = L²/9 and E =  9 h²/π²mL² and 27 h²/8π²mL². The first energy is 1.125 slightly above what we calculated above, while the second energy is 3 times the first. This second energy is the first excited state. As we divide the space further, we get slightly more accuracy and an additional level for every point. Generally we don’t need more than 2-4 levels, or points to get “engineering accuracy.”

For more complex problems, involving complex potential energy surfaces, or more points in space, you first calculate V each point, and estimate ψ at each so that the ∑ψ²=1. You can then use an iterative, relaxation procedure to find E and ψ. Using BASIC, FORTRAN, or PYTHON write a program to solve for E at each point using the estimate for d²ψ/dx² that we presented above. Unless you are very lucky, the values for E will be different at each point. Now tweak your values of ψ in whatever direction moves the E values closer together after you renormalize ψ so that ψ²=1 Continue tweaking the ψs this way until you relax to a set of ψ where E is the same for each point. Here’s an approach to tweaking things. 

Usually, you don’t need more than 2-4 space points, since your potential energy surface won’t be that clearly defined. If you find you need more than 2 energies to get a value in excess of kT you can assume the quantum stuff went away, and things behave pretty classically (Newtonian, non -quantum). Most of the time, reality looks Newtonian, and you don’t have to think quantum: when’s the last time someone’s diffracted through a door?

The above is how it’s all done for time independent cases; time-dependent are done similarly except you have two parts to ψ for every point for every time. The real and imaginary parts of Ψ are related to time as we discussed here; as time goes on the one gets bigger and the other gets smaller in a sin- cos way, like a clock dial moving around. Start with a value for ψ at each point and relax to your solution as above, realizing that each solution is only applicable at a certain time. It’s some work to predict quantum reality this way, but a lot less work than with path integrals, in my opinion, and the solutions make more sense to me.

RE. Buxbaum, July 14, 2014. Philosophy and science should go together, if only to help you make better lemonade or reduce the heat loss from your house.