What is the upside down delta in the Schrodinger Equation?
I'm going to abbreviate the "upside-down triangle" symbol as this: V, simply because I have no better way to do so. The V -- actually, it includes a squaring symbol (the '2') -- in the time-dependent SE is a mathematical operator called the "Laplacian." Basically, the Laplacian (the V2) is the divergence of the gradient.
Confused? You should be!! Both the gradient (sometimes called the "grad") and the divergence ("div") ARE difficult to understand -- I'll do my best.
In its most basic form, V is mathematical shorthand for (d /dx + d /dy + d /dz). Thus
Vf(x,y,z) = df/dx + df/dy + df/dz
It's a bit more complicated than that, but I'm trying to keep it simple.
The grad basically takes a scalar quantity in space and changes it into a vector quantity, with the direction of the vectors at each point being in the direction of increase of that scalar, and the size of the vector dependent on how rapidly the scalar is increasing.
Example: let's say our scalar quantity is air temperature ('T'), and we measure T just over a hot plate. Since T will be increasing as we go down towards the plate, the VT will, at each point will be a vector pointing down. Since the CHANGE in T over space gets larger as we get closer to the plate, then VT will also increase as the position gets closer to the plate.
The div basically takes a vector quantity at a point and changes it into a scalar quantity. If the vector "flow" at a specific point is "outward," then the div will positive; if inward, the div will be negative.
Let's take this hot plate again, and look at air flow. Because air expands from hotter regions, the air flow from hotter regions will be "outward." Thus, the div of air flow will be a positive number.
Now recall that a grad takes a scalar quantity at a point and makes it a vector at that point; and the div takes a vector quantity at a point and makes it a scalar. So the Laplacian -- the div of the grad -- starts with a scalar and ends up with a scalar.
When you do all the work, you find that the Laplacian of Φ is related to the velocity of a particle, and the change in Φ is related to its energy. Thus, the time dependent SE gives
ih(dΦ/dt) = [-h^2 x V2/2m + PE(x,y,z,t)]Φ(x,y,z,t)
where 'i' is the square root of -1, h is h-bar (Planck's constant divided by 2π), V2 is the Laplacian, 'm' is the mass of the particle, and PE is the potential energy function of the particle at a specific point and time. I'd prefer not to explain all the meaning of this equation, but if you have SPECIFIC questions after reading
I can try to help.