The "Derivation" of Schroedinger equation

So, let's say it again that it's principally impossible to write down fundamental principles in form of basic equations.  One can (and surely must) try to draw conclusions using heuristic way of thinking and then try to compare a number of solutions of "invented" equation with experimental measurements. 

We have seen earlier it's possible to describe observations using sum of probability amplitude. One can write down the following expression for one-dimensional wave: 

φ = A e−iwt+ikx    where   k = /λ

according to de Broglie we can write down the impulse p: λ = h/p and energy E = hν = hw of a particle. Substituting expressions for impulse and energy in the first equation and replacing φ by the probability amplitude y one can obtain the following expression:

y = Ae−iE/ht+ip/hx


  ¶y/∂t  =  -iE/h· Ae-iE/ht+ip/h    ®    (ih/t)y  =  E · y
  ¶y/∂x  =  ip/h· Ae-iE/ht+ip/h   ®    (h/i/x)y  =  p · y

One can see that there are E and p on the right sides of either equation and they certainly can be measured in experiments. On the left side of each expression there are kind of instruction how one can obtain energy or impulse: differentiating y on x (and multiplying it by ih) one can obtain Ey; differentiating y on x (and multiplying it by h/i) one can obtain py. This mathematical instructions ate also referred to as operators. We have introduced them at the end of last chapter but now we can draw a parallel between physical quantities and operators:
 

Energy E ih /∂t Energy operator ≡ H
Impulse p h/i/∂x Impulse operator ≡ p
Coordinate x  x     Characterizing operators we use "bold" symbols



Short note: according to our present knowledge of  x, p, H operators we can now write down the expression for kinetic moment L in classical meaning L = r xp
 
Classical Operator
p  =  (px, py, pz) p  =  h/i(/∂x, /∂y, /∂z)
E  =  /2m + V H  =  −h²/2mΔ + V 
Lx  =  ypz − zpy Lx  = h/i(y/∂z− z/∂y)
Ly  =  zpx − xpz Ly  = h/i(z/∂x− x/∂z)
Lz  =  xpy − ypx Lz  = h/i(x/∂y− y/∂x)

 Δ is the Laplace operator: Δ  =  ∂²/∂x² + ∂²/∂y² + ∂²/∂z².



To the best of our knowledge according to energy conservation law it should be E = Ekin + Epot. If one write down directly potential V(x) for potential energy and  ½mv2 = /2m for kinetic energy and multiply it by y:

/2my + V(x)y = Ey

and then expand expression for impulse using operator conception p²y = p(py) = p·(h/i·/∂xy) = − h² (/∂x)(/∂x)y = - h² (∂²/∂x²)y, and also use energy operator  ih /∂t, one can obtain the following
 

time dependent Schroedinger equation:

[h²/2m∂²/x² + V]y  =  ih /ty

Then let's try to spread the Schroedinger equation on three dimensions x, y, z:

(px, py, pz)h/i(/∂x, /∂y, /∂z)

p. =  p2  =  (px2 + py2 + pz2)→ − h2(∂²/∂x² + ∂²/∂y² + ∂²/∂z²)  =  − h2Δ

[h²/2mΔ + V] y  =  ih/∂t y
½¾¾¾¯¾¾½
 H Hamilton operator
 So, and in a more short form it will be
time dependent Schroedinger equation:

H y=  ih /∂ty

If potential (or generally H) doesn't depend on time and energy E is almost independ of time we can get rid of time here in the following way:
 

y(x,y,z,t)  = yu(x,y,z) e−iE/h t
ODE:
Hy  =  e−iE/ht.Hyu  =  ih /∂t(yue−iE/ht)  =  E · yue−iE/ht
And finally we will have:
time independent Schroedinger equation:

Hyu  =  Eyu

This ODE gives us values of stationary energy states.

If we know the time independent solution yu we can readily write down the solution of time dependent equation unless H is not a function of t and yu describe stationary time independent energy states E:
 

y  =yue−iE/h t

The normalization condition:
 

 -¥ò|y(x,y,z)|2dx dy dz = 1 

The expression |y(x,y,z)|2dxdydz gives us the possibility P(x,y,z) to detect particle in the location (x,y,z) with the range [x,x+dx], [y,y+dy] and [z,z+dz]. P(x,y,z) has unlike y(x,y,z) clear physical meaning of probability. The probability amplitude describe considered system completely and one can obtain all information about the system according to probability amplitude.