The "Derivation" of the Schroedinger equation

Let it be clear that it's practically impossible to describe fundamental principles in the form of basic equations.  One can (and surely must) try to draw conclusions using a heuristic way of thinking and then compare the solutions of "invented" equations with experimental measurements. 

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

φ = A e−iwt+ikx    where   k = 2π/λ

According to de Broglie we can write the momentum as p: λ = h/p and energy as E = hν = hw for a particle. By substituting expressions for momentum 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/hx     ®    (ih∂/∂t)y  =  E · y
  ¶y/∂x  =  ip/h· Ae-iE/ht+ip/hx    ®    (h/i∂/∂x)y  =  p · y

Notice that E and p are on the right sides of both equations. These two quantities can be measured experimentally. On the left side of each expression there is a clue as to how one could obtain energy or momentum: 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. These mathematical instructions are also referred to as operators. We introduced them at the end of the last chapter but now were able to draw parallels between physical quantities and operators:
 

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


Short note: according to our present knowledge of  x, p and H operators, we can write 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  =  p²/2m + 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.


With respect to the law of conservation of energy, we should have E = Ekin + Epot. If we use V(x) for potential energy and  ½mv2 = p²/2m for kinetic energy and multiply this by y we have:

p²/2my + V(x)y = Ey

we can then expand the expression for impulse by using the concept of an operator p²y = p(py) = p·(h/i·∂/∂xy) = − h² (∂/∂x)(∂/∂x)y = - h² (∂²/∂x²)y, and by using the energy operator  ih ∂/∂t, we obtain the following
 

time dependent Schroedinger equation:

[− h²/2m∂²/∂x² + V]y  =  ih ∂/∂ty

Next, let's try to expand the Schroedinger equation to three dimensions x, y, z:

p  =  (px, py, pz) →h/i(∂/∂x, ∂/∂y, ∂/∂z)

p. 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
  In a more concise form we have
time dependent Schroedinger equation:

H y=  ih ∂/∂ty

If potential (or generally H) doesn't depend on time and energy E is almost independent of time we can get rid of time 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 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 easily write the solution of the time dependent equation unless H is not a function of t and yu describes 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 probability P(x,y,z) of locating the 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 describes the considered system completely and we can obtain all information about the system according to the probability amplitude.

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