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Re: Schrodinger equation origins

A further point:
Note that the necessity of using i and a complex PHI in the Shrodinger
(free particle) wave equation,
(d/dx)^2 PHI(x,t) = i*Const* d/dt PHI(x,t),
really means that two real wave functions are needed to describe the quantum
particle, and that they obey a set of two coupled wave equations (containing
no i's):

Define PHI(x,t) = U(x,t) + iV(x,t), where U and V are real.
Sub this into the above wave equation and get the (completely real) coupled

(d/dx)^2 U(x,t) = -Const*d/dt V(x,t) and
(d/dx)^2 V(x,t) = Const*d/dt U(x,t)

Note how neatly these are satisfied by a COSINE/SINE for U/V respectively.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
----- Original Message -----
Some addendum thoughts: shows how the Schrodinger eq
emerges from
Einstein: E=hf; DeBroglie: p=h/lamda; and p^2/2m + U(x) = E.

1)Why did S use a first order time derivative and not a 2nd order time
derivative, as in the "standard" wave equation: (d/dx)^2 PHI(x,t) =
Const*(d/dt)^2 PHI(x,t)? and

2) Why is the imaginary (i) necessary?

1) The wave function PHI(x,t) is to be a complete description of the
particle's state at any time t. This means that the governing differential
equation must be able to develop PHI(x,t) solely from a knowledge of
PHI(x,0), where t=0 is any convenient "starting" time. This requires the
governing differential (Wave) equation to be first order in time
derivatives. A second order time derivative in the wave equation would
require a knowledge of both PHI(x,0) and (d/dt) PHI(x,0) as initial
conditions. PHI(x,0) would not alone be a complete state description.
(In the same way, the second order N2: F=m*(d/dt)^2 x(t) requires a
knowledge of both the position x and the velocity dx/dt as initial
conditions to specify and develop a particle state - given the environment,

2) Now SIN[k(x-ut)] is not a solution of the first (time) order wave
equation: (d/dx)^2 PHI(x,t) = Const* d/dt PHI(x,t) :
Sines and Cosines repeat only after 2 differentiations. But the exponential
combination wave of real cosine and imaginary sine
exp[ik(x-ut)] IS a solution if we choose an imaginary Const in the wave
equation. (We speak here of a free particle.)

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)