The Oscillator Propagator, Newton's Law and Schrodinger Equation
| Main Author: | Francesco R. Ruggeri |
|---|---|
| Format: | info publication-preprint Journal |
| Terbitan: |
, 2021
|
| Subjects: | |
| Online Access: |
https://zenodo.org/record/4637515 |
Daftar Isi:
- The oscillator propagator exp(i Action) = Q(t) exp[i mw/2 { cos(wt)/sin(wt) (XX + YY) - 2XY / sin(wt)} ] = Q(t) exp( i (-m/2 dA/dt / A) (XX +YY) - (mw/2) 2XY/A ) (where XA(t) is a classical solution with X as amplitude) is a solution of the time-dependent Schrodinger equation. Inserting the form with A(t) into the time-dependent Schrodinger equation yields Newton’s second law (with two other terms which cancel) i.e. m d/dt dA/dt = - k A(t). Newton’s second law may be multiplied by dA/dt and integrated, yielding the well-known conservation of energy equation: Energy = XX dA/dt dA/dt (m/2) + (k/2) XX AA. Thus, the two are equivalent as is known from classical mechanics. One may consider solving the time dependent Schrodinger equation in a different manner, namely postulating a separable solution of the form: exp(-i C t) H(x). Substituting into the oscillator Schrodinger equation yields C = -1/2m d/dx d/dx W + .5 k xx W. Mathematically, this equation may be solved, but only a discrete set of C values yield solutions, namely C= .5 w(n+.5) with corresponding Wn(x). We argue that this mathematical equation may be linked to a physical one, namely the conservation of energy equation if C is associated with energy. In such a case, one must associate -1/2m d/dx dW/dx with classical kinetic energy at a x. At this point, we suggest one must establish a physical interpretation for -1/2m d/dx dW/dx if W. We argue that X in exp(i Action) is an amplitude. Allowing all values of X (e.g. -1/2m d/dX d/dX) means allowing all amplitudes for a particular spring. Thus it seems that one is considering a fluctuating amplitude picture in which these fluctuations can yield an average energy associated with a particular En. This average equation follows classical mechanics in terms of average KE(x) + V(x) = En.