The Time-Independent Schrodinger Equation as Free Energy Equal to Zero for the Ground State Oscillator
| Main Author: | Francesco R. Ruggeri |
|---|---|
| Format: | info publication-preprint |
| Terbitan: |
, 2019
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| Subjects: | |
| Online Access: |
https://zenodo.org/record/2600497 |
Daftar Isi:
- In a previous note, it was shown that for a free quantum particle in a box with only kinetic energy, one could apply the thermodynamic relationship dU=TdS to obtain fp*fp=exp(-p*p/2mT) i.e. the Maxwell-Boltzmann factor. In this case, a wavefunction of W(x)= Sum over p fp sin(px) was actually used (instead of the usual approach of density = Sum over p exp(-p*p/2mT) sin(px)sin(px)). It was noted that if the box length becomes large, W(x) approaches a Gaussian which is the ground state solution of the oscillator problem. The oscillator ground state solution, however, is obtained solving the time-independent Schrodinger equation for V(x)=Cx*x. This raises a question. Why should the thermodynamic equation dU=TdS (for zero work) have any connection with the time-independent Schrodinger equation? In another previous note, it was shown that for the oscillator, kinetic energy density in momentum space summed over p is proportional to Shannon’s entropy for momentum i.e. Sum over p 2fp*fp ln(fp). It was also shown that W(x)W(x)V(x) for the oscillator is proportional to spatial entropy d(x) ln[d(x)], where d(x)=W(x)W(x) using Shannon’s entropy. Thus, a connection between classical statistical mechanics and quantum mechanics seems to exist for the oscillator ground state. In this note, we wish to show that the Schrodinger equation in momentum space is equivalent to a free energy equal to zero for the case of pure kinetic motion. This explains, it seems, why the temperature dependent wavefunction for a free gas without a potential is equivalent to the ground state of an oscillator.