Link Between Classical Statistical Mechanics and Quantum Mechanics for the Ground State Oscillator
Main Author: | Francesco R. Ruggeri |
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Format: | info publication-preprint |
Terbitan: |
, 2019
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Subjects: | |
Online Access: |
https://zenodo.org/record/2593441 |
Daftar Isi:
- In this note, we investigate two velocities present in a quantum bound state. The first is the root mean square velocity, which appears in the time independent Schrodinger equation as (-1/2m(1/W) d/dx d/dx W)= .5m v(x)*v(x) where W is the wavefunction and v(x) is equal to the classical velocity. The second is (1/W) d/dx W(x) which is an average momentum flux related to the spatial derivative of the density. From Newton’s conservation of energy, one has Mvdv = -d/dx V(x). This formula can always be applies to the root mean square velocity in quantum mechanics, but usually not to the flux velocity. In the case of the oscillator, however, the equation holds for both velocities (up to a factor of 2 related to d(x)=WW). This is important as the flux velocity is related to d/dx d(x), the change in density. In classical statistical mechanics, the spatial density d(x)=C exp(-V(x)/T) also leads to a relationship for d/dx d(x), thus one can use this as a point for comparing quantum mechanics and statistical mechanics at least for the oscillator case. It can be noted that exp(-.5mv*v/T), the Maxwell-Boltzmann distribution implies a spatial density of C exp(-V(x)/T). The velocity and spatial distributions are not independent. Similarly, in quantum mechanics, spatial density is W(x)W(x), but W(x)=Sum over p fp sin(px) so given a wavefunction, its Fourier transform is fp, the momentum distribution. The two are not independent. Finally, we try to show that for the case of the oscillator, one can obtain an expression for thermodynamic free energy in quantum mechanics, and show that exp(-ap*p) is an extremum for free energy.