Quantum Entropy and Thermodynamics
| Main Author: | Francesco R. Ruggeri |
|---|---|
| Format: | info publication-preprint |
| Terbitan: |
, 2020
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| Subjects: | |
| Online Access: |
https://zenodo.org/record/4035160 |
Daftar Isi:
- Correction Sept. 18, 2020 The last sentence of the section Temperature for a Quantum Bound State, namely: One may note that even though W(x)W(x) is of the same form as the classical statistical result exp(-.5kx*x/T) for an oscillator, the classical and quantum temperature values differ. should read: "do not differ." The temperature from the hydrodynamic equaiton and the Maxwell-Boltzmann form are the same. In the literature, one often sees entropy density expressions of the form: - [W*W] ln[W*W] - a(p)a(p) ln[a(p)a(p)] ((1)) (1). In a recent note (2), we suggested entropy density may be written as: -W x d/dx W - a(p)a(-p) ln[a(p)a(-p)] ((2)). In this note, we wish to compare these two expressions with the thermodynamic equation: dE = TdS - PdV. Thus, an expression for temperature must first be found. We use the hydrodynamical equation: T= m/3 ( |v-u|*|v-u|> and find that T(x) = 1/m { -d/dx d/dx W / W + [dW/dx / W][dW/dx /W] } in one-dimension. Thus, temperature is a function of x except for the ground state oscillator case i.e. W(x)=sqrt(B) exp(-ax*x) where a=.5 sqrt(km) and B= sqrt( km/3.14). In this case, temperature is constant in space. We consider this situation, and find that expression ((1)) yields dE/T not equal to dS and ((2)) dE = T dS for the ground state oscillator, if one varies k, the spring constant.