I Spectra in the thermal interpretation

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It is demonstrated how the thermal interpretation explains the spectra of atoms and molecules.
In any quantum system, the differences of the energy levels (the eigenvalues of the Hamiltonian ##H##) are in principle directly observable, since they represent excitable oscillation frequencies of the system and thus can be probed by coupling the system to a harmonic oscillator with adjustable frequency. Thus the observed spectral properties of quantum systems appear in the thermal interpretation as natural resonance phenomena.

To see this, we shall assume for simplicity a quantum system whose Hamiltonian has a purely discrete spectrum. We work in the Heisenberg picture in a basis of eigenstates of the Hamiltonian, such that ##H|k\rangle =E_k|k\rangle ## for certain energy levels ##E_k##. The q-expectation
$$\langle A(t)\rangle =Tr ~\rho A(t)=\sum_{j,k}\rho_{jk}A_{kj}(t)$$
is a linear combination of the matrix elements
$$A_{kj}(t)=\langle k|A(t)|j\rangle =\langle k|e^{iHt/\hbar}Ae^{-iHt/\hbar}|j\rangle=e^{iE_kt/\hbar}\langle k|A|j\rangle e^{-iE_jt/\hbar}=e^{i\omega_{kj}t}\langle k|A|j\rangle ,$$
where
$$\omega_{kj} = \frac{E_k-E_j}{\hbar}.$$
Thus every q-expectation exhibits multiply periodic oscillatory behavior whose frequencies ##\omega_{jk}## are scaled differences of energy levels. This relation is the modern general form of the Rydberg--Ritz combination principle.

Linear coupling of ##A## to a macroscopic (essentially classical, large ##m##) harmonic oscillator leads after standard approximations to a forced damped classical harmonic oscillator dynamics
$$m\ddot q + c\dot q + kq =F(t),$$ where ##F(t)=\langle A(t)\rangle## has the above form. If ##k## is adjusted according to ##k=\omega^2m,## standard analysis gives approximately amplitudes depending on ##\omega## in an approximately Lorentz-shaped way.

This is indeed what is observed in actual measured spectra.
 
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Typo ?

##\langle A(t)\rangle =Tr ~\rho A(t)=\sum_{j,k}\rho_{jk}A_{kj}(t)##
 
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Mentz114 said:
Typo ?

##\langle A(t)\rangle =Tr ~\rho A(t)=\sum_{j,k}\rho_{jk}A_{kj}(t)##
Thanks, corrected!
 
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