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The first-order energy shift is not well defined, since there is no unique way to choose a basis of eigenstates for the unperturbed system. The various eigenstates for a given energy will perturb with different energies, or may well possess no continuous family of perturbations at all.
Using perturbation theory, the first-order energy shift can be calculated as = >, which requires the knowledge of accurate many-electron wave function. Due to the 1 / M N {\displaystyle 1/M_{N}} term in the expression, the specific mass shift also decrease as 1 / M N 2 {\displaystyle 1/M_{N}^{2}} as mass of nucleus increase, same as normal mass ...
If g = 1 (as is often the case for electronic states of molecules) the first-order energy becomes proportional to the expectation (average) value of the dipole operator , = | | = . Because the electric dipole moment is a vector ( tensor of the first rank), the diagonal elements of the perturbation matrix V int vanish between states that have a ...
More precisely, if , each of these three components is actually a group of several transitions due to the residual spin–orbit coupling and relativistic corrections (which are of the same order, known as 'fine structure'). The first-order perturbation theory with these corrections yields the following formula for the hydrogen atom in the ...
The fine structure energy corrections can be obtained by using perturbation theory.To perform this calculation one must add three corrective terms to the Hamiltonian: the leading order relativistic correction to the kinetic energy, the correction due to the spin–orbit coupling, and the Darwin term coming from the quantum fluctuating motion or zitterbewegung of the electron.
A key example of this phenomenon is the spin–orbit interaction leading to shifts in an electron's atomic energy levels, due to electromagnetic interaction between the electron's magnetic dipole, its orbital motion, and the electrostatic field of the positively charged nucleus.
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where ln denotes the natural logarithm, is the thermodynamic equilibrium constant, and R is the ideal gas constant.This equation is exact at any one temperature and all pressures, derived from the requirement that the Gibbs free energy of reaction be stationary in a state of chemical equilibrium.