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Mean-Square Fluctuation

One physical quantity of great interest is the variance in the position of atoms at equilibrium, $\langle (\Delta x)^2 \rangle$. For our model oscillator, $\langle x \rangle = 0$; so $\langle (\Delta x)^2 \rangle = \langle x^2 \rangle$. This mean-square fluctuation about the average position is related to the B factors of crystallography and is also measurable by neutron scattering [3] and by Mössbauer spectroscopy [4]. It is one of the most important quantities to keep an eye on in molecular dynamics simulations as well. What is this fluctuation for the harmonic oscillator in equilibrium at constant $N\cal{V}\it {T}$ according to classical and quantum mechanics? We again use equations 2, 4, and 5, now with $O = x^2$, and consider the same three frequencies of proton vibration. Because $V = kx^2/2$ and $\langle V \rangle = \langle K \rangle = \langle E \rangle/2$ for harmonic oscillation, the quantum and classical results are proportional to those obtained above for the average energy. That is, $\langle x^2 \rangle = U/k = U/m(2 \pi f)^2$. Again, m is taken as the mass of a proton. (To plot the three frequencies on one scale, results have been scaled to the f = 100/ps values (green): Results for f = 10/ps (blue) were scaled by 0.01 ; results for f = 1/ps (cyan) were scaled by 0.0001. Consequently, the three classical curves (red), $\langle x^2 \rangle = k_B T/k$, coincide.)

Mean-square fluctuation of H atom undergoing harmonic oscillation.


next up previous
Next: Overall Comparison - What Up: Classical vs. Quantum Mechanics: Previous: Average Energy U and
Peter J. Steinbach 2010-11-15