Near a stationary configuration of a two-atom system

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Linear algebra and vector spaces · Linear algebra and vector spaces

Conditions of Validity, Ch The spectral theorem, degeneracy and the matrices that will not diagonalise

Problem

Near a stationary configuration (x,y)=(0,0)(x,y) = (0,0) of a two-atom system, the potential energy is well approximated by V(x,y)=12rTHrV(x,y) = \tfrac12\mathbf r^{\mathrm T} H \mathbf r, where r=(x,y)\mathbf r = (x,y) are small displacements in metres from the stationary point and H=(375−125−125375) N m−1H = \begin{pmatrix}375&-125\\-125&375\end{pmatrix}\,\mathrm{N\,m^{-1}} is the Hessian (second-derivative) matrix of VV there. Take the Boltzmann constant kB=1.381×10−23 J K−1k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}. ((a)) Find the eigenvalues of HH and classify the stationary point (minimum, maximum or saddle). ((b)) Give the singular value decomposition of HH, treated as a matrix in its own right, and its condition number κ(H)\kappa(H) in the Euclidean norm. ((c)) Using the principal axes found in part (a) and the two-dimensional Gaussian integral, evaluate Z=∬−∞∞exp⁡ ⁣(−V(x,y)kBT)dx dyZ = \iint_{-\infty}^{\infty} \exp\!\left(-\frac{V(x,y)}{k_BT}\right)dx\,dy at T=300 KT = 300\,\mathrm K.

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