3502_18_octo25

# 3502_18_octo25 - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 18, October 25, 2010 Solved Homework If the approximate energy is given by < H > = 3 a 4 – 4 a 3 –36 a 2 + 10 where a is a variational parameter, local minima and maxima of the energy will correspond to stationary points of < H > with respect to a , i.e., points where d H da = 0 Setting the derivative equal to zero gives 0 = 12 a 3 – 12 a 2 –72 a which may be simplified as 0 = a ( a 2 a – 6) = a ( a –3)( a + 2) The three solutions to this cubic equation are a = –2, 0, and 3. It is a trivial matter to plug these values back into the expression for the energy and find that the corresponding values of < H > = E are –54, 10, and –219. Analysis of the sign of the second derivative of < H > with respect to a indicates these values to be a local minimum, a local maximum, and a local minimum, respectively (although their characters are obvious in this case). The variational principle states that our lowest energy (–219) will be an upper bound to the true lowest energy (which would be the ground-state energy), so the highest energy the system can possibly have in its ground state is –219 (in whatever units a is expressed). Some Simple Variational Calculations Last lecture, we considered the variational principle in the context of basis functions and their uses for molecular calculations. The variational approach provides a prescription for computing molecular orbitals as linear combinations of atomic orbitals,

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18-2 where the coefficients of the linear combinations are the variational parameters subject to optimization. However, to better appreciate some of the concepts associated with the variational principle, let’s now take a step back and focus on a vastly simpler system with only a single variational parameter. Consider the ground state for a particle having mass 1 a.u. in a box of length L = 1 a.u. We know the eigenfunction and eigenvalue (in a.u.) for this particle-in-a-box system exactly, namely Ψ 1 x ( ) = sin π x ( ) 0 x 1 E 1 = π 2 2 (18-1) Now, let us imagine that we had no idea how to derive the correct eigenfunction. But, we certainly imagine that the function should have no nodes (it’s the ground state, after all), and we also know from the boundary conditions it must be zero at both ends of the box. One very simple choice for a trial function might be ξ x ( ) = x 1 x ( ) (18-2) This function has the correct behavior at x = 0 and x = 1 and it is everywhere positive in between. Of course, there’s no variational parameter in ξ , so there is nothing there to optimize. A simple way to introduce such a parameter is to choose instead ξ x ; a ( ) = x a 1 x ( ) (18-3) where a is the variational parameter to be optimized. The optimization condition minimizes the energy of the trial wave function, which from the variational principle establishes an upper bound on the “true” energy. The variational condition is 0 = d da
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## This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.

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3502_18_octo25 - Chem 3502/5502 Physical Chemistry...

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