Quantum Chemistry Exam 2 Take-home Solutions

Transcription

1 Cheistry 60 Fall 07 Dr Jean M Standard Nae KEY Quantu Cheistry Exa Take-hoe Solutions 5) (0 points) In this proble, the nonlinear variation ethod will be used to deterine an approxiate solution for the quartic oscillator The Hailtonian operator for the oscillator can be written where is the quartic force constant Ĥ(x)! d dx x, A linear variational calculation eploying the following orthonoral basis functions will be carried out: φ (x) / e α x / ; φ ( x) / ( αx ) e α x / In the basis functions shown above, α is a fixed paraeter For this proble, use atoic units (so that! au and au) Also use values of the paraeters such that au and α au a) Plot the quartic potential energy as a function of x and turn in this graph Use the paraeters given above, with a range for x of to au You ay use Excel or another graphing progra to do this The quartic potential is shown below Note that it is fairly flat in the vicinity of x0, but rises steeply

2 5) continued b) Using the basis functions given above, deterine and report the Hailtonian integrals H, H, H, and H Your results ay depend on the paraeters α and (with! and in atoic units) Do not use the nuerical values of the paraeters; work these out analytically first Show all your work To work out the Hailtonian integrals, the operation of the Hailtonian operator on the basis functions φ (x) and φ (x) ust be deterined The Hailtonian operator consists of two parts: the kinetic energy part involves a second derivative operator, while the potential energy part involves a "ultiply by" operator Thus, the second derivative of the basis functions ust be worked out For the two basis functions (excluding the noralization constants), we have d dx d dx eα x / ( ) d dx x αx / eα ( ) α e α x / α x e α x / eα x / ( ) ( α α x ) e α x / d dx d dx ( αx ) e α x / d dx d dx d dx α x / αx e ( αx ) ( αx) e α x / ( αx α x 3 αx) e α x / ( 5αx α x 3 ) e α x / ( ) e α x / ( 5αx α x 3 ) αx 5α 6α x ( 5α 6α x 5α x α 3 x ) e α x / ( αx ) e α x / ( 5α α x α 3 x ) e α x / ( ) e α x / The first of the Hailtonian integrals ay now be worked out, H φ Ĥ φ H!!! α e αx /! d dx x e αx / dx e αx / d dx e αx / e αx / α α x e αx dx! α ( ) dx α ( )e αx / dx e αx / x e αx / dx α x e αx dx x e αx dx α x e αx dx

3 3 5 b) continued The integrals ay be evaluated fro integral tables, e bx dx e bx dx 0 π b x e bx dx x e bx dx 0 π b b x e bx dx x e bx dx 0 3 π b b Substituting, H H! α! α! α e αx dx! α π! α! α 6α! α 6α x e αx dx π α α α x e αx dx 3 π α The next Hailtonian integral is H H φ Ĥ φ!!!! H H ( αx ) e αx /! d dx x e αx / dx ( αx ) e αx / d dx e αx / ( ) dx α ( αx )e αx / ( α α x )e αx / dx ( α x α 3 x α α x ) e αx dx ( ) 3α x α 3 x α e αx dx 3! α x e αx dx! α 3 α α ( αx )e αx / x e αx / dx ( αx ) x e αx dx ( αx 6 x ) e αx dx x 6 e αx dx x e αx dx x 6 e αx dx! α e αx dx x e αx dx x e αx dx

4 5 b) continued The sae integrals used for H ay be used here, and in addition the following integral is required, x 6 e bx dx x 6 e bx dx 0 5 π b 3 b Substituting, H H 3! α 3! α H H 3! α! α x e αx dx π α 3! α! α α! α 3! α 3 x e αx dx x 6 e αx dx 3 π α 5 5 6α π α 3 6α! α! α π x 3 π α e αx dx e αx dx The final Hailtonian integral is H φ Ĥ φ!!!! ( αx ) e αx /! d dx x ( αx )e αx / dx ( αx ) e αx / d ( dx αx )e αx / { } dx ( αx ) e αx / 5α α x α 3 x ( ) e αx / dx ( 0α x α 3 x α x 6 5α α x α 3 x ) H! α x e αx dx ( ) α x α 3 x α x 6 5α e αx dx x e αx dx x e αx dx 6! α 3 x e αx dx x 6 e αx dx x 6 e αx dx x e αx dx! α x 6 e αx dx e αx dx ( αx )e αx / x ( αx )e αx / dx ( α x αx ) x e αx dx x x e αx dx e αx dx x 6 e αx dx 5! α x e αx dx e αx dx

5 5 5 b) continued The sae integrals used previously ay be used here, and in addition the following integral is required, x e bx dx x 6 e bx dx 0 05 π 6b b Substituting, H! α H! α! α 5! α x e αx dx 6! α 3 π α 9! α 39 6α x e αx dx 05 6! α 3 π 6α 5! α 5! α x e αx dx! α 3 π α x 6 e αx dx α! α π α 3 5 6α 5 x 6 e αx dx 5! α π α 3 3α 3 π α x e αx dx 5! α π e αx dx To suarize, the Hailtonian integrals are H! α 6α H H H 5! α! α 39 6α α In atoic units, the Hailtonian integrals becoe, H α H H H 5α 6α α 39 6α α

6 6 5) continued c) Construct the secular deterinant for solution of the linear variation proble Your secular deterinant will depend on the paraeters α and Setting up the secular deterinant for the linear variation ethod with two basis functions gives H ES H ES H ES H ES The basis functions are orthonoral; thus, the overlap integrals are S S and S S 0 Substituting the results for the Hailtonian integrals, the secular deterinant becoes α E 6α α α α 5α α 39 6α E d) Solve the secular deterinant using the paraeter values au and α au Deterine the approxiate energies of the ground and first excited states Report these results in atoic units Setting the secular deterinant fro part (c) equal to zero yields, α 6α E α α α 5α α 39 6α E 0 Expanding, 6α E 5α 39 6α E α α 0 After ultiplying out the ters, the equation becoes 5α α α E 5 6α 7 56α 6α E 5α E 39 6α E E α α 9 3α 0

7 7 5 d) continued Collecting ters and siplifying the expression gives E α 6α 5α E 3α α E 39 6α 3α 6 E 5α α 5 56α 39 6α 0 5 6α 7 56α α α 9 3α 0 Solving the quadratic equation, we have E 3α α ± 3α α 3α α 5 56α 3α 6α ± 9α 6 α 6α 3α 39 α 5 6α E 3α 6α ± 3α α 99 6α Using the paraeter values α au and au, this equation becoes E 3α 6α ± 3α α 99 6α 3 6 ± E 3606 ± 5966 au Thus, the ground and first excited state approxiate energies are E,approx 0630 au, E,approx au

8 5) continued e) Obtain the linear variation coefficients c and c for the ground state of the quartic oscillator The secular equations of the linear variation ethod for two basis functions are: c ( H ES ) c ( H ES ) 0 c ( H ES ) c ( H ES ) 0 More specifically for this case, the secular equations siplify due to the orthonorality of the basis functions, c ( H E) c H 0 c H c ( H E) 0 In atoic units, the Hailtonian integrals are, H α H H H 5α 6α α 39 6α α Using the paraeter values α au and au, these integrals becoe H 03 au H H 379 au H au Substituting these values for the Hailtonian integrals, the secular equations are therefore given by c ( 03 E) 379 c c c ( E) 0 Substituting the approxiate ground state energy, E,approx 0630 au, into the secular equations yields c ( ) 379 c c c ( ) 0

9 9 5 e) continued Siplifying, the secular equations are, 0390 c 379 c c 6709 c 0 The first equation yields 0390 c 379 c c 379 c c 360 c The second equation yields 379 c 6709 c c 6709 c c 360 c Both equations give the sae result So while we know the relationship between the coefficients, the actual values are undeterined To coplete the deterination of the coefficients, we ust use the noralization condition For two basis functions, this condition is ψ approx ψ approx c φ ( x) c φ x ( ) c φ ( x) c φ ( x) c φ φ c c φ φ c c φ φ c φ φ c c c 0 c c 0 c c c Note that in the expressions above, the coefficients c and c have been assued to be real (which is generally a valid assuption) In addition, the integrals involving the basis functions have been siplified as a result of the orthonorality of the functions

10 0 5 e) continued Substituting the relation fro the secular equations, c 360 c, gives c c ( 360 c ) c 00c c 300c c c Then the coefficient c ay be deterined fro the secular equation result, c 360 c c Thus, the best approxiate ground state wavefunction of the quartic oscillator is ψ approx ( x) c φ x ( ) c φ ( x) φ ( x) φ ( x)

11 5) continued ( ) c φ ( x), using the the linear f) Make a plot of the approxiate wavefunction, ψ approx c φ x variation coefficients c and c that you deterined in part (e) Use a range for x fro to au and paraeter values au and α au Turn in this graph The approxiate ground state wavefunction is ψ approx ( x) c φ x ( ) c φ ( x) φ ( x) φ ( x) Substituting the for of the basis functions yields, ψ approx ( x) φ x α ( ) φ ( x) / e α x / α π / ( αx ) e α x / A graph of this function is shown below