Sample Quantum Chemistry Exam 2 Solutions

Transcription

1 Chemistry 46 Fall 7 Dr. Jean M. Standard Name SAMPE EXAM Sample Quantum Chemistry Exam Solutions.) ( points) Answer the following questions by selecting the correct answer from the choices provided. a.) Which of the following statements are false about the Variation Method? ) Setting the secular determinant equal to zero in the inear Variation Method yields a polynomial equation for the approximate energy. ) In the Nonlinear Variation Method, the energy expectation value is minimized with respect to the nonlinear parameter to obtain the best solution. 4) In the inear Variation Method, the set of functions used to represent the approximate wavefunction is referred to as the basis set. 5) All of the above. 6) None of the above. 3) The energy expectation value of an approximate wavefunction is always less than or equal to the exact eigenvalue. b.) Which of the statements below are true regarding the quantum mechanical harmonic oscillator? ) The quantized energy levels are evenly spaced. ) The eigenfunction corresponding to quantum number v= has odd symmetry. 3) The ground state wavefunction exhibits quantum mechanical tunneling. 4) The integral between eigenfunctions ψ and ψ 4 equals. 5) Both () and (3). 6) Both () and (4). d A c.) Consider the Schrödinger equation for a system given by the form E dx x ψ ( x ) =, where E is the energy eigenvalue and A is a real constant and the range of the coordinate x is x. Which of the following statements about the use of the Sommerfeld Polynomial Method for this equation are true? ) The first step in the method is determination of the asymptotic solution for x =. ) The asymptotic solution has the form e bx, where b = E. 4) The recursion relation is used to determine the asymptotic solution. 5) Both () and (3). 6) Both () and (4). 3) The full solution is written as the asymptotic solution times a trigonometric function.

2 . Continued d.) Which of the following statements are false about the Heisenberg Uncertainty Principle? ) The minimum uncertainty product for position and momentum (both in the x-direction) is!. ) If the position of a particle is known exactly, then its momentum is completely unknown. 3) If two operators commute, their uncertainty product is identically zero. 4) The quantum mechanical uncertainty is similar in form to the standard deviation from statistics. 5) All of the above. 6) None of the above..) ( points) Briefly explain and discuss the significance in quantum chemistry of each of the following topics or concepts. a.) Overlap integral The overlap integral S ij is defined between two functions, φ i and φ j, such that S ij = φ i φ j. The overlap integral provides a measure of how similar two functions are and ranges from to. If the overlap is zero the functions have nothing in common, and if the overlap is one the functions are identical. The overlap integral appears in the secular equations (and secular determinant) of the inear Variation Method and also in the definition of the Franck-Condon Factor for electronic transitions. b.) Correspondence Principle The Correspondence Principle states that as the energy of a quantum mechanical system increases, its behavior becomes more like a classical system. This can be observed in many systems, but the example we saw was in the harmonic oscillator, where the behavior of the ground state (v=) system was furthest away from the classical oscillator with respect to probability distribution (maximum at x= and close to zero near the turning points), whereas for higher quantum numbers (such as v= or higher) the probability distribution begins to more closely mimic the classical distribution (maxima near the turning points and low probability near x=). c.) Commutator The commutator of two quantum mechanical operators  and ˆB is defined as Â, ˆB =  ˆB ˆBÂ. If two operators are said to commute, the commutator equals zero. Otherwise, for non-commuting operators, the commutator is non-zero. This operator plays an important role in the Heisenberg Uncertainty Principle. Only physical observables corresponding operators that commute may be measured exactly and simultaneously. This is because only commuting operators may share a set of simultaneous eigenfunctions.

3 .) Continued 3 d.) Variation Principle The Variation Principle states that the expectation value of the energy, E, for any approximate wavefunction is always greater than or equal to the exact ground state energy eigenfunction, E E exact. This principle has important consequences in quantum chemistry since so many problems cannot be solved exactly. It allows us to determine the best approximate solution by determining the minimum energy expectation value with respect to some parameter.

4 3.) ( points) Determine the average value of x for the harmonic oscillator in the v th state. It may be helpful to use the recursion relation for Hermite polynomials, H v z that the harmonic oscillator eigenfunctions ψ v ( x) have the form = z H v ( z) v H v- ( z). Recall 4 ψ v ( x) = N v H v x e x /, with N v = /4 v v! / and = mk! /. For a normalized wavefunction, the expectation value x is x = ψ * v ( x) ˆx ψ v ( x) dx. Here, we know that coordinate operators are equivalent to "multiply by" operators, so that ˆx = x. Substituting, x = ψ * v ( x) x ψ v ( x) dx. Using the form of the harmonic oscillator wavefunctions, we have x = N v e x / x H v ( x) e x / dx H v x. In order to evaluate this integral, we must develop an expression for the product x H v ( x). To begin, we can use the recursion relation, H v ( x) = x H v ( x) v H v- ( x), and solve it for x H v ( x), x H v ( x) = H v( x) v H v-( x). Multiplying this expression by x to get x H v ( x) yields x H v ( x) = x H v x = x v H v-( x) x H v( x) v x H v-( x). The first term on the right involves x H v ( x), while the second term on the right involves x H v- x We can obtain expressions for these by shifting the index in the recursion formula either by or. The results are: x H v ( x) =.x H v- ( x) = and H v ( x) v H v ( x) H v ( x) v- H v- ( x)..

5 3.) Continued 5 Substituting these relations into the expression for x H v ( x) gives x H v ( x) = = x H v ( x) = x H v( x) = 4 H v ( x) v x H v-( x) H v ( x) v 4 H v ( x) H v ( x) v H v ( x) v H v ( x) v v H v ( x) v( v-) H v- H v ( x) v- H v- ( x) v( v-) ( x). H v- ( x) Now we can put this expression for x H v ( x) back into the expectation value and break it up into three separate integrals, x = N v = N v = N v 4 H v H v H v ( x) e x / x H v x e x / dx ( x) e x / 4 H v ( x) ( x) e x / H v x N v v N v v v- H v e x / dx ( x) e x / H v x v H v ( x) e x / dx v( v-) e x / H v- ( x) e x / dx H v x. H v- ( x) e x / dx Notice that the terms H v ( x) e x / and H v- x e x / on the right in the first and third integrals correspond to harmonic oscillator eigenfunctions, except that the normalization constants N v and N v- are missing. The next step is therefore to include the appropriate normalization constants for the v and v wavefunctions by multiplying and dividing by the requisite factors (that is, we are really just multiplying by ), and also to put the normalization factor N v back with the rest of the vth wavefunctions: x = 4 N v N v H v x N v e x / { } N v H v ( x) e x / v v v- { } dx N v H v ( x) e x / { } N v H v x e x / { } dx e x / { } N v- H v- ( x) e x / { } dx N v N v H v x. N v-

6 6 3.) Continued The terms that appear in brackets in each integral are harmonic oscillator eigenfunctions. Identifying them yields x = = 4 N v N v H v x N v e x / { } N vh v ( x) e x / v v v- { } dx N v H v ( x) e x / { } N vh v x N v N v H v x N v- 4 N v ψ v ( x) ψ v ( x) dx N v e x / { } dx e x / { } N v-h v- ( x) e x / v ψ v v v- ( x) ψ v x { } dx dx N v ψ v ( x) ψ v- ( x) dx. N v- Note that there are no complex conjugates shown because the harmonic oscillator eigenfunctions are real. Because of orthogonality, the first integral on the right is zero and the third integral on the right is also zero. The second integral on the right equals one due to normalization. Substituting, we have x = = x = 4 N v ψ v ( x) ψ v ( x) dx N v 4 N v N v v. v v v- v ψ v v v- N v N v- ( x) ψ v x dx N v ψ v x N v- ψ v- ( x) dx

7 4.) ( points) A typical angular wavefunction representing an electron in a p x -type orbital has the form 7 ψ px ( θ,φ) = 3 / cosφ. a.) Show that this function is normalized. The normalization integral is * θ,φ dθ dφ. Y px Y px ( θ,φ) Substituting the wavefunction yields * θ,φ dθ dφ = Y px Y px ( θ,φ) 3 4 ( cosφ) ( cosφ) dθ dφ. Simplifying, * θ,φ dθ dφ = Y px Y px ( θ,φ) 3 4 sin 3 θ dθ cos φ dφ. The integrals involving θ and φ must be evaluated; they may be found in integral tables. The result for the intergral involving θ is: sin 3 θ dθ = cosθ sin θ 3 = cos sin 3 cos 3 = 3 [ ] [ ] 3 sin = 3 3 sin 3 θ dθ = 4 3. The result for the intergral involving φ is: cos φ dφ = = cos φ dφ =. φ sinφ 4 sin4 4 sin 4 = { } { }

8 8 4 a.) Continued Substituting these into the normalization integral yields, * ψ px ( θ,φ) ψ px ( θ,φ) dθ dφ = = 3 4 sin 3 θ dθ cos φ dφ * ψ px ( θ,φ) ψ px ( θ,φ) dθ dφ =. Therefore, the angular function for the px-type orbital, ψ px ( θ,φ), is normalized. b.) Determine by explicit operation whether or not this function is an eigenfunction of the angular momentum z-component operator, ˆ z. If it is an eigenfunction, determine the eigenvalue. Recall that ˆ z = i! φ. To show that the function ψ px ( θ,φ) is an eigenfunction of ˆ z, we apply ˆ z to the function, ˆ z ψ px ( θ,φ) = i! 3 φ / cosφ = i! 3 = i! 3 ˆ z ψ px ( θ,φ) = i! 3 / / / cosφ φ { } φ cosφ { } sinφ. Since the righthand side does not equal a constant times ψ px ( θ,φ), the function ψ px θ,φ eigenfunction of the ˆ z operator. is not an

9 4.) Continued 9 c.) Demonstrate by explicit operation whether or not this function is an eigenfunction of the angular momentum squared operator, ˆ. If it is an eigenfunction, determine the eigenvalue. Recall that ˆ =! θ cotθ θ sin θ φ. To show that the function ψ px ( θ,φ) is an eigenfunction of ˆ, we apply ˆ to the function, ˆ ψ px ( θ,φ) =! θ cotθ θ 3 sin θ φ =! 3 =! 3 / / θ θ / cosφ cotθ θ sin θ φ cosφ ( cosφ) cotθ ( cosφ) θ ( sin θ φ cosφ). Evaluating the derivatives separately gives cosφ θ = cosφ θ = cosφ cosθ. θ φ ( cosφ) = cosφ θ = cosφ. ( cosφ) = φ ( ) ( cosφ ) = cosφ. Substituting, ˆ ψ px ( θ,φ) =! 3 / θ ( cosφ) cotθ ( cosφ) θ ( sin θ φ cosφ) =! 3 / =! 3 ˆ ψ px ( θ,φ) =! 3 / / cosφ cotθ cosφ cosθ cosφ cosθ cos θ ( cosθ cosφ) cosφ. ( cosφ) sin θ cosφ

10 4 c.) Continued The term in brackets can be simplified by use of a common denominator, cos θ = sin θ cos θ = sin θ cos θ = sin θ sin θ = sin θ =. Note that the trigonometric identity cos θ = sin θ was used in order to simplify the above expression. Substituting, ˆ ψ px ( θ,φ) =! 3 / cos θ cosφ =! 3 =! 3 / / ˆ ψ px ( θ,φ) =! ψ px θ,φ ( )cosφ cosφ. So, we see that the function ψ px ( θ,φ) is an eigenfunction of the ˆ operator. The eigenvalue is!.

11 Chemistry 46 Dr. Jean M. Standard Quantum Chemistry Exam Take-home Portion Name SAMPE EXAM This portion of the exam is take-home. You may consult your notes and any other written or online references. Please work alone. You may discuss this problem ONY with Dr. Standard. 5.) ( points) In this problem, the linear variation method will be employed to determine approximations to the ground state and first excited state of a particle in a uneven box. The Hamiltonian for this onedimensional system is given by ˆ H (x) = p ˆ x m V (x), where V (x) =, x <, x < V, x, x >, and V is a constant. A linear variational calculation employing two basis functions will be carried out. The basis functions that you will use are the first two wavefunctions of the particle in a one-dimensional infinite box (of width ), φ (x) = sin x, φ (x) = x. a.) Construct the secular determinant for solution of the linear variation problem. Use atomic units with! = and m =. Your secular determinant will depend on the parameters and V. The first step in getting the secular determinant is to calculate the Hamiltonian matrix elements. The Hamiltonian matrix element H is defined as H = φ =! m ˆ H φ = sin x! d m dx sin x dx sin x dx d V ( x ) V sin x dx sin x dx. / Note that the limits for the integral involving the potential energy are / to, since the potential energy is only non-zero in that interval.

12 5 a.) Continued Evaluating the second derivative, we have d dx sin x = sin x. Substituting, H =! m d sin x dx sin x dx =! m sin x dx From integral tables, the integral appearing above is V V / sin x dx. / sin x dx sin bx dx = x sin bx 4b. Substituting and evaluating at the limits yields, H =! m sin x dx =! m 3 =! m 3 x sin x / 4( /) V / V 4 sin 4 sin =! m 3 V 4 H =! m V. V sin x dx x sin x / 4( /) / 4 sin 4 4 sin The other Hamiltonian matrix elements are evaluated in a similar fashion. The Hamiltonian matrix element H is defined as H = φ =! m H =! m 4 ˆ H φ = sin x! d m dx V ( x ) sin x x dx dx d sin x sin x dx V V / x dx sin x sin x dx sin x sin x dx. /

13 5 a.) Continued 3 From integral tables, the integral appearing above is sin ax sinbx dx = sin( a b)x a b sin( a b)x. a b For the integral above, this becomes sin x sin x dx = = x sin x 3 x 6 x. Substituting and evaluating at the limits yields, H = 4! m 3 = 4! m 3 = 4! m 3 sin x sin x dx sin x = 4! m 3 H = 4V 3. 3 x 6 sin sin 3 6 V V 6 V / sin x sin x dx V sin 6 sin sin sin 3 6 sin x 3 x 6 sin 6 sin 3 / Because the Hamiltonian operator is Hermitian, we also have H = H = 4V 3. Finally, the Hamiltonian matrix element H is defined as H = φ =! m H =! m 4 ˆ H φ = x! d m dx V ( x ) sin x x dx dx d sin x sin x dx V V / x dx sin x sin x dx sin x sin x dx. /

14 5 a.) Continued 4 From integral tables, the integral appearing above is sin bx dx = x sin bx 4b. Substituting and evaluating at the limits yields, H = 4! sin x m 3 dx = 4! m 3 = 4! m 3 x sin 4 x / 4( /) V sin x dx / V 8 sin4 8 sin V = 4! m 3 V 4 H =! m V. x sin 4 x / 4( /) / 8 sin4 4 8 sin The secular determinant of the linear variation method has the general form, H ES H ES H ES H ES =. Substituting the values of the Hamiltonian matrix elements, and using the information that the basis functions are orthonormal (because they are eigenfunctions of a hermitian operator, so that S ij = δ ij ), the secular determinant takes the form,! m V E 4V 3 4V 3! m V E =. In atomic units with with! = and m =, the secular determinant simplifies to V E 4V 3 4V 3 V E =.

15 5 5.) Continued b.) Solve the secular determinant using = and V = 8 (in atomic units). Determine the approximate energies of the ground and first excited states. Compare these results to the energies of a particle in an infinite box of width (that is, compare to the exact result that you would get if you set V = ). Do the energies shift in the way that you would expect? Explain. In atomic units with with = and V =, the secular determinant becomes 8 4 E E =. Expanding the secular determinant gives Solving the quadratic equation, 8 4 E 4 E 3 3 = 3 3 ( E) ( E).58 = E E.58 = E 4.685E =. [ ( ) ]. E = ± ( 4.685) 4 = ± or, E = 3.739,.95 a.u. To compare with the energies of a particle in an infinite box of width, note that the energies of this system are given by E n = n! m. In atomic units with! =, m =, and =, the infinite box energies for the ground and first excited states are E = 8 E = =.337 a.u., = a.u. We see that the energy levels of the infinite box are much lower than those predicted for the uneven box. This is expected, since the added potential on the right side of the uneven box increases the energy of the particle.

16 5 b.) Continued 6 As the uneven potential gets larger, the energies of the uneven box increase. In the limit as V, the system becomes equivalent to an infinite box of width /. The energies of this system are given by E n = n! m( /) = n! m. In atomic units with! =, m =, and =, the energies for the ground and first excited states of an infinite box of width / are E = = a.u., E = = a.u. We see that the energy levels of the uneven box with V = 8 a.u. are lower than those predicted for the limit V (as might be expected). However, they are significantly higher than the energies calculated for the uneven box with V = a.u. c.) Obtain the linear variation coefficients c and c for the ground state of the uneven particle in a box. In matrix form, the secular equations are H ES H ES H ES H ES More specifically for this case, the secular equations are c c E E =. c c =. Substituting the ground state energy, E = a.u., into the secular equations yields c c = c c =. Multiplying the matrices out leads to the two secular equations,.63c c = c 5.774c =.

17 5 c.) Continued 7 The first equation yields.63c c =, or c =.6839c. The second equation yields The results are identical for the two equations c 5.774c =, or c =.6839c. To complete the determination of the coefficients, we must use the normalization condition. For two orthonormal basis functions, this condition is c c =. Substituting c =.6839c, we have c c = (.6839c ) c.8356c c c = c = c =.56. = = Then c =.6839c =.6839 (.56) c = Thus, the best approximate ground state wavefunction ψ is given by ψ ( x) = c φ ( x) c φ ( x) =.8598 sin x.56 x. For = a.u., the ground state wavefunction is =.8598 x ψ x.56sin x.

18 5.) Continued 8 d.) Plot the approximate wavefunction of the ground state of the uneven particle in a box. Compare this function to the exact wavefunction for the V = case. Explain why the approximate wavefunction has the shape that it does. The wavefunction for the particle in an infinite box of width a.u. is perfectly symmetric with a maximum at x=. a.u. Compared with the infinite box wavefunction, the uneven box wavefunction is shifted to the left. This occurs in order to move probability out of the region of high potential on the right side of the uneven box. ittle probability density remains in the region from x=/ to, particularly closer to x=.