( ) = 9φ 1, ( ) = 4φ 2.

Transcription

1 Chemistry 46 Dr Jean M Standard Homework Problem Set 6 Solutions The Hermitian operator A ˆ is associated with the physical observable A Two of the eigenfunctions of A ˆ are and These eigenfunctions are normalized and orthogonal The eigenvalues of and are and, so that ˆ A ˆ A Suppose that a state ψ has a form given by ψ Determine the uncertainty ΔA for the state ψ The uncertainty ΔA is defined as ΔA A A / In order to calculate the uncertainty, the expectation values must first be evaluated The expectation value A is A ψ ψ ˆ A ψ ψ ψ ˆ A ψ, as long as the wavefunction is normalized Evaluating the expectation value, A ψ ˆ A ψ Applying the operator twice, ˆ A A ˆ ˆ A A ˆ A ˆ A ˆ A ˆ 9, 4 A ˆ + 9 ˆ A

2 continued Substituting, The expectation value A is A A 45 A ψ ψ ˆ A ψ ψ ψ ˆ A ψ Evaluating the expectation value, A ψ ˆ A ψ ˆ A ˆ A ˆ A A ˆ + 9 ˆ A Using the eigenvalue equations, A A Substituting the expectation values into the expression for the uncertainty, ΔA A A ΔA / / /

3 Evaluate the uncertainties Δx and Δp x for the ground state of the harmonic oscillator The uncertainty Δx is defined as The expectation values above must be evaluated Δx [ x x ] / For x, we have x / / α x / e x x e α x dx e α x / dx From integral tables, x e bx dx π 4b b Since this integral is defined only from to, we have since the integrand is an even function, x e bx dx x e bx dx π b b Substituting this result into the expression for x, x x α / / π α α x α x e dx / For x, we have x / / α x / e x e α x / dx x e α x dx

4 4 continued Since the integrand is an odd function, the integral over the entire range must be zero, so x / x e α x dx Alternately, you can look the integral up in a table From integral tables, x e bx dx b Note however that this integral is defined only from to and not from to If the integrand f ( x) is an odd function, then f ( x) f ( x), and therefore So, the integral becomes x f ( x) dx f ( x) dx x / / / α x x e dx x e α x { dx + x e α x dx } α + α Knowing the expectation values, we can next calculate the uncertainty, Δx x x / Δx α α /

5 5 continued The uncertainty Δp x is defined as Δp x [ p x p x ] / The expectation values p x and p x must be evaluated For p x, we have p x / e α x / ˆp x e α x / dx / e α x /! d dx e α x / dx p x! / dx e α x / d dx e α x / Evaluating the second derivative, Substituting, the expectation value becomes d dx e α x / ( α + α x ) e α x / p x p x! /!!! / / / e α x / d e α x / dx e α x / dx ( α + α x ) e α x / dx ( α + α x ) e α x dx α e α x dx + α { x e α x dx } From integral tables, the first integral is e α x dx e α x dx From integral tables, the second integral is π α / x e α x dx x e α x dx π α α /

6 6 continued Substituting these results into the expression for p x we have, p x!! / / α e α x dx + α { x e α x dx } α π α / + α π α α /!! α / π / α α + α p x! α For p x we have, p x / / p x i! α e α x / e α x / / ˆp x e α x / dx i! d dx e α x / dx dx e α x / d dx e α x / Evaluating the first derivative, d dx e α x / α x e α x / Substituting, the expectation value becomes From integral tables, p x i! α / i! α / p x iα! α / e α x / e α x / / dx d dx e α x x e α x dx ( α x) e α x / dx x e bx dx b

7 7 continued Note that this integral is defined only from to and not from to If the integrand f ( x) is an odd function, then f ( x) f ( x), and therefore So, the integral becomes f ( x) dx f ( x) dx p x iα! iα! iα! p x / / / α x x e dx x e α x { dx + x e α x dx } α + α Knowing the expectation values, we can next calculate the uncertainty, Δp x p x p x /! α / Δp x! α

8 8 Prove the commutator relation A ˆ, B ˆ C ˆ [ ] B ˆ A ˆ [, C ˆ ] + A ˆ [, B ˆ ] ˆ This relation can be proved simply by verification that the left and right sides are equal The left side is A ˆ [, B ˆ C ˆ ] A ˆ B ˆ C ˆ B ˆ C ˆ A ˆ C Working out the right side, [ ] + A ˆ [, B ˆ ] ˆ B ˆ A ˆ, C ˆ [ ] + A ˆ [, B ˆ ] ˆ B ˆ A ˆ, C ˆ + A ˆ ( B ˆ B ˆ A ˆ ) C ˆ C B ˆ A ˆ C ˆ C ˆ A ˆ B ˆ A ˆ C ˆ B ˆ C ˆ A ˆ + A ˆ B ˆ C ˆ B ˆ A ˆ C ˆ C A ˆ B ˆ C ˆ B ˆ C ˆ A ˆ Thus, it is clear that the left and right sides are equal and so the identity is verified

9 9 4 Approximate the exact solution for the ground state wavefunction of a particle in a one-dimensional inifinite box using the function (x) x a ( x ) The parameter a is the nonlinear variational parameter and is the width of the box Note that this function is not normalized but it does satisfy the proper boundary conditions at x and x ; recall that the wavefunction equals zero outside the box a) Find the value of the energy expectation value for the approximate wavefunction Your answer will depend on the variational parameter a as well as on the constants and parameters in the Hamiltonian operator,!, m, and The energy expectation value is E ˆ H In order to evaluate this expectation value, we must operate the Hamiltonian operator for the particle in an infinite box on the approximate wavefunction Ĥ x!! m! m! m d m dx x d { } dx x a x d { dx x a+ x a } d dx! m a a + {( a +) x a ax a } { x a a( a )x a } Ĥ ( x)! m a ( a + ) x a +! m a a x a Substituting this into the expression for the energy expectation value yields, E E Ĥ x a ( x )! m a a + x a x a + ( x ) x a x x a+ x a! m a ( a + ) x a +! m a ( a )x a! m a ( a )x a x a+ x a x a+ x a

10 4 a) continued Working out the integrals in the denominator, we have x a+ x a x a+ x a x a+ x a+ + x a+ x a + x a x a+ + x a x a x a+ x a+ x a+ x a x a x a+ + x a x a x a+ dx x a+ dx x a+ dx + x a dx x a+ dx x a+ dx + x a dx Working out these integrals and evaluating at the limits yields: x a+ dx x a+ dx x a dx a + xa+ a + xa+ a + xa+ a + a+ a + a+ a + a+ Substituting, the denominator becomes x a+ dx x a+ dx + x a dx a + a+ a + a+ a + a+ + a + a+ a + a+ + a + a + + a + a+ a + a+ ( a + ) a + ( 4a + 6a + ) 4a +8a + ( a + ) a + ( a + ) a + ( a + ) ( a +) + ( a + ) ( a + ) ( a + ) ( a + ) ( a +) ( a +) + ( 4a + a + 6) ( a +) a+ a+ a+

11 4 a) continued Working out the integrals in the numerator, Ĥ xa+ x a! m a ( a + ) x a +! m a ( a )x a x a+! m a ( a + ) x a + x a+! m a ( a )x a + x a! m a ( a + ) x a + x a! m a ( a )x a! m a ( a + ) x a+ x a +! m a ( a ) x a+ x a! m a a + Ĥ! m a a + +! m a ( a + ) x a x a! m a ( a ) x a x a x a dx +! m a ( a ) x a dx +! m a ( a + ) x a dx! m a ( a ) x a dx x a dx +! m a Working out these integrals and evaluating at the limits yields: x a dx! m a ( a ) x a dx x a dx x a dx x a dx a + xa+ a xa a xa a a a + a+ a a

12 4 a) continued Substituting, the numerator becomes Ĥ! m a a +! m a a + x a dx! a( a +) m a + a+ +! a m! m a a + a +! a m Ĥ! a m +! m a x a dx! m a ( a ) x a dx a + a+ +! m a a a! m a ( a ) a a! m a a+! a( a ) m a a+ + a a ( a ) a a+ ( a +) ( a ) + ( a +) ( a ) ( a ) ( a +) ( a +) ( a ) + ( 4a ) ( a a ) a + a a + a + ( a ) ( a ) a+ a+ a+ Putting the numerator and denominator together, we have an expression for the expectation value, E E Ĥ! a m a + ( a + ) a + ( a )! a m a + ( a +) ( a ) ( a + )! a a + 4m a a+ a+ a + ( a + ) ( a +)

13 4 continued b) Minimize the energy expectation value in order to determine the best possible value of the variational parameter a Substitute this into the energy expectation value to determine the best approximation to the ground state energy of the particle in a box Compare with the exact result for an electron in a box of width au For a minimum, d E da From part 4a, the expectation value is E ( a + )! a a + 4m a! 4m ( 4a + a + 6a) ( a ) Evaluating the derivative, d E da! 4m ( a + a + 6) ( a )! 4a 4m ( + a + 6a) ( a ) Setting this equal to zero and solving for a,! 4m ( a + a + 6) ( a ) ( a + a + 6) a! 4a 4m ( + a + 6a) ( a ) 4a + a + 6a a ( a ) ( a + a + 6) ( 4a + a + 6a) ( 4a + 4a + a a a 6) 8a + a + a 6a + 8a a 6 8a + 4a a The cubic equation listed above may be solved using Solver in Microsoft Excel or a programmable calculator A graph of the equation is helpful in showing the approximate positions of the roots If we let the left side of the equation, which is a function of the variational parameter a, then we can plot the function f a 8a + 4a a on the y-axis versus the parameter a on the x-axis This graph is shown on the next page

14 4 4 b) continued Notice that the function exhibits roots at around a,, and For the approximation to the particle in a box problem, we do not want the parameter a to take a negative value If it does, the wavefunction will blow up at x Thus, we seek the positive root at around From solver, the root is a 47 Substituting, the energy expectation value for the best approximation is given by E ( a + )! a a + 4m a + +! m ( 47) E ! m Plugging in the values given for the problem, m and au, the numerical result for the energy of the best approximation is E ! m ( au ) ( au ) au E 474 au The exact solution for the ground state of the particle in a box is E,exact h 8m, or π in atomic m units With m and au, the numerical value for the exact ground state result is 7 au Therefore, the variational approximation in this case has only about a % error

15 5 4 b) continued While not required, it is interesting to plot the approximate and exact wavefunctions for comparison The approximate wavefunction has the form (x) x a ( x ) Here, for the best approximation, we have shown that the parameter a is equal to 47 This wavefunction is not normalized, however, so in order to compare it to the exact result for the ground state of the particle in a box, we must include the normalization constant Writing the approximation wavefunction as a normalization constant N times the original function yields, norm ( x) N x N x a ( x ) For normalization, we must have norm norm or N The integral has been previously evaluated in part (a) Substituting, N a + ( a + ) ( a +) N a+ Solving for the normalization constant N yields N a+ / ( a + ) a + ( a +) / Therefore, the normalized approximate wavefunction has the form norm ( x) a+ / ( a + ) a + ( a +) / x a ( x ) Plots of this approximate wavefunction and the exact ground state particle in a box wavefunction, ψ ( x) sin π x, are shown in the figure on the next page for the case when m au and au

16 6 4 b) continued Notice that the approximate and exact wavefunctions differ by a sign This is an indication that the phase of the wavefunction is arbitrary Because the observable probability density is related to the square of the wavefunction (or the absolute square if the wavefunction is complex), we can multiply any wavefunction by an overall sign and not change any observable properties Thus, we can arbitrarily multiply the approximate wavefunction by to compare with the exact wavefunction and give both functions the same overall phase Thus, the new normalized approximate wavefunction is: norm ( x) a+ / ( a + ) a + ( a +) / x a ( x ) This function and the exact particle in a box ground state wavefunction are plotted in the figure below Now we see that the approximate and exact wavefunctions agree quite well in this case, with the approximate wavefunction just shifted slightly to the right relative to the exact solution

17 7 5 The Hamiltonian operator in atomic units for the hydrogen atom in its ground state is given by H ˆ r + r r r The radial coordinate r is the distance between the electron and the nucleus with the nucleus placed at the origin Approximate the exact solution for the ground state of the hydrogen atom using the normalized function (r) α / 4 e α r The parameter α is the nonlinear variational parameter a) Find the value of the energy expectation value of (r) Your answer will depend on the variational parameter α When calculating the expectation value and integrating over the coordinate r, the volume element is r dr and the range of the radial coordinate is from to Also, the integrals over the angular coordinates θ and must be included; for these coordinates, the range of the polar angle θ is to π, and the volume element is sinθ dθ, ; the range of the azimuthal angle is to π, and the volume element is d The energy expectation value is E ˆ H In order to evaluate this expectation value, we must operate the Hamiltonian operator on the approximate wavefunction ˆ H r ˆ H r r + r r r r r + r r r r r r r ( r ) r r r r

18 8 5 a) continued Substituting the expression for the approximate wavefunction, this becomes ˆ H r ˆ H r r r r / 4 α r e α r α / 4 r r r r r r α / 4 e α r r r e α r r r e α r r e α r α π / 4 e α r Next, we can work out the derivatives, r e α r α r e α r Substituting the derivatives yields ˆ H r ˆ H r α α α α / 4 / 4 r e α r α e α r + 4α r e α r r e α r r r e α r r e α r α e α r + 4α r e α r r α r e α r r e α r / 4 / 4 α e α r α r e α r + α e α r r e α r α e α r α r e α r r e α r Inserting this expression into the Hamiltonian gives ˆ H α α / / Next, the integrals can be written out explicitly e α r α e α r α r e α r r e α r α e α r e α r α e α r r e α r e α r r e α r ˆ H α / α 4π α e α r e α r α e α r r e α r / α e α r r dr α e α r r 4 dr e α r r e α r e α r r dr

19 9 5 a) continued In the previous equation, the factor of 4π comes from the integrals over the angular coordinates θ and The three radial integrals may be separately evaluated using integral tables, e α r r dr π 4 α e α r r 4 dr π 8 α e α r r dr Substituting, the Hamiltonian integral becomes 4α 5 / / H ˆ α 4π / α α 4π α π α α π 4 α π H ˆ α α α π / / / / α π 8 α 5 / 4α Substituting into the expression for the energy eigenvalue, E ˆ H α α E α α / /

20 5) continued b) Determine the value of the variational parameter α that minimizes the approximate energy Use this value of α to compute the best approximation to the ground state energy (in atomic units) Compare this with the exact hydrogen atom ground state energy of 5 hartrees For a minimum, d E dα Evaluating the derivative and using the result from part 5a, d E dα / α / d E dα / α / Setting this equal to zero and solving for α, / α / α / / 8 α 9π Substituting this value of α into the energy expectation value yields the best approximation, E 8 9π 4 π 8 π 4 π E 44 au / 8 9π / The exact solution for the ground state of the hydrogen atom is E exact 5 au Therefore, the variational approximation in this case has only about a 5% error While not required, it is instructive to compare the approximate wavefunction to the exact wavefunction for the ground state of the hydrogen atom The exact ground state hydrogen atom wavefunction in atomic units is: ψ(r) / e r A graph of the approximate wavefunction (with the best value of α) along with the exact wavefunction is shown in the graph on the next page

21 5 b) continued Note that the approximate wavefunction is a gaussian-type function, so that it flattens out near r On the other hand, the exact wavefunction is simply a decaying exponential, which has a cusp at the origin Therefore, there is a significant amount of deviation in the approximate and exact wavefunctions as r approaches This is an example of the error in the wavefunction being fairly high, while the error in the energy is lower (due to the quadratic error scaling of the energy compared with the wavefunction) For larger values of r, the agreement between the approximate and exact wavefunctions is much better