Sample Quantum Chemistry Exam 1 Solutions

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1 Chemistry 46 Fall 217 Dr Jean M Standard September 27, 217 Name SAMPE EXAM Sample Quantum Chemistry Exam 1 Solutions 1 (24 points Answer the following questions by selecting the correct answer from the choices provided a Consider a particle of mass m trapped in a one-dimensional box of width with infinite potential barriers on both sides Which of the statements below are true regarding this system? 1 The average value of the position x is always /2 2 The most probable value of the position x is always /2 3 When the width of the box is doubled, the quantized energies all decrease by a factor of 2 4 When the mass of the particle is doubled, the quantized energies all decrease by a factor of 2 5 Both (1 and (3 6 Both (1 and (4 b Which of the following statements are false about the "free" particle in one dimension? 1 The Hamiltonian operator is Ĥ =!2 d 2 2m dx 2 2 One general solution is given by = Asinkx + Bcoskx, with k = 3 Its energy levels are quantized 2mE! 4 Its wavefunction may be considered to have one component corresponding to particles moving to the right and another component corresponding to particles moving to the left 5 All of the above 6 None of the above c Consider an operator  defined as  = x d Which of the following statements are false about this dx operator? 1 It is a linear operator 2 It does not commute with the operator ˆB = x 4 The function ( x = xe 2x is an eigenfunction of  sin x 3 Operation of  on the function x the result sin x + cos x x yields 5 All of the above 6 None of the above

2 1 Continued 2 c Which of the following statements about a proper quantum mechanical wavefunction are true? 1 It must be single-valued 2 It may take complex values 3 It must not blow up as the coordinate approaches ± 4 It must be continuous 5 All of the above 6 None of the above 2 (16 points Explain or provide an example illustrating each of the following concepts a Born probability interpretation The Born probability interpretation states that the probability of finding a particle in the volume dτ = dx dy dz between x, y, z and ( x + dx, y + dy, z + dz is given by * ( x, y,z ( x, y,z dτ d Tunneling Tunneling is the penetration of the wavefunction into a classically forbidden region (ie, a region in which E < V c Orthogonal functions Two functions f and g are said to be orthogonal if the integral of their product (with one function as the complex conjugate over all space equals, f * g dτ = d Eigenvalue equation An eigenvalue equation is a special type of operator equation in which the result is a constant times the original function For example, the equation  f = c f is an eigenvalue equation for the operator eigenvalue  in which f is the eigenfunction and c is the constant (called the

3 3 3 (2 points Consider two operators A ˆ and B ˆ which are Hermitian Prove that their product A ˆ B ˆ is Hermitian only if A ˆ and B ˆ commute; that is, only if A ˆ B ˆ = B ˆ A ˆ Recall that for an operator O ˆ to be Hermitian, it must satisfy the relation or in Dirac notation, * ( x O ˆ φ ( x dx = O ˆ ( x dx, * φ x O ˆ φ = O ˆ φ We must show that the product operator ˆ A ˆ B satisfies the Hermitian relation, * ( x A ˆ B ˆ φ ( x dx = A ˆ B ˆ ( x dx To begin, we can start with the left side of the Hermitian relation, * ( x A ˆ B ˆ φ ( x dx, * φ x and then manipulate it to try to produce the right side of the Hermitian relation In the integral above, recall that the operator nearest the function operates first to produce a new function Therefore, B ˆ operates on φ( x to produce the new function B ˆ φ( x Grouping terms, the integral above becomes dx * ( x A ˆ B ˆ φ ( x This looks like the left side of the Hermitian relation for the operator A ˆ if the function on the right is B ˆ φ( x Using the Hermitian character of A ˆ gives dx * ˆ dx * ˆ * ( x ˆ ( x = A ˆ ( x B φ( x = A ˆ ( x B φ( x dx The right side integral now looks like an integral with one function on the left, A ˆ ( x, the operator B ˆ in the middle, and another function, φ( x, on the right Using the Hermitian property of B ˆ, we have A ˆ ( ( x * B ˆ φ( x dx = B ˆ ( A ˆ ( x * φ( x dx Equating the integral we started with to the right side of the equation above gives * ( x A ˆ B ˆ φ ( x dx = B ˆ A ˆ ( x dx * φ x

4 3 Continued 4 The relation for the product operator ˆ A ˆ B to be Hermitian is * ( x A ˆ B ˆ φ ( x dx = A ˆ B ˆ ( x dx * φ x From this, we see that the left sides of the previous two equations are the same; however, the right sides are the same only if A ˆ B ˆ = B ˆ A ˆ ; thus, the product operator A ˆ B ˆ is Hermitian only if A ˆ B ˆ = B ˆ A ˆ * * * * * * * * * * * * * * * * * * * * * * * * Carrying out the same proof using Dirac notation, for the product operator ˆ A ˆ B to be Hermitian the relation is A ˆ B ˆ φ = A ˆ B ˆ φ We may again start with the left side, manipulate it, and try to get the right side The left side is A ˆ B ˆ φ = A ˆ ˆ B φ, where the only thing done on the right is to group terms Using the Hermitian character of ˆ A gives A ˆ B ˆ φ = A ˆ ˆ B φ Then, using the Hermitian property of ˆ B, we have A ˆ B ˆ φ = A ˆ ˆ B φ = ˆ B ˆ A φ The relation for the product operator ˆ A ˆ B to be Hermitian is A ˆ B ˆ φ = A ˆ B ˆ φ From this, we see that the left sides of the previous two equations are the same; however, the right sides are the same only if A ˆ B ˆ = B ˆ A ˆ ; thus, the product operator A ˆ B ˆ is Hermitian only if A ˆ B ˆ = B ˆ A ˆ

5 4 (2 points Determine the average value of the kinetic energy T for the one-dimensional particle in an infinite box Recall that the wavefunctions of the one-dimensional particle in an infinite box have the form 5 n ( x = + -, $ nπ x ' / sin&, x % ( - -, x <, x > 1 -, where n is the quantum number and is the width of the box The average (or expectation value of the kinetic energy T for the particle in an infinite box is defined as T = * n ( x T ˆ n ( x dx Note that the integration limits are to since the wavefunction for the particle in an infinite box is zero outside that range We have also used the fact that the particle in a box wavefunction is normalized Substituting the wavefunction into the expression for the average value yields T = = 2 =!2 2m 2 =!2 m 2 # nπ x & sin% ( ˆ 2 # nπ x & T sin% ( dx $ ' $ ' # sin nπ x & # % (!2 d 2 & # % $ ' $ 2m dx 2 ( sin nπ x & % ( dx ' $ ' # sin nπ x & # d 2 & # % ( % $ ' $ dx 2 ( sin nπ x & % ( dx ' $ ' # sin nπ x & # % ( n 2 π 2 # sin nπ x && % $ ' 2 % (( dx $ $ '' T =!2 n 2 π 2 sin 2 # nπ x & m 3 % ( dx $ ' To evaluate the integral, we can look it up in a table From the integral handout, sin 2 bx dx = x 2 sin2bx 4b By making the substitution b = nπ, the integral becomes sin 2 bx dx = x 2 % 2nπx ( sin' * 4nπ & Substituting, the average value expression becomes T =!2 n 2 π 2 m 3 * x 2 $ 2nπx '-, sin& / + 4nπ % (

6 4 Continued 6 Evaluating at the limits yields T =!2 n 2 π 2 * $ m 3 & 2 sin 2nπ % 4nπ ' $ & ( % 4nπ sin( '-, / + ( =!2 n 2 π 2 m 3 T =!2 n 2 π 2 2m 2 * +, / This result is the same as the energy eigenvalue Since the potential energy inside the box equals, and the total energy is just the sum of the kinetic and potential energies, it makes sense that the average kinetic energy would equal the total energy

7 5 (2 points In this problem, you will consider a particle trapped between an infinite wall (at x= and a finite potential barrier, shown below and defined by 7 V (x = $ & % & '&, x <, x V, < x Q, x > Q ( & & *& V=V I II III IV x= x= x=q a Write down appropriate solutions for the Schrödinger equation in regions I, II, III, and IV for E < V Assume that the particle starts in the trapped region (II; that is, it is impinging on the barrier (region III from the left If necessary, eliminate any terms in the solutions that would lead to unacceptable wavefunctions in the limit that x ± Make sure that you define any constants that you use in defining your solutions The general solutions for the wavefunctions in regions I-IV for the case E < V are ( x = ( x = A sin kx + B cos kx ( x = Ce λx + De λx ( x = Fe ikx, I II III IV where k for both regions II and IV and λ for region III are defined as k = 2mE!, λ = 2m( V E! In region I, the wavefunction must equal because the potential is infinite None of the terms in these solutions need to be eliminated because they go to infinity as x ±

8 5 continued 8 b Apply the boundary conditions for the wavefunction at x = How does this simplify the solution for the wavefunction in region II? The boundary condition at x = is I ( = II ( Substituting the forms of the wavefunctions leads to the equation = A sin + B cos Since sin = and cos = 1, the equation becomes = B Therefore, the solution for the wavefunction in region II simplifies to II ( x = A sin kx c Apply the boundary conditions for the wavefunction and the first derivative at both x = and x = Q You should then take the ratio of the derivative and wavefunction equations at x = and x = Q This leads to two equations Report the two equations, making sure to eliminate any constants that cancel, but do not do anything further with them The boundary conditions at x = are II ( = III and # II ( = # III ( Substituting the forms of the wavefunctions and their first derivatives leads to the equations II ( = III & II A sin k = Ce λ + De λ = & ( Ak cos k = λce λ λde λ III Taking the ratio of these equations yields " II II = " III III

9 5 c Continued 9 Substituting, ka cos k A sin k = λ ( Ceλ De λ Ce λ + De λ, or k cot k = λ ( Ceλ De λ Ce λ + De λ The boundary conditions at x = Q are III ( Q = IV ( Q and # III ( Q = # IV ( Q Substituting the forms of the wavefunctions and their first derivatives leads to the equations III ( Q = IV ( Q Ce λq + De λq = Fe ikq & III ( Q = & ( Q λce λq λde λq = ikfe ikq IV Taking the ratio of these equations gives " III Q III Q = " IV Q Q IV Substituting, λ( Ce λq De λq Ce λq + De λq = ikfeiλq Fe iλq, or λ( Ce λq De λq Ce λq + De λq = ik Summary The two equations found from matching at x = and x = Q are k cot k = λ ( Ceλ De λ, Ce λ + De λ λ( Ce λq De λq Ce λq + De λq = ik These equations or their inverses are acceptable solutions

10 5 Continued 1 d Explain, without solving the equations, how you would go about calculating the transmission coefficient T for this system In this case, the transmission coefficient may be defined as T = Coefficient of wave in region IV Coefficient of wave in region II 2 In this case, the transmission coefficient is defined as T = F A 2 Therefore an expression for the ratio of coefficients F/A must be obtained The two matching equations involving ratios from part (c do not contain the coefficients F or A; therefore those equations may not be used directly to determine the transmission coefficient ooking at the matching equations without taking ratios, we see that the wavefunction matching equation at x = includes the coefficient A, A sin k = Ce λ + De λ, or A = Ceλ + De λ sin k In addition, the wavefunction matching equation at x = Q includes the coefficient F, Ce λq + De λq = Fe ikq, or F = CeλQ + De λq e ikq These two equations may be combined to obtain the ratio F/A, F A sin k Ce λq + De λq = e ikq Ce λ + De λ This equation includes the constants C and D These may be determined using the ratios of the matching equations determined in part (c by solving the two equations for the two unknowns C and D Once the constants C and D are determined, they may be used to calculate the ratio F/A and thus the transmission coefficient

11 5 continued 11 e Without solving any equations, sketch the expected form of the ground state wavefunction for the system given above for E < V Qualitatively compare and contrast this wavefunction with the ground state wavefunction for the particle in a half-infinite well of the same width Pay particular attention to the shape of the wavefunctions in the various regions Example graphs are shown below for the ground state wavefunctions for the system given in this problem (with a finite width barrier and the particle in a half-infinite well These wavefunctions were constructed using a box width of =3 bohr for both systems, and a barrier width of 15 bohr for the finite barrier (ie, =3 and Q=45 bohr In both cases the barrier height V is 45 hartrees This Problem Ground State (n=1 Half-Infinite Well (x x (bohr 1 (x x (bohr In general, the shapes of the wavefunctions for the two systems are similar (note that the wavefunctions are not normalized, so the scale on the y-axis for each is arbitrary Since the wavefunctions correspond to ground states, there are no nodes in the wavefunctions within the well or barrier regions In addition, both wavefunctions exhibit similar tunneling behavior in classically forbidden region III The differences between the two functions are most distinct in region IV Here, the barrier in the halfinfinite well continues to infinity, whereas in this problem the finite width of the barrier means that the potential drops to in region IV (for x>45 bohr in the example, or x>q in general The particle in the half-infinite well has a wavefunction that continues to die off monotonically to as x increases On the other hand, the particle trapped behind the finite barrier becomes free in region IV, and the wavefunction therefore exhibits free particle-like behavior in this region The oscillations of the free particle-like wavefunction in region IV are low amplitude as a result of the small probability density for tunneling through the barrier However, the oscillations in the wavefunction continue undamped with wavelength 2π k for all x>q