1. Transform (using the coordinate system provided below) the following functions accordingly: Z Θ

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1 Section Exercises, Problems, and Solutions Review Exercises. Transform (using the coordinate system provided below) the following functions accordingly: Z Θ r Y X φ a. from cartesian to spherical polar coordinates 3x + y - 4z = b. from cartesian to cylindrical coordinates y + z = 9 c. from spherical polar to cartesian coordinates r = Sinθ Cosφ. Perform a separation of variables and indicate the general solution for the following expressions: a. 9x + 6y y x = b. y + y x + 6 = 3. Find the eigenvalues and corresponding eigenvectors of the following matrices: a. - b. - -

2 4. For the hermitian matrix in review exercise 3a show that the eigenfunctions can be normalized and that they are orthogonal. 5. For the hermitian matrix in review exercise 3b show that the pair of degenerate eigenvalues can be made to have orthonormal eigenfunctions. 6. Solve the following second order linear differential equation subject to the specified "boundary conditions": d x dt Exercises + k x(t) =, where x(t=) =, and dx(t=) dt. Replace the following classical mechanical expressions with their corresponding quantum mechanical operators. =. a. K.E. = mv in three-dimensional space. b. p = mv, a three-dimensional cartesian vector. c. y-component of angular momentum: y = zp x - xp z.. Transform the following operators into the specified coordinates: a. x = h i y z - z y b. z = hī φ from cartesian to spherical polar coordinates. from spherical polar to cartesian coordinates. 3. Match the eigenfunctions in column B to their operators in column A. What is the eigenvalue for each eigenfunction? Column A Column B i. (-x ) d dx - x d dx ii. d dx 5x 4 iii. x d dx 4x 4 - x + 3 e 3x + e -3x iv. d dx - x d dx v. x d dx + (-x) d dx 4x 3-3x x - 4x + 4. Show that the following operators are hermitian. a. P x b. x 5. For the following basis of functions (Ψ p-, Ψ p, and Ψ p+ ), construct the matrix representation of the x operator (use the ladder operator representation of x ). Verify that

3 the matrix is hermitian. Find the eigenvalues and corresponding eigenvectors. Normalize the eigenfunctions and verify that they are orthogonal. Ψ p- = 8π / Z a 5/ re -zr/a Sinθ e -iφ Ψ po = π / Z a 5/ re -zr/a Cosθ Ψ p = 8π / Z a 5/ re -zr/a Sinθ e iφ 6. Using the set of eigenstates (with corresponding eigenvalues) from the preceding problem, determine the probability for observing a z-component of angular momentum equal to h - if the state is given by the x eigenstate with h - x eigenvalue. 7. Use the following definitions of the angular momentum operators: x = h i y z - z y z = h i x y - y x and the relationships:, y = h i z x - x z, and = x + y + z, [x,p x ] = ih, [y,p y ] = ih, and [z,p z ] = ih, to demonstrate the following operator identities: a. [ x, y ] = ih z, b. [ y, z ] = ih x, c. [ z, x ] = ih y, d. [ x, ] =, e. [ y, ] =, f. [ z, ] =. 8. In exercise 7 above you determined whether or not many of the angular momentum operators commute. Now, examine the operators below along with an appropriate given function. Determine if the given function is simultaneously an eigenfunction of both operators. Is this what you expected? a. z,, with function: Y (θ,φ) =. 4π b. x, z, with function: Y (θ,φ) =. 4π c. z, 3, with function: Y (θ,φ) = 4π Cosθ.,

4 d. x, z, with function: Y (θ,φ) = 3 4π Cosθ. 9. For a "particle in a box" constrained along two axes, the wavefunction Ψ(x,y) as given in the text was : in x πx -in x πx in y πy -in y πy Ψ(x,y) = x y e x - e x e y - e y, with n x and n y =,,3,... Show that this wavefunction is normalized.. Using the same wavefunction, Ψ(x,y), given in exercise 9 show that the expectation value of p x vanishes.. Calculate the expectation value of the x operator for the first two states of the harmonic oscillator. Use the v= and v= harmonic oscillator wavefunctions given below which are normalized such that Ψ(x) dx =. Remember that Ψ = α π /4 e -αx / and Ψ = 4α 3 π /4 xe -αx /.. For each of the one-dimensional potential energy graphs shown below, determine: a. whether you expect symmetry to lead to a separation into odd and even solutions, b. whether you expect the energy will be quantized, continuous, or both, and c. the boundary conditions that apply at each boundary (merely stating that Ψ and/or Ψ is continuous is all that is necessary). x

5 3. Consider a particle of mass m moving in the potential: V(x) = for x < Region I V(x) = for x Region II V(x) = V(V > ) for x > Region III a. Write the general solution to the Schrödinger equation for the regions I, II, III, assuming a solution with energy E < V (i.e. a bound state). b. Write down the wavefunction matching conditions at the interface between regions I and II and between II and III. c. Write down the boundary conditions on Ψ for x ±. d. Use your answers to a. - c. to obtain an algebraic equation which must be satisfied for the bound state energies, E.

6 e. Demonstrate that in the limit V, the equation you obtained for the bound state energies in d. gives the energies of a particle in an infinite box; E n = n h π m ; n =,,3,... Problems. A particle of mass m moves in a one-dimensional box of length, with boundaries at x = and x =. Thus, V(x) = for x, and V(x) = elsewhere. The normalized eigenfunctions of the Hamiltonian for this system are given by Ψ n (x) = / nπx Sin, with E n = n π h, where the quantum number n can take on the values n=,,3,... m a. Assuming that the particle is in an eigenstate, Ψ n (x), calculate the probability that the particle is found somewhere in the region x 4. Show how this probability depends on n. b. For what value of n is there the largest probability of finding the particle in x 4? c. Now assume that Ψ is a superposition of two eigenstates, Ψ = aψ n + bψ m, at time t =. What is Ψ at time t? What energy expectation value does Ψ have at time t and how does this relate to its value at t =? d. For an experimental measurement which is capable of distinguishing systems in state Ψ n from those in Ψ m, what fraction of a large number of systems each described by Ψ will be observed to be in Ψ n? What energies will these experimental measurements find and with what probabilities? e. For those systems originally in Ψ = aψ n + bψ m which were observed to be in Ψ n at time t, what state (Ψ n, Ψ m, or whatever) will they be found in if a second experimental measurement is made at a time t' later than t? f. Suppose by some method (which need not concern us at this time) the system has been prepared in a nonstationary state (that is, it is not an eigenfunction of H). At the time of a measurement of the particle's energy, this state is specified by the normalized wavefunction Ψ = 3 / x(-x) for x, and Ψ = elsewhere. What is the 5 probability that a measurement of the energy of the particle will give the value E n = n π h m for any given value of n? g. What is the expectation value of H, i.e. the average energy of the system, for the wavefunction Ψ given in part f?. Show that for a system in a non-stationary state,

7 Ψ = C j Ψ j e -ie j t/h-, the average value of the energy does not vary with time but the j expectation values of other properties do vary with time. 3. A particle is confined to a one-dimensional box of length having infinitely high walls and is in its lowest quantum state. Calculate: <x>, <x >, <p>, and <p >. Using the definition Α = (<A > <A> ) /, to define the uncertainty, A, calculate x and p. Verify the Heisenberg uncertainty principle that x p h /. 4. It has been claimed that as the quantum number n increases, the motion of a particle in a box becomes more classical. In this problem you will have an oportunity to convince yourself of this fact. a. For a particle of mass m moving in a one-dimensional box of length, with ends of the box located at x = and x =, the classical probability density can be shown to be independent of x and given by P(x)dx = dx regardless of the energy of the particle. Using this probability density, evaluate the probability that the particle will be found within the interval from x = to x = 4. b. Now consider the quantum mechanical particle-in-a-box system. Evaluate the probability of finding the particle in the interval from x = to x = 4 for the system in its nth quantum state. c. Take the limit of the result you obtained in part b as n. How does your result compare to the classical result you obtained in part a? 5. According to the rules of quantum mechanics as we have developed them, if Ψ is the state function, and φ n are the eigenfunctions of a linear, Hermitian operator, A, with eigenvalues a n, Aφ n = a n φ n, then we can expand Ψ in terms of the complete set of eigenfunctions of A according to Ψ = c n φ n, where c n = φ n * Ψ dτ. Furthermore, the n probability of making a measurement of the property corresponding to A and obtaining a value a n is given by c n, provided both Ψ and φ n are properly normalized. Thus, P(a n ) = c n. These rules are perfectly valid for operators which take on a discrete set of eigenvalues, but must be generalized for operators which can have a continuum of eigenvalues. An example of this latter type of operator is the momentum operator, p x, which has eigenfunctions given by φ p (x) = Ae ipx/h- where p is the eigenvalue of the p x operator and A is a normalization constant. Here p can take on any value, so we have a continuous spectrum of eigenvalues of p x. The obvious generalization to the equation for Ψ is to convert the sum over discrete states to an integral over the continuous spectrum of states: Ψ(x) = C(p)φ p (x)dp = C(p)Ae ipx/h- dp

8 The interpretation of C(p) is now the desired generalization of the equation for the probability P(p)dp = C(p) dp. This equation states that the probability of measuring the momentum and finding it in the range from p to p+dp is given by C(p) dp. Accordingly, the probability of measuring p and finding it in the range from p to p is given by p p P(p)dp = C(p) * C(p)dp. C(p) is thus the probability amplitude for finding the particle p p with momentum between p and p+dp. This is the momentum representation of the wavefunction. Clearly we must require C(p) to be normalized, so that C(p) * C(p)dp =. With this restriction we can derive the normalization constant A =, giving a direct πh relationship between the wavefunction in coordinate space, Ψ(x), and the wavefunction in momentum space, C(p): Ψ(x) = C(p)e ipx/h- dp, πh and by the fourier integral theorem: C(p) = Ψ(x)e ipx/h- dx. πh ets use these ideas to solve some problems focusing our attention on the harmonic oscillator; a particle of mass m moving in a one-dimensional potential described by V(x) = kx. a. Write down the Schrödinger equation in the coordinate representation. b. Now lets proceed by attempting to write the Schrödinger equation in the momentum representation. Identifying the kinetic energy operator T, in the momentum representation is quite straightforward T = p m = - Error!. Writing the potential, V(x), in the momentum representation is not quite as straightforward. The relationship between position and momentum is realized in their commutation relation [x,p] = ih, or (xp - px) = ih This commutation relation is easily verified in the coordinate representation leaving x untouched (x = x. ) and using the above definition for p. In the momentum representation we want to leave p untouched (p = p. ) and define the operator x in such a manner that the commutation relation is still satisfied. Write the operator x in the momentum representation. Write the full Hamiltonian in the momentum representation and hence the Schrödinger equation in the momentum representation. c. Verify that Ψ as given below is an eigenfunction of the Hamiltonian in the coordinate representation. What is the energy of the system when it is in this state?

9 Determine the normalization constant C, and write down the normalized ground state wavefunction in coordinate space. Ψ(x) = C exp (- mk x h ). d. Now consider Ψ in the momentum representation. Assuming that an eigenfunction of the Hamiltonian may be found of the form Ψ(p) = C exp (-αp ), substitute this form of Ψ into the Schrödinger equation in the momentum representation to find the value of α which makes this an eigenfunction of H having the same energy as Ψ(x) had. Show that this Ψ(p) is the proper fourier transform of Ψ(x). The following integral may be useful: e -βx π Cosbxdx = β e-b /4β. Since this Hamiltonian has no degenerate states, you may conclude that Ψ(x) and Ψ(p) represent the same state of the system if they have the same energy. 6. The energy states and wavefunctions for a particle in a 3-dimensional box whose lengths are,, and 3 are given by E(n,n,n 3 ) = h 8m n + n + n 3 and 3 Ψ(n,n,n 3 ) = Sin n πx 3 Sin n πy Sin n 3 πz. 3 These wavefunctions and energy levels are sometimes used to model the motion of electrons in a central metal atom (or ion) which is surrounded by six ligands. a. Show that the lowest energy level is nondegenerate and the second energy level is triply degenerate if = = 3. What values of n, n, and n 3 characterize the states belonging to the triply degenerate level? b. For a box of volume V = 3, show that for three electrons in the box (two in the nondegenerate lowest "orbital", and one in the next), a lower total energy will result if the box undergoes a rectangular distortion ( = 3 ). which preserves the total volume than if the box remains undistorted (hint: if V is fixed and =, then 3 = V and is the only "variable"). c. Show that the degree of distortion (ratio of 3 to ) which will minimize the total energy is 3 =. How does this problem relate to Jahn-Teller distortions? Why (in terms of the property of the central atom or ion) do we do the calculation with fixed volume? d. By how much (in ev) will distortion lower the energy (from its value for a cube, = = 3 ) if V = 8 Å 3 and h 8m = 6. x -7 erg cm. ev =.6 x - erg 7. The wavefunction Ψ = Ae -a x is an exact eigenfunction of some one-dimensional Schrödinger equation in which x varies from to. The value of a is: a = (Å) -. For

10 now, the potential V(x) in the Hamiltonian (H = - h d m + V(x)) for which Ψ(x) is an dx eigenfunction is unknown. a. Find a value of A which makes Ψ(x) normalized. Is this value unique? What units does Ψ(x) have? b. Sketch the wavefunction for positive and negative values of x, being careful to show the behavior of its slope near x =. Recall that x is defined as: x = x if x > -x if x < c. Show that the derivative of Ψ(x) undergoes a discontinuity of magnitude (a) 3/ as x goes through x =. What does this fact tell you about the potential V(x)? d. Calculate the expectation value of x for the above normalized wavefunction (obtain a numerical value and give its units). What does this expectation value give a measure of? e. The potential V(x) appearing in the Schrödinger equation for which Ψ = Ae -a x is an exact solution is given by V(x) = h a m δ(x). Using this potential, compute the expectation value of the Hamiltonian (H = - h d m + V(x)) for your normalized dx wavefunction. Is V(x) an attractive or repulsive potential? Does your wavefunction correspond to a bound state? Is <H> negative or positive? What does the sign of <H> tell you? To obtain a numerical value for <H> use h m = 6.6 x -8 erg cm and ev =.6 x - erg. f. Transform the wavefunction, Ψ = Ae -a x, from coordinate space to momentum space. g. What is the ratio of the probability of observing a momentum equal to ah to the probability of observing a momentum equal to -ah? 8. The π-orbitals of benzene, C 6 H 6, may be modeled very crudely using the wavefunctions and energies of a particle on a ring. ets first treat the particle on a ring problem and then extend it to the benzene system. a. Suppose that a particle of mass m is constrained to move on a circle (of radius r) in the xy plane. Further assume that the particle's potential energy is constant (zero is a good choice). Write down the Schrödinger equation in the normal cartesian coordinate representation. Transform this Schrödinger equation to cylindrical coordinates where x = rcosφ, y = rsinφ, and z = z (z = in this case). Taking r to be held constant, write down the general solution, Φ(φ), to this Schrödinger equation. The "boundary" conditions for this problem require that Φ(φ) = Φ(φ + π). Apply this boundary condition to the general solution. This results in the quantization of the energy levels of this system. Write down the final expression for the normalized wavefunction and quantized energies. What is the physical significance of these quantum

11 numbers which can have both positive and negative values? Draw an energy diagram representing the first five energy levels. b. Treat the six π-electrons of benzene as particles free to move on a ring of radius.4 Å, and calculate the energy of the lowest electronic transition. Make sure the Pauli principle is satisfied! What wavelength does this transition correspond to? Suggest some reasons why this differs from the wavelength of the lowest observed transition in benzene, which is 6 Å. 9. A diatomic molecule constrained to rotate on a flat surface can be modeled as a planar rigid rotor (with eigenfunctions, Φ(φ), analogous to those of the particle on a ring) with fixed bond length r. At t =, the rotational (orientational) probability distribution is 4 observed to be described by a wavefunction Ψ(φ,) = 3π Cos φ. What values, and with what probabilities, of the rotational angular momentum, -ih φ, could be observed in this system? Explain whether these probabilities would be time dependent as Ψ(φ,) evolves into Ψ(φ,t).. A particle of mass m moves in a potential given by V(x,y,z) = k (x + y + z ) = kr. a. Write down the time-independent Schrödinger equation for this system. b. Make the substitution Ψ(x,y,z) = X(x)Y(y)Z(z) and separate the variables for this system. c. What are the solutions to the resulting equations for X(x), Y(y), and Z(z)? d. What is the general expression for the quantized energy levels of this system, in terms of the quantum numbers n x, n y, and n z, which correspond to X(x), Y(y), and Z(z)? e. What is the degree of degeneracy of a state of energy E = 5.5h k m for this system? f. An alternative solution may be found by making the substitution Ψ(r,θ,φ) = F(r)G(θ,φ). In this substitution, what are the solutions for G(θ,φ)? g. Write down the differential equation for F(r) which is obtained when the substitution Ψ(r,θ,φ) = F(r)G(θ,φ) is made. Do not solve this equation.. Consider an N molecule, in the ground vibrational level of the ground electronic state, which is bombarded by ev electrons. This leads to ionization of the N molecule to form N +. In this problem we will attempt to calculate the vibrational distribution of the newly-formed N + ions, using a somewhat simplified approach. a. Calculate (according to classical mechanics) the velocity (in cm/sec) of a ev electron, ignoring any relativistic effects. Also calculate the amount of time required for a ev electron to pass an N molecule, which you may estimate as having a length of Å. b. The radial Schrödinger equation for a diatomic molecule treating vibration as a harmonic oscillator can be written as:

12 - h µr r r Ψ r + k (r - r e) Ψ = E Ψ, Substituting Ψ(r) = F(r) r, this equation can be rewritten as: - h µ r F(r) + k (r - r e) F(r) = E F(r). The vibrational Hamiltonian for the ground electronic state of the N molecule within this approximation is given by: H(N ) = - h d µ dr + k N (r - r N ), where r N and k N have been measured experimentally to be: r N =.9769 Å; k g N =.94 x 6 sec. The vibrational Hamiltonian for the N + ion, however, is given by : H(N ) = - h d µ dr + k N + (r - r N + ), where r N + and k N + have been measured experimentally to be: r N + =.64 Å; k N + =.9 x 6 g sec. In both systems the reduced mass is µ =.64 x -3 g. Use the above information to write out the ground state vibrational wavefunctions of the N and N+ molecules, giving explicit values for any constants which appear in them. Note: For this problem use the "normal" expression for the ground state wavefunction of a harmonic oscillator. You need not solve the differential equation for this system. c. During the time scale of the ionization event (which you calculated in part a), the vibrational wavefunction of the N molecule has effectively no time to change. As a result, the newly-formed N+ ion finds itself in a vibrational state which is not an eigenfunction of the new vibrational Hamiltonian, H(N + ). Assuming that the N molecule was originally in its v= vibrational state, calculate the probability that the N + ion will be produced in its v= vibrational state.. The force constant, k, of the C-O bond in carbon monoxide is.87 x 6 g/sec. Assume that the vibrational motion of CO is purely harmonic and use the reduced mass µ = amu. a. Calculate the spacing between vibrational energy levels in this molecule, in units of ergs and cm -. b. Calculate the uncertainty in the internuclear distance in this molecule, assuming it is in its ground vibrational level. Use the ground state vibrational wavefunction (Ψ v= ), and calculate <x>, <x >, and x = (<x > - <x> ) /.

13 c. Under what circumstances (i.e. large or small values of k; large or small values of µ) is the uncertainty in internuclear distance large? Can you think of any relationship between this observation and the fact that helium remains a liquid down to absolute zero? 3. Suppose you are given a trial wavefunction of the form: φ = Z e 3 πa 3 exp -Z e r a exp -Z e r a to represent the electronic structure of a two-electron ion of nuclear charge Z and suppose that you were also lucky enough to be given the variational integral, W, (instead of asking you to derive it!): W = Z e - ZZ e Z e e a. a. Find the optimum value of the variational parameter Z e for an arbitrary nuclear charge Z by setting dw dz =. Find both the optimal value of Z e and the resulting value of e W. b. The total energies of some two-electron atoms and ions have been experimentally determined to be: Z = H ev Z = He ev Z = 3 i ev Z = 4 Be ev Z = 5 B ev Z = 6 C ev Z = 7 N ev Z = 8 O ev Using your optimized expression for W, calculate the estimated total energy of each of these atoms and ions. Also calculate the percent error in your estimate for each ion. What physical reason explains the decrease in percentage error as Z increases? c. In 98, when quantum mechanics was quite young, it was not known whether the isolated, gas-phase hydride ion, H -, was stable with respect to dissociation into a hydrogen atom and an electron. Compare your estimated total energy for H - to the ground state energy of a hydrogen atom and an isolated electron (system energy = -3.6 ev), and show that this simple variational calculation erroneously predicts H - to be unstable. (More complicated variational treatments give a ground state energy of H - of ev, in agreement with experiment.) 4. A particle of mass m moves in a one-dimensional potential given by H = - h d m dx + a x, where the absolute value function is defined by x = x if x and x = -x if x. a. Use the normalized trial wavefunction φ = b 4 e -bx to estimate the energy of π the ground state of this system, using the variational principle to evaluate W(b).

14 b. Optimize b to obtain the best approximation to the ground state energy of this system, using a trial function of the form of φ, as given above. The numerically calculated exact ground state energy is.8866 h3 m - 3 a - 3. What is the percent error in your value? 5. The harmonic oscillator is specified by the Hamiltonian: H = - h d m dx + kx. Suppose the ground state solution to this problem were unknown, and that you wish to approximate it using the variational theorem. Choose as your trial wavefunction, 5 φ = 6 a-5 (a - x ) for -a < x < a φ = for x a where a is an arbitrary parameter which specifies the range of the wavefunction. Note that φ is properly normalized as given. a. Calculate φ * Hφdx and show it to be given by: φ * Hφdx = 5 h 4 ma + ka 4. b. Calculate φ * Hφdx for a = b h km 4 with b =.,.4,.6,.8,.,.5,.,.5, 3., 4., and 5., and plot the result. c. To find the best approximation to the true wavefunction and its energy, find the minimum of φ * Hφdx by setting d da φ * Hφdx = and solving for a. Substitute this value into the expression for φ * Hφdx given in part a. to obtain the best approximation for the energy of the ground state of the harmonic oscillator. d. What is the percent error in your calculated energy of part c.? 6. Einstein told us that the (relativistic) expression for the energy of a particle having rest mass m and momentum p is E = m c 4 + p c. a. Derive an expression for the relativistic kinetic energy operator which contains terms correct through one higher order than the "ordinary" E = mc + p m

15 b. Using the first order correction as a perturbation, compute the first-order perturbation theory estimate of the energy for the s level of a hydrogen-like atom (general Z). Show the Z dependence of the result. 3 Note: Ψ(r) s = Z a e -Zr a and E π s = - Z me 4 h c. For what value of Z does this first-order relativistic correction amount to % of the unperturbed (non-relativistic) s energy? 7. Consider an electron constrained to move on the surface of a sphere of radius r. The Hamiltonian for such motion consists of a kinetic energy term only H = m e r, where is the orbital angular momentum operator involving derivatives with respect to the spherical polar coordinates (θ,φ). H has the complete set of eigenfunctions ψ () lm = Y l,m (θ,φ). a. Compute the zeroth order energy levels of this system. b. A uniform electric field is applied along the z-axis, introducing a perturbation V = -eεz = -eεr Cosθ, where ε is the strength of the field. Evaluate the correction to the energy of the lowest level through second order in perturbation theory, using the identity (l+m+)(l-m+) Cosθ Y l,m (θ,φ) = (l+)(l+3) Y l+,m (θ,φ) + (l+m)(l-m) (l+)(l-) Y l-,m(θ,φ). Note that this identity enables you to utilize the orthonormality of the spherical harmonics. c. The electric polarizability α gives the response of a molecule to an externally applied electric field, and is defined by α = - E ε where E is the energy in the presence ε= of the field and ε is the strength of the field. Calculate α for this system. d. Use this problem as a model to estimate the polarizability of a hydrogen atom, where r = a =.59 Å, and a cesium atom, which has a single 6s electron with r.6 Å. The corresponding experimental values are α H =.6668 Å 3 and α Cs = 59.6 Å An electron moving in a conjugated bond framework can be viewed as a particle in a box. An externally applied electric field of strength ε interacts with the electron in a fashion described by the perturbation V = eε x -, where x is the position of the electron in the box, e is the electron's charge, and is the length of the box. a. Compute the first order correction to the energy of the n= state and the first order wavefunction for the n= state. In the wavefunction calculation, you need only () compute the contribution to Ψ made by Ψ (). Make a rough (no calculation needed) () () sketch of Ψ + Ψ as a function of x and physically interpret the graph. b. Using your answer to part a. compute the induced dipole moment caused by the polarization of the electron density due to the electric field effect µ induced = - e Ψ * x - Ψdx. You may neglect the term proportional to ε ; merely obtain the term linear in ε.

16 c. Compute the polarizability, α, of the electron in the n= state of the box, and explain physically why α should depend as it does upon the length of the box. Remember that α = µ ε ε=. Review Exercises Solutions. The general relationships are as follows: Z Θ r Y X φ x = r Sinθ Cosφ r = x + y + z y = r Sinθ Sinφ Sinθ = z = r Cosθ Cosθ = Tanφ = y x a. 3x + y - 4z = 3(rSinθCosφ) + rsinθsinφ - 4(rCosθ) = r(3sinθcosφ + SinθSinφ - 4Cosθ) = b. x = rcosφ r = x +y y = rsinφ z = z Tanφ = y x x + y x + y + z z x + y + z

17 y + z = 9 r Sin φ + z = 9 c. r = SinθCosφ r = x r r = x x +y + z = x x - x +y + z = x - x + + y + z = (x - ) + y + z =. a. 9x + 6y y x = 6ydy = -9xdx 6 y = - 9 x + c 6y = -9x + c' y 9 + x 6 = c'' (general equation for an ellipse) b. y + y x + 6 = y + 6 = - dy dx y + 3 = - dy dx -dx = dy y + 3 -x = ln(y + 3) + c c'e -x = y + 3 y = c'e -x a. First determine the eigenvalues: det - - λ - λ = (- - λ)( - λ) - = - + λ - λ + λ - 4 = λ - λ - 6 = (λ - 3)(λ + ) = λ = 3 or λ = -. Next, determine the eigenvectors. First, the eigenvector associated with eigenvalue -: - C C = - C C

18 -C + C = -C C = -C (Note: The second row offers no new information, e.g. C + C = -C ) C + C = (from normalization) (-C ) + C = 4C + C = 5C = C =. C =., and therefore C = -.. For the eigenvector associated with eigenvalue 3: - C C = 3 C C -C + C = 3C -4C = -C C =.5C (again the second row offers no new information) C + C = (from normalization) (.5C ) + C =.5C + C =.5C = C =.8 C =.8 =., and therefore C =.. Therefore the eigenvector matrix becomes: b. First determine the eigenvalues: det - - λ - - λ - λ det [ - - λ ] det - - λ - λ = From 3a, the solutions then become -, -, and 3. Next, determine the eigenvectors. First the eigenvector associated with eigenvalue 3 (the third root): - C - C C C = 3 C 3 C 3 - C 3 = 3C 3 (row one) C 3 = -C 3 + C 33 = 3C 3 (row two) C 33 = 4C 3 C 33 = C 3 (again the third row offers no new information) C 3 + C 3 + C 33 = (from normalization) + C 3 + (C 3 ) = =

19 5C 3 = C 3 =., and therefore C 33 =.. Next, find the pair of eigenvectors associated with the degenerate eigenvalue of -. First, root one eigenvector one: -C = -C (no new information from row one) -C + C 3 = -C (row two) C = -C 3 (again the third row offers no new information) C + C + C 3 = (from normalization) C + (-C 3 ) + C 3 = C + 5C 3 = C = - 5C 3 (Note: There are now two equations with three unknowns.) Second, root two eigenvector two: -C = -C (no new information from row one) -C + C 3 = -C (row two) C = -C 3 (again the third row offers no new information) C + C + C 3 = (from normalization) C + (-C 3 ) + C 3 = C + 5C 3 = C = - 5C 3 (Note: Again there are now two equations with three unknowns) C C + C C + C 3 C 3 = (from orthogonalization) Now there are five equations with six unknowns. Arbitrarily choose C = C = = - 5C 3 5C 3 = C 3 =. C = -. C C + C C + C 3 C 3 = (from orthogonalization) + -.(-C 3 ) +. C 3 = 5C 3 = C 3 =, C =, and C = Therefore the eigenvector matrix becomes: Show: <φ φ > =, <φ φ > =, and <φ φ > = <φ φ > =? (-. ) + (. ) =? 4(.) +. =?

20 .8 +. =? = <φ φ > =? (. ) + (. ) =?. + 4(.) =?. +.8 =? = <φ φ > = <φ φ > =? =? -(.) + (.) =? =? = 5. Show (for the degenerate eigenvalue; λ = -): <φ φ > =, <φ φ > =, and <φ φ > = <φ φ > =? + (-. ) + (. ) =? 4(.) +. =?.8 +. =? = <φ φ > =? + + =? = <φ φ > = <φ φ > =? ()() + (-. )() + (. )() =? = 6. Suppose the solution is of the form x(t) = e αt, with α unknown. Inserting this trial solution into the differential equation results in the following: d dt eαt + k e αt = α e αt + k e αt = (α + k ) x(t) = (α + k ) = α = -k

21 α = -k α = ± ik Solutions are of the form e ikt, e -ikt, or a combination of both: x(t) = C e ikt + C e -ikt. Euler's formula also states that: e ±iθ = Cosθ ± isinθ, so the previous equation for x(t) can also be written as: x(t) = C {Cos(kt) + isin(kt)} + C {Cos(kt) - isin(kt)} x(t) = (C + C )Cos(kt) + (C + C )isin(kt), or alternatively x(t) = C 3 Cos(kt) + C 4 Sin(kt). We can determine these coefficients by making use of the "boundary conditions". at t =, x() = x() = C 3 Cos() + C 4 Sin() = C 3 = at t =, dx() dt = d dt x(t) = d dt (C 3Cos(kt) + C 4 Sin(kt)) d dt x(t) = -C 3kSin(kt) + C 4 kcos(kt) d dt x() = = -C 3kSin() + C 4 kcos() C 4 k = C 4 = The solution is of the form: x(t) = Cos(kt) Exercises. a. K.E. = mv = m mv m K.E. = m (p x + p y + p z ) = (mv) m = p m K.E. = m h i + x h i + y h i z K.E. = -h m x + y + z b. p = mv = ip x + jp y + kp z h p = i i x + j h iy + k h iz where i, j, and k are unit vectors along the x, y, and z c. y = zp x - xp z h y = z i x - x h i z axes.

22 . First derive the general formulas for x, y, z in terms of r,θ, and φ, and r, θ, and φ in terms of x,y, and z. The general relationships are as follows: x = r Sinθ Cosφ r = x + y + z y = r Sinθ Sinφ Sinθ = z = r Cosθ Cosθ = Tanφ = y x x + y x + y + z z x + y + z First x, y, and from the chain rule: z x = r x y,z r + θ x y,z θ + φ x y,z φ, y = r y x,z r + θ y x,z θ + φ y x,z φ, z = r z x,y r + θ z x,y θ + φ z x,y φ. Evaluation of the many "coefficients" gives the following: r x = Sinθ Cosφ, y,z θ x Cosθ Cosφ = y,z r, φ x = - Sinφ y,z r Sinθ, r y = Sinθ Sinφ, x,z θ y Cosθ Sinφ = x,z r, φ y = Cosφ x,z r Sinθ, r z = Cosθ, x,y θ z = - Sinθ x,y r, and φ z =. x,y Upon substitution of these "coefficients": Cosθ Cosφ + r θ - Sinφ r Sinθ φ, x = Sinθ Cosφ r y = Sinθ Sinφ r + Cosθ Sinφ r z = Cosθ r - Sinθ r θ + φ. Next r, θ, and from the chain rule: φ r = x r θ,φ x + y r θ,φ y + z r θ + Cosφ r Sinθ θ,φ z, φ, and

23 θ = x θ r,φ x + y θ r,φ y + z θ r,φ z, and φ = x φ r,θ x + y φ r,θ y + z φ r,θ z. Again evaluation of the the many "coefficients" results in: x r = x, θ,φ x + y + z y r = y, θ,φ x + y + z z r = z, θ,φ x + y + z x θ = x z, r,φ x + y y θ = y z, r,φ x + y z θ = - r,φ x + y, x φ = -y, r,θ y φ = x, and r,θ z φ = r,θ Upon substitution of these "coefficients": r = x x + y + z x + y x + y + z y z + x + y + z z θ = x z x + y x + y z x + y y - x + y z φ = -y x + x y + z. Note, these many "coefficients" are the elements which make up the Jacobian matrix used whenever one wishes to transform a function from one coordinate representation to another. One very familiar result should be in transforming the volume element dxdydz to r Sinθdrdθdφ. For example: f(x,y,z)dxdydz = f(x(r,θ,φ),y(r,θ,φ),z(r,θ,φ)) a. x = h i y z - z y x r θφ x θ rφ x φ rθ y r θφ y θ rφ y φ rθ z r θφ z θ rφ z φ rθ x = h i rsinθsinφ Cosθ r - Sinθ r θ drdθdφ

24 b. z = h i - h i rcosθ SinθSinφ r + CosθSinφ r x = - h i Sinφ z = h i φ = - ih φ -y x + x y θ + CotθCosφ φ θ + Cosφ rsinθ φ 3. B B' B'' i. 4x 4 - x + 3 6x 3-4x 48x - 4 ii. 5x 4 x 3 6x iii. e 3x + e -3x 3(e 3x - e -3x ) 9(e 3x + e -3x ) iv. x - 4x + x - 4 v. 4x 3-3x x - 3 4x B(v.) is an eigenfunction of A(i.): (-x ) d dx - x d dx B(v.) = (-x ) (4x) - x (x - 3) 4x - 4x 3 - x 3 + 3x -36x 3 + 7x -9(4x 3-3x) (eigenvalue is -9) B(iii.) is an eigenfunction of A(ii.): d dx B(iii.) = 9(e 3x + e -3x ) (eigenvalue is 9) B(ii.) is an eigenfunction of A(iii.): x d dx B(ii.) = x (x 3 ) x 4 4(5x 4 ) (eigenvalue is 4) B(i.) is an eigenfunction of A(vi.): d dx - x d dx B(i) = (48x - 4) - x (6x 3-4x) 48x - 4-3x x -3x x - 4-8(4x 4 - x + 3) (eigenvalue is -8) B(iv.) is an eigenfunction of A(v.):

25 x d dx + (-x) d dx B(iv.) = x () + (-x) (x - 4) x + x x + 4x -x + 8x - 4 -(x - 4x +) (eigenvalue is -) 4. Show that: f * Agdτ = g(af) * dτ a. Suppose f and g are functions of x and evaluate the integral on the left hand side by "integration by parts": so: f(x) * (-ih x )g(x)dx f(x) * (-ih let dv = x g(x)dx and u = -ih f(x) * v = g(x) du = -ih x f(x)* dx Now, udv = uv - vdu, x )g(x)dx = -ih f(x) * g(x) + ih g(x) x f(x)* dx. Note that in, principle, it is impossible to prove hermiticity unless you are given knowledge of the type of function on which the operator is acting. Hermiticity requires (as can be seen in this example) that the term -ih f(x) * g(x) vanish when evaluated at the integral limits. This, in general, will occur for the "well behaved" functions (e.g., in bound state quantum chemistry, the wavefunctions will vanish as the distances among particles approaches infinity). So, in proving the hermiticity of an operator, one must be careful to specify the behavior of the functions on which the operator is considered to act. This means that an operator may be hermitian for one class of functions and non-hermitian for another class of functions. If we assume that f and g vanish at the boundaries, then we have f(x) * (-ih x )g(x)dx = g(x) -ih x f(x) * dx b. Suppose f and g are functions of y and z and evaluate the integral on the left hand side by "integration by parts" as in the previous exercise: f(y,z) * -ih y z - z y g(y,z)dydz = f(y,z) * -ih y z g(y,z)dydz - f(y,z) * -ih z y g(y,z)dydz For the first integral, f(z) * -ih y z g(z)dz,

26 let dv = z g(z)dz u = -ih yf(z) * v = g(z) du = -ih y z f(z)* dz so: f(z) * (-ih y z )g(z)dz = -ih yf(z) * g(z) + ih y g(z) z f(z)* dz = g(z) -ih y z f(z) * dz. For the second integral, f(y) * -ih z y g(y)dy, so: let dv = y g(y)dy u = -ih zf(y) * v = g(y) du = -ih z y f(y)* dy f(y) * (-ih z y )g(y)dy = -ih zf(y) * g(y) + ih z g(y) y f(y)* dy = g(y) -ih z y f(y) * dy f(y,z) * -ih y z - z y g(y,z)dydz = g(z) -ih y z f(z) * dz - g(y) -ih z y f(y) * dy = g(y,z) -ih y z - z y f(y,z) * dydz. Again we have had to assume that the functions f and g vanish at the boundary = x + i y - = x - i y, so = x, or x = ( ) + Y l,m = l(l + ) - m(m + ) h Y l,m+ - Y l,m = l(l + ) - m(m - ) h Y l,m- Using these relationships: - Ψ p- =, - Ψ p = h Ψ p-, - Ψ p+ = h Ψ p

27 + Ψ p- = h Ψ p, + Ψ p = h Ψ p+, + Ψ p+ =, and the following x matrix elements can be evaluated: This matrix: x (,) = < Ψ p- > x (,) = < Ψ p- > x (,3) = < Ψ p- > x (,) = < Ψ p > x (,) = < Ψ p > x (,3) = < Ψ p > x (3,) = < Ψ p+ > x (3,) = < Ψ p+ > x (3,3) = - λ h h - λ - λ ( ) Ψ p- = ( ) Ψ p = ( ) Ψ p+ = ( ) Ψ p- = ( ) Ψ p = ( ) Ψ p+ = h ( ) Ψ p- = ( ) Ψ p = h h h h h - λ h (-λ) - h - λ Expanding these determinants yields: (λ - h )(-λ) - h (-λ) h h h h h, can now be diagonalized: = = h h - λ h =

28 -λ(λ - h ) = -λ(λ - h )(λ + h ) = with roots:,h, and -h Next, determine the corresponding eigenvectors: For λ = : h h h h C C C 3 = h C = (row one) C = h C + h C 3 = (row two) C + C 3 = C = -C 3 C + C + C 3 = (normalization) C + (-C ) = C = C =, C =, and C 3 = - C C C 3 For λ = h : h h h h C C C 3 = h h C = h C (row one) C = C h C + h C 3 = h C (row two) C + C 3 = C C + C 3 = C C C C 3

29 For λ = -h : C 3 = C C 3 = C C + C + C 3 = (normalization) C + C + C = C + C + C = C = C = C =, C =, and C 3 = h h h h h C 3 = -h C 3 (row one) C 3 = - C 3 h C C 3 + C 3 C 3 C 33 = -h h C 33 = -h C 3 (row two) C 33 = -C 3 - C 3 + C 33 = -C 3 C 33 = - C 3 C 33 = - C 3 C 3 + C 3 + C 33 = (normalization) - C 3 + C C 3 = C 3 + C 3 + C 3 = C 3 = C 3 C 3 C 33

30 C 3 = C 3 = -, C 3 =, and C 33 = - Show: <φ φ > =, <φ φ > =, <φ 3 φ 3 > =, <φ φ > =, <φ φ 3 > =, and <φ φ 3 > =. <φ φ > =? + + -? = + =? = <φ φ > =? + +? = =? = <φ 3 φ 3 > =? ? = =? = <φ φ > = <φ φ > =? + ( ) + - =? 4 = - 4 =? <φ φ 3 > = <φ 3 φ > =? - + ( ) =? = 4 =? <φ φ 3 > = <φ 3 φ > =? =?

31 =? = 6. P p+ = Ψ h x P p+ = < φ > p+ Ψ h x = φ p - - φ p + - < > φ p+ φ p+ = (or 5%) 7. It is useful here to use some of the general commutator relations found in Appendix C.V. a. [ x, y ] = [yp z - zp y, zp x - xp z ] = [yp z, zp x ] - [yp z, xp z ] - [zp y, zp x ] + [zp y, xp z ] = [y,z]p x p z + z[y,p x ]p z + y[p z,z]p x + yz[p z,p x ] - [y,x]p z p z - x[y,p z ]p z - y[p z,x]p z - yx[p z,p z ] - [z,z]p x p y - z[z,p x ]p y - z[p y,z]p x - zz[p y,p x ] + [z,x]p z p y + x[z,p z ]p y + z[p y,x]p z + zx[p y,p z ] As can be easily ascertained, the only non-zero terms are: [ x, y ] = y[p z,z]p x + x[z,p z ]p y = y(-ih )p x + x(ih )p y = ih (-yp x + xp y ) = ih z b. [ y, z ] = [zp x - xp z, xp y - yp x ] = [zp x, xp y ] - [zp x, yp x ] - [xp z, xp y ] + [xp z, yp x ] = [z,x]p y p x + x[z,p y ]p x + z[p x,x]p y + zx[p x,p z ] - [z,y]p x p x - y[z,p x ]p x - z[p x,y]p x - zy[p x,p x ] - [x,x]p y p z - x[x,p y ]p z - x[p z,x]p y - xx[p z,p y ] + [x,y]p x p z + y[x,p x ]p z + x[p z,y]p x + xy[p z,p x ] Again, as can be easily ascertained, the only non-zero terms are: [ y, z ] = z[p x,x]p y + y[x,p x ]p z = z(-ih )p y + y(ih )p z = ih (-zp y + yp z ) = ih x c. [ z, x ] = [xp y - yp x, yp z - zp y ] = [xp y, yp z ] - [xp y, zp y ] - [yp x, yp z ] + [yp x, zp y ] = [x,y]p z p y + y[x,p z ]p y + x[p y,y]p z + xy[p y,p z ] - [x,z]p y p y - z[x,p y ]p y - x[p y,z]p y - xz[p y,p y ] - [y,y]p z p x - y[y,p z ]p x - y[p x,y]p z - yy[p x,p z ] + [y,z]p y p x + z[y,p y ]p x + y[p x,z]p y + yz[p x,p y ] Again, as can be easily ascertained, the only non-zero terms are: [ z, x ] = x[p y,y]p z + z[y,p y ]p x

32 = x(-ih )p z + z(ih )p x = ih (-xp z + zp x ) = ih y d. [ x, ] = [ x, x + y + z ] = [ x, x ] + [ x, y ] + [ x, z ] = [ x, y ] + [ x, z ] = [ x, y ] y + y [ x, y ] + [ x, z ] z + z [ x, z ] = (ih z ) y + y (ih z ) + (-ih y ) z + z (-ih y ) = (ih )( z y + y z - y z - z y ) = (ih )([ z, y ] + [ y, z ]) = e. [ y, ] = [ y, x + y + z ] = [ y, x ] + [ y, y ] + [ y, z ] = [ y, x ] + [ y, z ] = [ y, x ] x + x [ y, x ] + [ y, z ] z + z [ y, z ] = (-ih z ) x + x (-ih z ) + (ih x ) z + z (ih x ) = (ih )(- z x - x z + x z + z x ) = (ih )([ x, z ] + [ z, x ]) = f. [ z, ] = [ z, x + y + z ] = [ z, x ] + [ z, y ] + [ z, z ] = [ z, x ] + [ z, y ] = [ z, x ] x + x [ z, x ] + [ z, y ] y + y [ z, y ] = (ih y ) x + x (ih y ) + (-ih x ) y + y (-ih x ) = (ih )( y x + x y - x y - y x ) = (ih )([ y, x ] + [ x, y ]) = 8. Use the general angular momentum relationships: J j,m> = h (j(j+)) j,m> J z j,m> = h m j,m>, and the information used in exercise 5, namely that: x = ( ) + Y l,m = l(l + ) - m(m + ) h Y l,m+ - Y l,m = l(l + ) - m(m - ) h Y l,m- Given that: Y, (θ,φ) = 4π =,>

33 3 Y, (θ,φ) = Cosθ =,>. 4π a. z,> =,> = Since and z commute you would expect,> to be simultaneous eigenfunctions of both. b. x,> = z,> = x and z do not commute. It is unexpected to find a simultaneous eigenfunction (,>) of both... for sure these operators do not have the same full set of eigenfunctions. c. z,> =,> = h,> Again since and z commute you would expect,> to be simultaneous eigenfunctions of both. d. x,> = h,-> + h,> z,> = Again, x and z do not commute. Therefore it is expected to find differing sets of eigenfunctions for both. 9. For: Ψ(x,y) = x y e in x πx/ x - e -in x πx/ x e in y πy/ y - e -in y πy/ y <Ψ(x,y) Ψ(x,y)> =? et: a x = n xπ, and a y = n yπ x and using Euler's formula, expand the exponentials into Sin y and Cos terms. Ψ(x,y) = x [Cos(a y x x) + isin(a x x) - Cos(a x x) + isin(a x x)] [Cos(a y y) + isin(a y y) - Cos(a y y) + isin(a y y)] Ψ(x,y) = x isin(a y x x) isin(a y y) Ψ(x,y) = - x Sin(a y x x) Sin(a y y) <Ψ(x,y) Ψ(x,y)> = - x y Sin(a x x) Sin(a y y) dxdy = x y Sin (a x x) Sin (a y y) dxdy Using the integral:

34 Sin nπx dx =, <Ψ(x,y) Ψ(x,y)> = x y x y =. x y <Ψ(x,y) p x Ψ(x,y)> = y Sin (a y y)dy x Sin(a x x)(-ih x )Sin(a xx)dx But the integral: = -ih x a x x Sin(a x x)cos(a x x)dx x Cos(a x x)sin(a x x)dx =, <Ψ(x,y) p x Ψ(x,y)> =. <Ψ x Ψ > = α π e -αx/ ( x ) e -αx/ dx = α π x e -αx dx Using the integral: x n e -βx dx = (n-) n+ π β n+, <Ψ x Ψ > = α π π α 3 <Ψ x Ψ > = α <Ψ x Ψ > = 4α 3 π xe -αx/ ( x ) xe -αx/ dx

35 = 4α 3 π x 4 e -αx/ dx Using the previously defined integral: <Ψ x Ψ > = 4α 3 π 3 3 π α 5 <Ψ x Ψ > = 3 α.

36 3.

37 a. Ψ I (x) = Ψ II (x) = Ae i me/h x + Be -i me/h x Ψ III (x) = A'e i m(v-e)/h x + B'e -i m(v-e)/h x b. I II Ψ I () = Ψ II () Ψ I () = = Ψ II () = Ae i me/h () -i + Be me/h () = A + B B = -A Ψ ' I() = Ψ ' II() II III Ψ II () = Ψ III () (this gives no useful information since Ψ ' I(x) does not exist at x = ) Ae i me/h + Be -i me/h = A'e i m(v-e)/h + B'e -i m(v-e)/h Ψ ' II() = Ψ ' III() A(i me/h )e i me/h - B(i me/h )e -i me/h = A'(i m(v-e)/h )e i m(v-e)/h - B'(i m(v-e)/h )e -i m(v-e)/h c. as x, Ψ I (x) = as x, Ψ III (x) = A' =

38 d. Rewrite the equations for Ψ I () = Ψ II (), Ψ II () = Ψ III (), and Ψ ' II() = Ψ ' III() using the information in 3c: B = -A (eqn. ) Ae i me/h + Be -i me/h = B'e -i m(v-e)/h (eqn. ) A(i me/h )e i me/h - B(i me/h )e -i me/h = - B'(i m(v-e)/h )e -i m(v-e)/h (eqn. 3) substituting (eqn. ) into (eqn. ): Ae i me/h - Ae -i me/h = B'e -i m(v-e)/h A(Cos( me/h ) + isin( me/h )) - A(Cos( me/h ) - isin( me/h )) = B'e -i m(v-e)/h AiSin( me/h ) = B'e -i m(v-e)/h Sin( me/h ) = B' Ai e-i m(v-e)/h (eqn. 4) substituting (eqn. ) into (eqn. 3): A(i me/h )e i me/h + A(i me/h )e -i me/h = - B'(i m(v-e)/h )e -i m(v-e)/h A(i me/h )(Cos( me/h ) + isin( me/h )) + A(i me/h )(Cos( me/h ) - isin( me/h )) = - B'(i m(v-e)/h )e -i m(v-e)/h Ai me/h Cos( me/h ) = - B'i m(v-e)/h e -i m(v-e)/h Cos( me/h ) = - B'i m(v-e)/h -i m(v-e)/h e Ai me/h Cos( me/h ) = - B' V-E A E e-i m(v-e)/h (eqn. 5)

39 Dividing (eqn. 4) by (eqn. 5): Sin( me/h ) = B' -A E e -i Cos( me/h Ai ) B' V-E e -i Tan( me/h ) = - E / V-E m(v-e)/h m(v-e)/h e. As V, Tan( me/h ) So, E n = n π h m me/h = nπ Problems. a. Ψ n (x) = Sin nπx P n (x)dx = Ψ n (x) dx The probability that the particle lies in the interval x 4 4 P n = P n (x)dx = 4 Sin nπx dx is given by: This integral can be integrated to give (using integral equation with θ = nπx ): nπ 4 P n = nπ Sin nπx d nπx nπ 4 P n = nπ Sin θdθ P n = nπ - 4 Sinθ + θ nπ 4 = nπ - 4 Sinnπ nπ 4 + ()(4)

40 = 4 - πn Sin nπ nπ b. If n is even, Sin = and P n = 4. nπ If n is odd and n =,5,9,3,... Sin = and P n = 4 - πn nπ If n is odd and n = 3,7,,5,... Sin = - and P n = 4 + πn The higher P n is when n = 3. Then P n = 4 + π3 P n = 4 + 6π =.33 -iht -ie n t -ie m t c. Ψ(t) = e h [ aψ n + bψ m ] = aψ n e h + bψ m e h -ie n t ie m t HΨ = aψ n E n e h + bψ m E m e < Ψ H Ψ > = a E n + b E m + a * be -i(e m -E n) t h i(e n -E m) t h < Ψ n H Ψ m > + b * ae h < Ψ m H Ψ n > Since < Ψ n H Ψ m > and < Ψ m H Ψ n > are zero, < Ψ H Ψ > = a E n + b E m (note the time independence) d. The fraction of systems observed in Ψ n is a. The possible energies measured are E n and E m. The probabilities of measuring each of these energies is a and b. e. Once the system is observed in Ψ n, it stays in Ψ n. < > f. P(E n ) = Ψ n Ψ = c n c n = Sin nπx 3 5 x(-x)dx = 6 nπx 6 x(-x)sin dx

41 6 = 6 nπx xsin dx - x nπx Sin dx These integrals can be evaluated from integral equations 4 and 6 to give: 6 c n = 6 n π Sin nπx - x nπ Cos nπx - c n = 6 6 x n π Sin nπx 6 3 6{ n π ( Sin(nπ) - Sin() ) - nπ ( Cos(nπ) - Cos ) ) - ( n π ( Sin(nπ) - Sin() ) - ( n π - ) - n π x - 3 n 3 π 3Cos nπx 3 n 3 π 3Cos(nπ) + n π () - 3 3Cos())} n 3 π c n = -3 6 {- 3 nπ Cos(nπ) + ( n π - ) 3 n 3 π 3Cos(nπ) + 3 n 3 π 3} c n = 6 - nπ (-)n + ( n π - ) n 3 π 3(-)n + c n = 6 - nπ + nπ - n 3 π 3 (-)n + n 3 π 3 c n = 6 n 3 π 3 )( -(-)n + ) n 3 π 3 c n = 4(6) n 6 π 6 )( -(-)n + ) If n is even then c n = If n is odd then c n = (4)(6)(4) n 6 π 6 = 96 n 6 π 6 The probability of making a measurement of the energy and obtaining one of the eigenvalues, given by: E n = n π h m is: P(E n ) = if n is even

42 P(E n ) = 96 n 6 if n is odd π6 Ψ H Ψ = 3 5 x(-x) -h d m dx 3 5 x(-x)dx g. < > = 3 -h 5 m x(-x) d dx ( x-x) dx. < Ψ H Ψ > = ij = -5h m 5 x(-x)(-)dx = 3h m 5 x-x dx = 3h m 5 x -x3 3 = 3h m = 3h m - 3 = 3h 6m = 5h m ie i t C i * e Since < Ψ i H Ψ j > = E j δ ij < Ψ H Ψ > = j < Ψ H Ψ > = For other properties: < Ψ A Ψ > = ij j h < Ψ i H Ψ j > e i(e j -E j )t C j * C j E j e h -ie j t h C j C j * C j E j (not time dependent) ie i t C i * e h < Ψ i A Ψ j > e but, < Ψ i A Ψ j > does not necessarily = a j δ ij. -ie j t h C j

43 This is only true if [A,H] =. i(e i -E j )t < < Ψ A Ψ > = C i * Ψ i A Ψ j > C j e h ij Therefore, in general, other properties are time dependent. 3. For a particle in a box in its lowest quantum state: Ψ = Sin πx < x > = Ψ * xψdx = xsin πx dx Using integral equation 8: = x 4 - x 4π Sin πx - 8π Cos πx < > = Cos()) 8π(Cos(π) = 4 = x = Ψ * x Ψdx = x Sin πx dx Using integral equation 9: = x3 6 - x 4π - 3 8π 3 Sin πx - x 4π Cos πx = ()Cos()) 4π(Cos(π) = π = 3 - π

44 < p > = Ψ * pψdx = πx Sin h i d dx Sin πx dx = h π i πx Sin Cos πx dx = h i πx Sin Cos πx d πx Using integral equation 5 (with θ = πx ): < > = h i - Cos (θ) π p = Ψ * p Ψdx = πx Sin -h = π h 3 Sin πx dx = = πh Sin πx d πx d dx Sin πx dx Using integral equation (with θ = πx ): = πh - 4 Sin(θ) + θ π = πh π = π h

45 x = < x > - < x> = 3 - π - 4 = - π p = < p > - < p> = π h - = πh x p = πh - π = h 4π - 4 = h π 3 - h Finally, π 3 - > h (3) 3 - = h x p > h /4 /4 4. a. P(x)dx = dx = x /4 = 4 = 4 = 5% P classical = 4 (for interval - /4) b. This was accomplished in problem a. to give: P n = 4 - πn Sin nπ (for interval - /4) c. imit n P quantum = imit n 4 - πn Sin nπ imit n P quantum = 4 Therefore as n becomes large the classical limit is approached.

46 5. a. The Schrödinger equation for a Harmonic Oscillator in -dimensional coordinate representation, H Ψ(x) = E x Ψ(x), with the Hamiltonian defined as: H = -h d m dx + kx, becomes: -h d m dx + kx Ψ(x) = E x Ψ(x). b. The transformation of the kinetic energy term to the momentum representation is trivial : T = p x m. In order to maintain the commutation relation [x,p x] = ih and keep the p operator unchanged the coordinate operator must become x = ih d dp. The Schrödinger x equation for a Harmonic Oscillator in -dimensional momentum representation, H Ψ(p x ) = Ep x Ψ(p x ), with the Hamiltonian defined as: H = m p x - kh d dp x, becomes: m p x - kh d dp x Ψ(p x ) = Ep x Ψ(p x ). c. For the wavefunction Ψ(x) = C exp (- mk x h ), mk let a = h, and hence Ψ(x) = C exp (-ax ). Evaluating the derivatives of this expression gives: d dx Ψ(x) = d dx C exp (-ax ) = -axc exp (-ax ) d d Ψ(x) = dx dx C exp (-ax ) = d dx -axc exp (-ax ) = (-axc) (-ax exp (-ax )) + (-ac) (exp (-ax )) = (4a x - a) Cexp (-ax ). H Ψ(x) = E x Ψ(x) then becomes: H Ψ(x) = -h m (4a x - a) + kx Ψ(x). Clearly the energy (eigenvalue) expression must be independent of x and the two terms containing x terms must cancel upon insertion of a: E x = -h m 4 mk h x - mk h + kx = -h m 4mkx 4h = - kx + h mk m + h mk m h + kx + kx

47 = h mk m. Normalization of Ψ(x) to determine the constant C yields the equation: C exp (- mk x ) dx =. h Using integral equation () gives: C π mk - h = C πh mk = C mk = πh mk C = 4 πh mk Therefore, Ψ(x) = 4 πh exp (- mk x h ). d. Proceeding analogous to part c, for a wavefunction in momentum space Ψ(p) = C exp (-αp ), evaluating the derivatives of this expression gives: d dp Ψ(p) = d dp C exp (-αp ) = -αpc exp (-αp ) d d Ψ(p) = dp dp C exp (-αp ) = d dp -αpc exp (-αp ) = (-αpc) (-αp exp (-αp )) + (-αc) (exp (-αp )) = (4α p - α) Cexp (-αp ). H Ψ(p) = E p Ψ(p) then becomes: H Ψ(p) = m p - kh (4α p - α) Ψ(p) Once again the energy (eigenvalue) expression corresponding to E p must be independent of p and the two terms containing p terms must cancel with the appropriate choice of α. We also desire our choice of α to give us the same energy we found in part c (in coordinate space). E p = m p - kh (4α p - α) Therefore we can find α either of two ways: