University of Illinois at Chicago Department of Physics. Quantum Mechanics Qualifying Examination. January 7, 2013 (Tuesday) 9:00 am - 12:00 noon

Transcription

1 University of Illinois at Chicago Department of Physics Quantum Mechanics Qualifying Examination January 7, 13 Tuesday 9: am - 1: noon Full credit can be achieved from completely correct answers to 4 questions If the student attempts all 5 questions, all of the answers will be graded, and the top 4 scores will be counted toward the exam s total score Formulas x n e ax dx = n!/a n+1, valid for complex a as long as Rea > e λx dx = π/λ, valid for complex a as long as Reλ x e λx dx = 1 π λ λ, Fourier transform: ψk = 1 σ x = 1 1 O =, σ y = x e λx dx eλx dx = 1 λ e ikx ψxdx, ψx = 1 Ψ xoψxd 3 x i i, σ z = 1 1 σ x σ y = iσ z = σ y σ x, σ y σ z = iσ x = σ z σ y, σ z σ x = iσ y = σ x σ z ψke ikx dk 1

2 1 Gaussian Wave Packet A Gaussian wave packet describes the initial amplitude of a free non-relativistic one-dimensional quantum particle of mass m = 1 in units h = 1, ψx, t = = N exp3x + 5x + i1x a By completing the square of the real part of the exponent, determine the wave function normalization factor N b Find the mean position x and the mean wave number p of the particle c What is the wave-number amplitude ψk in the wave-number k-representation or the momentum representation? d Find the uncertainties of the position and the momentum, x x, p p, e What is the group velocity of the packet? f How is this wave-number amplitude ψ changed in time? g Find ψx, t Your result can be in an integral form Describe qualitatively how the wave function evolves By completing the square for the real part of the exponent, we derive ψx, = N exp 3x i1x It is easy to claim x = 5 The Gaussian integral gives 6 1/N = e 6x 5 6 dxe 5 6, N = e π Let x = x 5 6 ψx, = e i 5 6 exp 3x 5 π 6 + i1x 5 6 be the new coordinate Then in primed variables, ψ x, = e i 5 6 exp 3x + i1x π However, the physics is just a translation, so we drop the primes unless we specify the variables of the original coordinate Therefore, x = ψ x, xψx, dx = in the translated coordinate The Gaussian standard deviation gives x = ψ x, x ψx, dx = = 1 1 The Fourier analysis gives the wave amplitude in the momentum representation, ψk = 1 ψx, e ikx dx = e i π }{{} C e 3x k1ix dx

3 The exponent in the integrand can be in the complete square form, which is 3x + i 6 k k 1, and it gives ψk = C π 3 e 1 1 k1 It is again a Gaussian, centered at k = p = 1 The Gaussian standard deviation gives k 1 = 1 6 = 3, k = 3 Combining earlier result in the position uncertainty, x = 1, we observe the 1 saturation of the Heisenberg s uncertainty inequality, k x = 1, 4 The time evolution in the momentum representation ψ is simply given by attaching the time-energy factor e ikt in our units, m = 1, h = 1 ψk, t = C π 1 3 e 1 k1 ik t, ψx, t = 1 ψk, te +ikx dk In the narrow k approximation, the wave packet travels at the group velocity k = during the beginning short interval However, when the time is much m longer, the wave will disperse Hydrogen bound states In units m = 1, h = 1, the radial Schrödinger equation of the hydrogen atom is given by [ d g + ] ll+1 dr r r ur = εur The lowest eigenstate of a given l is known to have the form, u lr = C l r l+1 expr/a l a For a given l, determine the eigenvalue ε l and the size parameter a l, in terms of the Coulomb strength g b The initial 3-dimensional wave function at t = is the superposition of the above states l =, 1 r ψx, = D e g + g r re g 4 cos θ Determine the quantum expectation average of cos θ as a function of time d d dr u lr = C l [l + 1r l 1 a l r l+1 ] expr/a l, u dr lr = C l [ll + 1/r r [ d dr r l+1 a l l+1 a l + 1 ]r l+1 expr/a a l l + ll+1 r ] u l r = 1 a l u lr Comparing it with the Schrödinger equation, we identify the relations, g = l + 1 a l, so the ground state energy is ɛ l = 1 a l = g 4 /[l + 1] Now the initial state is given by ψx, = D r e g + gr r e g 4 cos θ 3

4 The first term corresponds to the ground s-wave, and the second one corresponds to the lowest p-wave Their energies are g 4 /4 and g 4 /16 respectively r ψx, t = D e g g4 +i 4 t + g r re g g4 +i 4 16 t cos θ The normalization can be determined at t = 1/D = 4π e gr + 13 g4 r e g r r dr = 4π! = 13/g 6 g 3 3 Above, the cross term is understood to be zero However, the average cos θ is from the cross term, +1 cos θ = g8 cos θd cos θ e 3 4 gr r 3 dr cos g4 t 1 cos θ = cos 3 16 g4 t 3 Planar Rotor and Perturbation A permanent planar dipole p, which lies on the x-y plane, is described by the rotation Hamiltonian H R = h d, where φ is the angle of p with respect to the x I dφ axis a Write down the three lowest energy eigenvalues and their corresponding eigenstates Arrange these states to be eigenstates of the angular momentum operator L z = i h d dφ b A weak electric field E along the y axis is turned on The interaction is given by p E Find all matrix elements of this perturbed energy operator between H R eigenstates in a The result is expressed in terms of p,e, h and I c Determine the perturbed energies to the second order effect for the lowest lying states The equation h ψφ = ɛψφ implies the eigenvalues to be quantized as ɛ = I dφ h m with the integer magnetic number m =, ±1, ±, because of the periodic I condition in the angle φ When we label the eigenstates ψ m by the m index, they are d ψ φ = 1, ψ ±1 φ = 1 e ±iφ, ψ ± φ = 1 e ±iφ, ψ ±m φ = 1 e ±imφ, They are also eigenstates of the angular momentum operator L z, whose eigenvalues are, ±1, ±,, ±m, Now the perturbed Hamiltonian is H = p E = pe sin φ [ m H m = pe e imm φ sin φdφ = pe 1 i e imm +1φ e ] imm 1φ dφ 4

5 m H m = pe for m = m + 1, i for m = m 1, i otherwise pe The corrected energies up to the second order are E = + p E 1 h + = p E I 4 h +, I E 1 = h I + + p E 1 h 1 + = h 4 3 h I + p E I 3 h +, I I E = 4 h I + + p E = 4 h 4 3 h 5 h I + p E I 15 h +, I I Note that the degeneracy relation E m = E m still holds 4 Fermi-Golden rule, scattering length, Born approximation The asymptotic form of a scattering wave is given by ψx e ik x + fθ eikr for a r particle of mass m in a spherical potential V r The scattering amplitude is given by the Born approximation for a weak potential, fθ = m h e ik k r V r d 3 r a Carry out the angular integration in fθ so that only the radial integration remains Simplify the result in terms of q q = k k b Find the differential cross section dσ/dω = fθ for a weak delta-shell potential V r = g δr R, located at the radius R c On the other hand, the cross-section can be derived from the Fermi Golden Rule about the transition rate from an initial state i to the final states f, Γ = h f H i ρe f, where ρe f counts the final state density when the system is confined in a very large cube with the periodic boundary condition Derive the Born crosssection result from the Fermi golden rule by working out the state density, the solid angle differential, and the incident flux d The scattering problem can also be solved by the phase shift method In the low energy limit of a very small k, the s-wave outside the potential range becomes a straight line rψr = ur A r a Here A is an arbitrary multiplicative constant The parameter a, ie the extrapolated intercept of the outside wave, is called the scattering length As the s-wave effect dominates at the low k, we know that fθ a Determine the scattering length a for the above delta-shell potential, in terms of the shell radius R and the strength g, by solving the corresponding radial Schrödinger equation at k, d dr ur + gδr Rur = Confront your result with Born approximation 5

6 The Born scattering amplitude for a central force is given by fθ = m e ikk r V r d 3 r = m r V rdr h h +1 1 iqr cos θ dcos θ e [ = m h r iq e iq e iq] V rdr = m r sinqrv rdr, h q where the momentum transfer q = k k and q = k sin θ For the present case that V r = gδr R, fθ = mgr dσ sin qr, = h q dω fθ = mgr h q sin qr The density of state in the Fermi Golden Rule FGR is δρede = L3 d 3 p h = L3 4πp dp 3 h 3, δρe pdp m = L3 p dpδω h 3 We have δρe = L3 mp h 3 δω Since the transition rate Γ in the FGR, when normalized by the incoming flux I = 1 L 3 p, is the cross-section, we obtain m δσ = δγ I = h e ikk r V r 1 L 3 d 3 r L3 mp 1 L 3 p m h 3 δω Note that the artificial L 3 factors cancel completely We obtain the desirable Born formula dσ dω = m h e ikk r V r d 3 r The 1-d radial solution ur at the low energy k becomes a straight line, { Ar a for r > R, here a is the scattering length, ur = Br for r < R because u= The continuity of u at r = R relate A and B so that AR a = BR The delta potential gives rise to a kink, h m [u R > u R < ] + gur =, h A B + gar a = m Combining the matching conditions, we obtain a = gr/g + h The result is mr in agreement with the Born approximation for a weak coupling g at the very low energy k, where fθ = a 5 Coupled Angular Momenta We study the composite system of two localized spin-half particles, 1 and Their corresponding Pauli matrices are σ 1 i and σ i The spin-spin interaction among them is described by H = σ 1 x σ x + σ 1 y σ y + σ 1 z σ z 6

7 a Find the energy eigenstates and eigenvalues of H, by using the property of the angular momentum sum b Then find the energy eigenstates and eigenvalues of another Hamiltonian, H + = σ 1 y σ y + σ 1 x σ x c When the spin state ψ of one single spin-half particle is rotated about the y-axis by an angle β, the new state ψ = U ψ is described by the unitary transformation U = expiσ y β/ Work out the explicit entries in the matrix U in terms of β The Pauli matrices σ i i = x, y, z when transformed becomes σ i = Uσ i U Work out the explicit relation that σ x = c 1 σ x + c σ z and express the coefficients c 1, c in terms of the angle β Do the same calculation for σ z and σ y Explain the physical meaning of the transformation Show the result for the special case of β = π d Finally, If the relative sign of terms in H + is flipped to give the third Hamiltonian, H = σ y 1 σ y σ x 1 σ x How is H related to H + by a unitary transformation? What is the energy eigenstates and eigenvalues of H? Uβ = e iβσy/ = cos β iσ y sin β = cos β sin β sin β cos β It is easy to see σ y unchanged after transformation, Uβσ y U β = σ y However, Similarly, Uβσ x U β = cos β iσ y sin β = cos β sin β σx sin β cos β Uβσ z U β = cos β iσ y sin β = cos β sin β σz + sin β cos β σx cos β + iσ y sin β σz = cos βσ x sin βσ z σz cos β + iσ y sin β σx = cos βσ z + sin βσ z Therefore, the -component spinor is transformed by the half-angle formula However, the operators are transformed by the whole angle β, like rotating a vector In summary, σ y = σ y, σ x = cos βσ x sin βσ z, σ z = sin βσ x + cos βσ z In the special case, β = π, we have σ y = σ y, σ x = σ x, σ z = σ z Now we work on the spin-spin coupling in the problem H = σ 1 σ = σ 1 + σ σ 1 σ = S s 1 s 7

8 = 4SS The energy eigenvalues for the triplet and the singlet are E S = SS = { +1 for the triplet S = 1, 3 for the singlet S = Now we move to a less symmetric system, H + = H σ 1 z σ + z, E S,S z = E S σ z 1 σ z Therefore, E + S,S z = for the triplet S = 1, S z = ±1, or, for the triplet S = 1, S z =, 1 +, for the singlet S =, S z =, 1 The arrows in the first and the second positions represent the state configuration of the first and second spins Now we study the last system H, which is similar to H + with a relative sign flip The two Hamiltonians are related by a unitary transformation, U H U = H +, where we only rotate the second spin by the angle β = π The eigenvalues are unchanged We obtain E S,S z = for or, for 1, + for 1 + 8