Quantum Mechanics: Simple Harmonic Oscillator & Angular Momentum in Magnetic Fields, Exams of Quantum Physics

Download Quantum Mechanics: Simple Harmonic Oscillator & Angular Momentum in Magnetic Fields and more Exams Quantum Physics in PDF only on Docsity! L.J. Sham December 11, 2007 Physics 130B Final Examination PRINT your name and attach only the front page to the front of your bluebook. Name: Do problem 1 and choose ONLY THREE PROBLEMS from problem 2 to problem 5 . (Cross out the problem you do NOT wish to be graded.) Problem 1 2 3 4 5 Total Grade Out of 40 20 20 20 20 100 Useful information: Pauli spin matrices I = [ 1 0 0 1 ] , σx = [ 0 1 1 0 ] σy = [ 0 −i i 0 ] σz = [ 1 0 0 −1 ] . (1) Simple Harmonic Oscillator Ĥ = p̂2 2m + 1 2 mω2x̂2 = ~ω ( ĉ†ĉ + 1 2 ) , Ĥ|n〉 = ~ω(n + 1 2 )|n〉, n = 0, 1, 2, . . . (2) x̂ = a(ĉ + ĉ†), p̂ = ~ i2a (ĉ − ĉ†), where a = √ ~ 2mω ; (3) ĉ|n〉 = √ n|n − 1〉, ĉ†|n〉 = √ n + 1|n + 1〉, [ĉ, ĉ†] = 1. (4) Angular momentum states ℓ = 1: In polar coordinates (θ, φ), Y1,1 = − 1√ 2 sin θeiφ; Y1,0 = cos θ; Y1,−1 = 1√ 2 sin θe−iφ. (5) 1 1. Quickies. Each answer must be supported by brief reasoning. (a) An electron in the hydrogen atom has at a certain instant the same form of wave func- tion as the ground state except the Bohr radius is replaced by a length which is larger by 1%. The mean energy of the electron at that instant is different from -1 Ry by (I) 1%; (II) 10%; (III) 0.01%. Choose the correct one and reason. (b) Let T̂ be the operator representing a rotation of the ammonia molecule, NH3, through 120◦ about the nitrogen axis normal to the plane of the hydrogen atoms. Find the eigenvalues of the operator T̂ . Figure 1: The top view from the nitrogen atom (circle) above the normal axis to the centroid of the equilateral triangle of hydrogen atoms (black dots). (c) A hydrogen atom is excited by white light with the polarization vector (i.e., the electric field) in the x direction. Consider the possible transitions induced between energy eigenstates of the electron motion relative to the proton, denoted by the usual notation |n, ℓ, m, ms〉 for the principal n, orbital angular momentum (ℓ, m) and spin ms quantum numbers. Out of the eight states: |2, 0, 0, ms〉, |2, 1, 1, ms〉, |2, 1, 0, ms〉, |2, 1,−1, ms〉, with ms = ±12 , select the ones to which the transition from the ground state |1, 0, 0, +1 2 〉 is possible. (It might or might not be helpful to look at Problem 3 first but p. 1 is.) (d) A beam of light with a spectral distribution centered at the resonance frequency of two atomic levels is shone on the atom. Explain succinctly (i) the reason that the Fermi Golden Rule gives the same probability per unit time for the excitation from the lower to high energy level and for the reverse process from the higher to the lower; (ii) the reason that we are not happy with this result. (e) A beam of neutrons is sent through a Stern-Gerlach apparatus. The detector screen shows two lines of equal intensity. The starting condition is known to be one of the only two mutually exclusive possibilities: i. half of the neutrons in the beam are in the spin-up state |+〉 along the direction of the vertical magnetic field (the z axis) and half in the spin-down state |−〉; ii. every neutron in the beam is in the spin state along the positive x axis which is normal to the plane of motion of the neutrons (y − z plane), i.e., | + x〉 = 1√ 2 ( |+〉 + |−〉 ) . (6) Design an experiment to distinguish between only these two possibilities. 2 L.J. Sham December 11, 2007 Physics 130B Final Examination Solutions 1 Correction of Frequently Made Errors 1. (a) Use the variational principle. (b) (c) Find the selection rules from the matrix elements of x between two states. The mag- netic field term is weak by comparison with the electric field term but its consideration was given credit. (d) (e) It is encouraging that a surprising number of students figured out the right answer. 2. The state was given as a tensor product of three vector spaces x, y, z: |nx, ny, nz〉 ≡ |nx〉|ny〉|nz〉. Then, the operator in the x space acts only on the vector in x space, e.g., p̂x|nx, ny, nz〉 = [p̂x|nx〉]|ny〉|nz〉, so that, 〈mx, my, mz|p̂x|nx, ny, nz〉 = 〈mx|p̂x|nx〉〈my|ny〉〈mz|nz〉. 3. Be careful to pick the variational wave function which will contain information on the added term in the Hamiltonian. In this case, the SHO Hamiltonian in Eq. (2) is symmetry under mirror operation x → −x and so is the ground state wave function but the additional term, V̂ , is odd under the mirror operation. So an even variational wave function would give zero, which means not that the effect is zero but that your variational wave function contains no possibility to include the effect of the perturbation. Remember that for 〈Ψ| given by Eq. (11), take the complex conjugate of the elements. 4. A little nitpicking: the magnetic moment factor for proton is µp. Useful information: Pauli spin matrices I = [ 1 0 0 1 ] , σx = [ 0 1 1 0 ] σy = [ 0 −i i 0 ] σz = [ 1 0 0 −1 ] . (1) 1 Simple Harmonic Oscillator Ĥ = p̂2 2m + 1 2 mω2x̂2 = ~ω ( ĉ†ĉ+ 1 2 ) , Ĥ|n〉 = ~ω(n+ 1 2 )|n〉, n = 0, 1, 2, . . . (2) x̂ = a(ĉ + ĉ†), p̂ = ~ i2a (ĉ− ĉ†), where a = √ ~ 2mω ; (3) ĉ|n〉 = √ n|n− 1〉, ĉ†|n〉 = √ n+ 1|n+ 1〉, [ĉ, ĉ†] = 1. (4) Angular momentum states ℓ = 1: In polar coordinates (θ, φ), Y1,1 = − 1√ 2 sin θeiφ; Y1,0 = cos θ; Y1,−1 = 1√ 2 sin θe−iφ. (5) 1. Quickies. Each answer must be supported by brief reasoning. (a) An electron in the hydrogen atom has at a certain instant the same form of wave func- tion as the ground state except the Bohr radius is replaced by a length which is larger by 1%. The mean energy of the electron at that instant is different from -1 Ry by (I) 1%; (II) 10%; (III) 0.01%. Choose the correct one and reason. Solution – (III) is correct. Let the radius used in the 1s wave function be a = aB(1+ε) where aB is the true Bohr radius and ε is the variational parameter. Then the expectation value of the Hamiltonian is E(ε) = E(0) + E ′(0)ε+ 1 2 E ′′(0)ε2 + . . . , (6) for a small change ε. By the variational principle, the ground state energy at ε = 0 is a minimum. Hence, E ′(0) = 0, and the correction to the energy is O(ε2). (b) Let T̂ be the operator representing a rotation of the ammonia molecule, NH3, through 120◦ about the nitrogen axis normal to the plane of the hydrogen atoms. Find the eigenvalues of the operator T̂ . Figure 1: The top view from the nitrogen atom (circle) above the normal axis to the centroid of the equilateral triangle of hydrogen atoms (black dots). Solution – T̂ 3 = Î , T̂ |α〉 = α|α〉, α3 = 1, α = e−i2πk, (7) where k = 0, 1/3, 2/3. 2 (c) A hydrogen atom is excited by white light with the polarization vector (i.e., the electric field) in the x direction. Consider the possible transitions induced between energy eigenstates of the electron motion relative to the proton, denoted by the usual notation |n, ℓ,m,ms〉 for the principal n, orbital angular momentum (ℓ,m) and spinms quantum numbers. Out of the eight states: |2, 0, 0, ms〉, |2, 1, 1, ms〉, |2, 1, 0, ms〉, |2, 1,−1, ms〉, with ms = ±12 , select the ones to which the transition from the ground state |1, 0, 0,+1 2 〉 is possible. (It might or might not be helpful to look at Problem 3 first but p. 1 is.) Solution – Consider the electric perturbation only, He = eE x̂. From Eq.(5), we could make, |Y1,±1〉 = 1√ 2 (|x〉 ± i|y〉, and |Y1,0〉 = |z〉. (8) The the selection rules from the consideration of the matrix elements between |1, 0, 0,+1 2 〉 and the excited states are the change ∆ms = 0 (no spin-dependent perturbation is in- volved), ∆ℓ = 0, and ∆m = ±1. Hence, the only possible final states are |2, 1,±1, 1 2 〉. (d) A beam of light with a spectral distribution centered at the resonance frequency of two atomic levels is shone on the atom. Explain succinctly (i) the reason that the Fermi Golden Rule gives the same probability per unit time for the excitation from the lower to high energy level and for the reverse process from the higher to the lower; (ii) the reason that we are not happy with this result. Solution – i. Consider the three terms in the Golden rule, they are all the same for the absorption and the emission process. ii. The higher energy state can spontaneously decay to the lower energy state. This is not taken into account by the Fermi Golden rule. (e) A beam of neutrons is sent through a Stern-Gerlach apparatus. The detector screen shows two lines of equal intensity. The starting condition is known to be one of the only two mutually exclusive possibilities: i. half of the neutrons in the beam are in the spin-up state |+〉 along the direction of the vertical magnetic field (the z axis) and half in the spin-down state |−〉; ii. every neutron in the beam is in the spin state along the positive x axis which is normal to the plane of motion of the neutrons (y − z plane), i.e., | + x〉 = 1√ 2 ( |+〉 + |−〉 ) . (9) Design an experiment to distinguish between only these two possibilities. Solution – Repeat the experiment with the magnetic axis rotated about the path through π/4 radians. Scenario (i) still yields two lines on the detector screen but Scenario (ii) now yields only one line. 3 The exact solution. Note that the addition of a potential linear in x yields a displaced harmonic oscillator since the total potential is, V = 1 2 mω2x2 − eEx = 1 2 mω2(x− x0)2 − 1 2 mω2x2 0 , (25) where x0 = eE mω2 . (26) The displacement SHO gives the same ground state. Thus, the net change in energy is ∆E0 = − 1 2 mω2x2 0 = − (eE) 2 2mω2 . (27) The perturbation method. The first order change is zero by symmetry, 〈0|V |0〉 = 0. (28) The second order perturbation contribution is ∆E0 = ∞ ∑ n=1 |〈n|V |0〉|2 E0 −En = |〈1|V |0〉|2 E0 − E1 = − (eE) 2 2mω2 . (29) (c) Compare the accuracy of your two answers. Solution – All three methods give the same answer. The perturbation result is correct because the exact result is a power series which terminates at second order in E . The variational method is a bit of a fluke. Its result before expansion is not exactly correct but the choice of the wave function leads to the correct O(E)2 term. 4. A spin-less particle is confined to be in an ℓ = 1 orbital angular momentum state. The orthonormal basis set is given by the spherical harmonics |Y1,m〉, m = 1, 0,−1. The matrix representations of the angular momentum operators L̂ are given by Lz = 1   1 0 0 0 0 0 0 0 −1   ; Lx = 1√ 2   0 1 0 1 0 1 0 1 0   ; Ly = 1√ 2   0 −i 0 i 0 −i 0 i 0   . (30) (a) At time t = 0, the particle has the state vector Ψ(t = 0) =   − i 2 1√ 2 i 2   . (31) Find the expectation value and uncertainty of L̂y at t = 0. 6 Solution – Verify that LyΨ(t = 0) = Ψ(t = 0), i.e., the initial state is an eigenstate of L̂y. Thus, 〈Ψ(0)|L̂y|Ψ(0)〉 = 〈Ψ(0)|Ψ(0)〉 = 1, (32) and the variation is zero. (b) The Hamiltonian of the particle in a uniform static magnetic field B is given by H = µBB · L̂, (33) where µB is the Bohr magneton. If the magnetic field is fixed along the y direction, find the expectation value of L̂y at time t with the particle initial state given by Eq. (31). Solution – Since H = µBBL̂y, (34) , the initial state is also an eigenstate of the Hamiltonian and thus at time t, it becomes |Ψ(t)〉 = |Ψ(t = 0)〉e−iωt/~, where ω = µBB ~ . (35) The expectation value remains constant. (c) If the magnetic field is fixed along the z direction since t = 0, find the expectation value of L̂y at time t with the particle initial state given by Eq. (31). Solution – From H = µBB   1 0 0 0 0 0 0 0 −1   , (36) the state at time t is Ψ(t) =   − i 2 e−iωt 1√ 2 i 2 eiωt   . (37) Then 〈Ψ(t)|L̂y|Ψ(t)〉 = Ψ(t)†LyΨ(t) = cos(ωt). (38) 5. An immobilized proton was prepared with its spin pointing vertically up, thus in state |+ z〉, in zero magnetic field. We wanted to flip the spin to point vertically down, i.e., to flip to the state | − z〉. An electromagnet was installed to provide a horizontal magnetic field along the x axis with time dependence B(t) which may be controlled at will within a range of ±b. 7 (a) Find the Hamiltonian for the proton spin with the installed magnet. Solution – For the proton with magnetic dipole µp, the Hamiltonian is H = −µpB(t)σx. (39) (b) Design the time dependence of B(t) which will bring the proton spin to the down state and to stop there. Solution – Let B(t) = b for 0 < t < T and = 0 otherwise. (40) Then for ωT = π, or T = π~ µpB , (41) the proton spin is flipped from up to down. (c) Unfortunately, the professor bumped into the magnet and knocked the field axis to an angle φ from the x axis in the x − y plane. Find the Hamiltonian for the proton spin with the displaced magnet. Solution – H = −µpB · σ = −µpBσφ, where σφ = σx cosφ+ σy sin φ. (42) In matrix form, H = −µpB [ 0 e−iφ eiφ 0 ] . (43) (d) Moreover, exactly half way through the application of theB(t) field using your design, there was a blackout, stopping the experiment. What was the final state of the spin, in the basis of the two states | ± z〉? Solution – The spin vector will rotate about the axis n = (cosφ, sinφ, 0) through an an- gle of −π/2. It will end in the xy plane normal to n, i.e., along n′ = (− sinφ, cosφ, 0). From n ′ · σ [ 0 e−i(φ+ π 2 ) ei(φ+ π 2 ) 0 ] . (44) Ψfinal = [ e−i( φ 2 + π 4 ) ei( φ 2 + π 4 ) ] . (45) 8