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PHY4604–Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004 These problems are similar but not identical to the actual test. One or two parts will actually show up. 1. Short answer. (a) Recall the Bohr energy levels of the Hydrogen atom are En = − 2h̄m2 n2 ³ e2 4π²0 ´2 (1) Compare the wavelengths of the 2p → 1s transitions in i. hydrogen The energy which must be carried away by a photon in the transition is E2 − E1 = hc/λ, so λH = hc/(E2 − E1 ) = 8h̄3 3m µ 4π²0 e2 ¶2 ii. deuterium (mass of nucleus 2× that of H). The mass which appears in the Bohr formula is the electron mass m, which is replaced by the reduced mass µ = mmp /(m+mp ) if one allows for the proton motion. Changing mp → 2mp doesn’t change much, so to a good approximation the photon wavelength doesn’t change, λD ' λH . iii. positronium. In this case the positron is not heavier than the electron, but in fact has the same mass. The reduced mass of 2 identical particles of mass m is m/2, so we obtain the correct answer from the Hydrogen expression by substituting m → m/2, or λpos = 2λH . (b) An electron is in the ground state of tritium, a form of heavy hydrogen where the nucleus has two neutrons in addition to a proton. A nuclear reaction now changes the nucleus instantaneously to 3 He, i.e. two protons 1 and a neutron. Calculate the probability that theqelectron remains in the ground state of 3 He. You may need ψ100 = (2/ 4πa30 ) exp −r/a0 , with a0 = h̄2 /(mZe2 ), where Z is the nuclear charge. Z, the nuclear charge or number of protons, is 1 for tritium and 2 for 3 He. Thus the Bohr radius for 3 He is twice as small as for tritium or hydrogen, 3 a0He ' 12 at0 . The electron in the 3 He ion is more tightly bound to the nucleus. We’re told the system starts out in the tritium ground state, 2 t = q ψ100 t 3 4π(aH 0 ) e−r/a0 , which can be expressed as a linear combination of any complete set of states in Hilbert space, for example the eigenstates of the 3 He Hamiltonian, 3 3 3 He He t He + ... + cψ210 ψ100 = aψ100 + bψ200 meaning the electron in the tritium ground state will be found after a measurement in the 3 He ground state with probability amplitude 3 t He a = hψ100 |ψ100 i µ Z ¶ 1 + 31 at a He 0 0 4 1 q d3 r e 4π (at )3 (a3 He )3 0 0 4 √ Z∞ t 8 dr r2 e−3r/a0 = t 3 (a0 ) 0 √ Z √ 16 2 2 −3y = = 0.838 = 4 8 dy y e 27 = −r and probability = |a|2 = 0.702. (c) An electron moving in the Coulomb field of a proton is in a state described by the wave function √ 1 Ψ = [4ψ100 + 3ψ211 − ψ210 + 10ψ21−1 ] 6 (2) i. What is the expectation value of the energy? 1 (16h100|H|100i + 9h211|H|211i + h210|H|210i 36 + 10h21 − 1|H|21 − 1i) 1 1 = (16E1 + 9E2 + 1E2 + 10E2 ) = (16E1 + 20E2 ). 36 36 hΨ|H|Ψi = P Note this is an example of general rule hψ|O|ψi = n on |cn |2 , where the on are the eigenvalues of the Hermitian operator O, and the cn the expansion coefficients of ψ in the basis of O eigenstates. 2 ii. What is the expectation value of L̂2 ? X 2 hψ|L2 |ψi = h̄ `(` + 1)|cn`m |2 = n`m h̄2 10h̄2 (16 · 0 + 9 · 2 + 1 · 2 + 10 · 2) = 36 9 iii. What is the expectation value of L̂z ? X hψ|Lz |ψi = h̄m|cn`m |2 = n`m h̄ h̄ (16 · 0 + 9 · 1 + 1 · 0 + 10 · (−1)) = − 36 36 (d) How large would a constant magnetic field have to be to split two H-atom states which are degenerate in zero field by an amount so as to maximally absorb light of wavelength λ? e H = −µ · B = S · B m e e = = Sz Bz = h̄ms Bz , m m where the last step where the operator is replaced by its eigenvalues holds only when applied to Sz eigenstates, and where I have used ms for the Sz quantum number. The two Sz states have a difference of ∆ms = 1, so the energy of the photon produced must be me h̄Bz ≡ hc , or λ = (2πmc/(eBz )). λ (e) For two particles a and b such that `a = 1 and `b = 1, argue that it must be true that |` = 2, m = 1i = α|ma = 1, mb = 0i + β|ma = 0, mb = 1i, where `, m are the quantum numbers corresponding to total angular momentum Lq= La + Lb , and find the coefficients α and β. Hint: use L± |`mi = `(` + 1) − m(m ± 1)|`m ± 1i. a) states given are the only possible ones with m = ma + mb and ma ≤ `a , mb ≤ `b . b) Start at top of angular momentum ladder, where we know there is only one possible |`m = `i state, equal to the one possible |ma = `a , mb = `b i, then apply lowering operator as in HW, remembering square root factors to keep states normalized: √ L− |22i`m = (La + Lb )|11ima mb √ 2 · 3 − 2 · 1|21i = 2(|01i + |10i) 3 So divide by √ 2 to get |21i = √1 (|10i 2 + |01i). 2. Angular momentum 1. Consider an angular momentum 1 system. (a) What are the possible eigenvalues of L2 and Lz corresponding to the eigenvectors |` = 1, mi? L2 |1mi = 2h̄2 Lz |1mi = h̄m (b) In the basis where the eigenvectors |` = 1, mi of the operator L̂z are given by (1, 0, 0), (0, 1, 0) and (0, 0, 1), construct the matrix representation of the operator L̂x (Hint: you will need to calculate the matrix elements hm|L̂x |m0 i.) Write Lx = (L+ + L− )/2, find, e.g. L+ + L− |1i = 0 2 √ 2 L+ + L− h1| |0i = h̄ 2 2 L+ + L− | − 1i = 0... h1| 2 h1| where I’ve used the effect of L± acting on |mi, and the orthonormality of the |mi. It is a little tedious to do them all, but eventually we find Lx 0 1 0 h̄ √ = 1 0 1 2 0 1 0 (c) Find the eigenvectors of L̂x . Now we just have a matrix eigenvalue problem, and I assume you can find the eigenvectors, then normalize them. The results are √ 1/2 1/2 −1/ 2 √ √ 0√ |1i = 1/ 2 |0i = | − 1i = −1/ 2 1/2 1/2 1/ 2 4 (d) If a system is prepared in the state vector 1 1 |ψi = √ 4 , 26 −3 (3) what is the probability that a measurement of L̂x yields the value 0? P = |h0|ψi|2 = 4/13 3. Pauli principle. Consider two electrons described by the Hamiltonian H= where p̂21 p̂2 + 2 + V (x1 ) + V (x2 ) 2m 2m ∞ x < −a/2 V (x) = 0 −a/2 ≤ x ≤ a/2 ∞ x > a/2 Assume both electrons are in same spin state. (4) (5) (a) What is the lowest (ground state) energy? (b) What is the energy eigenfunction for this ground state? Parts a) and b): Denote the single-particle eigenfunctions of the ordinary infinite square well by ψ0 , ψ1 , .... The two-particle ground state must be antisymmetric under exchange of all particle labels according to Pauli, but we are told that the spins are both the same, e.g. up. The wave function must be Ψ0 (1, 2) = (ψ0 (x1 )ψ1 (x2 ) − ψ1 (x1 )ψ0 (x2 ))χ↑↑ From our theorem about additive Hamiltonians, the energy of this state is the sum of the energies of the 2 single particle states 0 and 1: Eground = E0 + E1 = h̄2 (π/a)2 h̄2 (2π/a)2 5h̄2 (π/a)2 + = 2m 2m 2m (c) What is the energy and the wave function of the first excited state (still with equal-spin condition!)? Ψ1 = (ψ0 ψ2 − ψ2 ψ0 )χ↑↑ Eexc,1 = E0 + E2 = 5h̄2 (π/a)2 h̄2 (π/a)2 h̄2 (3π/a)2 + = 2m 2m m 5