PHY4604 Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004


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1 PHY464 Introduction to Quantum Mechanics Fall 4 Practice Test 3 November, 4 These problems are similar but not identical to the actual test. One or two parts will actually show up.. Short answer. (a) Recall the Bohr energy levels of the Hydrogen atom are E n = m h n ( e 4πɛ ) () Compare the wavelengths of the p s transitions in i. hydrogen The energy which must be carried away by a photon in the transition is E E = hc/λ, so λ H = hc/(e E ) = 8 h3 3m ( 4πɛ e ) ii. deuterium (mass of nucleus that of H). The mass which appears in the Bohr formula is the electron mass m, which is replaced by the reduced mass µ = mm p /(m+m p ) if one allows for the proton motion. Changing m p m p doesn t change much, so to a good approximation the photon wavelength doesn t change, λ D λ H. iii. positronium. In this case the positron is not heavier than the electron, but in fact has the same mass. The reduced mass of identical particles of mass m is m/, so we obtain the correct answer from the Hydrogen expression by substituting m m/, or λ pos = λ H. (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
2 and a neutron. Calculate the probability that the electron remains in the ground state of 3 He. You may need ψ = (/ 4πa 3 ) exp r/a, with a = h /(mze ), where Z is the nuclear charge. Z, the nuclear charge or number of protons, is for tritium and for 3 He. Thus the Bohr radius for 3 He is twice as small as for tritium or hydrogen, a 3 He at. 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, ψ t = e r/at 4π(a H ) 3 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, ψ t = aψ 3 He + bψ 3 He + cψ 3 He +... meaning the electron in the tritium ground state will be found after a measurement in the 3 He ground state with probability amplitude a = ψ t ψ 3 He = 4 4π = (a t ) 3 (a 3 He ) 3 d 3 r e r 4 8 dr r e 3r/at (a t ) 3 = 4 8 and probability = a =.7. dy y e 3y = 6 7, ( a t + a 3 He =.838 (c) An electron moving in the Coulomb field of a proton is in a state described by the wave function Ψ = 6 [4ψ + 3ψ ψ + ψ ] () i. What is the expectation value of the energy? ) Ψ H Ψ = (6 H + 9 H + H 36 + H ) = 36 (6E + 9E + E + E ) = 36 (6E + E ). Note this is an example of general rule ψ O ψ = n o n c n, where the o n are the eigenvalues of the Hermitian operator O, and the c n the expansion coefficients of ψ in the basis of O eigenstates.
3 ii. What is the expectation value of ˆL? ψ L ψ = h l(l + ) c nlm = h h ( ) = nlm 36 9 iii. What is the expectation value of ˆL z? ψ L z ψ = hm c nlm = h nlm 36 ( ( )) = h 36 (d) How large would a constant magnetic field have to be to split two Hatom states which are degenerate in zero field by an amount so as to maximally absorb light of wavelength λ? H = µ B = e m S B = = e m S zb z = e m hm sb z, where the last step where the operator is replaced by its eigenvalues holds only when applied to S z eigenstates, and where I have used m s for the S z quantum number. The two S z states have a difference of m s =, so the energy of the photon produced must be e m hb z hc λ, or λ = (πmc/(eb z)). (e) For two particles a and b such that l a = and l b =, argue that it must be true that l =, m = = α m a =, m b = + β m a =, m b =, where l, m are the quantum numbers corresponding to total angular momentum L = L a + L b, and find the coefficients α and β. Hint: use L ± lm = l(l + ) m(m ± ) lm ±. a) states given are the only possible ones with m = m a + m b and m a l a, m b l b. b) Start at top of angular momentum ladder, where we know there is only one possible lm = l state, equal to the one possible m a = l a, m b = l b, then apply lowering operator as in HW, remembering square root factors to keep states normalized: L lm = (L a + L b ) mam b 3 = ( + ) 3
4 So divide by to get = ( + ).. Angular momentum. Consider an angular momentum system. (a) What are the possible eigenvalues of L and L z corresponding to the eigenvectors l =, m? L m = h L z m = hm (b) In the basis where the eigenvectors l =, m of the operator ˆL z are given by (,, ), (,, ) and (,, ), construct the matrix representation of the operator ˆL x (Hint: you will need to calculate the matrix elements m ˆL x m.) Write L x = (L + + L )/, find, e.g. L + + L = L + + L = h L + + L =... where I ve used the effect of L ± acting on m, and the orthonormality of the m. It is a little tedious to do them all, but eventually we find (c) Find the eigenvectors of ˆL x. L x = h Now we just have a matrix eigenvalue problem, and I assume you can find the eigenvectors, then normalize them. The results are = / / / = / / = / / / 4
5 (d) If a system is prepared in the state vector ψ = 6 4 3, (3) what is the probability that a measurement of ˆL x yields the value? P = ψ = 4/3 3. Pauli principle. Consider two electrons described by the Hamiltonian where H = ˆp m + ˆp m + V (x ) + V (x ) (4) x < a/ V (x) = a/ x a/ x > a/ Assume both electrons are in same spin state. (a) What is the lowest (ground state) energy? (b) What is the energy eigenfunction for this ground state? Parts a) and b): Denote the singleparticle eigenfunctions of the ordinary infinite square well by ψ, ψ,... The twoparticle 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 (5) Ψ (, ) = (ψ (x )ψ (x ) ψ (x )ψ (x ))χ From our theorem about additive Hamiltonians, the energy of this state is the sum of the energies of the single particle states and : E ground = E + E = h (π/a) m + h (π/a) m = 5 h (π/a) m (c) What is the energy and the wave function of the first excited state (still with equalspin condition!)? Ψ = (ψ ψ ψ ψ )χ E exc, = E + E = h (π/a) m + h (3π/a) m = 5 h (π/a) m 5
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