Solutions to exam : 1FA352 Quantum Mechanics 10 hp 1

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Lawrence Russell
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1 Solutions to exam 6--6: FA35 Quantum Mechanics hp Problem (4 p): (a) Define the concept of unitary operator and show that the operator e ipa/ is unitary (p is the momentum operator in one dimension) (b) Show that e ipa/ x is an eigenstate to the position operator What is the eigenvalue? Hint: Consider [x, e ipa/ ] (a) An operator U is unitary if it satisfies U U = UU = With U = e ipa/ where p is Hermitian, it follows that U = e ipa/ and, therefore, U U = e ipa/ e ipa/ = = e ipa/ e ipa/ = UU (b) With the position operator x x = x x, we should demonstrate that x e ipa/ x = C e ipa/ x where C is the eigenvalue: x e ipa/ x = {[x, e ipa/ ] + e ipa/ x} x = ( using [x i, F ( p)] = i F p i from the collection of formulae ) = i eipa/ x + e ipa/ x x = i ia p eipa/ x + e ipa/ x x = ( a + x )e ipa/ x, ie eigenstate of x with eigenvalue x a This means a translation by a, which is in accordance with e ipa/ being the translation operator Problem (4 p): Consider the one-dimensional harmonic oscillator: states n, energies E n = ω(n + ) (a) Show that the expectation value n x n = for all n (b) Construct the state α = c + c that makes α x α as large as possible (c, c can be chosen real) (c) With the oscillator in state α at time t =, derive the state vector (in Schrödinger picture) for t > and calculate the expectation value α x α as a function of time (a) Using x = mω (a+a ) and a n = n n, a n = n + n+ (from collection of formulae) one obtains n x n = mω n a+a ( ) n = mω n n n + n + n n + = due to orthonormality m n = δ mn of the energy eigenstates of the harmonic oscillator In words, operator x is given by creation and annihilation operators that shift the state n to n ± which is orthogonal to n, giving zero expectation value (b) x = α x α = (c + c )x(c + c ) = c c x + c c x + c c x + c c x Using x, a, a as in (a) one obtains x = mω (c c + c c ) = c mω c, where in the last step the normalisation condition c + c = has been used together with c, c real since a complex phase does not affect the following Maximum of x is then obtained by d x /dc = giving c = / (maximum since d x /dc < for c = / ), and then c = / Hence x max = /mω for α = ( + )/ (up to an overall phase factor) (c): Applying the time evolution operator e iht/, and using e iht/ n = e iω(n+/)t n, one gets α, t = e iht/ α, t = = e iht/ ( + )/ = (e iωt/ + e 3iωt/ )/ as the state vector for t > in the Schrödinger picture The expectation value is then α, t x α, t = / (e iωt/ +e 3iωt/ )x(e iωt/ +e 3iωt/ )/ = ( x +e iωt x + e iωt x + x )/ = (again using x a + a as in (a)) = /mω cos ωt

2 Solutions to exam 6--6: FA35 Quantum Mechanics hp Problem 3 (4 p): (a) Derive the relation m m j j ; m m j j ; jm = for Clebsch-Gordan coefficients and comment on its physical interpretation (b) An electron is in a p-orbital (l = ) Give, with motivations/explanations, all possible states jm of total angular momentum j of the electron, expressed in terms of its spin and orbital angular momentum (a) When combining two angular momenta j, j, the new states with total angular momentum j can be expressed (using the completeness relation) as j j ; j, m = m m j j ; m m j j ; m m j j ; j, m in the old direct-product basis Multiplying with j j ; jm gives the left-hand side j j ; jm j j ; jm = and the right-hand side m m j j ; jm j j ; m m j j ; m m j j ; jm = m m j j ; m m j j ; jm, ie the desired relation is obtained Since the Clebsch-Gordan coefficients are the expansion coefficients of the new j j ; jm basis states in the old direct-product basis, the given relation expresses the conservation of probability, ie the sum of squared amplitude coefficients should be unity (b) Add orbital angular momentum l = (m l =,,,) and spin angular momentum s = / (m s = /, /) to total angular momentum j = l s,, l + s = /, 3/ (m j = j, j +, j, j for j = / and j = 3/, resp) and m j = m l + m s Express new states (eigenstates to L, S, J, J z ) as linear combinations in old direct product basis (eigenstates to L, S, L z, S z ) using change of basis via completeness relation: l, s; j, m j = m l m s l, s; m l, m s l, s; m l, m s l, s; j, m j, where the scalar products are the expansion coefficients called Clebsch-Gordan coefficients Reading them from the table gives (suppressing l, s): j = 3, m j = + 3 = m l = +, m s = + j = 3, m j = + = +, +, + j =, m 3 3 j = + = +,, j = 3, m j = =, +, + j =, m 3 3 j = =,, j = 3, m j = 3 = m l =, m s = Here one should note the orthogonality between all states, in particular those on the same line due to the relative + and signs To derive all these states, one starts with the extreme state j = 3, m j = + 3 = m l = +, m s = + with coefficient (chosen real by convention) since only possibility is maximum m l and m s Then one operates with ladder operators J = L +S on respective sides giving j = 3, m j = + = above Iterating with the ladder operators gives the following states with j = 3/ The state j =, m j = + is then constructed to be orthogonal to j = 3, m j = + and have sum of squared coefficients equal unity The ladder operators are then iteratively applied to get the remaining j = / states Problem 4 (5 p): A hydrogen atom in its ground state nlm = is at time t = placed in a time-dependent electric field E(t) = E ẑ exp( t/τ) along ẑ direction, with E the field strength (a) In first-order time-dependent perturbation theory, which ones of the excited states lm are accessible in a direct transition? Motivate your answer (b) Calculate the transition probability from the ground state to these excited states You do not need to perform the radial integral, but perform all other integrals Hint: Energy levels: E n = e a ; Bohr radius: a n = ; Y me (θ, φ) =

3 Solutions to exam 6--6: FA35 Quantum Mechanics hp 3 (a) The perturbation is an electric field E(t) along the ẑ direction, which gives a perturbation potential dependent on z, V (t) = ee z exp( t/τ), using the potential in electrodynamics The initial state is and the final states are lm, with l =, and m =,, + The transition amplitudes in the zero th and first order are c () (t) = δ, c () lm (t) = i t with the difference in the energy levels is ω = E E elements of the perturbation dt e iω t V lm, (t ), = e ( 3e a ) = 4 8a V lm, (t ) = lm ( ee ) ze t /τ = ee e t /τ lm z and the matrix The possible transitions from to lm are the ones where the matrix elements of the operator z are non-zero The parity of the z operator is odd and lm states (or Yl m ) have parity ( ) l, which excludes transition from to states with even l Moreover, z is the zero-component of a rank- spherical tensor (z Y ), so the Wigner-Ekhardt theorem applies and gives the selection rules l = l ± and m = m Therefore, the only non-zero matrix element is z (which can also be obtained by detailed inspection of the polar angle integrals) Thus, the only direct transition in first-order perturbation theory is (b) We calculate the matrix element z using the position representation and write the wave function in spherical coordinates Ψ nlm (x) = x nlm = R nl (r) Yl m (θ, φ), z = d 3 x Ψ (x) z Ψ (x) = d 3 x R (r)y (θ, φ) z R (r)y (θ, φ) = r drdω R (r)y (θ, φ) r cos θ R (r)y (θ, φ) = dr r R (r)r (r) dω Y (θ, φ) cos θy (θ, φ) where we have separated the radial part and the angular part The radial integral does not need to be estimated and we can denote it I r = dr r R (r)r (r) = = 4 6 a ( 3) 5 The angular integral can be estimated using the spherical harmonics Y and Y π 3 I θ,φ = dω Y (θ, φ) cos θy (θ, φ) = dφ d(cos θ) cos θ cos θ 3 = π d(cos θ) cos θ = 3 The transition amplitude to first order is c () (t) = i = i ( ee ) t t dt e iω t V, (t ) = i t dt e iω t ( ee ) e t /τ z dt e iω t e t /τ I r I θ,φ = iee exp(iω t t ) τ iω I r I θ,φ τ The transition probability for is finally given by ( ) P (t) = c () (t) ee [ ] = + e t τ e t τ cos(ω t) ω + Ir 3 τ

4 Solutions to exam 6--6: FA35 Quantum Mechanics hp 4 Problem 5 (4 p): Consider the elastic scattering of particles, with mass m and wave vector k, on an atom represented by a 3D δ-potential V (x) = V δ(x), with V > (a) Calculate the differential scattering cross-section and the total scattering cross-section using the Born approximation (b) Perform the same calculations for a molecule represented by a double δ-potential V (x) = V (δ(x + R) + δ(x R)), where R k (a) The scattering amplitude in Born approximation is given by f () (k, k) = m (π) 3 k V k = m d 3 x e i(k k) x V (x ) Using the single δ-potential V (x) = V δ(x), the expression becomes f () (k, k) = m d 3 x e i(k k) x ( V )δ(x ) = m V k) ei(k = mv π The differential scattering cross-section and the total scattering cross-section become ( ) dσ dω = f () (k, k) mv =, σ π tot = dω dσ ( ) dω = mv σ π atom (b) Using the double δ-potential V (x) = V (δ(x + R) + δ(x R)), the amplitude is f () (k, k) = m d 3 x e i(k k) x ( V )(δ(x +R)+δ(x R)) = m V ( ) e i(k k) R + e i(k k) R With the notation that θ = (k, k) and considering that R k, the scattering amplitude becomes f () (k, k) = mv π ( e ikr( cos θ) + e ikr( cos θ)) The differential scattering cross-section becomes ( ) dσ dω = f () (k, k) mv ( = + e ikr( cos θ) + e ikr( cos θ)), π and the total scattering cross-section after performing the angular integrals is σ tot = dω dσ dω = π d(cos θ) dσ ( ) ( dω = mv + sin(4kr) ) π 4kR Finally, we note that ( σ molecule = σ atom + sin(4kr) ) 4kR

5 Solutions to exam 6--6: FA35 Quantum Mechanics hp 5 Problem 6 (3 p): To determine possible violation of the Bell inequality, one needs to evaluate expectation values of operators on the form a σ b σ, where a = (a x, a y, a z ) and b = (b x, b y, b z ) are unit vectors and σ= (σ x, σ y, σ z ) are Pauli matrices Suppose Alice and Bob share the Bell state Φ + = ( + ) (a) Calculate the expectation value Φ + a σ b σ Φ + (b) Suppose these are spin / states and Alice decides to measure S z, ie she chooses her vector a = (,, ) Which measurement must Bob choose to get a maximum expectation value? (a) With the notation a = (a x, a y, a z ) and b = (b x, b y, b z ) we can calculate the expectation value using the properties of the Pauli matrices σ x, σ y, σ z Φ + a σ b σ Φ + = ( + )(a σ b σ)( + ) = ( a σ b σ + a σ b σ ) + a σ b σ + a σ b σ = [ a z b z + (a x ia y ) (b x ib y ) ] + (a x + ia y ) (b x + ib y ) + ( a z ) ( b z ) = a x b x a y b y + a z b z Here we made use of a σ = a x σ x + a y σ y + a z σ z = a z, and so on (b) If Alice measures S z, then a x =, a y =, a z =, in which case the expectation value becomes Φ + a σ b σ Φ + = b z The maximum value is, when Bob also measures S z, ie b = (,, ) Simply put, this corresponds to the case where Alice and Bob measure S z on entangled spins and discover their measurements are fully correlated

Quantum Mechanics Solutions

Quantum Mechanics Solutions (a (i f A and B are Hermitian, since (AB = B A = BA, operator AB is Hermitian if and only if A and B commute So, we know that [A,B] = 0, which means that the Hilbert space H