Selection rules - electric dipole As an example, lets take electric dipole transitions; when is j, m z j 2, m 2 nonzero so that j 1 = 1 and m 1 = 0. The answer is equivalent to the question when can j 1, j 2, j; m j 1, j 2, 0, m 2 be nonzero. This means that j j 2 1 and m = m 2. However, it turns out that 1, j 2, j; m 1, j 2, 0, m must be zero by parity, so we have in fact only j = j 1 ± 1. j, m T j 1 m 1 j 2, m 2 = N j1,j 2,j j 1 j 2, jm j 1 j 2, m 1 m 2 1
Electric quadrupole As a second example, lets take electric quadrupoletransitions; j, m x 2 y 2 j 2, m 2 nonzero. (I have not bothered working out exactly what value of m 1, the operator x 2 y 2 corresponds to.) The answer is equivalent to the question when can 2, j 2, j; m 2, j 2, m 1, m 2 be nonzero. This means that j j 2 2 and m m 2 2. However 2, j 2, j; m 2, j 2, m 1, m 2 is zero unless j + j 1 + 2 is even, so in fact only j = j 2 ± 2 or j = j 2 is allowed and j = j 2 ± 1 is forbidden. j, m T j 1 m 1 j 2, m 2 = N j1,j 2,j j 1 j 2, jm j 1 j 2, m 1 m 2 2
Lecture 11, Density matrices, entanglement Density matrix, pure and mixed ensembles Density submatrix and unitary invariance Bell states and entanglement 3
Recall the Stern-Gerlach experiment... out of the oven comes unpolarized atoms 4
Mixtures We need to define the expectation of an experiment done on a mixture of quantum states: Let [ A ] be the average of a quantum measurment and let w i denotes the probability of being in state "i". let ρ = i i w i i [ A ] = i = i w i i A i i ρa i = tr ρ A 5
Properties of ρ Note that ρ is symmetric and Hermitian, hence can be diagonalized. We have tr ρ = 1 and for an ensemble consisting of a pure state 1 0 0... ρ =... 0 so that ρ 2 = ρ and tr ρ = 1 is necessary and sufficient for ρ to correspond to a pure state. In this case, ρ = α α. 6
Examples, spin 1/2 (a) Let +z polarized state: w = 1 and ρ = = 1 0 0 0 (b) Let z polarized state: w = 1 and ρ = = 0 0 0 1 7
Examples... (c) Let +x polarized state: w x = 1. Then ρ = x x = ( + ) ( + ) /2 = 1 1 1 2 1 1 (d) Let x polarized state: w x = 1. Then ρ = x x = ( )( )/2 = 1 1 1 2 1 1 ρ 2 = ρ ( check ) ( check ρ 2 = ρ ) 8
Mixtures (f) Mixture +x polarized and x polarized: ρ = ρ +x + ρ x = = 1 1 1 + 1 1 4 1 1 1 1 = 1 1 0 = ρ 2 +z + ρ z 0 1 i.e. this uniform mixture is rotation invariant. (Completely unpolarized) 9
Time evolution Note that Hence [A(t)] = tr ρ A(t) = tr ρ U (t)au(t) = tr U(t)ρ U (t)a = tr U ( t)ρ U( t)a = tr ρ( t) A i ρ t = [ ρ, H ] i.e. it transform opposite to an operator! 10
Entropy We define the entropy S of a mixture as S = k i w i log w i = k tr ρ log(ρ) (which base does not matter since tr ρ = 1 which just shifts the entropy by a constant) Assertion: maximum S implies w i = constant i.e. w i = 1/N where N is the number of states. Proof: Introduce Lagrange multiplier λ. ( ) 0 = ( k w j log(w j ) λw j ) w i j 0 = k log(w i ) + 1 + λ. p.8/16 11
Thermal equilibrium Thermal equilibrium implies ρ = const i.e. [ ρ, H ] = 0 Hence, simultaneous eigenvalues of H and ρ exist. Now maximize entropy subject to fixed energy expectation. 12
Partition function Introduce LaGrange multipliers for i w i and energy j e jw j : 0 = w i j ( k w j log(w j ) λw j γw j e j ) = k(1 + log(w i )) λ γe i i.e. defining Z = e 1+λ/k and identify γ = 1/T : w i = e e i/(kt ) /Z Z = i e e i/(kt ) 13
Statistical mechanics With the identification of the probability distribution simply from the constancy of the density matrix, statistical mechanics follow. Thermodynamics allows us to identify Z = e F/(kT ) 14
Density submatrices Consider two subsystems A and B and a combined wave function, for example the ground state. ψ = αβ C αβ α β Let A be an operato acting "only on A" A = A 1. A B 15
Density submatrices A = αβα β C αβ β α A α β C α β = αβα β C αβc α β α A α β β = αβα C α βc βα α A α = tr ρ A where ρ = β C α βc βα CC (note: C is not a square matrix!) 16
Entanglement The density submatrix: ρ A = β C α βc βα a entropy: The entanglement S = tr ρ A log ρ A = tr ρ B log ρ B where the last equality is not obvious! (Prove it!) a Bennet et al, Phys. Rev. A 54, 3824 (1996) 17
Consistency check What do we need of sensible definition of entanglement: It should be invariant under separate choices of bases of A and B. Lets check that this works. Let α U A α and β U B β Then i.e. C αβ (U A ) αα (U B ) ββ C α β C U A CU T B ) CC ( U A CU T B) ( U BC U A Now we have U T B U B = (U B U B) T = 1 ρ U A ρu A. p.15/1 Since all the eigenvalues are preserved, so is the entropy 18
Bell states In the two spin 1 problem, there are four "Bell states" with 2 maxmimal entropy that have nice properties a e 1 = 1 ( + ) 2 e 2 = 1 i( ) 2 e 3 = 1 i( + ) 2 e 4 = 1 ( ) 2 Note that e 4 is the singlet Bell state used in the EPR discussion. a S. Hill and W.K. Wooters, Phys. Rev. Lett. 78, 5022 (1997) 19
Symmetries and conservation laws 20
Continuous symmetries Assume we have an operator which commutes with the Hamiltonian [ G, H ] = 0. This generates a symmetry operation: where by symmetry we mean G = 1 iɛ G G HG = H 21
Conservation laws Multiply this out (1 + iɛ ) G H (1 iɛ G ) i.e. This in turn implies iɛ [ G, H ] = 0 dg dt = [ G, H ] = 0 i.e. conservation law. 22
Degeneracies Since [ G, H ] = 0, G and H can be simultaneously diagonalized HG ɛ = GH ɛ H(G ɛ ) = ɛ(g ɛ ) (1) i.e. if G ɛ is not invariant, the energy eigenvalue ɛ is degenerate. 23
Parity Example 1: Let π be the parity operator α π α We define the parity operator to have the action π rπ = r i.e. it takes space to minus itself. If the spatial dimension is odd, parity is disconnected from identity. Otherwise it is a rotation by (the angle) π. 24
parity... continued π = 1 0 0 1... 1 i.e. det π = ( 1) d. Hence it cannot be part of the rotations in odd dimension since det R = 1. 25
Momemtum r p p r = i Multiply left by π and right by π: π (rp)π π (pr)π = i (π rπ)(π pπ) (π pπ)(π rπ) = i r(π pπ) + (π pπ)r = i i.e. π pπ = p so that p is also a vector. 26
Angular momentum Since π 2 = 1 we must have eigenvalues of π = ±1. Pseudovectors and vectors are even or odd under parity. π rπ = r hence r is odd and a vector and same for p. Note that L = r p so that π Lπ = ( r) ( p) = r p = L i.e. the angular momemtum is a positive eigenoperator and is called a pseudovector. 27
Scalars and pseudoscalars scalars: π r 1 r 2 π = ( r 1 ) ( r 2 ) = r 1 r 2 Hence dot product of two pseudovectors or two vectors is a a scalar. Similarly r L is a pseudoscalar, which is odd under parity. 28
Wavefunctions Wavefunctions under parity Note that L± commutes with parity. Hence the entire multiplet l, m for l m l has to have the same transformation. Note that l, l (x + iy) l hence π l, m = ( 1) l l, m 29
Degeneracies Suppose [ H, π ] = 0. Then nondegenerate eigenstates are also eigenstates of parity. The usual argument: H(π α ) = πh α H(π α ) = e α (π α ) i.e. if e α is nondegenerate, π α must be the same state. 30
Selection rules (Mini Wigner-Eckart theorem) Assume π α = ɛ α and π β = ɛ β. Then α (π xπ) β = ( α π )x(π β ) α x β = ɛ α ɛ β α x β Hence α x β = 0 unless ɛ α ɛ β = 1. Similar arguments for other operators with definite signatures under parity. 31