# PS3sol - Solutions to Problem Set 3 Physics 341 by Callum...

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Unformatted text preview: Solutions to Problem Set 3 Physics 341 by: Callum Quigley 1 Some Nice Matrix Relations In the first part, we can save a bit of tedious algebra by first showing [ AB,C ] = A [ B,C ] + [ A,C ] B (1.1) which also implies [ A,BC ] = B [ A,C ] + [ A,B ] C . Applying this relation, we have [ AB,CD ] = A [ B,CD ] + [ A,CD ] B = A [ B,C ] D + AC [ B,D ] + C [ A,D ] B + [ A,C ] DB (1.2) For the second part, if e C = e λA e λB then (at least formally) we have C = log ( e λA e λB ) . (1.3) To make sense of the above statement, we must Taylor expand (being careful about the ordering of A s and B s): C = log " X m,n ≥ 1 m ! 1 n ! λ n + m A m B n # = log " 1 + X k ≥ 1 λ k k ! Y k # with Y k = k X p ≥ k p A k- p B p = X N> (- 1) N +1 N X k ≥ 1 λ k k ! Y k ! N (1.4) = λY 1 + 1 2 λ 2 ( Y 2- Y 2 1 ) + 1 12 λ 3 (2 Y 3- 3 Y 1 Y 2- 3 Y 2 Y 1 + 4 Y 3 1 ) + + 1 24 λ 4 ( Y 4- 2( Y 1 Y 3 + Y 3 Y 1 )- 3 Y 2 2 + 4( Y 2 1 Y 2 + Y 1 Y 2 Y 1 + Y 2 Y 2 1 )- 6 Y 4 1 ) + ... At O ( λ ), we simply have Y 1 = A + B . At O ( λ 2 ), we have Y 2- Y 2 1 = A 2 + 2 AB + B 2- ( A + B )( A + B ) = A 2 + { A,B } + [ A,B ] + B 2- ( A 2 + { A,B } + B 2 ) (1.5) = [ A,B ] 1 At O ( λ 3 ) and higher, the algebra gets a little nasty. Let me spare you the details, and just present the final result up to O ( λ 4 ) C = log ( e λA e λB ) = λ ( A + B ) + 1 2 λ 2 [ A,B ] + 1 12 λ 3 [ A- B, [ A,B ]]- 1 24 λ 4 [ A, [ B, [ A,B ]]] + ... (1.6) In the case A = X and B = P , note that [ X,P ] = i ~ commutes with everything, so the expansion stops at O ( λ 2 ), and we simply have e λX e λP = e λ ( X + P )+ i ~ λ 2 / 2 (1.7) 2 Density Matrices Recall that if a system has probability p k of being in the state | ψ k i , then the density matrix of the system is defined by ρ = X k p k | ψ k ih ψ k | . (2.1) (We’re assuming the {| ψ k i} form a complete orthonormal basis, so ∑ k p k = 1.) Since p k ∈ R , then ρ † = X p k | ψ k ih ψ k | † = X p * k ( h ψ k | ) † ( | ψ k i ) † = X p k | ψ k ih ψ k | = ρ. (2.2) Also, using the completeness and orthonormality of {| ψ k i} , we see that Tr ρ = X k h ψ k | ρ | ψ k i = X k,‘ p ‘ |h ψ k | ψ ‘ i| 2 = X k,‘ p ‘ δ k‘ = X k p k = 1 , (2.3) while Tr ρ 2 = X k h ψ k | ρ 2 | ψ k i = X k,‘,m p ‘ p m h ψ k | ψ ‘ ih ψ ‘ | ψ m ih ψ m | ψ k i = X k,‘,m p ‘ p m δ k‘ δ ‘m δ mk = X k p 2 k = ( X k p k ) 2- X k 6 = ‘ p k p ‘ = 1- 2 X k<‘ p k p ‘ ≤ 1 . (2.4) For a pure ensemble, we have ρ = | ψ ih ψ | for some state | ψ i = ∑ k c k | ψ k i and so ρ 2 = ( | ψ ih...
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PS3sol - Solutions to Problem Set 3 Physics 341 by Callum...

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