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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 3: Solutions Fundamentals of Quantum Mechanics 1. [10pts] The trace of an operator is defined as Tr { A } = ∑ m ( m  A  m ) , where { m )} is a suitable basis set. (a) Prove that the trace is independent of the choice of basis. Answer: Let { m )} and { e m )} be two independent basis sets for our Hilbert space. We must show that ∑ m ( e m  A  e m ) = ∑ m ( m  A  m ) . Proof: summationdisplay m ( e m  A  e m ) = summationdisplay mm ′ m ′′ ( e m  m ′ )( m ′  A  m ′′ )( m ′′  e m ) (1) = summationdisplay mm ′ m ′′ ( m ′′  e m )( e m  m ′ )( m ′  A  m ′′ ) (2) = summationdisplay m ′ m ′′ ( m ′′  m ′ )( m ′  A  m ′′ ) (3) = summationdisplay m ′ m ′′ δ m ′ m ′′ ( m ′  A  m ′′ ) (4) = summationdisplay m ′ ( m ′  A  m ′ ) (5) = summationdisplay m ( m  A  m ) (6) (b) Prove the linearity of the trace operation by proving Tr { aA + bB } = aTr { A } + bTr { B } . Answer: Tr { aA + bB } = summationdisplay m ( m  aA + bB  m ) (7) = summationdisplay m ( a ( m  A  m ) + b ( m  B  m ) ) (8) = a summationdisplay m ( m  A  m ) + b summationdisplay m ( m  B  m ) (9) = aTr { A } + bTr { B } (10) 1 (c) Prove the cyclic property of the trace by proving Tr { ABC } = Tr { BCA } = Tr { CAB } . Answer: First, if Tr { ABC } = Tr { BCA } then it follows that Tr { BCA } = Tr { CAB } , so we need only prove the first identity. Tr { ABC } = summationdisplay m ( m  ABC  m ) (11) = summationdisplay mm ′ m ′′ ( m  A  m ′ )( m ′  B  m ′′ )( m ′′  C  m ) (12) = summationdisplay mm ′ m ′′ ( m ′′  C  m )( m  A  m ′ )( m ′  B  m ′′ ) (13) = Tr { CAB } (14) 2 2. Consider the system with three physical states { 1 ) ,  2 ) ,  3 )} . In this basis, the Hamiltonian matrix is: H = 1 2 i 1 − 2 i 2 − 2 i 1 2 i 1 (15) Find the eigenvalues { ω 1 ,ω 2 ,ω 3 } and eigenvectors { ω 1 ) ,  ω 2 ) ,  ω 3 )} of H . Assume that the initial state of the system is  ψ (0) ) =  1 ) . Find the three components ( 1  ψ ( t ) ) , ( 2  ψ ( t ) ) , and ( 3  ψ ( t ) ) . Give all of your answers in proper Dirac notation. Answer: The eigenvalues are solutions to det  H − planckover2pi1 ωI  = 0 (16) Taking the determinate in Mathematica gives 4 ω + 4 ω 2 − ω 3 = 0 (17) which factorizes as ω ( ω 2 − 4 ω − 4) = 0 (18) which has as its solutions ω 1 = 2(1 − √ 2) (19) ω 2 = 0 (20) ω 3 = 2(1 + √ 2) (21) the corresponding eigenvectors are  ω 1 ) = 1 2 (  1 ) + √ 2 i  2 ) +  3 ) ) (22)  ω 2 ) = 1 √ 2 ( − 1 ) +  3 ) ) (23)  ω 3 ) = 1 2 (  1 ) − √ 2 i  2 ) +  3 ) ) (24) The components of  ψ ( t ) ) are found via  ψ ( t ) ) = e − iHt  ψ (0) ) , giving ( 1  ψ ( t ) ) = 1 4 parenleftBig 2 + e − i 2(1 − √ 2) t + e − i 2(1+ √ 2) t parenrightBig (25) ( 2  ψ ( t ) ) = i 2 √ 2 parenleftBig e −...
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This note was uploaded on 12/02/2010 for the course PHYSICS 851 taught by Professor M.moore during the Fall '09 term at Michigan State University.
 Fall '09
 M.Moore
 mechanics, Work

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