851HW2_09_Solutions - PHYS851 Quantum Mechanics I Fall 2009...

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PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 2: Postulates of Quantum Mechanics 1. [10 pts] Assume that A | φ n ) = a n | φ n ) but that ( φ n | φ n ) negationslash = 1. Prove that | a n ) = c | φ n ) is also an eigenstate of A . What is its eigenvalue? What should c be so that ( a n | a n ) =1? Solution: A | a n ) = Ac | φ n ) = cA | φ n ) = ca n | φ n ) = a n ( c | φ n ) ) = a n | a n ) . ( a n | a n ) = | c | 2 ( φ n | φ n ) . Setting equal to unity requires c = 1 radicalbig ( φ n | φ n ) . 2. [10 pts] Assume that | φ n ) and | a n ) are degenerate eigenstates of A with eigenvalue a n . They sat- isfy ( φ n | φ n ) = ( a n | a n ) = 1 and ( φ n | a n ) negationslash = 1. Show that the states | a n , 1 ) = | a n ) and | a n , 2 ) = | φ n )−| a n )( a n | φ n ) ||| φ n )−| a n )( a n | φ n )|| are both eigenstates of A with eigenvalue a n , and are mutually orthogonal. Now assume that the system is in an arbitrary state | ψ ) when A is measured. Write an expression for the probability to obtain the result a n . Write down the state of the system immediately after the measurement, assuming that a n was obtained by random chance. Solution: A | a n , 1 ) = A | a n ) = a n | a n ) Let N = ||| φ n ) − | a n )( a n | φ n )|| , A | a n , 2 ) = A | φ n ) − | a n )( a n | φ n ) N = A | φ n ) − A | a n )( a n | φ n ) N = a n | φ n ) − a n | a n )( a n | φ n ) N = a n | φ n ) − | a n )( a n | φ n ) N = ( a n , 1 | a n , 2 ) = ( a n | | φ n ) − | a n )( a n | φ n ) N = ( a n | φ n ) − ( a n | a n )( a n | φ n ) N = ( a n | φ n ) − ( a n | φ n ) N = 0 . P ( a n ) = ( ψ | parenleftBig | a n , 1 )( a n , 1 | + | a n , 2 )( a n , 2 | parenrightBig | ψ ) = |( a n , 1 | ψ )| 2 + |( a n , 2 | ψ )| 2 | ψ ) = 1 radicalbig P ( a n ) parenleftBig | a n , 1 )( a n , 1 | + | a n , 2 )( a n , 2 | parenrightBig | ψ ) = 1 radicalbig P ( a n ) parenleftBig | a n , 1 )( a n , 1 | ψ ) + | a n , 2 )( a n , 2 | ψ ) parenrightBig 1
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3. [20 pts] Consider the wavefunction φ ( x ) ≡ ( x | φ ) = N e ( x x 0 ) 2 2 σ 2 . What should the value of N be so that ( φ | φ ) = 1?
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