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Unformatted text preview: CHEM 120A Problem Set 3 Solutions David Hoffman , Tara Yacovitch , Doran Bennett 2/12/2010 1. The solution can be found in the McQuarrie Solutions manual. 2. Before we begin this problem it would be good to remember that when a given operator, ˆ A , is expressed in a basis, { n i} , it’s matrix elements are A mn = h m  ˆ A  n i . In addition the vector representation of the basis is e n = (0 ,..., 1 ,..., 0) where the 1 appears at the n th position. This will all be made clearer below. (a) I will demonstrate this problem using both BraKet notation and vector notation. From looking at H we can see that, A = h A H A i (1a) B = h B H B i (1b) K * = h B H A i (1c) K = h A H B i (1d) From this we can determine the effect of H operating on  A i and  B i which is, H A i = A  A i  K *  B i (1e) H B i = B  B i  K  A i (1f) Remember that if  φ i is a stationary state then H φ i = E  φ i . This does not hold for  A i and  B i and thus they are not stationary states. When K = 0,  A i and  B i are stationary states and the expectation value for the energy in each of those states is, h E A i = h A H A i = A (1g) h E B i = h B H B i = B (1h) In vector notation we would have  A i = 1 and  B i = 1 (2a) So H A i = A K K * B 1 = A K * = A 1 K * 1 (2b) And again neither  A i or  B i are eigenvectors of the Hamiltonian. 1 (b) This is an eigenvalue problem. First we must find the characteristic polynomial, det( A λ I ), A λ K K * B λ = ( A λ )( B λ ) K 2 = λ 2 λ ( A + B )+( A B K 2 ) = 0 (3a) The roots to this equation are, λ ± = ( A + B ) ∓ p ( A + B ) 2 4( A B K 2 ) 2 (3b) = ( A + B ) ∓ p ( A B ) 2 + 4 K 2 2 (3c) Which are the eigenvalues. The eigenvectors are, v ± = ( B A ) ± √ ( A B ) 2 +4 K 2 2 K 1 ! (3d) Note that these eigenvectors have not been normalized. If we say that A = B = then the eigenvalues/eigenvectors become, λ ± = ∓ K (3e) v ± = 1 √ 2 1 ± 1 (3f) Which in BraKet notation becomes, ±i = 1 √ 2 (  A i± B i ) (3g) A diagram of the energy levels of this system is drawn in figure 1 . (c) Here our eigenvalues and vectors are the ones shown in equations ( 3e ) and ( 3g ). We can expand our initial ket  Ψ(0) i =  A i in terms of our eigenkets  + i and i ,  Ψ(0) i =  A i = 1 √ 2 (  + i + i ) (4a) Therefore our timedependent wave function is,  Ψ( t ) i = 1 √ 2 (  + i e iE + t/ ~ + i e iE t/ ~ ) (4b) If we choose the zero of energy to be equal to A then our energies become E ± = ∓ K and our timedependent wave function is,  Ψ( t ) i = 1 √ 2 (  + i e iKt/ ~ + i e iKt/ ~ ) = e iKt/ ~ 1 √ 2 (  + i e iωt + i e iωt ) (4c) 2 E A Δ m + √ Δ 2 + K 2 B Δ m √ Δ 2 + K 2 Δ m 2 √ Δ 2 + K 2 (a) Energy level diagram for the two level system Minus 4 Minus 2 2 4 CapDelta Slash1 K K K E Slash1 K E A E B E Plus E Minus (b) A depiction of the splitting induced by the coupling of the two levels as a function of Δ Figure 1: We have defined...
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 Spring '10
 CHandler
 Cos, wave function, A

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