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PS 3 Sol - CHEM 120A Problem Set 3 Solutions David Homan...

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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 Bra-Ket 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 0 and | B i = 0 1 (2a) So H| A i = A - K - K * B 1 0 = A - K * = A 1 0 - K * 0 1 (2b) And again neither | A i or | B i are eigenvectors of the Hamiltonian. 1
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(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 Bra-Ket 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 time-dependent 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 time-dependent 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
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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 Δ = ( B - A ) / 2 and Δ m = ( B + A ) / 2 . Notice how as the energy difference between the two levels, Δ , gets larger E - approaches A and E + approaches B . 3
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Where on the last line we have used the fact that a ket multiplied by a complex number of modulus 1 represents the same state, you can see that this overall complex phase factor disappears whenever we calculate an expectation value. We find that the angular frequency is ω = K/ ~ so the frequency is ω/ 2 π = K/h .
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