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

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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, A , is expressed in a basis, {| n i} , its matrix elements are A mn = h m | 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 | i is a stationary state then H| i = E | 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- I ), A- - K- K * B- = ( A- )( B- )- K 2 = 2- ( A + B )+( A B- K 2 ) = 0 (3a) The roots to this equation are, = ( 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, = 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 | (0) i = | A i in terms of our eigenkets | + i and |-i , | (0) i = | A i = 1 2 ( | + i + |-i ) (4a) Therefore our time-dependent wave function is, | ( 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, | ( t ) i = 1 2 ( | + i e iKt/ ~ + |-i e- iKt/ ~ ) = e iKt/ ~ 1 2 ( | + i e it + |-i e- it ) (4c) 2 E A m + 2 + K 2 B m- 2 + K 2 m 2 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 Figure 1: We have defined...
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PS 3 Sol - CHEM 120A Problem Set 3 Solutions David Hoffman...

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