This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 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  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 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  (0) i =  A i in terms of our eigenkets  + i and i ,  (0) i =  A i = 1 2 (  + i + i ) (4a) Therefore our timedependent 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 timedependent 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...
View
Full
Document
 Spring '10
 CHandler

Click to edit the document details