hw4sol_fa07

# hw4sol_fa07 - C/CS/Phy191 Problem Set 4 Solutions Out 1...

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Unformatted text preview: C/CS/Phy191 Problem Set 4 Solutions Out: October 14, 2007 1. Show that the trace of an operator is independent of the basis in which it is evaluated. Answer : Let {| i i} and {| j i} be two orthonormal bases. Then the trace of an operator A in the first basis is tr A = i i A | i i = i i A j | j ih j | ! | i i = i , j i A | j ih j | i i = i , j h j | i i i A | j i = j j A | j i , where we used the resolution of the identity, I = | j ih j | = | i ih i | , twice. 2. When is e ( A + B ) = e A e B ? In the case that they are not equal, estimate the difference to first order in the commutator [ A , B ] . Show all reasoning explicitly. Answer : If [ A , B ] = 0, then certainly e A + B = e A e B , and the proof is identical to the proof that e a + b = e a e b for scalars a and b : e A e B = m = n = A m m ! B n n ! = k = 1 k ! m + n = k ( k m ) A m B k- m = k = 1 k ! ( A + B ) k = e A + B . In the third step, we used the binomial identity ( A + B ) k = k m = ( k m ) A m B k- m , which holds if A and B commute (proof by induction), but not in general. For the second part of the question, well expand each of e A e B and e A + B to the quadratic terms: e A e B = ( 1 + A + 1 2 A 2 + cubic terms )( 1 + B + 1 2 B 2 + cubic terms ) = 1 + A + B + AB + 1 2 ( A 2 + B 2 )+ cubic terms , while e A + B = 1 +( A + B )+ 1 2 ( A + B ) 2 + cubic terms = 1 + A + B + 1 2 ( A 2 +...
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## This note was uploaded on 12/06/2009 for the course PHYSICS 191 taught by Professor Birgittawhaley during the Fall '07 term at Berkeley.

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hw4sol_fa07 - C/CS/Phy191 Problem Set 4 Solutions Out 1...

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