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Unformatted text preview: muting operators and Î¾ is a parameter, then exp (Î¾A ) B exp ( Î¾A ) = B + Î¾ 1! [ B,A ] + Î¾ 2 2! [[ B,A ] ,A ] + Î¾ 3 3! [[[ B,A ] ,A ] ,A ] + Â·Â·Â·Â·Â·Â· . [ Hint : Expand exp (Î¾A ) B exp ( Î¾A ) in a power series of Î¾ .] 4. Theorem 2 : If A and B are two noncommuting operators and Î¾ is a parameter, then exp (Î¾A ) B n exp ( Î¾A ) = [exp (Î¾A ) B exp ( Î¾A )] n for some integer n , and exp (Î¾A ) F ( B ) exp ( Î¾A ) = F (exp (Î¾A ) B exp ( Î¾A )) where F ( B ) is an arbitrary function of B . 1 5. Theorem 3 : If the operators a and b satisfy the commutation relation [ a,b ] = 1, then show that exp (Î¾ba ) a exp ( Î¾ba ) = a exp ( Î¾ ) exp (Î¾ba ) b exp ( Î¾ba ) = b exp (Î¾ ) exp Â±1 2 Î¾ Â² b 2a 2 Â³ Â´ a exp Â± 1 2 Î¾ Â² b 2a 2 Â³ Â´ = a cosh ( Î¾ ) + b sinh ( Î¾ ) exp Â±1 2 Î¾ Â² b 2a 2 Â³ Â´ b exp Â± 1 2 Î¾ Â² b 2a 2 Â³ Â´ = b cosh ( Î¾ ) + a sinh ( Î¾ ) . â€”â€” End â€”â€” 2...
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 Spring '11
 CFLO
 Charge, Mass, Î¾A)

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