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Unformatted text preview: muting operators and ξ is a parameter, then exp (ξA ) B exp ( ξA ) = B + ξ 1! [ B,A ] + ξ 2 2! [[ B,A ] ,A ] + ξ 3 3! [[[ B,A ] ,A ] ,A ] + ······ . [ Hint : Expand exp (ξA ) B exp ( ξA ) in a power series of ξ .] 4. Theorem 2 : If A and B are two noncommuting operators and ξ is a parameter, then exp (ξA ) B n exp ( ξA ) = [exp (ξA ) B exp ( ξA )] n for some integer n , and exp (ξA ) F ( B ) exp ( ξA ) = F (exp (ξA ) B exp ( ξA )) 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 (ξba ) a exp ( ξba ) = a exp ( ξ ) exp (ξba ) b exp ( ξba ) = b exp (ξ ) exp ±1 2 ξ ² b 2a 2 ³ ´ a exp ± 1 2 ξ ² b 2a 2 ³ ´ = a cosh ( ξ ) + b sinh ( ξ ) exp ±1 2 ξ ² b 2a 2 ³ ´ b exp ± 1 2 ξ ² b 2a 2 ³ ´ = b cosh ( ξ ) + a sinh ( ξ ) . —— End —— 2...
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.
 Spring '11
 CFLO
 Charge, Mass

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