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Unformatted text preview: Physics 139B Solutions to Homework Set 1 Fall 2009 1. Liboff, problem 11.27 on page 498. (a) Let A be an hermitian operator. We first demonstrate that ( e iA ) † = e iA . To prove this, we use the series expansion that defines the exponential, e iA = ∞ summationdisplay n =0 ( iA ) n n ! . The sum converges for any operator A . Then, e iA = ∞ summationdisplay n =0 ( − iA ) n n ! = ∞ summationdisplay n =0 [( iA ) † ] n n ! = bracketleftBigg ∞ summationdisplay n =0 ( iA ) n n ! bracketrightBigg † = ( e iA ) † , where we have used the fact that ( iA ) † = − iA † = − iA , since A is hermitian. In the above derivation, we have also used the fact that ( A + B ) † = A † + B † , which also holds for the sum of an infinite number of terms, assuming that the sum converges. A unitary operator hatwide U is defined as an operator that satisfies hatwide U hatwide U † = hatwide U † hatwide U = I , where I is the identity operator. Defining hatwide U ≡ e iA , we see that: hatwide U hatwide U † = e iA ( e iA ) † = e iA e iA = e i ( A A ) = e = I . (1) In this derivation, I used the fact that for any two operators A and B , 1 e A e B = e B e A = e A + B , if and only if AB = BA . (2) This result applies in eq. (1) since iA and − iA clearly commute. Finally, we use the fact that e = I , where 0 is the zero operator, which is again a consequence of the definition of the exponential of an operator via its series expansion. Setting A = − hatwide Ht/ planckover2pi1 , we note that if hatwide H is hermitian, then so is A . It then follows that hatwide U = exp( − i hatwide Ht/ planckover2pi1 ) is unitary. (b) Given  ψ ( t ) ) = hatwide U  ψ (0) ) , it follows that ( ψ ( t )  = ( ψ (0)  hatwide U † . Hence, using the results of part (a), ( ψ ( t )  ψ ( t ) ) = ( ψ (0)  U † U  ψ (0) ) = ( ψ (0)  I  ψ ) = ( ψ (0)  ψ (0) ) . That is, the normalization of ψ is independent of the time t . 1 For a proof of eq. (2), see, e.g., Jacob T. Schwartz, Introduction to Matrices and Vec tors (Dover Publications, Inc., Mineola, NY, 2001) pp. 157–159, which can be viewed at http://books.google.com . 2. Liboff, problem 11.39 on page 512. (a) We are given a wave function of a rigid rotator, ψ ( t ) = 1 √ 14 1 2 3 e iEt/ planckover2pi1 , E ≡ planckover2pi1 2 /I . Using the results of Liboff, problem 11.38, we can write at time t = 0: 1 √ 14 1 2 3 = − 1 √ 7 ξ (0) x + 1 √ 7 ( √ 2 − 1) ξ ( 1) x + 1 √ 7 ( √ 2 + 1) ξ (1) x , where, ξ (0) x ≡ 1 √ 2 1 − 1 , ξ ( 1) x ≡ 1 2 1 − √ 2 1 , ξ (1) x ≡ 1 2 1 √ 2 1 , are eigenstates of hatwide L x , with eigenvalues, 0, − planckover2pi1 and + planckover2pi1 , respectively. Note that the ξ ( m ) x are all eigenstates of vector L 2 with eigenvalue 2 planckover2pi1 2 . Thus, ψ ( t ) = bracketleftbigg − 1 √ 7 ξ (0) x + 1 √ 7 ( √ 2 − 1) ξ ( 1) x...
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 Spring '09
 Mucciolo
 mechanics, Work, Eigenvalue, eigenvector and eigenspace, Hilbert space, Identity function, spinor

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