4604_Solutions_Set4

4604_Solutions_Set4 - PHY4604 Problem Set 4 Solutions PHY...

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PHY4604 Problem Set 4 Solutions Department of Physics Page 1 of 11 PHY 4604 Problem Set #4 Solutions Problem 1 (25 points): . (a) (2 points) Show that the sum of two hermitian operators is hermitian. Solution: If (A op ) = A op and (B op ) = B op then (A op +B op ) = (A op ) +(B op ) = (A op +B op ). (b) (2 points) Suppose that H op is a hermitian operator, and α is a complex number. Under what condition (on α ) is α H op hermitian? Answer: α is real Solution: ( α H op ) = α ( H op ) = α H op . Thus, α H op hermitian provided α = α ( i.e. α is real). (c) (2 points) When is the product of two hermitian operators hermitian? Answer: When the two operators commute. Solution: If (A op ) = A op and (B op ) = B op then (A op B op ) = (B op ) (A op ) = B op A op = A op B op + [B op, A op ] and (A op B op ) = A op B op provided [B op, A op ] = 0. Note that I used (e). (d) (2 points) If dx d O op = , what is op O ? Answer: dx d O op = Solution: We know that [] Ψ Ψ = Ψ Ψ dx t x O t x dx t x t x O op op ) , ( ) , ( ) , ( ) , ( 1 * 2 1 * 2 We see that Ψ Ψ = Ψ Ψ dx dx t x d t x dx t x dx t x d ) , ( ) , ( ) , ( ) , ( 1 * 2 1 2 where I integrated by parts and dropped the boundary term. Thus, dx d O op = . (e) (2 points) Show that = op op op op A B B A ) (. Solution: We know that < ψ 1 |(A op B op )| ψ 2 >* = < ψ 2 |(A op B op ) | ψ 1 >, but < ψ 1 |(A op B op )| ψ 2 >* = < ψ 1 |A op |B ψ 2 >* = < B ψ 2 |(A op ) | ψ 1 > = < ψ 2 |(B op ) (A op ) | ψ 1 >, where I used |B ψ 2 > = B op | ψ 2 > and <B ψ 2 | = < ψ 2 |(B op ) . (f) (2 points) Prove that [AB,C] = A[B,C] + [A,C]B, where A, B, and C are operators. Solution: We see that [AB,C] = ABC – CAB A[B,C] + [A,C]B = A(BC-CB) + (AC-CA)B = ABC - CAB (g) (2 points) Show that dx df i x f p op x h = )] ( , ) [( , for any function f(x). Solution: We see that ) ( ) ( ) ( )] ( , ) [( x dx df i dx d f i dx d f i dx df i dx d f i f dx d i x x f p op x ψ h h h h h h = + = + =

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PHY4604 Problem Set 4 Solutions Department of Physics Page 2 of 11 (h) (2 points) Show that the anti-hermitian operator, I op , has at most one real eigenvalue (Note: anti-hermitian means that op op I I = ). Solution: We see that if I op | ψ > = λ | ψ > then λ = < ψ |I op | ψ > and λ = ( < ψ |I op | ψ >)* = < ψ |(I op ) | ψ > = -< ψ |I op | ψ > = - λ. If we let λ = x+iy then λ = x-iy and λ = - λ implies that x-iy = -x-iy or 2x= 0. The only real eigenvalue is λ = 0. (i) (2 points) If A op is an hermitian operator, show that 0 2 >≥ < op A . Solution: The norm of any allowed state is positive definite. Namely, 0 <A ψ |A ψ > = < ψ |(A op ) A op | ψ > = < ψ |(A op ) 2 | ψ > = <(A op ) 2 >, where I used |A ψ > = A op | ψ > and <A ψ | = < ψ |(A op ) and (A op ) = A op . (j) (2 points) The parity operator, P op , is defined by P op Ψ (x,t) = Ψ (-x,t). Prove that the parity operator is hermitian and show that 1 2 = op P , where 1 is the identity operator. Compute the eigenvalues of the parity operator. Solution: An operator is hermitian if [] Ψ Ψ = Ψ Ψ dx t x O t x dx t x t x O op op ) , ( ) , ( ) , ( ) , ( 1 * 2 1 * 2 If we let O op = P op we get Ψ Ψ = Ψ Ψ = Ψ Ψ = Ψ Ψ dx t x P t x dx t x t x dx t x t x dx t x t x P op op ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( 1 * 2 1 * 2 1 * 2 1 * 2 where I changed variables x -x in the integration.
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This note was uploaded on 04/19/2008 for the course PHY 4604 taught by Professor Field during the Spring '07 term at University of Florida.

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4604_Solutions_Set4 - PHY4604 Problem Set 4 Solutions PHY...

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