exam_solutions_1

# exam_solutions_1 - 1(50 points You have a particle of mass...

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Unformatted text preview: 1. (50 points) You have a particle of mass m confined to a 1D box bewteen- a 2 < x < a 2 . Take the set of states {| n i} to refer to the eigenstates of the Hamiltonian; ˆ H | n i = n 2 E 1 | n i , for n = 1 , 2 , 3 ,... . You already know the eigenfunctions h x | n i . (a) You can write the energy representations of the operators ˆ x and ˆ p , by saying ˆ x = X mn | m ih m | ˆ x | n ih n | = X mn x mn | m ih n | where x mn = h m | ˆ x | n i ; and ˆ p = X mn | m ih m | ˆ p | n ih n | = X mn p mn | m ih n | where p mn = h m | ˆ p | n i . Find expressions for the “matrix elements” x mn and p mn , for arbitrary m and n . Simplify the expressions as much as possible (look up some trigonometric identities). Just to make it easy for me to check your answer, write down the a-dependent values for x mn and p mn for m,n = 1 , 2 , 3 , 4 (the 4 × 4 sub-matrices). Hint: Paying attention to parity will simplify many of your integrals. Answer: Convert to wave function integrals: h m | ˆ x | n i = Z dxϕ * m xϕ n If m and n are of the same parity, the integral will be zero, since x is an odd function. Also, since the ϕ n are real, and h m | ˆ x | n i = h n | ˆ x | m i * = h n | ˆ x | m i , this means x mn = x nm . So we only need do the integral for even n and odd m (or vice versa). In this case, x nm = 2 a Z a/ 2- a/ 2 dxx sin nπ a x cos mπ a x After use of integral tables and some algebra, x nm = 2 a π 2 " 1 ( n + m ) 2- 1 ( n- m ) 2 # (- 1) ( n + m- 1) / 2 The ˆ p matrix elements are also 0 for states of the same parity, since the derivative reverses parity. Now, however,the derivative reverses parity....
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exam_solutions_1 - 1(50 points You have a particle of mass...

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