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Midterm 7 Solutions

# Midterm 7 Solutions - Intermediate QMI Chapter 7 Midterm...

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Intermediate QMI Chapter 7 Midterm Solutions Problem 1 ü Part a We have H ` = p ` x 2 2 m + 1 2 m w 2 x ` 2 = 1 2 m w 2 J x ` 2 + 1 m 2 w 2 p ` x 2 N = 1 2 m w 2 AI x ` - Â m w p ` x M I x ` + Â m w p ` x M - Â m w A x ` , p ` x EE ¬ A x ` , p ` x E = Â = 1 2 m w 2 AI x ` - Â m w p ` x M I x ` + Â m w p ` x M + 1 m w E H ` = w A m w 2 I x ` - Â m w p ` x M I x ` + Â m w p ` x M + 1 2 E Let m w 2 I x ` - Â m w p ` x M = a ` m w 2 I x ` + Â m w p ` x M = a ` Thus H ` = w J a ` a ` + 1 2 N and we have B a ` , a ` F = B m w 2 I x ` + Â m w p ` x M , m w 2 I x ` - Â m w p ` x MF to which we apply linearity of the commutator in both arguments, together with the fact that an operator commutes with itself, to obtain B a ` , a ` F = I m w 2 M 9 - Â m w A x ` , p ` x E + Â m w A p ` x , x ` E= = I m w 2 MA - Â m w H Â L + Â m w H LE therefore B a ` , a ` F = 1 Printed by Mathematica for Students

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ü Part b From (a) H ` = w J N ` + 1 2 N where N ` = a ` a ` , so the eigenstates 8 n \< of N ` are also eigenstates of H ` , and the energy eigenvalues are given by E n = w I n + 1 2 M To solve for the eigenstates of N ` (and hence of H ` ), we use B N ` , a ` F = B a ` a ` , a ` F = a ` A a ` , a ` E + B a ` , a ` F a ` B N ` , a ` F = B a ` a ` , a ` F = a ` B a ` , a ` F + B a ` , a ` F a ` B N ` , a ` F = - a ` B N ` , a ` F = a ` The latter of these implies B N ` , H a ` L n F = H a ` L n - 1 B N ` , a ` F + B N ` , H a ` L n - 1 F a ` = H a ` L n + B N ` , H a ` L n - 1 F a ` B N ` , H a ` L n F = H a ` L n + :H a ` L n - 1 + B N ` , H a ` L n - 2 F a ` > a ` Continuing this recursion relation, we find that B N ` , H a ` L n F = n H a ` L n Therefore, the eigenstates of N ` are of the form n \ = H a ` L n n ! 0 \ where N ` 0 ^ = 0 as N ` H a ` L n n ! 0 \ = B n H a ` L n n ! 0 \ + H a ` L n n ! N ` 0 \F = 0 N ` H a ` L n n ! 0 \ = n H a ` L n n ! 0 \ 2 MidtermExam7Solutions.nb Printed by Mathematica for Students
ü Part c The condition that we have equal probability of measuring E = 1 2 w or E = 3 2 w implies (from (b)) y H 0 L\ = 1 2 B 0 \ + ‰ Â f 1 \F

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Midterm 7 Solutions - Intermediate QMI Chapter 7 Midterm...

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