ProblemSet2Solutions

# ProblemSet2Solutions - Chem 120A Problem Set 2 Solutions 1...

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Chem 120A Problem Set 2 Solutions 2/22/06 1. a) We are given that operator A = | α i ± β ² ² , and we want to know what A , the operator adjoint to A, is. Let A act on an arbitrary state | a i A | a i = | b i | α ih β | a i = | b i Taking the inner product with another arbitrary state, ± d ² ² we get: h d | α ih β | a i = h d | b i If we take the complex conjugate of both sides ( h d | α ih β | a i ) * = ( h d | b i ) * h a | β ih α | d i = h b | d i Now the deFnition of A is ± a ² ² A = ± b ² ² (This is true whether the operator is Hermitian or not) Therefore, h a | β ih α | d i = ± a ² ² A | d i Thus we have shown that A = | β i ± α ² ² b) Let A and B be two Hermitian operators, i.e. A = A and B = B . We want to know under what circumstances AB = ( AB ) . To solve this, you need to use the fact that AB = B A (see proof below). Since A and B are Hermitian, then AB = ( AB ) = B A = BA . So what this problem is really asking us is when is AB = BA ? This is true when the two operators commute, i.e. [ A, B ] = AB - BA = 0. So in short form: AB = ( AB ) AB = B A AB = BA [ A, B ] = 0 Proof that ( AB ) = B A Chem 120A, Spring 2006 1

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Use A = | α i ± β ² ² and B = | m i ± n ² ² From a) we know that: A = | β i ± α ² ² and B = | n i ± m ² ² so AB = | α ih β | m i ± n ² ² and B A = | n ih m | β i ± α ² ² Let (AB) act on an arbitrary state, | y i ( AB ) | y i = | z i , then by defnition ± y ² ² ( AB ) = ± z ² ² Taking the inner product, as in part a) ± x ² ² ( AB ) | y i = h x | z i h x | α ih β | m ih n | y i = h x | z i And taking the complex conjugate o± both sides ( h x | α ih β | m ih n | y i ) * = ( h x | z i ) * h y | n ih m | β ih α | x i = h z | x i h x | α ih β | m ih n | y i = ± y ² ² ( AB ) | x i Thus | n ih m | β i ± α ² ² = ( AB ) But ±rom our defnition o± B A above, we conclude that ( AB ) = B A 2. The energy di²erence between the ground state (lowest energy level) and frst excited state (next level up) in a hydrogen atom is about 10 eV (eV = electron-volt). The diameter o± a hydrogen atom is 1 Angstrom = 10 - 10 meters. I± we model a hydrogen atom as a 1-D box with hard walls, then what is the length o± a the box to get the same energy level spacing as hydrogen? Answer : The energy eigenvalues ±or a particle in a 1-D box are given by E n = ~ 2 n 2 π 2 2 mL 2 . The energy di²erence between the frst excited state and the ground state is given by: Chem 120A, Spring 2006 2
Δ E = E 2 - E 1 = ~ 2 (2) 2 π 2 2 mL 2 - ~ 2 (1) 2 π 2 2 mL 2 = ~ 2 π 2 2 mL 2 ( 2 2 - 1 2 ) = 3 ~ 2 π 2 2 mL 2 In the case of hydrogen, the particle is an electron which has a mass m = 9.11 × 10 - 31 kg. Continuing in MKS units, we have 1 eV = 1.60 × 10 - 19 J, and ~ = 1 . 05 × 10 - 34 J. We can solve the previous equation for L and plug in the appropriate quantities: L = s 3 ~ 2 π 2 2 m e Δ E = s 3 (1 . 05 × 10 - 34 J · s ) 2 π 2 2 (9 . 11 × 10 - 31 kg ) (1 . 60 × 10 - 19 J ) = 3 . 36 × 10 - 10 m

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ProblemSet2Solutions - Chem 120A Problem Set 2 Solutions 1...

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