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# hw2add - PHYS 507 Answers to Homework II 1(a i i T(a)|p =...

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PHYS 507 Answers to Homework II 1. (a) T ( a ) | p 0 i = exp - i ¯ h pa | p 0 i = exp - i ¯ h p 0 a | p 0 i . (b) There are different ways of doing this. This is a longer way. Start with ˜ φ ( p 0 ) = h p 0 | φ i = h p 0 | T ( a ) | ψ i . As a result, we first need to calculate h p 0 | T ( a ). The dual complement of this expression is T ( a ) | p 0 i = T ( - a ) | p 0 i = exp + i ¯ h p 0 a | p 0 i , as a result h p 0 | T ( a ) = exp - i ¯ h p 0 a h p 0 | . This leads to ˜ φ ( p 0 ) = exp - i ¯ h p 0 a h p 0 | ψ i = exp - i ¯ h p 0 a ˜ ψ ( p 0 ) . A shorter way: First expand the kets in terms of momentum eigenkes using the momentum space wavefunction. | ψ i = Z | p 0 ih p 0 | ψ i dp 0 = Z ˜ ψ ( p 0 ) | p 0 i dp 0 ⇒ | φ i = T ( a ) | ψ i = Z ˜ ψ ( p 0 ) T ( a ) | p 0 i dp 0 = Z ˜ ψ ( p 0 ) exp - i ¯ h p 0 a | p 0 i dp 0 ˜ φ ( p 0 ) = exp - i ¯ h p 0 a ˜ ψ ( p 0 ) In any case, we see that the effect of translation in momentum space is a phase change where the amount of the phase depends on the particular value of mo- mentum. An immediate conclusion is that the distribution of momentum does not change by translation, i.e., | ˜ φ ( p 0 ) | 2 = | ˜ ψ ( p 0 ) | 2 . 2. (a) S ( μ 1 ) S ( μ 2 ) = S ( μ 1 + μ 2 ) and S ( μ ) = S ( - μ ). Since S ( μ ) S ( μ ) = S ( μ ) S ( - μ ) = S (0) = 1, S ( μ ) is a unitary operator. (b) These commutators will be useful for part (c). i ¯ h [ A, x ] = i h [ px + xp, x ] = i h ([ p, x ] x + x [ p, x ]) = i h (2 ¯ h i x ) = x . i ¯ h [ A, p ] = i h [ px + xp, p ] = i h ( p [ x, p ] x + [ x, p ] p ) = i h ( - 2 ¯ h i p ) = - p . 1

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(c) Using shorthand notation B = iA/ ¯ h , we have S ( μ ) = exp( μB ) and S ( μ ) xS ( μ ) = x + μ [ B, x ] + μ 2 2! [ B, [ B, x ]] + · · · + μ n n ! [ B, [ B, ... [ B, x ]]] + · · · From part (b) we can easily see that [ B, [ B, ... [ B, x ]]] = x . Therefore S ( μ ) xS ( μ ) = 1 + μ + μ 2 2! + · · · + μ n n ! + · · · x = e μ x . Similarly, S ( μ ) pS ( μ ) = p + μ [ B, p ] + μ 2 2! [ B, [ B, p ]] + · · · + μ n n ! [ B, [ B, ... [ B, p ]]] + · · · = 1 - μ + μ 2 2! - · · · + ( - 1) n μ n n ! + · · · p = e - μ p . (d) h x i φ = h φ | x | φ i = h ψ | S ( μ ) xS ( μ ) | ψ i = e μ h ψ | x | ψ i = e μ h x i ψ h p i φ = h φ | p | φ i = h ψ | S ( μ ) pS ( μ ) | ψ i = e - μ h ψ | p | ψ i = e - μ h p i ψ 3. (a) h ψ | ψ i = Z + -∞ dx 0 | ψ ( x 0 ) | 2 = Z a 0 dx 0 | N | 2 = a | N | 2 = 1 . we choose N = 1 / a . (b) ˜ ψ ( p 0 ) = Z dx 0 h p 0 | x 0 ih x 0 | ψ i = 1 2 π ¯ h Z dx 0 exp - i ¯ h p 0 x 0 ψ ( x 0 ) = N 2 π ¯ h Z a 0 dx 0 exp - i ¯ h ( p 0 - ¯ hk ) x 0 = N 2 π ¯ h ¯ h ( - i ) 1 p 0 - ¯ hk exp - i ¯ h ( p 0 - ¯ hk ) a - 1 Here we use e - - 1 = e - iθ/ 2 ( e - iθ/ 2 - e iθ/ 2 ) = e - iθ/ 2 ( - 2 i sin θ/ 2). Then, ˜ ψ ( p 0 ) = N 2 π ¯ h h p 0 -
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