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hw6a - PHYS 507 Answers to Homework VI(Fall'05 1(a,b The...

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PHYS 507 Answers to Homework VI (Fall ’05) 1. (a,b) The gauge transformation is ˜ ~ A = ~ A + ~ Λ , ˜ φ = φ - 1 c Λ ∂t , and we want ˜ φ = 0. Therefore, φ = Ze r = 1 c Λ ∂t -→ Λ = Zect r . From here we find ~ ˜ A ( ~ r,t ) = ~ Λ = - Zect r 2 ˆ r . (c) ˜ ψ ( ~ r 0 ,t ) = exp ± i ( - e ) ¯ hc Λ( ~ r 0 ,t ) ψ ( ~ r 0 ,t ) = exp ± - i Ze 2 t ¯ hr ψ ( ~ r 0 ,t ) . 2. (a) The electric field is E = - ∂φ ∂x - 1 c ∂A ∂t . In the first gauge A = 0 and hence φ = - Ex . In the second gauge ˜ φ = 0 and hence ˜ A = - cEt . (b) Λ = - cExt . (c) ˜ ψ ( x 0 ,t ) = exp ± - i QE ¯ h x 0 t ψ ( x 0 ,t ) . 3. (a) [ π x y ] = h p x + e c A x ,p y + e c A y i = e c ([ p x ,A y ] + [ A x ,p y ]) = e c ¯ h i ( x A y - y A x ) = e c ¯ h i B = - i ¯ hmω . (b) [ a,a ] = D 2 [ π x - y x + y ] = 2 iD 2 [ π x y ] = 2¯ hmωD 2 = 1 . We can set D = 1 / hmω . (c)
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hw6a - PHYS 507 Answers to Homework VI(Fall'05 1(a,b The...

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