s11f1037

# s11f1037 - m . [ L i , L j ] = X k X l X m X n X r ± ikl...

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Solution for assignment 37 Poisson Brackets (21a) Let us consider functions g = g ( q k , p k , t ) and h = h ( q k , p k , t ). The Poisson bracket is deﬁned by [ g, h ] def = X k ± ∂g ∂q k ∂h ∂p k - ∂h ∂q k ∂g ∂p k ! . The properties of the assignment are shown in the following: 1. Resulting in the deﬁnition of the total time derivative: dg dt = [ g, H ] + ∂g ∂t = X k ± ∂g ∂q k ∂H ∂p k - ∂H ∂q k ∂g ∂p k ! + ∂g ∂t = X k ± ∂g ∂q k ˙ q k + ∂g ∂p k ˙ p k ! + ∂g ∂t . 2. Using ∂q j /∂p k = 0: ˙ q j = [ q j , H ] = X k ± ∂q j ∂q k ∂H ∂p k - ∂H ∂q k ∂q j ∂p k ! = X k δ jk ˙ q k = ˙ q j , 3. Using ∂p j /∂q k = 0: ˙ p j = [ p j , H ] = X k ± ∂p j ∂q k ∂H ∂p k - ∂H ∂q k ∂p j ∂p k ! = X k δ jk ˙ p k = ˙ p j , 4. Commutators of generalized coordinates and momenta: [ q i , q j ] = X k ± ∂q i ∂q k ∂q j ∂p k - ∂q j ∂q k ∂q i ∂p k ! = 0 as ∂q j /∂p k = 0 and ∂q i /∂p k = 0. Similarly [ p i , p j ] = X k ± ∂p i ∂q k ∂p j ∂p k - ∂p j ∂q k ∂p i ∂p k ! = 0 , while [ q i , p j ] = X k ± ∂q i ∂q k ∂p j ∂p k - ∂p j ∂q k ∂q i ∂p k ! = X k ( δ ik δ j k - 0) = δ ij . (21b) Angular momentum commutator: [ x i , L j ] = X k X l X m ± jlm ± ∂x i ∂x k ( x l p m ) ∂p k - ( x l p m ) ∂x k ∂x i ∂p k ! = X k X l X m ± jlm δ ik x l δ mk = X l X m ± jlm δ im x l = X l ± jli x l = X l ± ijl x l .

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[ p i , L j ] = X k X l X m ± jlm ± ∂p i ∂x k ( x l p m ) ∂p k - ( x l p m ) ∂x k ∂p i ∂p k ! = - X k X l X m ± jlm δ lk p m δ ik = - X l X m ± jlm δ li p m = - X m ± jim p m = X m ± ijm x
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Unformatted text preview: m . [ L i , L j ] = X k X l X m X n X r ± ikl ± jmn ± ∂ ( x k p l ) ∂x r ∂ ( x m p n ) ∂p r-∂ ( x m p n ) ∂x r ∂ ( x k p l ) ∂p r ! = X k X l X m X n X r ± ikl ± jmn ( δ kr δ nr x m p l-δ mr δ lr x k p n ) = X k X l X m X n ± ikl ± jmn ( δ kn x m p l-δ ml x k p n ) = X k X l ± X m ± ikl ± jmk x m p l-X n ± ikl ± jln x k p n ! = X k X l ± X m ± kli ± kjm x m p l-X n ± lik ± lnj x k p n ! = X l X m ( δ lj δ im-δ lm δ ij ) x m p l-X k X n ( δ in δ kj-δ ij δ kn ) x k p n = x i p j-δ ij ~x · ~ p-x j p i + δ ij ~x · ~ p = x i p j-x j p i . This agrees with X k ± ijk L k = X k X l X m ± ijk ± klm x l p m = X k X l X m ± kij ± klm x l p m = X l X m ( δ il δ jm-δ im δ jl ) x l p m = x i p j-x j p i ....
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## This note was uploaded on 12/11/2011 for the course PHY 4936 taught by Professor Berg during the Fall '11 term at FSU.

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s11f1037 - m . [ L i , L j ] = X k X l X m X n X r ± ikl...

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