p268r-ps4-solutions-part2 - Phys 268r Fall 2013 Problem Set 4 PART II Solutions B Halperin Solutions to Problem Set 4 Part II Problem 5 a We have from

p268r-ps4-solutions-part2 - Phys 268r Fall 2013 Problem Set...

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Phys 268r Fall 2013 Problem Set 4 - PART II - Solutions B. Halperin Solutions to Problem Set 4 - Part II Problem 5. a) We have from the definitions: Z dxdx 0 ψ ( x 0 ) h Ψ( x 0 ) , Ψ ( x ) ˆ H Ψ( x ) i = EX, where ˆ H = h vp x σ z + { p x , ~ f ( s ) } · + V ( x ) + h ( x ) σ x i . We compute the commutator, showing the first and third terms of ~ H (the others follow similarly). For the third term, we have: Ψ( x 0 ) , Ψ ( x ) V ( x )Ψ( x ) = V ( x ) ( Ψ( x 0 ( x )Ψ( x ) - Ψ ( x )Ψ( x )Ψ( x 0 ) ) = V ( x ) ( δ ( x - x 0 ) - Ψ ( x )Ψ( x 0 ) Ψ( x ) - Ψ ( x )Ψ( x )Ψ( x 0 ) ) = V ( x ) ( δ ( x - x 0 )Ψ( x ) - Ψ ( x )Ψ( x 0 )Ψ( x ) - Ψ ( x )Ψ( x )Ψ( x 0 ) ) = V ( x ) ( δ ( x - x 0 )Ψ( x ) + Ψ ( x )Ψ( x )Ψ( x 0 ) - Ψ ( x )Ψ( x )Ψ( x 0 ) ) = V ( x ) δ ( x - x 0 )Ψ( x ) . For the first term (different because of the p x ), we have: Ψ( x 0 ) , Ψ ( x ) vp x σ z Ψ( x ) = v ( Ψ( x 0 ( x ) p x σ z Ψ( x ) - Ψ ( x ) p x σ z Ψ( x )Ψ( x 0 ) ) = v ( δ ( x - x 0 ) - Ψ ( x )Ψ( x 0 ) p x σ z Ψ( x ) - Ψ ( x ) p x σ z Ψ( x )Ψ( x 0 ) ) = v ( δ ( x - x 0 ) p x σ z Ψ( x ) - Ψ ( x )Ψ( x 0 ) p x σ z Ψ( x ) - Ψ ( x ) p x σ z Ψ( x )Ψ( x 0 ) ) = v ( δ ( x - x 0 ) p x σ z Ψ( x ) - Ψ ( x )Ψ( x 0 ) p x σ z Ψ( x ) - Ψ ( x ) p x σ z Ψ( x )Ψ( x 0 ) ) = v δ ( x - x 0 ) p x σ z Ψ( x ) - Ψ ( x ) ( x - x 0 )
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  • Fall '13
  • BertrandI.Halperin
  • Physics, Equals sign, plane wave, Ψ, B. Halperin, px σz

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