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where I’ve let c=1.

Problem 5-3b Solution. where I’ve let c=1. †. †. . i. Recalling that  is a function of k ,. it should be primed when k is. Again, I’ve adopted the ħ , c = 1. So the left-hand-side of. †. †. By taking the Fourier transform of both sides:.  d 3 re - i K ·r.  d 3 re - i K ·r.

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where I’ve let c=1.

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  1. Problem 5-3b Solution where I’ve let c=1. † †  i Recalling that  is a function of k, it should be primed whenk is. Again, I’ve adopted the ħ,c = 1. So the left-hand-side of † † By taking the Fourier transform of both sides: d3re-iK·r d3re-iK·r e-ik·r I’ve already cancelled 1/(2p)3 from both sides (2)33(k-K)  (K) After integrating over d3k, the right-hand-side simplifies to: † † Next do the Fourier transform: d3r'eiK'·r' d3r'e-iK'·r' e-ik·r Again already canceling 1/(2p)3 from both sides (2)33(K' -K)  '  (K' )

  2. Once the dust settles: † † If this yields 0. But since so for all surviving NON-ZERO terms . So † † † † Notice that the right-hand-side is completely independent of time! For the full statement to be true for all times, it must be true that † † and so † † † [d-K' ,d-K] † Which means both terms on the left are of the form: For the whole statement ot be true, we must have †

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