- #1

- 716

- 49

∫ <x | p | p><p | o > dp = ∫ p<x | p><p | o > dp = -iħ d/dx ∫ <x | p><p | 0>dp

I am confused about how the -iħ d/dx arises. I thought the p produced when the p operator acts on |p> is an eigenvalue not an operator ? And it also seems as though the equation is in the momentum representation not the position one and -iħ d/dx is the momentum operator in the position representation ?

Thanks