Question

Does ddx⟨u|=(ddx|u⟩)†? Here |u⟩ is any vector in a Hilbert space and †

is the adjoint/conjugate-transpose.It seems to me that this should not be the case since (ddx|u⟩)†=⟨u|(ddx)†=−ddx⟨u|. Here we used that (ddx)†=−ddx
since the momentum operator p=−iddx is Hermitian so p†=p
and p†=i(ddx)†
.Related to Is the time derivative of the adjoint equal to the adjoint of the time derivative? , but possibly with a different conclusion for the space derivative and time derivative.
Does my logic make sense for why the derivative of the adjoint is not the adjoint of the derivative, or where did I make a mistake?

Answer 1

No, the derivative of the adjoint is not the adjoint of the derivative in this case. The derivative of the adjoint is -i(p†|u> = -i(-i(d/dx)|u>) = d/dx|u>.

Step-by-step explanation:

No, the derivative of the adjoint is not the adjoint of the derivative in this case. While it is true that the adjoint of the momentum operator p is -i(d/dx), the adjoint of the derivative operator (d/dx) is -i(p), not p. This can be seen by taking the adjoint of both sides of the equation (d/dx)|u> = p|u> and using the property of the adjoint (D†)|v> = <u|(D|v>)†. Therefore, (d/dx)|u>† = (<u|(d/dx)†)† = -i<u|p†)† = -i<u|p) = -i(p†|u> = -i(-i(d/dx)|u>) = d/dx|u>.