The Hermitian conjugate of operator $\mathrm{d} / \mathrm{dx}$, called $(\mathrm{d} / \mathrm{dx})^{\dagger}$, is actually equal to
(a)-d/dx
(b)$\mathrm{d} / \mathrm{dx}$
(c)$\mathrm{i}(\mathrm{d} / \mathrm{dx})$
(d)$-\mathrm{i}(\mathrm{d} / \mathrm{dx})$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Integration by parts gives ∫f*(dg/dx)dx = −∫(df/dx)*g dx for vanishing boundary terms, so (d/dx)† = −d/dx (a). d/dx is anti-Hermitian, and −i d/dx is Hermitian.