Discussion
The result of the following operation
\(\widehat{A}\,(e^{ikx})\) = ∵ \(\widehat{A} = -i\hbar\,\dfrac{d}{dx}\)
is :
Answer: C
The result is \(\hbar k\,e^{ikx}\), option C.
Apply the momentum operator to the function:
\[ \begin{aligned} \widehat{A}\,e^{ikx} &= -i\hbar\,\frac{d}{dx}e^{ikx} \\ &= -i\hbar\,(ik)\,e^{ikx} \\ &= \hbar k\,e^{ikx} \end{aligned} \]
since \(-i \times i = 1\). So eikx is an eigenfunction of the momentum operator with eigenvalue \(\hbar k\). Option A would mean the operator had no effect, B leaves out the factor \(\hbar\), and D changes the exponent to kx2, which differentiation of eikx never produces.
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