Question
Let us consider the wave function of a free electron:
$$
\psi (x,t) = \sin ( k x – \omega t )
$$
I am trying to calculate its de Broglie wavelength in terms of $ k $.
What I did
I tried to use the formula $ \lambda = h/p $ ($ p $ is the momentum). I calculated $ p $ as follows:
$$
\begin{align}
p &= -i \hbar \frac {d \psi} {dt} \\
&= -i \hbar k \cos ( k x – \omega t ) \\
\\
\implies \
\lambda &= h/p \\
&= \boxed { \left( \frac { 2 \pi } {k} \right) \color{blue} { \left( \frac {i} { \cos ( k x – \omega t) } \right) } }
\end{align}
$$
Problem
Where is the extra $ \color{blue} { ( i / \cos ( k x – \omega t) ) } $ factor coming from? Isn’t the de Broglie wavelength just $ 2 \pi / k $ ?
Best Answer
You calculated the result of applying the momentum operator on the wave function. That's not the same as calculating the expectation value of the operator.
You'll have an easier time of it if you use the exponential version of the wavefunction $$e^{i(kx-\omega t)}$$
BTW, I think there's a typo in your momentum operator.