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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 $ ?

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    $\begingroup$ $$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$ $\endgroup$ – Andrei Geanta Jan 18 '18 at 19:19
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First, the momentum operator is actually a derivative in x, not in time.

$p = -i\hbar \frac{\partial}{\partial x}$, with apologies to mathematicians, but it is understood.

If you want the mean value, you must do

$\int_{-\infty}^{+\infty} \psi^* (p\psi)dx$

So you will have

$\int_{-\infty}^{+\infty} sin(kx-\omega t)\cdot k \cdot cos(kx-\omega t)dx$

This integral does not converge, because an harmonic wave is not integrable. BUT, if you "try" to normalize dividing by

$\int_{-\infty}^{+\infty} sin^2 (kx-\omega t)dx$

The answer gives 0. But that is the mean value!

That's because, as other answers point out, this state is a superposition of two plain waves, travelling in opposite directions, since $\frac{z-z^*}{2} = Im(z)=sin(kx-\omega t)$.

So you can have both momentums $k$ and $-k$. That's why the mean value is 0.

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  • $\begingroup$ Please do not post complete solutions to homework-like problems. You might consider deleting this answer. $\endgroup$ – garyp Jan 19 '18 at 0:52
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    $\begingroup$ I don't see this as a HW-like question, but rather a concrete example that the asker has tried, and I think this example is interesting enough and it provides a small sand grain to knowledge . I thought this site was precisely that: a site for knowledge. I don't know why people in general so allergic to HW questions, if they really contribute to global understanding. It's askign about a specific concept, he's showing effort and he has made an attempt. What else do you want? This is not going to become a prank-site for answering this. Anyways, mayeb @Karan Karan wants to clarify to us... $\endgroup$ – FGSUZ Jan 19 '18 at 13:03
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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.

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  • $\begingroup$ But that would be a different wavefunction. In Electromagnetism we use to work with exponentials, but being aware of taking only the real part. In QM the wavefunction IS actually a complex number, with its imaginary part. The one you wrote has a constant modulus, while the asker's one is a $sin^2$. $\endgroup$ – FGSUZ Jan 18 '18 at 23:51
  • $\begingroup$ @FGSUZ Good point. The problem as originally posed is more interesting. $\endgroup$ – garyp Jan 19 '18 at 0:51
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Since $\psi=\frac{1}{2i}\sum_\pm\pm\exp\pm i(kx-\omega t)$, the momentum is $\pm\hbar k$. The wavelength is $\frac{2\pi\hbar}{p}=\frac{2\pi}{k}$ as you said.

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