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I know from my quantum mechanics course that the moving particle moves with the group velocity of the wave packet $v_{g}$ and I also Know that the momentum of the particle is $\hbar k=p$

so it implies \begin{gather} \hbar k=p=mv_{g} \end{gather}

so the $v_{g}$ always depend on the wave vector $k$ , but I've studied some cases where the $v_{g}$ is zero while $k$ has some value (like in the solution of kroning penny model )

Can someone help explain me how can the $v_{g}$ is zero while the $k$ is not ? and also clarify about their relationship

amin
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  • Imagine a Gaussian distribution that gets wider over time, but stays centred at 0. Does that help? – Quantum Mechanic May 28 '23 at 23:34
  • @QuantumMechanic, I don't understand what you want to say ? – Mans May 28 '23 at 23:54
  • @Mans that's an example where the overall wave packet is moving nowhere (group velocity zero) but there is a distribution of nonzero momenta – Quantum Mechanic May 29 '23 at 00:20
  • @QuantumMechanic, But isn't the group velocity get calculated from the momentum or in other words how $\hbar k =mv_{g} $ can't be zero when the group velocity is also zero ? – Mans May 29 '23 at 00:56
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    In solids, $\hbar k$ isn't momentum, nor is it the expectation value of momentum. It is called crystal momentum. Does the first paragraph of this answer address your question? – Puk May 29 '23 at 02:15

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