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Suppose you are given a wave function $\psi(x, t) = C\exp(i(kx - \omega t))$; how can you calculate the group and particle velocity?

My intuition was to use group velocity: $v_g = \frac{\partial w}{\partial k}$. However, I cannot rearrange for $\omega$.

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The plane wave is a free particle, which is a solution to the Schrodinger equation with zero potential.

If you take the Schrodinger equation with zero potential \begin{equation} i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} \end{equation} and plug in your ansatz \begin{equation} \psi = e^{i (\omega t - k x)} \end{equation} you will find that $\psi$ solves the Schrodinger equation when \begin{equation} \omega = \frac{\hbar k^2}{2m} \end{equation} Note if we multiplied the left and right hand side by $\hbar$, and used $E=\hbar \omega$ and $p = \hbar k$, this equation is just the classical expression for the kinetic energy, $E = p^2/2m$.

Now we can get the phase velocity \begin{equation} v_p = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m} \end{equation} and the group velocity \begin{equation} v_g = \frac{\partial \omega}{\partial k} = \frac{\hbar k}{m} = \frac{p}{m} \end{equation} The group velocity is equal to the classical velocity of a particle with momentum $p$.

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The functional dependence of $\omega(k)$ (the dispersion relation) is not contained in the plane-wave expression for $\psi$. In general, it is either measured or obtained from the particular wave equation for $\psi$.

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Your wave function can't be normalized, because $\varphi(x,t)$ in this case doesn't belong to $L_2$ space. Physical meaning of this - particle can't stay in this state, or particle with this state can't exist and can't be localized (you fixated impulse of this particle and $\Delta x=\infty$), so:

  1. Group velocity can't be found like velocity of localized quasi-monochromatic wave. You can't localize maximum of envelope of a quasi-monochromatic wave packet. If you can't find maximum, you can't differentiate it by time. It makes no sense to look for the speed of propagation of energy because the energy is not located in any particular place
  2. But remember that you fixated impulse and so after any measurement you will get the same momentum value, so in terms of quantum physics $<p>=p_0$, where $p_0=\hbar k$. So, you can find particle velocity like $v=\dfrac{<p>}{m}$
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    $\begingroup$ "Physical meaning of this - particle can't stay in this state" -- this doesn't seem right to me, since $\varphi(x, t)$ is a stationary state for a free particle. $\endgroup$
    – Andrew
    Commented Feb 16, 2022 at 23:51
  • $\begingroup$ The wave function is not normalizable in a Euclidean space, monochromatic stationary states can not correspond to physical realizable states. You can't get really free particles and you can't make really monochromatic waves in real life, you will always get a wave package. It's obvious in optics but is also true for particles $\endgroup$ Commented Feb 17, 2022 at 0:34
  • $\begingroup$ Sure. But to me "particle can't stay in this state" sounds like you are saying "if the particle were in this state, it would transition to another one," which isn't right since (a) it is a stationary state and (b) as you pointed out, the particle would never actually be in that state. $\endgroup$
    – Andrew
    Commented Feb 17, 2022 at 0:38
  • $\begingroup$ No, I didn't mean that it would transition to another one, I just said that this is an impossible state, so there is no sense to determine "velocity" for this state $\endgroup$ Commented Feb 17, 2022 at 0:42

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