I've recently begun studying Quantum Mechanics and went through the descriptions of phase and group velocities and the fact the group velocity or the velocity of the resulting wave envelope represents the velocity of a free particle for a non relativistic system. I know that the superposition of plane harmonic waves of different wavelengths and travelling with different phase velocities can be given as,$$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \phi(k)e^{ikx}e^{-\omega(k)t}dk.$$ where $\Psi$ is the wave function representing a free particle travelling in one dimension. It's assumed that the $\phi(k)$ peaks at some $k_0$ and hence the assumption is made that the particle has a precise momentum whose momentum space wave function peaks at $k_0$ and $\omega$ is assumed to be a function of $k$ as it is in most cases of a dispersive medium. Hence approximating $\omega$ with Taylor's expansion in the vicinity of $k_0$ $$\omega(k) = \omega(k_0) + \frac{d \omega(k)}{dk}(k-k_0)+ \frac{d^2 \omega(k)}{dk^2}(k-k_0)^2+...$$ where derivatives are calculated at $k_0$ and neglecting quadratic term gives the final form of $\Psi$ as, after defining variable $\kappa := k-k_0$ we get $$\Psi(x,t) \approx \frac{1}{\sqrt{2 \pi}}e^{-i\omega(k_0)}e^{i2\omega(k_0)} \int^{\infty}_{-\infty}\phi(\kappa + k_0)e^{i(\kappa + k_0)(x-\frac{d \omega(k)}{dk})}dk.$$ hence the wave function has a velocity equal to the group velocity $$ v_g=\frac{d \omega(k)}{dk} $$ evaluated at $k_0$ and hence we know the particle velocity since the particle velocity $v_p$ is given as, $$ v_g = \frac{d \omega(k)}{dk} = \frac{d E(p)}{dp} = \frac{d (p^2/2m)}{dp} = \frac{p}{m} = v_p$$ So, even if the spatial wave function peaked sharply, the group velocity would still be $\frac{d\omega}{dk}$ and hence we will always know the group velocity (and hence the particle velocity) even if we also know its position quite accurately.

So, my question is does the knowledge of group velocity violate the uncertainty principle that the velocity and position can't be known simultaneously beyond a certain accuracy? Since we are able to measure the velocity of the particle as group velocity $v_g$ hence the momentum $mv_g$,the spread. The momentum $\Delta p =0$ and hence the spread in position wave function should be infinity to satisfy the uncertainty principle $\Delta p\cdot\Delta x \ge \frac{h}{4\pi}$. But the wave propagates with a finite spread in position making a wave envelope.


1 Answer 1


If the particle has a precise momentum, then its wavefunction doesn't just peak at $k_0$, it is a delta-function at $k_0$. In this case the wavefunction is monochromatic. The spatial wavefunction is a single frequency sinusoid extending infinitely in both directions.

This wavefunction does not violate the uncertainty relation. Taking the appropriate limits you would find $\Delta p \rightarrow 0$ and $\Delta x \rightarrow \infty$, while maintaining a finite $\Delta x \cdot \Delta p$.

If the spatial wavefunction is a Gaussian enveloped sinusoid, then the momentum wavefunction will not be a delta-function. The momentum wavefunction will also be a Gaussian enveloped sinusoid with a finite, non-zero spread.

The Taylor series approximation you employ assumes the wavefunction is monochromatic. Your $v_g$ is the group velocity for a monochromatic wave with wavenumber $k_0$. A particle defined by a monochromatic wavefunction has a well defined wavenumber and momentum and speed. It also has infinite spatial extent.

All quantum wavefunctions are dispersive, different wavenumbers/frequencies travel at different speeds. This results in the wavefuntion spreading out as it travels. There is not one group velocity for a dispersive wavepacket, a phenomenon called group velocity dispersion.

  • $\begingroup$ Thanks for the answer, if we assume that the momentum space wavefunction is a gaussian peaking at $$k_0$$ then the spatial wavefunction may have some group velocity that will always be equal to the particle velocity. Sharper the wavefunction peak gets in momentum space, wider the spatial envelope becomes. But still the envelope will have a group velocity and hence we can measure the momentum as mass times Vg precisely. I don't understand what I'm missing here? $\endgroup$ Sep 18, 2023 at 3:06
  • $\begingroup$ Do the edits to address dispersion help? $\endgroup$
    – Paul T.
    Sep 18, 2023 at 3:45

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