I am reading a lecture note (author: Pr. Barton Zwiebach from the physics department of M.I.T., Course: 8.04 'Quantum Physics I', title: de Broglie Wavelength and Galilean Transformations, Phase and Group Velocities, Choosing the Wavefunction for a Free Particle) and they say that in the expression $$\Psi = \int dk \Phi(k) e^{i(kx-\omega (k) t)}$$ taking $\Phi$ real or complex doesn't change the group velocity of the wavepacket.

They prove that the wavepacket moves at the velocity $$v_g = \frac{d \omega}{dk}_{|k_0}$$ for $\Phi$ real, using the stationary phase principle (and, in a second step, using a Taylor expansion of $\omega (k)$ around $k_0$), but not for $\Phi$ complex.

So I tried to do it on my own:

If $\Phi$ is complex then $\Phi (k) = |\Phi(k)|e^{iArg(\Phi(k))}$

Using the stationary phase principle, I get:

$$x-\frac{d \omega}{dk}_{|k_0}t +\frac{d (Arg(\Phi(k))}{dk}_{|k_0} =0$$

From there I don't know what to do...

Same for the Taylor expansion method, I add a complex phase to $e^{ikx}$ but then I don't know how to cancel that phase out to find the same result as for the real case.

  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    May 31, 2022 at 13:29
  • $\begingroup$ Okay I will do it for the next posts! These are lecture notes of Pr. Barton Zwiebach from the physics department of M.I.T. The course is called 8.04 'Quantum Physics I' $\endgroup$
    – niobium
    May 31, 2022 at 13:58
  • $\begingroup$ Please also do it for this post. $\endgroup$
    – Qmechanic
    May 31, 2022 at 14:02

1 Answer 1


Suppose that we have waves with dispersion equation $\omega =\omega(k)$. A right-going wave-packet of finite extent, and with initial profile $\varphi(x)$, can be Fourier analyzed to give $$ \varphi(x)= \int_{-\infty}^{\infty} \frac {dk}{2\pi} \Phi(k) e^{ikx}. $$ At later times this will evolve to $$ \varphi(x,t)= \int_{-\infty}^{\infty} \frac {dk}{2\pi} \Phi(k) e^{ikx-i\omega(k) t}. $$ Let us suppose that $\Phi(k)$ is non-zero only for a narrow band of wavenumbers around $k_0$, and that, restricted to this narrow band, we can approximate the full $\omega(k)$ dispersion equation by $$ \omega(k) \approx \omega_0+ U(k-k_0). $$ Thus $$ \varphi(x,t)= \int_{-\infty}^{\infty} \frac {dk}{2\pi} \Phi(k) e^{ik(x-Ut)-i(\omega_0-Uk_0)t}. $$ Comparing this with the Fourier expression for the initial profile, we find that
$$ \varphi(x,t)= e^{-i (\omega_0-Uk_0) t} \varphi(x-Ut). $$ The pulse envelope therefore travels at speed $U$. The individual wave crests, on the other hand, move at the phase velocity $\omega(k)/k$. The phase velocity is encoded in the $e^{-i (\omega_0-Uk_0) t}$ overall phase factor.

  • $\begingroup$ $φ(x,t)=e^{−i(ω_0−Uk_0)t}φ(x−Ut).$ So this holds for $ \Phi$ complex or real ? Doesn't matter if $\Phi$ is complex ? $\endgroup$
    – niobium
    May 31, 2022 at 13:06
  • 2
    $\begingroup$ Yes. I assume complex $\Phi$ for simplicity. If you want real $\varphi$ you must have $\Phi(-k)= [\Phi(k)]^*$ and so $\Phi$ is big near both $k_0$ and $-k_0$ and you must add the two contributions and then the overall phase factor becomes $\cos[(\omega_0-Uk_0)t]$. Usually complex expressions are understood to have a tacit "real part" understood. $\endgroup$
    – mike stone
    May 31, 2022 at 13:10

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