2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ $φ(x,t)=e^{−i(ω_0−Uk_0)t}φ(x−Ut).$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.