I am reading Zettili's QM concepts and applications (second edition), and in section 1.8.3 where he discusses the motion of wave packets, we consider a wave packet where angular frequency $\omega$ is a function of the wave number $k$: $\omega = \omega(k)$.

We assume that $\phi(k)$ is a function $g(k-k_0)$ that is narrow and peaks at $k=k_0$, so we can Taylor expand $\omega(k)$ about $k_0$ as: $$\omega(k) = \omega(k_0)+(k-k_0)\left.\frac{d\omega(k)}{dk}\right|_{k=k_0} + \frac{1}{2}(k-k_0)^2\left.\frac{d^2\omega(k)}{dk^2}\right|_{k=k_0} + ...$$ $$\omega(k) = \omega(k_0)+(k-k_0)v_g + (k-k_0)^2\alpha + ...$$

then all we would need to do is substitute $\omega(k)$ into the equation $\Psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}\phi(k)e^{i(kx-\omega t)}dk$ . In the book, this is shown as: $$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}e^{ik_0(x-v_{ph}t)} \int_{-\infty}^{+\infty} g(k-k_0)e^{i(k-k_0)(x-v_gt)} e^{-i(k-k_0)^2 \alpha t + ...} dk$$

I don't understand where the phase velocity came from at all, and how did he get $e^{i(k-k_0)(x-v_gt)}$? Shouldn't it simply be $e^{\omega(k_0)-i(k-k_0)v_gt}$?

Also a sub question: In the integral, $e^{i(k-k_0)(x-v_gt)} e^{-i(k-k_0)^2 \alpha t + ...}$ ends with a $+...$, but if you write out more terms, they would all start with a minus sign because of the minus sign in $-\omega t$, right?


1 Answer 1


Zitteli is simply writing the exponent as \begin{eqnarray} i(kx-\omega(k) t) &=& i(kx-[\omega(k_0)+(k-k_0)v_g +(k-k_0)^2\alpha + ...]t) \end{eqnarray} and then adding and subtracting $k_0 x$, he writes \begin{equation} -\omega(k_0) t = k_0\left (x-\frac{\omega(k_0)}{k_0} t\right ) - k_0 x \,. \end{equation} The phase velocity is defined to be $v_{ph} = \frac{\omega(k_0)}{k_0}$, so this is \begin{equation} -\omega(k_0) t = k_0(x-v_{ph} t) - k_0 x \,. \end{equation} Substituting gives Zettili's expression.

Often $+...$ just means there are more terms that are ignored and doesn't imply a sign (over even a phase).


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