D.44 Derivation of group velocity

The objective of this note is to derive the wave function for a wave packet if time is large.

To shorten the writing, the Fourier integral (7.64) for $\Psi$ will be abbreviated as:

\begin{displaymath}
\Psi = \int_{k_1}^{k_2} f(k) e^{{\rm i}\varphi t} {\,\rm d...
... \frac{x}{t} - v_{\rm {g}}
\quad \varphi'' = - v_{\rm {g}}'
\end{displaymath}

where it will be assumed that $\varphi$ is a well behaved functions of $k$ and $f$ at least twice continuously differentiable. Note that the wave number $k_0$ at which the group velocity equals $x$$\raisebox{.5pt}{$/$}$$t$ is a stationary point for $\varphi$. That is the key to the mathematical analysis.

The so-called “method of stationary phase” says that the integral is negligibly small as long as there are no stationary points $\varphi'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 in the range of integration. Physically that means that the wave function is zero at large time positions that cannot be reached with any group velocity within the range of the packet. It therefore implies that the wave packet propagates with the group velocity, within the variation that it has.

To see why the integral is negligible if there are no stationary points, just integrate by parts:

\begin{displaymath}
\Psi = \frac{f(k)}{{\rm i}\varphi't} e^{{\rm i}\varphi t}\...
...}{{\rm i}\varphi't}\right)'
e^{{\rm i}\varphi t} {\,\rm d}k
\end{displaymath}

This is small of order 1$\raisebox{.5pt}{$/$}$$t$ for large times. And if $\overline{\Phi}_0(p)$ is chosen to smoothly become zero at the edges of the wave packet, rather than abruptly, you can keep integrating by parts to show that the wave function is much smaller still. That is important if you have to plot a wave packet for some book on quantum mechanics and want to have its surroundings free of visible perturbations.

For large time positions with $x$$\raisebox{.5pt}{$/$}$$t$ values within the range of packet group velocities, there will be a stationary point to $\varphi$. The wave number at the stationary point will be indicated by $k_0$, and the value of $\varphi$ and its second derivative by $\varphi_0$ and $\varphi_0''$. (Note that the second derivative is minus the first derivative of the group velocity, and will be assumed to be nonzero in the analysis. If it would be zero, nontrivial modifications would be needed.)

Now split the exponential in the integral into two,

\begin{displaymath}
\Psi = e^{{\rm i}\varphi_0 t} \int_{k_1}^{k_2} f(k)
e^{{\rm i}(\varphi-\varphi_0)t} {\,\rm d}k
\end{displaymath}

It is convenient to write the difference in $\varphi$ in terms of a new variable $\overline{k}$:

\begin{displaymath}
\varphi-\varphi_0 = {\textstyle\frac{1}{2}} \varphi_0'' \o...
...\qquad \overline{k} \sim k - k_0\quad\mbox{for}\quad k\to k_0
\end{displaymath}

By Taylor series expansion it can be seen that $\overline{k}$ is a well behaved monotonous function of $k$. The integral becomes in terms $\overline{k}$:

\begin{displaymath}
\Psi = e^{{\rm i}\varphi_0 t} \int_{\overline{k}_1}^{\over...
...d g(\overline{k}) = f(k) \frac{{\rm d}k}{{\rm d}\overline{k}}
\end{displaymath}

Now split function $g$ apart as in

\begin{displaymath}
g(\overline{k}) = g(0) + [g(\overline{k})-g(0)]
\end{displaymath}

The part within brackets produces an integral

\begin{displaymath}
e^{{\rm i}\varphi_0 t} \int_{\overline{k}_1}^{\overline{k}...
... i}\frac12\varphi_0''\overline{k}^2t}
{\,\rm d}\overline{k}
\end{displaymath}

and integration by parts shows that to be small of order 1/$t$.

That leaves the first part, $g(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(k_0)$, which produces

\begin{displaymath}
\Psi = e^{{\rm i}\varphi_0 t} f(k_0) \int_{\overline{k}_1}...
...rm i}\frac12\varphi_0''\overline{k}^2t} {\,\rm d}\overline{k}
\end{displaymath}

Change to a new integration variable

\begin{displaymath}
u \equiv \sqrt{\frac{\vert\varphi_0''\vert t}{2}} \overline{k}
\end{displaymath}

Note that since time is large, the limits of integration will be approximately $u_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\infty$ and $u_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\infty$ unless the stationary point is right at an edge of the wave packet. The integral becomes

\begin{displaymath}
\Psi = e^{{\rm i}\varphi_0 t} f(k_0) \sqrt{\frac{2}{\vert\...
..._0''\vert t}}
\int_{u_1}^{u_2} e^{\pm{\rm i}u^2} {\,\rm d}u
\end{displaymath}

where $\pm$ is the sign of $\varphi_0''$. The remaining integral is a Fresnel integral that can be looked up in a table book. Away from the edges of the wave packet, the integration range can be taken as all $u$, and then

\begin{displaymath}
\Psi = e^{{\rm i}\varphi_0 t} e^{\pm{\rm i}\pi/4} f(k_0)
\sqrt{\frac{2\pi}{\vert\varphi_0''\vert t}}
\end{displaymath}

Convert back to the original variables and there you have the claimed expression for the large time wave function.

Right at the edges of the wave packet, modified integration limits for $u$ must be used, and the result above is not valid. In particular it can be seen that the wave packet spreads out a distance of order $\sqrt{t}$ beyond the stated wave packet range; however, for large times $\sqrt{t}$ is small compared to the size of the wave packet, which is proportional to $t$.

For the mathematically picky: the treatment above assumes that the wave packet momentum range is not small in an asymptotic sense, (i.e. it does not go to zero when $t$ becomes infinite.) It is just small in the sense that the group velocity must be monotonous. However, Kaplun’s extension theorem implies that the packet size can be allowed to become zero at least slowly. And the analysis is readily adjusted for faster convergence towards zero in any case.