Group Velocity

We have seen that Equation (4.40) is the wavefunction of a particle whose most probable position at time $t=0$ is $x=x_0$. According to Equations (4.37) and (4.46), the wavefunction evolves in time as

$$\psi(x,t) = \frac{1}{(2\pi)^{3/4}\,({\mit\Delta} k)^{1/2}}\int_{-... ...mega\,t-{\rm i}\,(k-k_0)\,x_0-\frac{(k-k_0)^2}{4\,({\mit\Delta k})^2}\right]dk.$$ (4.50)

Here, $\omega$ is related to $k$ via the dispersion relation (4.23); in other words, $\omega=\omega(k)$. Now, the integrand on the right-hand side of the previous expression is strongly peaked at $k=k_0$. It follows that the only significant contribution to the corresponding integral comes from a small region of $k$-space centered on $k=k_0$. Let us Taylor expand the dispersion relation, $\omega=\omega(k)$, about $k=k_0$. Neglecting second-order terms in the expansion, we obtain

$$\omega \simeq \omega_0+ v_g\,(k-k_0),$$ (4.51)

where

$$\omega_ 0 = \omega(k_0),$$ (4.52)

$$v_g = \frac{d\omega(k_0)}{dk}.$$ (4.53)

Thus, we get

$$\psi(x,t) = \frac{1}{(2\pi)^{3/4}\,({\mit\Delta} k)^{1/2}}\exp[\,{\rm i}\,(... ...m i}\,(k-k_0)\,(x-x_0-v_g\,t) -\frac{(k-k_0)^2}{4\,({\mit\Delta k})^2}\right]dk$$

$$= \frac{2\,{\mit\Delta} k}{(2\pi)^{3/4}\,({\mit\Delta} k)^{1/2}}\... ...2}}{4\,({\mit\Delta}x)^{\,2}}\right]\int_{-\infty}^{\infty} {\rm e}^{-y^2}\,dy,$$ (4.54)

where $y=(k-k_0)/(2\,{\mit\Delta} k)-{\rm i}\,{\mit\Delta} k\,(x-x_0-v_g\,t)$, and use has been made of Equation (4.49). The previous equation reduces to

$$\psi(x,t) = \frac{1}{[2\pi\,({\mit\Delta} x)^2]^{1/4}}\, \exp\lef... ...,x-\omega_0\,t) - \frac{(x-x_0-v_g\,t)^{\,2}}{4\,({\mit\Delta}x)^{\,2}}\right],$$ (4.55)

where use has been made of Equations (4.43) and (4.49). Hence, the probability of finding the particle between $x$ and $x+dx$ at time $t$ is $\vert\psi(x,t)\vert^2\,dx$, where

$$\vert\psi(x,t)\vert^2 = \frac{1}{[2\pi\,({\mit\Delta} x)^2]^{1/2}}\, \exp\left[- \frac{(x-x_0-v_g\,t)^{\,2}}{2\,({\mit\Delta}x)^{\,2}}\right].$$ (4.56)

It can be seen that the particle's most probable location at time $t$ is

$$x = x_0 + v_g\,t.$$ (4.57)

If, as seems reasonable, we identify the velocity of the particle with the velocity of its most probable location then we deduce that the particle effectively moves at the so-called group velocity,

$$v_g = \frac{d\omega}{dk},$$ (4.58)

rather than the phase velocity,

$$v_p = \frac{\omega}{k}.$$ (4.59)

Incidentally, the distinction between these two velocities is as follows. The phase velocity is the propagation velocity of an individual wave maximum, whereas the group velocity is the propagation velocity of an interference peak.

We have seen that a spatially localized particle moves at the group velocity, (4.58), rather than the phase velocity, (4.59). Making use of the matter-wave dispersion relation, (4.23), the group velocity is

$$v_g = \frac{\hbar\,k}{m} = \frac{p}{m},$$ (4.60)

where use has been made of Equation (4.12). This velocity is identical to the classical velocity of a (non-relativistic) massive particle. We conclude that the matter-wave dispersion relation (4.23) is perfectly consistent with classical physics, as long as we recognize that particles must be identified with wave packets (which propagate at the group velocity) rather than plane waves (which propagate at the phase velocity).