# Joan Lindsay Orr


# Phase and Group Velocity

The most general (1-dimensional) wave function is a function of the form $$f(kx - \omega t)$$, where $$f:\RR\rightarrow\CC$$. Any particular value of $$f$$, say $$f(c)$$, moves to the right at a speed $$\omega/k$$, because this is how $$x$$ and $$t$$ evolve on the line $$kx - \omega t = c$$.

In particular a plane wave, $$e^{i (kx - \omega t)}$$, propagates to the right with speed $$\omega/k$$. In this case, $$k$$ is the wave number, the number of cycles in a spatial interval of length $$2\pi$$, and $$\omega$$ is the (normalized) frequency, the number of cycles in a temporal interval of length $$2\pi$$. From this we have the wavelength $$\lambda = 2\pi/k$$ and the frequency $$\nu = \omega / 2\pi$$ and recover the formula velocity = frequency $$\times$$ wavelength.

Physically, the refractive index of a medium will vary with the wavelength; this is the phenomenon which enables a prism to split a beam of white light. Mathematically we typically make $$\omega$$ depend on $$k$$ (called the dispersion relation__), and we can assemble a _wave packet of the form: $\Psi(x, t) = \int_{-\infty}^\infty c(k) e^{i (kx - \omega(k) t)} dk$

Now suppose that $$c(k)$$ is narrowly peaked about some value $$k_0$$. If the peak is narrow enough then $$\omega(k)$$ is well-approximated by the first two terms of its Taylor series for all values of $$k$$ for which $$c(k)$$ is non-zero. In this case \begin{align} \Psi(x, t) &= \int_{-\infty}^\infty c(k) e^{i (kx - (\omega(k_0) + \omega'(k_0)(k-k_0) )t)} dk \\ &= e^{i (k_0x - \omega(k_0)t} \int_{-\infty}^\infty c(k) e^{i ((k-k_0)x - \omega'(k_0)(k-k_0)t)} \\ &= e^{i (k_0x - \omega(k_0)t} \int_{-\infty}^\infty c(k) e^{i (x - \omega'(k_0)t)(k-k_0)} \\ &= e^{i (k_0x - \omega(k_0)t} F(x - \omega'(k_0)t) \end{align} The first term of these is a plane wave which propagates with velocity $$\omega(k_0) / k_0$$, and this is the phase velocity. This is modulated by the seond term, which propagates with velocity $$\omega'(k_0) / 1 = \omega'(k_0)$$. This is the group velocity.

## Application to the Schrödinger Equation

For a free particle (i.e. $$V(x) = 0$$ for all $$x$$) the time-independent Schrödinger Equation has solutions of the form $\Psi(x,t) = e^{-iEt/\hbar} \psi(x)$ where $$\psi$$ is a solution of $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi$ and so $\frac{d^2\psi}{dx^2} = -\frac{2mE}{\hbar^2}\psi = - k^2 \psi$ where $k^2 = 2mE/\hbar^2.$

Thus $$\psi(x) = e^{ikx}$$ (traveling in the positive direction; there's also a solution traveling in the negative direction of course), and $\Psi(x, t) = e^{i(kx - Et/\hbar)}$ It follows that $\omega(k) = \frac{E}{\hbar} = \frac{\hbar k^2}{2m}$

Thus the phase velocity is $\frac{\omega(k)}{k} = \frac{\hbar k}{2m}$ and the group velocity is $\omega'(k) = \frac{\hbar k}{m}$

Now the classical (macro) velocity $$v$$ satisfies $$E = \frac{1}{2}mv^2$$, and so $v^2 = \frac{2E}{m} = \frac{\hbar^2 k^2}{m^2}$ Thus we see the group velocity corresponds to the classical velocity of the particle/wave group, while the phase velocity is off by a factor of $$\sqrt{2}$$.