Wave Packets

As we have seen, the wavefunction of a massive particle of momentum $p$ and energy $E$, moving in free space along the $x$-axis, can be written

$$\psi(x,t) = \bar{\psi}\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)},$$ (4.36)

where $k= p/\hbar$, $\omega = E/\hbar$, and $\bar{\psi}$ is a complex constant. Here, $\omega$ and $k$ are linked via the matter-wave dispersion relation (4.23). Expression (4.36) represents a plane wave that propagates in the $x$-direction with the phase velocity $v_p=\omega/k$. However, it follows from Equation (4.24) that this phase velocity is only half of the classical velocity of a massive particle.

According to the discussion in the previous section, the most reasonable physical interpretation of the wavefunction is that $\vert\psi(x,t)\vert^{2}\,dx$ is proportional to (assuming that the wavefunction is not properly normalized) the probability of finding the particle between $x$ and $x+dx$ at time $t$. However, the modulus squared of the wavefunction (4.36) is $\vert\bar{\psi}\vert^{2}$, which is a constant that depends on neither $x$ nor $t$. In other words, the previous wavefunction represents a particle that is equally likely to be found anywhere on the $x$-axis at all times. Hence, the fact that this wavefunction propagates at a phase velocity that does not correspond to the classical particle velocity has no observable consequences.

How can we write the wavefunction of a particle that is localized in $x$? In other words, a particle that is more likely to be found at some positions on the $x$-axis than at others. It turns out that we can achieve this goal by forming a linear combination of plane waves of different wavenumbers; that is,

$$\psi(x,t) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk.$$ (4.37)

Here, $\bar{\psi}(k)$ represents the complex amplitude of plane waves of wavenumber $k$ within this combination. In writing the previous expression, we are relying on the assumption that matter waves are superposable. In other words, it is possible to add two valid wave solutions to form a third valid wave solution. The ultimate justification for this assumption is that matter waves satisfy the linear wave equation (4.22).

There is a fundamental mathematical theorem, known as Fourier's theorem, that states that if

$$f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \bar{f}(k)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk,$$ (4.38)

then

$$\bar{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.$$ (4.39)

Here, $\bar{f}(k)$ is known as the Fourier transform of the function $f(x)$. We can use Fourier's theorem to find the $k$-space function $\bar{\psi}(k)$ that generates any given $x$-space wavefunction $\psi(x)$ at a given time.

Figure 4.2: A one-dimensional Gaussian probability distribution.

For instance, suppose that at $t=0$ the wavefunction of our particle takes the form

$$\psi(x,0) =\frac{1}{[2\pi\,({\mit\Delta} x)^2]^{1/4}}\, \exp\left[{\rm i}\,k_0\,x - \frac{(x-x_0)^{\,2}}{4\,({\mit\Delta}x)^{\,2}}\right].$$ (4.40)

Thus, the initial probability distribution for the particle's $x$-coordinate is

$$\vert\psi(x,0)\vert^{2} =\frac{1}{[2\pi\,({\mit\Delta} x)^2]^{1/2}}\, \exp\left[- \frac{(x-x_0)^{2}}{2\,({\mit\Delta}x)^{2}}\right].$$ (4.41)

This particular distribution is called a Gaussian distribution (see Section 5.1.7), and is plotted in Figure 4.2. It can be seen that a measurement of the particle's position is most likely to yield the value $x_0$, and very unlikely to yield a value which differs from $x_0$ by more than $3\,{\mit\Delta} x$. Thus, Equation (4.40) is the wavefunction of a particle that is initially localized in some region of $x$-space, centered on $x=x_0$, whose width is of order ${\mit\Delta} x$. This type of wavefunction is known as a wave packet.

It is easily demonstrated that the wavefunction (4.40) is properly normalized. In fact,

$$\int_{-\infty}^{\infty} \vert\psi(x,0)\vert^{2}\,dx = \frac{1}{[2\pi\,({\mit\Delta} x)^2]^{1/2}}\int_{-\infty}^{\infty}\exp\left[- \frac{(x-x_0)^{2}}{2\,({\mit\Delta}x)^{2}}\right]\,dx$$

$$= \frac{2^{1/2}\,({\mit\Delta}x)}{[2\pi\,({\mit\Delta} x)^2]^{1/2... ... e}^{-y^2}\,dy= \frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty {\rm e}^{-y^2}\,dy=1.$$ (4.42)

Here, $y=(x-x_0)/(2^{1/2}{\mit\Delta x})$, and use has been made of the standard result

$$\int_{-\infty}^\infty {\rm e}^{-y^2}\,dy =\pi^{1/2}.$$ (4.43)

According to Equation (4.37),

$$\psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk.$$ (4.44)

Hence, we can employ Fourier's theorem to invert this expression to give

$$\bar{\psi}(k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi(x,0)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.$$ (4.45)

Making use of Equation (4.40), we obtain

$$\bar{\psi}(k) = \frac{1}{(2\pi)^{3/4}\,({\mit\Delta} x)^{1/2}} \int_{-\infty}^\... ...xp\left[-{\rm i}\,(k-k_0)\,x - \frac{(x-x_0)^2}{4\,({\mit\Delta} x)^2}\right]dx$$

$$= \frac{1}{(2\pi)^{3/4}\,({\mit\Delta} x)^{1/2}}\,\exp\left[-{\rm... ...x-x_0}{2\,{\mit\Delta} x} + {\rm i}\,{\mit\Delta} x\,(k-k_0)\right]^2\right\}dx$$

$$= \frac{2\,{\mit\Delta x}}{(2\pi)^{3/4}\,({\mit\Delta} x)^{1/2}}\... ...k_0)^2}{4\,({\mit\Delta k})^2}\right]\int_{-\infty}^{\infty} {\rm e}^{-y^2}\,dy$$

$$= \frac{1}{[2\pi\,({\mit\Delta} k)^2]^{1/4}}\,\exp\left[-{\rm i}\,(k-k_0)\,x_0-\frac{(k-k_0)^2}{4\,({\mit\Delta k})^2}\right],$$ (4.46)

where $y=(x-x_0)/(2\,{\mit\Delta} x)+{\rm i}\,{\mit\Delta} x\,(k-k_0)$,

$${\mit\Delta} k = \frac{1}{2\,{\mit\Delta} x},$$ (4.47)

and use has been made of Equation (4.43).

If $\vert\psi(x,0)\vert^{2}\,dx$ is the probability that a measurement of the particle's position yields a value in the range $x$ to $x+dx$ at time $t=0$ then it stands to reason that $\vert\bar{\psi}(k)\vert^{2}\,dk$ is the probability that a measurement of the particle's wavenumber yields a value in the range $k$ to $k+dk$. (Recall that $p = \hbar\,k$, so a measurement of the particle's wavenumber, $k$, is equivalent to a measurement of the particle's momentum, $p$.) According to Equation (4.46),

$$\vert\bar{\psi}(k)\vert^{2} = \frac{1}{[2\pi\,({\mit\Delta} k)^2]^{1/2}}\,\exp\left[- \frac{(k-k_0)^{2}}{2\,({\mit\Delta}k)^{2}}\right].$$ (4.48)

This probability distribution is a Gaussian in $k$-space. [See Equation (4.41) and Figure 4.2.] Hence, a measurement of $k$ is most likely to yield the value $k_0$, and very unlikely to yield a value that differs from $k_0$ by more than $3\,{\mit\Delta}k$. Note that the probability distribution (4.48) is properly normalized; that is, $\int_{-\infty}^\infty \vert\bar{\psi}(k)\vert^{2} \,dk = 1$.

We have just seen that a wave packet with a Gaussian probability distribution of characteristic width ${\mit\Delta} x$ in $x$-space [see Equation (4.41)] is equivalent to a wave packet with a Gaussian probability distribution of characteristic width ${\mit\Delta} k$ in $k$-space [see Equation (4.48)], where

$${\mit\Delta}x\,{\mit\Delta} k = \frac{1}{2}.$$ (4.49)

This illustrates an important property of wave packets. Namely, in order to construct a packet that is highly localized in $x$-space (i.e., with small ${\mit\Delta} x$) we need to combine plane waves with a very wide range of different $k$-values (i.e., with large ${\mit\Delta} k$). Conversely, if we only combine plane waves whose wavenumbers differ by a small amount (i.e., if ${\mit\Delta} k$ is small) then the resulting wave packet is highly extended in $x$-space (i.e., ${\mit\Delta} x$ is large).