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# 2.10: Wave-Packets

The previous discussion suggests that the wavefunction of a massive particle of momentum $$p$$ and energy $$E$$, moving in the positive $$x$$-direction, can be written $\label{e2.41} \psi(x,t) = \bar{\psi}\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)},$ where $$k= p/\hbar>0$$ and $$\omega = E/\hbar>0$$. Here, $$\omega$$ and $$k$$ are linked via the dispersion relation ([e2.38]). Expression ([e2.41]) represents a plane-wave whose maxima and minima propagate in the positive $$x$$-direction with the phase-velocity $$v_p=\omega/k$$. As we have seen, this phase-velocity is only half of the classical velocity of a massive particle.

From before, the most reasonable physical interpretation of the wavefunction is that $$|\psi(x,t)|^{\,2}$$ is proportional to the probability density of finding the particle at position $$x$$ at time $$t$$. However, the modulus squared of the wavefunction ([e2.41]) is $$|\bar{\psi}|^{\,2}$$, which depends on neither $$x$$ nor $$t$$. In other words, this wavefunction represents a particle that is equally likely to be found anywhere on the $$x$$-axis at all times. Hence, the fact that the maxima and minima of the wavefunction propagate at a phase-velocity that does not correspond to the classical particle velocity does not have any real physical consequences.

How can we write the wavefunction of a particle that is localized in $$x$$: that is, 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: in other words, $\label{e2.42} \psi(x,t) = \int_{-\infty}^{\infty} \bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk.$ Here, $$\bar{\psi}(k)$$ represents the complex amplitude of plane-waves of wavenumber $$k$$ in this combination. In writing the previous expression, we are relying on the assumption that particle waves are superposable: that is, that it is always possible to add two valid wave solutions to form a third valid wave solution. The ultimate justification for this assumption is that particle waves satisfy a differential wave equation that is linear in $$\psi$$. As we shall see, in Section 1.15, this is indeed the case. Incidentally, a plane-wave that varies as $$\exp[\,{\rm i}\,(k\,x-\omega\,t)]$$ and has a negative $$k$$ (but positive $$\omega$$) propagates in the negative $$x$$-direction at the phase-velocity $$\omega/|k|$$. Hence, the superposition ([e2.42]) includes both forward and backward propagating waves.

There is a useful mathematical theorem, known as Fourier’s theorem , which states that if $\label{e2.43} f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \bar{f}(k)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk,$ then $\label{e2.44} \bar{f}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.$ 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.

For instance, suppose that at $$t=0$$ the wavefunction of our particle takes the form $\label{e2.45} \psi(x,0) \propto \exp\left[{\rm i}\,k_0\,x - \frac{(x-x_0)^{\,2}}{4\,({\mit\Delta}x)^{\,2}}\right].$ Thus, the initial probability density of the particle is written $\label{e2.46} |\psi(x,0)|^{\,2} \propto \exp\left[- \frac{(x-x_0)^{\,2}}{2\,({\mit\Delta}x)^{\,2}}\right].$ This particular probability distribution is called a Gaussian distribution, and is plotted in Figure [f4]. 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 ([e2.45]) is the wavefunction of a particle that is initially localized around $$x=x_0$$ in some region whose width is of order $${\mit\Delta} x$$. This type of wavefunction is known as a wave-packet.

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According to Equation ([e2.42]), $\psi(x,0) = \int_{-\infty}^{\infty} \bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk.$ Hence, we can employ Fourier’s theorem to invert this expression to give $\label{e2.42a} \bar{\psi}(k)\propto \int_{-\infty}^{\infty} \psi(x,0)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.$ Making use of Equation ([e2.45]), we obtain $\bar{\psi}(k) \propto {\rm e}^{-{\rm i}\,(k-k_0)\,x_0}\int_{-\infty}^{\infty} \exp\left[ -{\rm i}\,(k-k_0)\,(x-x_0) - \frac{(x-x_0)^{\,2}}{4\,({\mit\Delta}x)^{\,2}}\right]dx.$ Changing the variable of integration to $$y=(x-x_0)/ (2\,{\mit\Delta} x)$$, this reduces to $\bar{\psi}(k) \propto {\rm e}^{-{\rm i}\,k\,x_0} \int_{-\infty}^{\infty}\exp\left(-{\rm i}\,\beta\,y - y^{\,2}\right) dy,$ where $$\beta = 2\,(k-k_0)\,{\mit\Delta}x$$. The previous equation can be rearranged to give $\bar{\psi}(k) \propto {\rm e}^{-{\rm i}\,k\,x_0 - \beta^{\,2}/4}\int_{-\infty}^{\infty} {\rm e}^{-(y-y_0)^{\,2}}\,dy,$ where $$y_0 = - {\rm i}\,\beta/2$$. The integral now just reduces to a number, as can easily be seen by making the change of variable $$z=y-y_0$$. Hence, we obtain $\label{e2.51} \bar{\psi}(k) \propto \exp\left[-{\rm i}\,k\,x_0 - \frac{(k-k_0)^{\,2}}{4\,({\mit\Delta}k)^{\,2}}\right],$ where ${\mit\Delta} k = \frac{1}{2\,{\mit\Delta} x}.$

If $$|\psi(x)|^{\,2}$$ is proportional to the probability density of a measurement of the particle’s position yielding the value $$x$$ then it stands to reason that $$|\bar{\psi}(k)|^{\,2}$$ is proportional to the probability density of a measurement of the particle’s wavenumber yielding the value $$k$$. (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 ([e2.51]), $\label{e2.53} |\bar{\psi}(k)|^{\,2} \propto \exp\left[- \frac{(k-k_0)^{\,2}}{2\,({\mit\Delta}k)^{\,2}}\right].$ Note that this probability distribution is a Gaussian in $$k$$-space. [See Equation ([e2.46]) and Figure [f4].] Hence, a measurement of $$k$$ is most likely to yield the value $$k_0$$, and very unlikely to yield a value which differs from $$k_0$$ by more than $$3\,{\mit\Delta}k$$. Incidentally, a Gaussian is the only simple mathematical function in $$x$$-space that has the same form as its Fourier transform in $$k$$-space.

We have just seen that a Gaussian probability distribution of characteristic width $${\mit\Delta} x$$ in $$x$$-space [see Equation ([e2.46])] transforms to a Gaussian probability distribution of characteristic width $${\mit\Delta} k$$ in $$k$$-space [see Equation ([e2.53])], where ${\mit\Delta}x\,{\mit\Delta} k = \frac{1}{2}.$ This illustrates an important property of wave-packets. Namely, if we wish to construct a packet that is very localized in $$x$$-space (i.e., if $${\mit\Delta}x$$ is small) then we need to combine plane-waves with a very wide range of different $$k$$-values (i.e., $${\mit\Delta}k$$ will be large). 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 will be very extended in $$x$$-space (i.e., $${\mit\Delta}x$$ will be large).

# Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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