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Next: Evolution of wave-packets Up: Wave-particle duality Previous: Quantum particles   Contents


Wave-packets

The above discussion suggests that the wave-function of a massive particle of momentum $p$ and energy $E$ can be written
\begin{displaymath}
\psi(x,t) = \bar{\psi}\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)},
\end{displaymath} (64)

where $k= p/\hbar$ and $\omega = E/\hbar$. Here, $\omega$ and $k$ are linked via the dispersion relation (61). Expression (64) represents a plane-wave which propagates in the $x$-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 wave-function is that $\vert\psi(x,t)\vert^{\,2}$ is proportional to the probability density of finding the particle at position $x$ at time $t$. However, the modulus squared of the wave-function (64) is $\vert\bar{\psi}\vert^{\,2}$, which depends on neither $x$ nor $t$. In other words, this wave-function represents a particle which is equally likely to be found anywhere on the $x$-axis at all times. Hence, the fact that the plane-wave wave-function (64) propagates at a phase-velocity which does not correspond to the classical particle velocity does not have any real physical consequences.

So, how can we write the wave-function of a particle which is localized in $x$: i.e., a particle which 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 wave-numbers: i.e.,

\begin{displaymath}
\psi(x,t) = \int_{-\infty}^{\infty} \bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,(k\,x-\omega\,t)}\,dk.
\end{displaymath} (65)

Here, $\bar{\psi}(k)$ represents the complex amplitude of plane-waves of wave-number $k$ in this combination. In writing the above expression, we are relying on the assumption that particle waves are superposable: i.e., it is 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 which is linear in $\psi$. As we shall see, in Sect. 3.12, this is indeed the case.

Now, there is a useful mathematical theorem, known as Fourier's theorem, which states that if

\begin{displaymath}
f(x) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^{\infty} \bar{f}(k)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk,
\end{displaymath} (66)

then
\begin{displaymath}
\bar{f}(k) = \frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^\infty f(x)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.
\end{displaymath} (67)

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 $\bar{\psi}(k)$ which generates any given $x$-space wave-function $\psi(x)$ at a given time.

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

\begin{displaymath}
\psi(x,0) \propto \exp\left[{\rm i}\,k_0\,x - \frac{(x-x_0)^{\,2}}{4\,({\mit\Delta}x)^{\,2}}\right].
\end{displaymath} (68)

Thus, the initial probability density of the particle is written
\begin{displaymath}
\vert\psi(x,0)\vert^{\,2} \propto \exp\left[- \frac{(x-x_0)^{\,2}}{2\,({\mit\Delta}x)^{\,2}}\right].
\end{displaymath} (69)

This particular probability distribution is called a Gaussian distribution, and is plotted in Fig. 4. 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, (68) is the wave-function of a particle which is initially localized around $x=x_0$ in some region whose width is of order ${\mit\Delta} x$. This type of wave-function is known as a wave-packet.

Figure 4: A Gaussian probability distribution in $x$-space.
\begin{figure}
\epsfysize =3.in
\centerline{\epsffile{gauss.eps}}
\end{figure}

Now, according to Eq. (65),

\begin{displaymath}
\psi(x,0) = \int_{-\infty}^{\infty} \bar{\psi}(k)\,{\rm e}^{\,{\rm i}\,k\,x}\,dk.
\end{displaymath} (70)

Hence, we can employ Fourier's theorem to invert this expression to give
\begin{displaymath}
\bar{\psi}(k)\propto \int_{-\infty}^{\infty} \psi(x,0)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.
\end{displaymath} (71)

Making use of Eq. (68), we obtain
\begin{displaymath}
\bar{\psi}(k) \propto
{\rm e}^{-{\rm i}\,(k-k_0)\,x_0}\int_{...
...0)\,(x-x_0) - \frac{(x-x_0)^2}{4\,({\mit\Delta}x)^2}\right]dx.
\end{displaymath} (72)

Changing the variable of integration to $y=(x-x_0)/ (2\,{\mit\Delta} x)$, this reduces to
\begin{displaymath}
\bar{\psi}(k) \propto {\rm e}^{-{\rm i}\,k\,x_0}
\int_{-\infty}^{\infty}\exp\left[-{\rm i}\,\beta\,y - y^2\right] dy,
\end{displaymath} (73)

where $\beta = 2\,(k-k_0)\,{\mit\Delta}x$. The above equation can be rearranged to give
\begin{displaymath}
\bar{\psi}(k) \propto {\rm e}^{-{\rm i}\,k\,x_0 - \beta^2/4}\int_{-\infty}^{\infty} {\rm e}^{-(y-y_0)^{\,2}}\,dy,
\end{displaymath} (74)

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
\begin{displaymath}
\bar{\psi}(k) \propto \exp\left[-{\rm i}\,k\,x_0 - \frac{(k-k_0)^{\,2}}{4\,({\mit\Delta}k)^2}\right],
\end{displaymath} (75)

where
\begin{displaymath}
{\mit\Delta} k = \frac{1}{2\,{\mit\Delta} x}.
\end{displaymath} (76)

Now, if $\vert\psi(x)\vert^{\,2}$ is proportional to the probability density of a measurement of the particle's position yielding the value $x$, then it stands to reason that $\vert\bar{\psi}(k)\vert^{\,2}$ is proportional to the probability density of a measurement of the particle's wave-number yielding the value $k$. (Recall that $p = \hbar\,k$, so a measurement of the particle's wave-number, $k$, is equivalent to a measurement of the particle's momentum, $p$). According to Eq. (75),

\begin{displaymath}
\vert\bar{\psi}(k)\vert^{\,2} \propto \exp\left[- \frac{(k-k_0)^{\,2}}{2\,({\mit\Delta}k)^{\,2}}\right].
\end{displaymath} (77)

Note that this probability distribution is a Gaussian in $k$-space--see Eq. (69) and Fig. 4. 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 mathematical function in $x$-space which has the same form as its Fourier transform in $k$-space.

We have seen that a Gaussian probability distribution of characteristic width ${\mit\Delta} x$ in $x$-space [see Eq. (69)] transforms to a Gaussian probability distribution of characteristic width ${\mit\Delta} k$ in $k$-space [see Eq. (77)], where

\begin{displaymath}
{\mit\Delta}x\,{\mit\Delta} k = \frac{1}{2}.
\end{displaymath} (78)

This illustrates an important property of wave-packets. If we wish to construct a packet which 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 wave-numbers 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).


next up previous contents
Next: Evolution of wave-packets Up: Wave-particle duality Previous: Quantum particles   Contents
Richard Fitzpatrick 2006-12-12