Subsections


7.10 Wave Packets

This section gives a full description of the motion of a particle according to quantum mechanics. It will be assumed that the particle is in free space, so that the potential energy is zero. In addition, to keep the analysis concise and the results easy to graph, it will be assumed that the motion is only in the $x$-​direction. The results may easily be extended to three dimensions by using separation of variables.

One thing that the analysis will show is how limiting the uncertainty in both momentum and position produces the various features of classical Newtonian motion. It may be recalled that in Newtonian motion through free space, the linear momentum $p$ is constant. In addition, since $p$$\raisebox{.5pt}{$/$}$$m$ is the velocity $v$, the classical particle will move at constant speed. So classical Newtonian motion would say:

\begin{displaymath}
v = \frac{p}{m} = \mbox{constant} \qquad x = v t + x_0\qquad
\mbox{for Newtonian motion in free space}
\end{displaymath}

(Note that $p$ is used to indicate $p_x$ in this and the following sections.)


7.10.1 Solution of the Schrö­din­ger equation.

As discussed in section 7.1, the unsteady evolution of a quantum system may be determined by finding the eigenfunctions of the Hamiltonian and giving them coefficients that are proportional to $e^{-{{\rm i}}Et/\hbar}$. This will be worked out in this subsection.

For a free particle, there is only kinetic energy, so in one dimension the Hamiltonian eigenvalue problem is:

\begin{displaymath}
-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi
\end{displaymath} (7.56)

Solutions to this equation take the form of exponentials

\begin{displaymath}
\psi_E = A e^{\pm{\rm i}\sqrt{2mE} x/\hbar}
\end{displaymath}

where $A$ is a constant.

Note that $E$ must be positive: if the square root would be imaginary, the solution would blow up exponentially at large positive or negative $x$. Since the square magnitude of $\psi$ at a point gives the probability of finding the particle near that position, blow up at infinity would imply that the particle must be at infinity with certainty.

The energy eigenfunction above is really the same as the eigenfunction of the $x$-​momentum operator ${\widehat p}_x$ derived in the previous section:

\begin{displaymath}
\psi_E = \frac{1}{\sqrt{2\pi\hbar}}
e^{{\rm i}p x/\hbar} \quad \mbox { with } p=\pm \sqrt{2mE}
\end{displaymath} (7.57)

The reason that the momentum eigenfunctions are also energy eigenfunctions is that the energy is all kinetic energy, and the kinetic operator equals ${\widehat T}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\widehat p}^2$$\raisebox{.5pt}{$/$}$$2m$. So eigenfunctions with precise momentum $p$ have precise energy $p^2$$\raisebox{.5pt}{$/$}$$2m$.

As shown by (7.55) in the previous section, combinations of momentum eigenfunctions take the form of an integral rather than a sum. In the one-di­men­sion­al case that integral is:

\begin{displaymath}
\Psi(x,t)= \frac{1}{\sqrt{2\pi\hbar}}
\int_{-\infty}^\infty \Phi(p,t) e^{{\rm i}p x/\hbar} { \rm d}p
\end{displaymath}

where $\Phi(p,t)$ is called the momentum space wave function.

Whether a sum or an integral, the Schrö­din­ger equation still requires that the coefficient of each energy eigenfunction varies in time proportional to $e^{-{{\rm i}}Et/\hbar}$. The coefficient here is the momentum space wave function $\Phi$, and the energy is $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p^2$$\raisebox{.5pt}{$/$}$$2m$, so the solution of the Schrö­din­ger equation must be:

\begin{displaymath}
\fbox{$\displaystyle
\Psi(x,t)= \frac{1}{\sqrt{2\pi\hbar...
...ac{p}{{\scriptscriptstyle 2}m}} t)/\hbar}
{ \rm d}p
$} %
\end{displaymath} (7.58)

Here $\Phi_0(p)\equiv\Phi(p,0)$ is determined by whatever initial conditions are relevant to the situation that is to be described. The above integral is the final solution for a particle in free space.


Key Points
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In free space, momentum eigenfunctions are also energy eigenfunctions.

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The one-di­men­sion­al wave function for a particle in free space is given by (7.58).

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The function $\Phi_0$ is still to be chosen to produce whatever physical situation is to be described.


7.10.2 Component wave solutions

Before trying to interpret the complete obtained solution (7.58) for the wave function of a particle in free space, it is instructive first to have a look at the component solutions, defined by

\begin{displaymath}
\psi_{\rm {w}} \equiv
e^{{\rm i}p (x - {\textstyle\frac{p}{{\scriptscriptstyle 2}m}} t)/\hbar}
\end{displaymath} (7.59)

These solutions will be called component waves; both their real and imaginary parts are sinusoidal, as can be seen from the Euler formula (2.5).

\begin{displaymath}
\psi_{\rm {w}} =
\cos\left(p \Big(x - \frac{p}{2m} t\Big...
... {\rm i}\sin\left(p \Big(x - \frac{p}{2m} t\Big)/\hbar\right)
\end{displaymath}

In figure 7.11, the real part of the wave (in other words, the cosine), is sketched as the red curve; also the magnitude of the wave (which is unity) is shown as the top black line, and minus the magnitude is drawn as the bottom black line.

Figure 7.11: The real part (red) and envelope (black) of an example wave.
\begin{figure}
\centering
\epsffile{wave.eps}
\end{figure}

The black lines enclose the real part of the wave, and will be called the envelope. Since their vertical separation is twice the magnitude of the wave function, the vertical separation between the black lines at a point is a measure for the probability of finding the particle near that point.

The constant separation between the black lines shows that there is absolutely no localization of the particle to any particular region. The particle is equally likely to be found at every point in the infinite range. This also graphically demonstrates the normalization problem of the momentum eigenfunctions discussed in the previous section: the total probability of finding the particle just keeps getting bigger and bigger, the larger the range you look in. So there is no way that the total probability of finding the particle can be limited to one as it should be.

The reason for the complete lack of localization is the fact that the component wave solutions have an exact momentum $p$. With zero uncertainty in momentum, Heisenberg's uncertainty relationship says that there must be infinite uncertainty in position. There is.

There is another funny thing about the component waves: when plotted for different times, it is seen that the real part of the wave moves towards the right with a speed $p$$\raisebox{.5pt}{$/$}$$2m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12v$, as illustrated in figure 7.12.

Figure 7.12: The wave moves with the phase speed.
 
wave

Move your mouse over the figure to see the animation. Javascript must be enabled on your browser. Give it a few seconds for the animation to load, especially on a phone line.

This is unexpected, because classically the particle moves with speed $v$, not $\frac12v$. The problem is that the speed with which the wave moves, called the “phase speed,” is not meaningful physically. In fact, without anything like a location for the particle, there is no way to define a physical velocity for a component wave.


Key Points
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Component waves provide no localization of the particle at all.

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Their real part is a moving cosine. Similarly their imaginary part is a moving sine.

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The speed of motion of the cosine or sine is half the speed of a classical particle with that momentum.

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This speed is called the phase speed and is not relevant physically.


7.10.3 Wave packets

As Heisenberg’s principle indicates, in order to get some localization of the position of a particle, some uncertainty must be allowed in momentum. That means that you must take the initial momentum space wave function $\Phi_0$ in (7.58) to be nonzero over at least some small interval of different momentum values $p$. Such a combination of component waves is called a wave packet.

The wave function for a typical wave packet is sketched in figure 7.13. The red line is again the real part of the wave function, and the black lines are the envelope enclosing the wave; they equal plus and minus the magnitude of the wave function.

Figure 7.13: The real part (red) and magnitude or envelope (black) of a wave packet. (Schematic).
\begin{figure}
\centering
\epsffile{packet.eps}
\end{figure}

The vertical separation between the black lines is again a measure of the probability of finding the particle near that location. It is seen that the possible locations of the particle are now restricted to a finite region, the region in which the vertical distance between the black lines is nonzero.

If the envelope changes location with time, and it does, then so does the region where the particle can be found. This then finally is the correct picture of motion in quantum mechanics: the region in which the particle can be found propagates through space.

The limiting case of the motion of a macroscopic Newtonian point mass can now be better understood. As noted in section 7.2.1, for such a particle the uncertainty in position is negligible. The wave packet in which the particle can be found, as sketched in figure 7.13, is so small that it can be considered to be a point. To that approximation the particle then has a point position, which is the normal classical description.

The classical description also requires that the particle moves with velocity $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p$$\raisebox{.5pt}{$/$}$$m$, which is twice the speed $p$$\raisebox{.5pt}{$/$}$$2m$ of the wave. So the envelope should move twice as fast as the wave. This is indicated in figure 7.14 by the length of the bars, which show the motion of a point on the envelope and of a point on the wave during a small time interval.

Figure 7.14: The velocities of wave and envelope are not equal.
 
wave packet

Move your mouse over the figure to see the animation. Javascript must be enabled on your browser. Give it a few seconds for the animation to load, especially on a phone line.

That the envelope does indeed move at speed $p$$\raisebox{.5pt}{$/$}$$m$ can be seen if you define the representative position of the envelope to be the expectation value of position. That position must be somewhere in the middle of the wave packet. The expectation value of position moves according to Ehrenfest's theorem of section 7.2.1 with a speed $\big\langle p\big\rangle $$\raisebox{.5pt}{$/$}$$m$, where $\big\langle p\big\rangle $ is the expectation value of momentum, which must be constant since there is no force. Since the uncertainty in momentum is small for a macroscopic particle, the expectation value of momentum $\big\langle p\big\rangle $ can be taken to be the momentum $p$.


Key Points
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A wave packet is a combination of waves with about the same momentum.

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Combining waves into wave packets can provide localization of particles.

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The envelope of the wave packet shows the region where the particle is likely to be found.

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This region propagates with the classical particle velocity.


7.10.4 Group velocity

As the previous subsection explained, particle motion in classical mechanics is equivalent to the motion of wave packets in quantum mechanics. Motion of a wave packet implies that the region in which the particle can be found changes position.

Motion of wave packets is not just important for understanding where particles in free space end up. It is also critical for the quantum mechanics of for example solids, in which electrons, photons, and phonons (quanta of crystal vibrations) move around in an environment that is cluttered with other particles. And it is also of great importance in classical applications, such as acoustics in solids and fluids, water waves, stability theory of flows, electromagnetodynamics, etcetera. This section explains how wave packets move in such more general systems. Only the one-di­men­sion­al case will be considered, but the generalization to three dimensions is straightforward.

The systems of interest have component wave solutions of the general form:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{component wave:}\quad \psi_{\rm{w}} = e^{{\rm i}(kx - \omega t)}
$} %
\end{displaymath} (7.60)

The constant $k$ is called the “wave number,” and $\omega$ the “angular frequency.” The wave number and frequency must be real for the analysis in this section to apply. That means that the magnitude of the component waves must not change with space nor time. Such systems are called nondissipative: although a combination of waves may get dispersed over space, its square magnitude integral will be conserved. (This is true on account of Parseval’s relation, {A.26}.)

For a particle in free space according to the previous subsection:

\begin{displaymath}
k=\frac{p}{\hbar} \qquad \omega = \frac{p^2}{2m\hbar}
\end{displaymath}

Therefore, for a particle in free space the wave number $k$ is just a rescaled linear momentum, and the frequency $\omega$ is just a rescaled kinetic energy. This will be different for a particle in a nontrivial surroundings.

Regardless of what kind of system it is, the relationship between the frequency and the wave number is called the

\begin{displaymath}
\fbox{$\displaystyle
\mbox{dispersion relation:}\quad \omega = \omega(k)
$} %
\end{displaymath} (7.61)

It really defines the physics of the wave propagation.

Since the waves are of the form $e^{{{\rm i}}k(x-\frac{\omega}{k}t)}$, the wave is constant if $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\omega/k)t$ plus any constant. Such points move with the

\begin{displaymath}
\fbox{$\displaystyle
\mbox{phase velocity:}\quad v_{\rm{p}} \equiv \frac{\omega}{k}
$} %
\end{displaymath} (7.62)

In free space, the phase velocity is half the classical velocity.

However, as noted in the previous subsection, wave packets do not normally move with the phase velocity. The velocity that they do move with is called the group velocity. For a particle in free space, you can infer that the group velocity is the same as the classical velocity from Ehrenfest's theorem, but that does not work for more general systems. The approach will therefore be to simply define the group velocity as

\begin{displaymath}
\fbox{$\displaystyle
\mbox{group velocity:}\quad v_{\rm{g}} \equiv \frac{{\rm d}\omega}{{\rm d}k}
$} %
\end{displaymath} (7.63)

and then to explore how the so-defined group velocity relates to the motion of wave packets.

Wave packets are combinations of component waves, and the most general combination of waves takes the form

\begin{displaymath}
\fbox{$\displaystyle
\Psi(x,t) = \frac{1}{\sqrt{2\pi}}
...
...overline{\Phi}_0(k) e^{{\rm i}(kx-\omega t)}{ \rm d}k
$} %
\end{displaymath} (7.64)

Here $\overline{\Phi}_0$ is the complex amplitude of the waves. The combination $\overline{\Phi}_0e^{-{\rm i}{\omega}t}$ is called the “Fourier transform” of $\Psi$. The factor $\sqrt{2\pi}$ is just a normalization factor that might be chosen differently in another book. Wave packets correspond to combinations in which the complex amplitude $\overline{\Phi}_0(k)$ is only nonzero in a small range of wave numbers $k$. More general combinations of waves may of course always be split up into such wave packets.

To describe the motion of wave packets is not quite as straightforward as it may seem: the envelope of a wave packet extends over a finite region, and different points on it actually move at somewhat different speeds. So what do you take as the point that defines the motion if you want to be precise? There is a trick here: consider very long times. For large times, the propagation distance is so large that it dwarfs the ambiguity about what point to take as the position of the envelope.

Finding the wave function $\Psi$ for large time is a messy exercise banned to derivation {D.44}. But the conclusions are fairly straightforward. Assume that the range of waves in the packet is restricted to some small interval $k_1$ $\raisebox{.3pt}{$<$}$ $k$ $\raisebox{.3pt}{$<$}$ $k_2$. In particular, assume that the variation in group velocity is relatively small and monotonous. In that case, for large times the wave function will be negligibly small except in the region

\begin{displaymath}
v_{{\rm {g}}1} t < x < v_{{\rm {g}}2} t
\end{displaymath}

(In case $v_{{\rm {g}}1}$ $\raisebox{.3pt}{$>$}$ $v_{{\rm {g}}2}$, invert these inequalities.) Since the variation in group velocity is small for the packet, it therefore definitely does move with the group velocity.

It is not just possible to say where the wave function is nonzero at large times. It is also possible to write a complete approximate wave function for large times:

\begin{displaymath}
\Psi(x,t)\sim \frac{e^{\mp{\rm i}\pi/4}}{\sqrt{\vert v_{{\...
...{{\rm i}(k_0x-\omega_0t)} \qquad v_{{\rm {g}}0} = \frac{x}{t}
\end{displaymath}

Here $k_0$ is the wave number at which the group speed is exactly equal to $x$$\raisebox{.5pt}{$/$}$$t$, $\omega_0$ is the corresponding frequency, $v_{{\rm {g}}0}'$ is the derivative of the group speed at that point, and $\mp$ stands for the sign of $-v_{{\rm {g}}0}'$.

While this precise expression may not be that important, it is interesting to note that $\Psi$ decreases in magnitude proportional to 1/$\sqrt{t}$. That can be understood from conservation of the probability to find the particle. The wave packet spreads out proportional to time because of the small but nonzero variation in group velocity. Therefore $\Psi$ must be proportional to 1$\raisebox{.5pt}{$/$}$$\sqrt{t}$ if its square integral is to remain unchanged.

One other interesting feature may be deduced from the above expression for $\Psi$. If you examine the wave function on the scale of a few oscillations, it looks as if it was a single component wave of wave number $k_0$ and frequency $\omega_0$. Only if you look on a bigger scale do you see that it really is a wave packet. To understand why, just look at the differential

\begin{displaymath}
{\rm d}(k_0x-\omega_0t)= k_0 {\rm d}x - \omega_0{\rm d}t + x{\rm d}k_0 - t{\rm d}\omega_0
\end{displaymath}

and observe that the final two terms cancel because ${\rm d}\omega_0$$\raisebox{.5pt}{$/$}$${\rm d}{k}_0$ is the group velocity, which equals $x$$\raisebox{.5pt}{$/$}$$t$. Therefore changes in $k_0$ and $\omega_0$ do not show up on a small scale.

For the particle in free space, the result for the large time wave function can be written out further to give

\begin{displaymath}
\Psi(x,t)\sim e^{-{\rm i}\pi/4} \sqrt{\frac{m}{t}}
\Phi_0\left(\frac{mx}{t}\right)
e^{{\rm i}mx^2/2\hbar t}
\end{displaymath}

Since the group speed $p$$\raisebox{.5pt}{$/$}$$m$ in this case is monotonously increasing, the wave packets have negligible overlap, and this is in fact the large time solution for any combination of waves, not just narrow wave packets.

In a typical true quantum mechanics case, $\Phi_0$ will extend over a range of wave numbers that is not small, and may include both positive and negative values of the momentum $p$. So, there is no longer a meaningful velocity for the wave function: the wave function spreads out in all directions at velocities ranging from negative to positive. For example, if the momentum space wave function $\Phi_0$ consists of two narrow nonzero regions, one at a positive value of $p$ and one at a negative value, then the wave function in normal space splits into two separate wave packets. One packet moves with constant speed towards the left, the other with constant speed towards the right. The same particle is now going in two completely different directions at the same time. That would be unheard of in classical Newtonian mechanics.


Key Points
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Component waves have the generic form $e^{{\rm i}(kx-{\omega}t)}$.

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The constant $k$ is the wave number.

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The constant $\omega$ is the angular frequency.

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The relation between $\omega$ and $k$ is called the dispersion relation.

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The phase velocity is $\omega$$\raisebox{.5pt}{$/$}$$k$. It describes how fast the wave moves.

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The group velocity is ${\rm d}\omega$$\raisebox{.5pt}{$/$}$${\rm d}{k}$. It describes how fast wave packets move.

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Relatively simple expressions exist for the wave function of wave packets at large times.


7.10.5 Electron motion through crystals

One important application of group velocity is the motion of conduction electrons through crystalline solids. This subsection discusses it.

Conduction electrons in solids must move around the atoms that make up the solid. You cannot just forget about these atoms in discussing the motion of the conduction electrons. Even semi-classically speaking, the electrons in a solid move in a roller-coaster ride around the atoms. Any external force on the electrons is on top of the large forces that the crystal already exerts. So it is simply wrong to say that the external force gives mass times acceleration of the electrons. Only the total force would do that.

Typically, on a microscopic scale the solid is crystalline; in other words, the atoms are arranged in a periodic pattern. That means that the forces on the electrons have a periodic nature. As usual, any direct interactions between particles will be ignored as too complex to analyze. Therefore, it will be assumed that the potential energy seen by an electron is a given periodic function of position.

It will also again be assumed that the motion is one-di­men­sion­al. In that case the energy eigenfunctions are determined from a one-di­men­sion­al Hamiltonian eigenvalue problem of the form

\begin{displaymath}
-\frac{\hbar^2}{2 m_{\rm e}} \frac{\partial^2\psi}{\partial x^2}
+ V(x) \psi = E \psi %
\end{displaymath} (7.65)

Here $V(x)$ is a periodic potential energy, with some given atomic-scale period $d$.

Three-di­men­sion­al energy eigenfunctions may be found as products of one-di­men­sion­al ones; compare chapter 3.5.8. Unfortunately however, that only works here if the three-di­men­sion­al potential is some sum of one-di­men­sion­al ones, as in

\begin{displaymath}
V(x,y,z) = V_x(x) + V_y(y) + V_z(z)
\end{displaymath}

That is really quite limiting. The general conclusions that will be reached in this subsection continue to apply for any periodic potential, not just a sum of one-di­men­sion­al ones.

The energy eigenfunction solutions to (7.65) take the form of Bloch waves:

\begin{displaymath}
\pp{k}/x/// = \pp{{\rm p},k}/x/// e^{{\rm i}k x} %
\end{displaymath} (7.66)

where $\pp{{\rm {p}},k}////$ is a periodic function of period $d$ like the potential.

The reason that the energy eigenfunctions take the form of Bloch waves is not that difficult to understand. It is a consequence of the fact that commuting operators have common eigenfunctions, chapter 4.5.1. Consider the “translation operator” ${\cal T}_d$ that shifts wave functions over one atomic period $d$. Since the potential is exactly the same after a wave function is shifted over an atomic period, the Hamiltonian commutes with the translation operator. It makes no difference whether you apply the Hamiltonian before or after you shift a wave function over an atomic period. Therefore, the energy eigenfunctions can be taken to be also eigenfunctions of the translation operator. The translation eigenvalue must have magnitude one, since the magnitude of a wave function does not change when you merely shift it. Therefore the eigenvalue can always be written as $e^{{{\rm i}}kd}$ for some real value $k$. And that means that if you write the eigenfunction in the Bloch form (7.66), then the exponential will produce the eigenvalue during a shift. So the part $\pp{{\rm {p}},k}////$ must be the same after the shift. Which means that it is periodic of period $d$. (Note that you can always write any wave function in Bloch form; the nontrivial part is that $\pp{{\rm {p}},k}////$ is periodic for actual Bloch waves.)

If the crystal is infinite in size, the wave number $k$ can take any value. (For a crystal in a finite-size periodic box as studied in chapter 6.22, the values of $k$ are discrete. However, this subsection will assume an infinite crystal.)

To understand what the Bloch form means for the electron motion, first consider the case that the periodic factor $\pp{{\rm {p}},k}////$ is just a trivial constant. In that case the Bloch waves are eigenfunctions of linear momentum. The linear momentum $p$ is then ${\hbar}k$. That case applies if the crystal potential is just a trivial constant. In particular, it is true if the electron is in free space.

Even if there is a nontrivial crystal potential, the so-called “crystal momentum” is still defined as:

\begin{displaymath}
\fbox{$\displaystyle
p_{\rm cm} = \hbar k
$} %
\end{displaymath} (7.67)

(In three dimensions, substitute the vectors ${\skew0\vec p}$ and ${\vec k}$). But crystal momentum is not normal momentum. In particular, for an electron in a crystal you can no longer get the propagation velocity by dividing the crystal momentum by the mass.

Instead you can get the propagation velocity by differentiating the energy with respect to the crystal momentum, {D.45}:

\begin{displaymath}
\fbox{$\displaystyle
v = \frac{{\rm d}{\vphantom' E}^{\rm p}}{{\rm d}p_{\rm cm}}
\qquad p_{\rm cm} = \hbar k
$} %
\end{displaymath} (7.68)

(In three dimensions, replace the $p$-​derivative by 1$\raisebox{.5pt}{$/$}$$\hbar$ times the gradient with respect to ${\vec k}$.) In free space, ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\omega$ and $p_{\rm {cm}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k$, so the above expression for the electron velocity is just the expression for the group velocity.

One conclusion that can be drawn is that electrons in an ideal crystal keep moving with the same speed for all times like they do in free space. They do not get scattered at all. The reason is that energy eigenfunctions are stationary. Each eigenfunction corresponds to a single value of $k$ and so to a corresponding single value of the propagation speed $v$ above. An electron wave packet will involve a small range of energy eigenfunctions, and a corresponding small range of velocities. But since the range of energy eigenfunctions does not change with time, neither does the range of velocities. Scattering, which implies a change in velocity, does not occur.

This perfectly organized motion of electrons through crystals is quite surprising. If you make up a classical picture of an electron moving through a crystal, you would expect that the electron would pretty much bounce off every atom it encountered. It would then perform a drunkard’s walk from atom to atom. That would really slow down electrical conduction. But it does not happen. And indeed, experimentally electrons in metals may move past many thousands of atoms without getting scattered. In very pure copper at very low cryogenic temperatures electrons may even move past many millions of atoms before getting scattered.

Note that a total lack of scattering only applies to truly ideal crystals. Electrons can still get scattered by impurities or other crystal defects. More importantly, at normal temperatures the atoms in the crystal are not exactly in their right positions due to thermal motion. That too can scatter electrons. In quantum terms, the electrons then collide with the phonons of the crystal vibrations. The details are too complex to be treated here, but it explains why metals conduct much better still at cryogenic temperatures than at room temperature.

The next question is how does the propagation velocity of the electron change if an external force $F_{\rm {ext}}$ is applied? It turns out that Newton’s second law, in terms of momentum, still works if you substitute the crystal momentum ${\hbar}k$ for the normal momentum, {D.45}:

\begin{displaymath}
\fbox{$\displaystyle
\frac{{\rm d}p_{\rm cm}}{{\rm d}t} = F_{\rm ext}
\qquad p_{\rm cm} = \hbar k
$} %
\end{displaymath} (7.69)

However, since the velocity is not just the crystal momentum divided by the mass, you cannot convert the left hand side to the usual mass times acceleration. The acceleration is instead, using the chain rule of differentiation,

\begin{displaymath}
\frac{{\rm d}v}{{\rm d}t}
= \frac{{\rm d}^2{\vphantom' E...
... d}^2{\vphantom' E}^{\rm p}}{{\rm d}p_{\rm cm}^2} F_{\rm ext}
\end{displaymath}

For mass times acceleration to be the force, the factor multiplying the force in the final expression would have to be the reciprocal of the electron mass. It clearly is not; in general it is not even a constant.

But physicists still like to think of the effect of force as mass times acceleration of the electrons. So they cheat. They ignore the true mass of the electron. Instead they simply define a new “effective mass” for the electron so that the external force equals that effective mass times the acceleration:

\begin{displaymath}
\fbox{$\displaystyle
m_{\rm eff} \equiv 1 \Bigg/ \frac{{...
...p}}{{\rm d}p_{\rm cm}^2}
\qquad p_{\rm cm} = \hbar k
$} %
\end{displaymath} (7.70)

Unfortunately, the effective mass is often a completely different number than the true mass of the electron. Indeed, it is quite possible for this mass to become negative for some range of wave numbers. Physically that means that if you put a force on the electron that pushes it one way, it will accelerate in the opposite direction! That can really happen. It is a consequence of the wave nature of quantum mechanics. Waves in crystals can be reflected just like electromagnetic waves can, and a force on the electron may move it towards stronger reflection.

For electrons near the bottom of the conduction band, the effective mass idea may be a bit more intuitive. At the bottom of the conduction band, the energy has a minimum. From calculus, if the energy ${\vphantom' E}^{\rm p}$ has a minimum at some wave number vector, then in a suitably oriented axis system it can be written as the Taylor series

\begin{displaymath}
{\vphantom' E}^{\rm p}= {\vphantom' E}^{\rm p}_{\rm min}
...
...rtial^2{\vphantom' E}^{\rm p}}{\partial k_z^2} k_z^2 + \ldots
\end{displaymath}

Here the wave number values are measured from the position of the minimum. This can be rewritten in terms of the crystal momenta and effective masses in each direction as
\begin{displaymath}
{\vphantom' E}^{\rm p}= {\vphantom' E}^{\rm p}_{\rm min}
...
...}{2}} \frac{1}{m_{{\rm eff},z}} p_{{\rm cm},z}^2
+ \ldots %
\end{displaymath} (7.71)

In this case the effective masses are indeed positive, since second derivatives must be positive near a minimum. These electrons act much like classical particles. They move in the right direction if you put a force on them. Unfortunately, the effective masses are not necessarily similar to the true electron mass, or even the same in each direction.

For the effective mass of the holes at the top of a valence band things get much messier still. For typical semiconductors, the energy no longer behaves as an analytic function, even though the energy in a specific direction continues to vary quadratically with the magnitude of the wave number. So the Taylor series is no longer valid. You then end up with such animals as heavy holes, light holes,” and “split-off holes. Such effects will be ignored in this book.


Key Points
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The energy eigenfunctions for periodic potentials take the form of Bloch waves, involving a wave number $k$.

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The crystal momentum is defined as ${\hbar}k$.

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The first derivative of the electron energy with respect to the crystal momentum gives the propagation velocity.

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The second derivative of the electron energy with respect to the crystal momentum gives the reciprocal of the effective mass of the electron.