Subsections


D.45 Motion through crystals

This note derives the semi-classical motion of noninteracting electrons in crystals. The derivations will be one-di­men­sion­al, but the generalization to three dimensions is straightforward.


D.45.1 Propagation speed

The first question is the speed with which a more or less localized electron moves. An electron in free space moves with a speed found by dividing its linear momentum by its mass. However, in a solid, the energy eigenfunctions are Bloch waves and these do not have definite momentum.

Fortunately, the analysis for the wave packet of a free particle is virtually unchanged for a particle whose energy eigenfunctions are Bloch waves instead of simple exponentials. In the Fourier integral (7.64), simply add the periodic factor $\pp{{\rm {p}},k}/x///$. Since this factor is periodic, it is bounded, and it plays no part in limit process of infinite time. (You can restrict the times in the limit process to those at which $x$ is always at the same position in the period.)

As a result the group velocity is again ${\rm d}\omega$$\raisebox{.5pt}{$/$}$${\rm d}{k}$. Since the energy is ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\omega$ and the crystal momentum $p_{\rm {cm}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k$, the velocity of a localized electron can be written as

\begin{displaymath}
v = \frac{{\rm d}{\vphantom' E}^{\rm p}}{{\rm d}p_{\rm cm}}
\end{displaymath}

In the absence of external forces, the electron will keep moving with the same velocity for all time. The large time wave function is

\begin{displaymath}
\Psi(x,t)\sim \frac{e^{\mp{\rm i}\pi/4}}{\sqrt{\vert v_{\r...
...e^{{\rm i}(k_0x-\omega_0t)} \qquad v_{\rm {g0}} = \frac{x}{t}
\end{displaymath}

where $k_0$ is the wave number at which the group speed equals $x$$\raisebox{.5pt}{$/$}$$t$. Note that the wave function looks locally just like a single Bloch wave for large time.


D.45.2 Motion under an external force

The acceleration due to an external force on an electrons is not that straightforward. First of all note that you cannot just add a constant external force. A constant force $F_{\rm {ext}}$ would produce an external potential of the form $V_{\rm {ext}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-F_{\rm {ext}}x$ and that becomes infinite at infinite $x$. However, it can be assumed that the force is constant over the nonzero range of the wave packet.

Next there is a trick. Consider the expectation value $\langle{\cal T}_d\rangle$ of the translation operator ${\cal T}_d$ that translates the wave function over one atomic cell size $d$. If the wave packet consisted of just a single Bloch wave with wave number $k_0$, the expectation value of ${\cal T}_d$ would be $e^{{{\rm i}}k_0d}$. A wave packet must however include a small range of $k$ values. Then $\langle{\cal T}_d\rangle$ will be an average of $e^{{{\rm i}}kd}$ values over the $k$ values of the wave packet. Still, if the range of $k$ values is small enough, you can write

\begin{displaymath}
\langle {\cal T}_d \rangle = A e^{{\rm i}k_0 d}
\end{displaymath}

where $k_0$ is a $k$ value somewhere in the middle of the wave packet and $A$ is a real number close to one. So $\langle{\cal T}_d\rangle$ still gives the typical $k$ value in the wave packet.

Moreover, its magnitude $\vert\langle{\cal T}_d\rangle\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $A$ is always less than one and the closer it is to one, the more compact the wave packet. That is because $\langle{\cal T}_d\rangle$ is an average of $e^{{{\rm i}}kd}$ values. These are all located on the unit circle in the complex plane, the plane with $\cos(kd)$ as the horizontal axis and $\sin(kd)$ as the vertical axis. If the wave packet would consist of just a single $k$ value $k_0$, then the average of $e^{{{\rm i}}kd}$ would be exactly $e^{{{\rm i}}k_0d}$, and be on the unit circle. If however the wave numbers spread out a bit around $k_0$, then the average moves inside the unit circle: if you average positions on a circle, the average is always inside the circle. In the extreme case that the $k$ values get uniformly distributed over the entire circle, the average position is at the origin. That would make $\vert\langle{\cal T}_d\rangle\vert$ zero. Conversely, as long as $\vert\langle{\cal T}_d\rangle\vert$ stays very close to one, the wave packet must be very compact in terms of $k$.

The time evolution of $\langle{\cal T}_d\rangle$ can be found using chapter 7.2:

\begin{displaymath}
\frac{{\rm d}\langle {\cal T}_d \rangle}{{\rm d}t} =
\fr...
...ar} \left\langle [H_0+V_{\rm {ext}},{\cal T}_d] \right\rangle
\end{displaymath} (D.28)

where $H_0$ is the Hamiltonian for the electron in the crystal, and $V_{\rm {ext}}$ the additional external potential. Now the commutator of $H_0$ and ${\cal T}_d$ is zero; the crystal Hamiltonian acts exactly the same on the wave function whether it is shifted one cell over or not. The remainder of the commutator gives, when applied on an arbitrary wave function,

\begin{displaymath}[V_{\rm {ext}},{\cal T}_d]\Psi
\equiv V_{\rm {ext}}{\cal T}_d\Psi - {\cal T}_d V_{\rm {ext}}\Psi
\end{displaymath}

Writing this out with the arguments of the functions explicitly shown gives:

\begin{displaymath}
V_{\rm {ext}}(x) \Psi(x+d) - V_{\rm {ext}}(x+d)\Psi(x+d)
= (V_{\rm {ext}}(x)-V_{\rm {ext}}(x+d)) {\cal T}_d\Psi(x)
\end{displaymath}

Now assume that the external force $F_{\rm {ext}}$ is constant over the extent of the wave packet. In that case the difference in the potentials is just $F_{\rm {ext}}d$, and that is a constant that can be taken out of the expectation value of the commutator. So:

\begin{displaymath}
\frac{{\rm d}\langle {\cal T}_d \rangle}{{\rm d}t} =
\frac{{\rm i}}{\hbar} F_{\rm {ext}} d \langle {\cal T}_d \rangle
\end{displaymath} (D.29)

The solution to this equation is:

\begin{displaymath}
\langle{\cal T}_d\rangle
= \langle{\cal T}_d\rangle_0 e^{{\rm i}F_{\rm {ext}} d t/\hbar}
\end{displaymath}

where $\langle{\cal T}_d\rangle_0$ is the value of $\langle{\cal T}_d\rangle$ at the starting time $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

It follows that the magnitude of the $\langle{\cal T}_d\rangle$ does not change with time. In view of the earlier discussion, this means that the wave packet maintains its compactness in terms of $k$. (In physical space the wave packet will gradually spread out, as can be seen from the form of the large-time wave function given earlier.)

It further follows that the average wave number $k_0$ in the wave packet evolves as:

\begin{displaymath}
\frac{{\rm d}h k_0}{{\rm d}t} = F_{\rm {ext}}
\end{displaymath}

Since the packet remains compact, all wave numbers in the wave packet change the same way. This is Newton’s second law in terms of crystal momentum.


D.45.3 Free-electron gas with constant electric field

This book discussed the effect of an applied electric field on free electrons in a periodic box in chapter 6.20. The effect was described as a change of the velocity of the electrons. Since the velocity is proportional to the wave number for free electrons, the velocity change corresponds to a change in the wave number. In this subsection the effect of the electric field will be examined in more detail. The solution will again be taken to be one-di­men­sion­al, but the extension to three dimensions is trivial.

Assume that a constant electric field is applied, so that the electrons experience a constant force $F_{\rm {ext}}$. The time-dependent Schrö­din­ger equation is

\begin{displaymath}
{\rm i}\hbar \frac{\partial \Psi}{\partial t}
= - \frac{...
...\frac{\partial^2 \Psi}{\partial x^2}
- F_{\rm {ext}} x \Psi
\end{displaymath}

Assume the initial condition to be

\begin{displaymath}
\Psi_0 = \sum_{k_0} c_{k_0} e^{{\rm i}k_0 x}
\end{displaymath}

in which a subscript 0 indicates the initial time.

The exact solution to this problem is

\begin{displaymath}
\Psi = \sum_{k_0} c(k_0,t) e^{{\rm i}k(t) x}
\qquad \frac{{\rm d}\hbar k}{{\rm d}t} = F_{\rm {ext}}
\end{displaymath}

where the magnitude of the coefficients $\vert c(k_0,t)\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert c_{k_0}\vert$ is independent of time. This exact solution is in terms of states $e^{{{\rm i}}k(t)x}$ that change in time. The probability of the particle being in those states does not change.

Unfortunately, this solution is only periodic with period equal to the length of the box $\ell$ for times at which $F_{\rm {ext}}t$$\raisebox{.5pt}{$/$}$$\hbar$ happens to be a whole multiple of the wave number spacing. At those times the Fermi sphere of occupied states has shifted the same whole multiple of wave number spacings to the right.

At intermediate times, the solution is not periodic, so it cannot be correctly described using the periodic box modes. The magnitude of the wave function is still periodic. However, the momentum has values inconsistent with the periodic box. The problem is that even though a constant force is periodic, the corresponding potential is not. Since quantum mechanics uses the potential instead of the force, the quantum solution is no longer periodic.

The problem goes away by letting the periodic box size become infinite. But that brings back the ugly normalization problems. For a periodic box, the periodic boundary conditions will need to be relaxed during the application of the electric field. In particular, a factor $e^{{{\rm i}}F_{\rm {ext}}{\ell}t/\hbar}$ difference in wave function and its $x$-​derivative must be allowed between the ends of the box. Since the periodic boundary conditions are artificial anyway for modeling a piece of electrical wire, this may not be a big concern. In any case, for a big-enough periodic box, the times at which the solution returns to its original periodicity become spaced very close together.