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

$$v = \frac{{\rm d}{\vphantom' E}^{\rm p}}{{\rm d}p_{\rm cm}}$$

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

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

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

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

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:

$$\frac{{\rm d}\langle {\cal T}_d \rangle}{{\rm d}t} = \frac... ...bar} \left\langle [H_0+V_{\rm {ext}},{\cal T}_d] \right\rangle$$ (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,

$$[V_{\rm {ext}},{\cal T}_d]\Psi \equiv V_{\rm {ext}}{\cal T}_d\Psi - {\cal T}_d V_{\rm {ext}}\Psi$$

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

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

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:

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

The so­lu­tion to this equa­tion is:

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

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:

$$\frac{{\rm d}h k_0}{{\rm d}t} = F_{\rm {ext}}$$

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

$${\rm i}\hbar \frac{\partial \Psi}{\partial t} = - \frac{\h... ...} \frac{\partial^2 \Psi}{\partial x^2} - F_{\rm {ext}} x \Psi$$

As­sume the ini­tial con­di­tion to be

$$\Psi_0 = \sum_{k_0} c_{k_0} e^{{\rm i}k_0 x}$$

in which a sub­script 0 in­di­cates the ini­tial time.

The ex­act so­lu­tion to this prob­lem is

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

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.