An electron in a three-dimensional lattice
Let’s look for a moment at how we could apply the same ideas to see what happens to an electron in three dimensions. The results turn out to be very similar. Suppose we have a rectangular lattice of atoms with lattice spacings of $a$, $b$, $c$ in the three directions. (If you want a cubic lattice, take the three spacings all equal.) Also suppose that the amplitude to leap in the $x$-direction to a neighbor is $(iA_x/\hbar)$, to leap in the $y$-direction is $(iA_y/\hbar)$, and to leap in the $z$-direction is $(iA_z/\hbar)$. Now how should we describe the base states? As in the one-dimensional case, one base state is that the electron is at the atom whose locations are $x$, $y$, $z$, where $(x,y,z)$ is one of the lattice points. Choosing our origin at one atom, these points are all at \begin{equation*} x=n_xa,\quad y=n_yb,\quad \text{and}\quad z=n_zc, \end{equation*}  where $n_x$, $n_y$, $n_z$ are any three integers. Instead of using subscripts to indicate such points, we will now just use $x$, $y$, and $z$, understanding that they take on only their values at the lattice points. Thus the base state is represented by the symbol $\ket{\text{electron at \(x,y,z\)}}$, and the amplitude for an electron in some state $\ket{\psi}$ to be in this base state is $C(x,y,z)=\braket{\text{electron at \(x,y,z\)}}{\psi}$. As before, the amplitudes $C(x,y,z)$ may vary with time. With our assumptions, the Hamiltonian equations should be like this:  \begin{align} i\hbar\dfrac{dC(x,y,z)}{dt}=E_0C(x,y,z) &-A_xC(x+a,y,z)-A_xC(x-a,y,z)\notag\\[-1pt] &-A_yC(x,y+b,z)-A_yC(x,y-b,z)\notag\\[4pt] \label{Eq:III:13:22} &-A_zC(x,y,z+c)-A_zC(x,y,z-c). \end{align} 
 \begin{align} i\hbar\dfrac{dC(x,y,z)}{dt}&=E_0C(x,y,z)\notag\\[-.3ex] -&A_xC(x\!+\!a,y,z)\!-\!A_xC(x\!-\!a,y,z)\notag\\[1ex] -&A_yC(x,y\!+\!b,z)\!-\!A_yC(x,y\!-\!b,z)\notag\\[.9ex] \label{Eq:III:13:22} -&A_zC(x,y,z\!+c\!)\!-\!A_zC(x,y,z\!-\!c). \end{align}  It looks rather long, but you can see where each term comes from. Again we can try to find a stationary state in which all the $C$’s vary with time in the same way. Again the solution is an exponential: \begin{equation} \label{Eq:III:13:23} C(x,y,z)=e^{-iEt/\hbar}e^{i(k_xx+k_yy+k_zz)}. \end{equation}  If you substitute this into (13.22) you see that it works, provided that the energy $E$ is related to $k_x$, $k_y$, and $k_z$ in the following way: \begin{equation} \label{Eq:III:13:24} E=E_0\!-\!2A_x\!\cos k_xa\!-\!2A_y\!\cos k_yb\!-\!2A_z\!\cos k_zc. \end{equation}   The energy now depends on the three wave numbers $k_x$, $k_y$, $k_z$, which, incidentally, are the components of a three-dimensional vector $\FLPk$. In fact, we can write Eq. (13.23) in vector notation as \begin{equation} \label{Eq:III:13:25} C(x,y,z)=e^{-iEt/\hbar}e^{i\FLPk\cdot\FLPr}. \end{equation}  The amplitude varies as a complex plane wave in three dimensions, moving in the direction of $\FLPk$, and with the wave number $k=(k_x^2+k_y^2+k_z^2)^{1/2}$. The energy associated with these stationary states depends on the three components of $\FLPk$ in the complicated way given in Eq. (13.24). The nature of the variation of $E$ with $\FLPk$ depends on relative signs and magnitudes of $A_x$, $A_y$, and $A_z$. If these three numbers are all positive, and if we are interested in small values of $k$, the dependence is relatively simple. Expanding the cosines as we did before to get Eq. (13.16), we can now get that \begin{equation} \label{Eq:III:13:26} E=E_{\text{min}}+A_xa^2k_x^2+A_yb^2k_y^2+A_zc^2k_z^2. \end{equation}  For a simple cubic lattice with lattice spacing $a$ we expect that $A_x$ and $A_y$ and $A_z$ would be equal—say all are just $A$—and we would have just \begin{equation} E=E_{\text{min}}+Aa^2(k_x^2+k_y^2+k_z^2),\notag \end{equation} or \begin{equation} \label{Eq:III:13:27} E=E_{\text{min}}+Aa^2k^2. \end{equation} This is just like Eq. (13.16). Following the arguments used there, we would conclude that an electron packet in three dimensions (made up by superposing many states with nearly equal energies) also moves like a classical particle with some effective mass. In a crystal with a lower symmetry than cubic (or even in a cubic crystal in which the state of the electron at each atom is not symmetrical) the three coefficients $A_x$, $A_y$, and $A_z$ are different. Then the “effective mass” of an electron localized in a small region depends on its direction of motion. It could, for instance, have a different inertia for motion in the $x$-direction than for motion in the $y$-direction. (The details of such a situation are sometimes described in terms of an “effective mass tensor.”)