In Kittel's Intro to solid state physics, when solving the schrodinger equation for a periodic potential, we begin by writing the potential and the wave function as fourier series of the form $$\psi = \sum_k C(k)\,e^{ikx} \qquad U(x) = \sum_G U_G\,e^{iGx}$$ where $k=n2\pi/L$ and $G$ is a reciprocal lattice vector. We then substitute these into the schrodinger equation and simplify until we get what Kittel refers to as the central equation which is $$\left(\frac{\hbar^2 k^2}{2m}-\epsilon\right)C(k)+ \sum_G U_G\,C(k-G)=0$$ Kittel then goes on to say that the above equation "represents a set of simultaneous linear equations that connect the coefficients for all reciprocal lattice vectors G" and that "there are as many equations as there are coefficients C". This I do not understand. In ashcroft and mermin, it is stated that central equation consists of $N$ equations where N is the number of vectors within the 1st Brillouin zone. But the sum $\sum_GU_G\,C(k-G)$ runs over all reciprocal vectors $G$ within reciprocal space and hence their should be an infinite amount of coefficents $C(k-G)$. So clearly their are an infinite amount of coefficients but not an infinite amount of equations. Is Kittel wrong when he states that "there are as many equations as there are coefficients". After this Kittel says that "these equations are consistent if the determinant of the coefficients vanishes" and then he writes that "a block of the determinant of the coeeficients is given by:

How does he get this matrix from the central equation? Any help on this would be most appreciated as its been driving me crazy.


1 Answer 1


I do not have Ashcroft and Mermin for reference, but as far as I can tell the central equation generally describes an infinite set of equations and coefficients. This also makes sense for the following reason: If we were dealing with phonons, we would only need to consider wave vectors $\mathbf{K}$ inside the first Brillouin Zone (BZ) to model the motion, but this is due to elastic waves being described in terms of the lattice constant. An elastic wave "exists" only at the ions as there is nothing in between them which can move, and as such, there is a minimum wavelength and a maximum wave vector required to describe the motion. This maximum wave vector lies at the edge of the first BZ. The wave function of a free electron (its probability distribution) exists also in between atoms, and we generally need to consider all possible wave vectors $\mathbf{k}$, even those outside the first BZ.

As for the determinant in question; it is derived under the simplifying assumption that the potential only has one (real) Fourier component; $U_{g} = U_{-g} \equiv U$ (the constant component $U_0 = 0$). Thus the sum over $G$ in the central equation

$$ \bigg( \frac{\hbar^2k^2}{2m} - \epsilon \bigg)C(k) + \sum_G U_GC(k-G) = 0 $$

contains two terms, $UC(k \pm g)$. Set $\lambda_k \equiv \hbar^2k^2/2m$ and write out the set of equations from $k-2g$ to $k+2g$;

\begin{equation} \begin{aligned} && \vdots \\ & (\lambda_{k-2g} - \epsilon)C(k-2g) && + \; U\big[C(k-g) + C(k-3g)\big] & = 0 \\ & (\lambda_{k-g} - \epsilon)C(k-g) && + \; U\big[C(k) + C(k-2g)\big] & = 0 \\ & (\lambda_{k} - \epsilon)C(k) && + \; U\big[C(k+g) + C(k-g)\big] & = 0 \\ & (\lambda_{k+g} - \epsilon)C(k+g) && + \; U\big[C(k) + C(k+2g)\big] & = 0 \\ & (\lambda_{k+2g} - \epsilon)C(k+2g) && + \; U\big[C(k+g) + C(k+3g)\big] & = 0 \\ && \vdots \end{aligned}. \end{equation}

These are part of an infinite system of equations which can equally well be written as a single matrix equation;

\begin{equation} \begin{bmatrix} \ddots & \vdots & \vdots & \vdots & \vdots & \vdots & ⋰ \\ \dots & \lambda_{k-2g} - \epsilon & U & 0 & 0 & 0 & \dots \\ \dots & U & \lambda_{k-g} - \epsilon & U & 0 & 0 & \dots \\ \dots & 0 & U & \lambda_{k} - \epsilon & U & 0 & \dots \\ \dots & 0 & 0 & U & \lambda_{k+g} - \epsilon & U & \dots \\ \dots & 0 & 0 & 0 & U & \lambda_{k+2g} - \epsilon & \dots \\ ⋰ & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix} \begin{bmatrix} \vdots \\ C(k-2g) \\ C(k-g) \\ C(k) \\ C(k+g) \\ C(k+2g) \\ \vdots \end{bmatrix} = \begin{bmatrix} \vdots \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \vdots \end{bmatrix}, \end{equation}

and for non-trivial solutions the determinant of the matrix on the left must then be zero. What is shown by Kittel is just a small central block of this determinant.

  • $\begingroup$ (1/n) Thanks! This is a great answer! I'm still confused about one thing though. We get that $\bigg( \frac{\hbar^2k^2}{2m} - \epsilon \bigg)C(k) + \sum_G U_GC(k-G) = 0$. After this result, you state that we should write out this out the set of equations from $k-2g$ to $k+2g$ with $g$ being a reciprocal lattice vector. It is my understanding that the central equation should hold for any $k$ within the first BZ and so why do we not write out the set of equations for any of the other possible $C(k)$ values. For example, say $k=1*2\pi/L$. You then explicitly write out equations for ... $\endgroup$ Jun 3, 2021 at 12:23
  • $\begingroup$ $k=1*2\pi/L-2g$ , $k=1*2\pi/L-g$, $k=1*2\pi/L$ , $k=1*2\pi/L+g$ and $k=1*2\pi/L+2g$ But what about the equations for when $k=n*2\pi/L$ where $n$ is now say $n=3$. Then we would need another list of equations just like yours but for each $k$ value and each list will produce a matrix similar to the matrix you derived. Am I correct? In essence, I am asking if each possible value for $k$ (which we can restrict to the 1st BZ) has its own matrix equation like the one you've derived? $\endgroup$ Jun 3, 2021 at 12:24
  • 1
    $\begingroup$ In essence, yes. Remember that $C(k)$ are the coefficients in the Fourier expansion of the wave function for a single electron. If a certain $k$ is present is this expansion, then all the other $k$'s in the expansion have the form $k+G$ where $G$ is any reciprocal lattice vector (I believe Kittel proves this somewhere). The starting $k$ can be any $2\pi n/L$ allowed by the periodic boundary conditions, but is usually taken to be within the first BZ. So in this case each starting $k$ represents a specific set of equations which determine the coefficients for the given electron. $\endgroup$
    – Engelmark
    Jun 3, 2021 at 18:27
  • $\begingroup$ Okay perfect. Thanks so much! So just to be totally clear, let us say that there are $N$ wave vectors within the 1st BZ. Then for each wave vector $k$ within the 1st BZ, we get a matrix of the form shown in your answer (so N matrices in total), each of which we can in theory solve to find the the set of coefficients for the corresponding wave function $\psi_k$? $\endgroup$ Jun 3, 2021 at 20:00
  • $\begingroup$ Yes, that's right. $\endgroup$
    – Engelmark
    Jun 3, 2021 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.