Unjustified claim in Kittel about Bloch functions [duplicate]

Question

In Kittel's Solid state physics, on page 171. He claimes

If one particular wavevector $k$ is contained in a $\psi$, then all other wavevectors in the Fourier expansion of this $\psi$ will have the form $k+G$, where $G$ is any reciprocal lattice vector.

He then continues with stating that a wavefunction labeled as $\psi_k$ may equally well be labeled as $\psi_{k+G}$ because if $k$ enters in the Fourier expansion, then $k+G$ may enter.

I am really wondering what is the proper argument for justifying this. Although I may have overlooked it when reading the text, I cannot find a good explaination.

His first statement is that $$ \psi = \sum_{k} C(k) e^{ikx},$$ but after discussing that all wavevectors have the form $k+G$, and deriving the central equation, he is able to rewrite this into $$ \psi_k = \sum_{G} C(k-G) e^{i(k-G)x} = \left( \sum_{G} C(k-G) e^{-iGx} \right) e^{ikx}, $$

which proves Bloch's theorem. Justifying the quoted claim therefore seems essential to proving the desired result. If this is not the case, I don't think I see the connection between the first and second expression for $\psi$ and would really like a explaination of the two expressions and their relation.

Answer 1

"...Justifying the quoted claim therefore seems essential to proving the desired result..."

quoted claim : "Not all wavevectors of the set $2\pi m/L$ enter the Fourier expansion of any one Bloch function. If one particular wavevector k is contained in a $\psi$, then all other wavevectors in the Fourier expansion of this $\psi$ will have the form k + G, where G is any reciprocal lattice vector. We prove this result in (29) below"

TL;DR: As you can see, Kittel explicitly says that he is going to prove the quoted claim. The "claim" is a consequence of what follows, not a prerequisite. But I agree, the wording is poor.

Long Version: The aim of this exercise is to find the form of eigenstates of the Hamiltonian, which has the form:

$\psi = \sum_{k} C_k e^{ikx}$ ; (eq (25)) ; $k = \frac{2\pi m}{L}$ , where $m$ $\epsilon$ $\mathbb{Z}$, and $L$ is the length of the crystal.

The sum over $k$ is not constrained to be in a single Brillouin Zone (covers all $m$ $\epsilon$ $\mathbb{Z}$).

Now, by explicitly plugging this into the wave equation for the periodic potential given as $U(x) = \sum_{G}U_Ge^{iGx}$ (sum over reciprocal lattice vectors $G$) we get to what he calls the Central equation (the eigenvalue equation), given as (eq(27)) :

$(\frac{(\hbar k)^2}{2m}-\epsilon)C_k = \sum_G U_G C_{k-G}$

The form of the above eigenvalue equation tells us that for any eigenstate,the presence of $U_G$'s mixes the $C$'s for $k$'s related by lattice translations. That is, $(I)$ for any solution of the above eigenvalue equation corresponding to some $k$, it is not just a single $C_k$ that is nonzero, but (potentially) all $C_{k-G}$.

We can also see that $(II)$ no eigensolution will have $C_k \neq 0$ and $C_{k'} \neq 0$ for $k \neq k' (mod G)$. So, the eigenstates only mix up $C_k$'s separated by reciprocal lattice vectors, and nothing else.

Note: $(I)$ and $(II)$ together constitute what Kittel is claiming. As you can see, these statements are merely a consequence of the form taken by the above eigenvalue equation.

Using these two observations, we are in a position to state Bloch's theorem : any eigenstate of the lattice Hamiltonian must be of form $\psi_k = \sum_G C_{k+G}e^{i(G+k)x}$ . Rearranging, we can write $\psi_k = u_k e^{ikx}$, where $u_k$ is periodic with the periodicity of the lattice.

Now, the part about the restriction to a single BZ. We have motivated that every solution of the above eigenvalue equation will obey property $(I)$.The implication is that we must label all eigensolutions referring to a single BZ (usually taken to be the first , $BZ_1$). There is no logical basis in assigning $k$ to one of the solutions, and $k-G$ to another, etc., as both solutions will have both $C_k \neq 0$ and $C_{k-G} \neq 0$.

However, we must still accommodate for the fact that we are not going to get a single solution for any given $k$, there are going to be multiple ; these different solutions corresponding to the same $k $ $\epsilon $ $BZ_1$ are said to belong to different bands. Therefore, to completely characterize the eigenstates of the Hamiltonian, we need another index, the Band index.

Finally, there is really a much simpler way to see all this:.

Consider the Hamiltonian of the system $\hat{H}$. The fact that we are on a lattice implies that $\hat{H}$ commutes with the lattice translation operator $\hat{T}(na)$, where $a$ is the lattice constant, and $n$ $\epsilon$ $\mathbb{Z}$. This lattice translation operator is generated by Crystal momentum $\hat{P}$ ; $\hat{T}(n) = e^{i\hat{P}na}$. We can see that same lattice translation is generated if we replace $\hat{P}$ by $\hat{P} + G$ for some reciprocal lattice vector $G = \frac{2\pi m}{a}$, that is, crystal momentum (by virtue of the fact that it is only required to perform lattice translations) is defined modulo the reciprocal lattice , ie, defined to have eigenvalues in one $BZ$ only.

We can write $[\hat{T}(n),\hat{H}] = 0$ $ \forall$ $n$ , which implies that $[\hat{P},\hat{H}] = 0$. That is, there exists simultaneous eigenstates of $\hat{P}$ and $H$ , $\psi_k$.

To see that these are indeed in the Bloch form, note that $\hat{P} \psi_k = k\psi_k$ implies that : $\left<r|\hat{T}(n)|\psi_k(r)\right> = e^{inak} \psi_k(r) = \psi_k(r+na) $ $\forall n$

Now, we define $u_k \equiv \psi_k e^{-ikr}$ . We can see that $u_k(r+na) = \psi_k(r+na) e^{-ik(r+na)} = \psi_k(r)e^{ikna} e^{-ik(r+na)} = u_k(r)$, which proves Bloch's theorem.