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I am a bit confused over Kittel's derivation of Bloch's theorem: enter image description here enter image description here enter image description here

My question lies in eq. 29. The k+G is only a subset of possible k value that satisfy the periodic boundary condition. Why is it proper to determine that there does not exist a $k'$ such that $k'\neq k+G$ and the corresponding coefficient for the series $C_{k'}$ would be non-zero?

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If I'm interpreting correctly, this is the troublesome sentence:

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.

Now, you ask why. The answer is that, well, Charles Kittel is lying here. They're not looking for just any solutions, they're looking for a helpful set of solutions. Let's take a step back and derive the central equation $$ (\lambda_k - \epsilon)C(k) + \sum_{G}U_GC(k-G) = 0. $$

Starting at the previous equation: $$ \sum_k \lambda_k C(k) e^{ikx} + \sum_G\sum_k U_GC(k)e^{i(k+G)x} = \epsilon\sum_kC(k)e^{ikx} $$ The second term on the right side can be rearranged: $$ \sum_G\sum_k U_GC(k)e^{i(k+G)x} = \sum_k\left(\sum_GU_gC(k-G)\right)e^{ikx} $$ and then we get to rearrange the previous equation to have a sum over $k$ on one side, zero on the other: $$ \sum_k \left((\lambda_k - \epsilon) C(k) + \sum_GU_gC(k-G)\right)e^{ikx} = 0 $$ Only way for this to be true is for the coefficients to each be zero: $$ (\lambda_k - \epsilon) C(k) + \sum_GU_gC(k-G) = 0 $$ Once we have a $k_0$ value with $C(k_0)\neq 0$, we have all of the $C(k_0-G)$ coefficients determined by the central equation. Moreover, saying that $C(k\neq k_0-G) = 0$ will yield a valid wavefunction with energy $\epsilon$.

But what if $C(k\neq k_0-G)\neq 0$? Well, the honest answer is that we end up with the same result: a wavefunction with energy $\epsilon$. There's nothing actually stopping us from having a wavefunction which includes all wavevectors $k$ with the right energy $\epsilon$.

But the helpful part of using $C(k\neq k_0-G) = 0$ is that it organizes the solutions into a set of $\psi_k$ with wonderful properties:

  1. The solutions $\psi_k$ form a basis.
  2. The solutions $\psi_k$ are mutually orthogonal.

Property 1 means that if $\psi$ satisfies the equation $\mathcal H\psi = \epsilon\psi$ then $\psi$ can be expressed as a combination (or superposition) of $\psi_k$s: $$ \psi(x) = \sum_k A_k\psi_k(x). $$ And property 2 means that if $k'\neq k+G$ then their inner product is zero: $$ \int_{-\infty}^\infty \psi_k(x)^* \psi_{k'}(x)\,\mathrm dx = 0. $$

The second property has some implications in quantum mechanics. In principle, we could measure the wavevector of any wavefunction $\psi$. If $\psi$ is one of our special solutions, $\psi_k$, then measuring the wavevector is guaranteed to yield $k$. The same cannot be said for an arbitrary $\psi$. For example, if we have wavevectors $k_1$ through $k_4$, then we can construct a wavefunction $\psi$: $$ \psi(x) = \frac{1}{2}\left(\psi_{k_1}(x) + \psi_{k_2}(x) + \psi_{k_3}(x) + \psi_{k_4}(x)\right) $$ If we try to measure the wavevector of $\psi$, there is a $1/4$ chance of measuring each $k_1$ through $k_4$.

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    $\begingroup$ "There's nothing actually stopping us from having all values of $k$ represented." Actually, just because you need a constant eigen-energy $\epsilon$, it is clear that there will only be a handful of $k$ points inside the 1BZ that would have the exact same eigen-energy. You can thus only pick from those points. $\endgroup$ Commented Jun 14, 2023 at 8:26
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    $\begingroup$ On the other hand, while Kittel is justified in looking for wavefunctions of this form, the statement I quoted is not true. The earlier proof was in the absence of degeneracy. In this proof, the statement that "if $k$ is in $\psi$ then all other wavevectors in the Fourier expansion will be of the form $k+G$" is not correct. You can have a wavevector with a well-defined energy which has $k$ and $k'\neq k+G$ in its Fourier expansion (like the $\psi$ I constructed at the end). $\endgroup$ Commented Jun 14, 2023 at 14:22
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    $\begingroup$ @蕭力諶 yes, that's correct $\endgroup$ Commented Jun 14, 2023 at 18:47
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    $\begingroup$ @蕭力諶 No, it will not be, and that is the point. If you want to be a Bloch state, in particular, if you want to be a simultaneous eigenfunction of Hamiltonian and Translation operators, then you need to only have one single $k$ in the 1BZ and no more. The concept that is appearing here is the same as that in single atoms: The different $m_j$ in H atom for the same $n,\ell,s,j$ have the same energy, but if you superpose them, they will no longer be $J_z$ eigenstates. Nobody is saying that every solution of TISE is in $J_z$ eigenstates, but rather that they span the whole Hilbert space. $\endgroup$ Commented Jun 15, 2023 at 1:37
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    $\begingroup$ @蕭力諶 Yes, and that is always the case, not just in the Bloch–translation case. We are almost always interested in some maximally commuting set of operators that include the Hamiltonian, and if you swap some of them for others, the old complete basis will span the same space as the new basis but will appear as superpositions that will violate the eigenvalue equation for the operators that are swapped out. $\endgroup$ Commented Jun 15, 2023 at 8:20

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