In my book (Ashcroft) it says:

"The ground state of N Bloch electrons is similarly constructed, except that the one-electron levels are now labeled by the quantum numbers n and k, $\varepsilon_n(\textbf k)$ (Energy) does not have the simple explicit free electron form, and k must be confined to a single primitive cell of the reciprocal lattice if each level is to be counted only once."

What do they mean by the last part: "and k must be confined to a single primitive cell of the reciprocal lattice if each level is to be counted only once." , I know that I get the same energy eigenvalue solution for multiple different wave vectors (k), I also know that electrons are fermions and thus two particles cannot occupy the same state. But if two states have the same energy and different k aren't those two different states ?

  • $\begingroup$ Are you talking about the eigenstate of independent electron's Hamiltonian? $\endgroup$ – Chuan Chen Aug 31 '17 at 12:13
  • $\begingroup$ Independent electrons with a weak potential. $\endgroup$ – Luka8281 Aug 31 '17 at 12:17
  • $\begingroup$ Just to be sure, are you asking why the quantum number "k" of the eigenstate of Hamiltonian can be confined to the 1BZ? $\endgroup$ – Chuan Chen Aug 31 '17 at 12:20
  • $\begingroup$ Why it must be confined $\endgroup$ – Luka8281 Aug 31 '17 at 12:32
  • $\begingroup$ Remember, the Bloch functions both extend through the crystal, and are the same at $k$ and $k+K$ where $K$ is a basis vector (this second part is a requirement of how they are defined). So, don't double count the $k+K$ part of the solution - it is still the one independent electron solution you are looking at. (At some level, you could consider the level at $k+K$ as the independent electron solution of some other electron in a different unit cell, but that isn't rigorous.) $\endgroup$ – Jon Custer Aug 31 '17 at 13:04

The best way I know to show this is actually from the section called the "second proof of Bloch theorem" in the Ashcroft and Mermin's book.

One can try to use the plane wave basis (I will just use $| k \rangle$ to denote the state with wave vector $k$) to represent the Hamiltonian (so does the eigenstate), assuming there is the Born-von Karman boundary condition, so the wave vector $k$ of the plane waves are quantized: $$\mathbf{k}=\sum_i\frac{m_i}{N_i}\mathbf{b_i} \quad (m_i \in \mathbb{Z}) $$ where the $\mathbf{b_i}$'s are the basis of the reciprocal lattice vector (up to now the $k$ vector does not have to be confined into the 1BZ), and it can be easily shown that the Hamiltonian of the single electron in the periodic potential is block diagonal with this basis. The non-zero matrix elements of $H$ are $\langle k|H|k' \rangle$ where $\mathbf{k}=\mathbf{k'}+\mathbf{K}$ (and $\mathbf{K}$ can be any reciprocal lattice vector). In other words, we can write the Hamiltonian in a form like: $$\begin{pmatrix} H_{k_1,k_1} & H_{k_1,k_1+K}& \cdots& 0& \\ H_{k_1+K,k_1} & H_{k_1+K,k_1+K} & \cdots & \\ \cdots & \cdots & \cdots & & & \\ 0 & & & H_{k_2,k_2} & H_{k_2,k_2+K'} &\cdots & 0 \\ & & & H_{k_2+K',k_2} & H_{k_2+K',k_2+K'} & \cdots \\ & & & \cdots & \cdots & \cdots & \\ &&&0\\ &&&&&&\cdots\\ &&&&&&&\cdots \end{pmatrix}$$ $k_i$'s are wave vectors within the 1BZ (of course if you want can be some larger one), so the difference between the $k_i$'s are smaller than a reciprocal lattice vector. As a result, the eigenstate (Bloch state) of the Hamiltonian can be given by diagonalise each of the blocks.

For example, one can denote the eigenstates from the block of $k_1$ as: $$| \psi_{k_1,n} \rangle = \sum_{K_j} c_{k_1+K_j,n} |k_1+K_j \rangle$$ and the $n$ here stands for different bands, and one can easily show that this solution satisfy the Bloch form.

To summarize, Bloch theorem tells us we can lable all the eigenstates of the Hamiltonian for a particle in a periodic potential with a quantum number $k$, and the value of $k$ can always be chosen within the 1BZ. And as mentioned in Ashcroft and Mermin's book (page 140)

Although full set of levels can be described with $k$ restricted to a single primitive unit cell, it is often useful to allow $k$ to range through all of $k$-space, even though this is a highly redundant description.

  • $\begingroup$ Exactly - and this just shows that Ashcroft and Mermin generally has it all... $\endgroup$ – Jon Custer Sep 1 '17 at 1:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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