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My lecturer is teaching the Bloch theorem, which I saw many years ago in Griffiths textbook on Quantum mechanics. I cannot recognise it in the form that my lecturer is using.

We are studying a tight-binding model for a one-dimensional chain of atoms. This is what my lecturerer writes:

Consider a one-dimensional chain of (sodium) atoms, each one represented by a fix point charge of +1. In the search for the eigenstates for an electron in interaction with these charges we resort to a basis of one s-like orbital per atom, that we will call $X$ , indicating where it is located, $X = j a$, with $a$ the lattice parameter and $j$ an integer indicating the lattice site. Let us make an additional approximation: The Hamiltonian matrix elements will be considered non-zero only when involving neighbouring sites, the nearest-neighbour approximation, i.e., $\langle X | H | X \rangle = \varepsilon$ and $\langle X +a | H | X \rangle = -t$. Therefore the standard matrix of the hamiltonian in this basis is tridiagonal: $$ H = \begin{bmatrix} \ddots& \\ &\varepsilon & -t & 0 \\ &-t& \varepsilon & -t & \\ & 0 &-t&\varepsilon&\\ &&&& \ddots \end{bmatrix} $$ The solution is trivial if we construct Bloch states as linear combinations of the basis states: $$ |k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle $$ For the basis set we have, with one single basis state per unit cell, there is only one such Bloch state per k value. Our Hamiltonian will thus be a block diagonal matrix with blocks 1x1: It will be directly diagonal!

I have some questions about this:

• How can my lecturer assume that if $|k \rangle$ is a bloch function, that there exists a formula like this: $|k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle$? The bloch theorem says that there exists a fuction $u$ that can take the role of $|X \rangle$ in this sum, if I (falsely) assume that $|X\rangle$ is a basis, that means (at best) that $ u = \sum_i \alpha_i |X \rangle$. I assume the lecturer means that $\langle x | X \rangle = 2 \alpha^{3/2} e^{-|x - ja|/\alpha}$ where $\alpha$ is the Bohr radius. Does he arrive at this in some other way?
• My lecturer says that there is only one Bloch state per k value. Is this to say that there is exactly one band? If so, how can we prove this?

My lecturer is teaching the Bloch theorem, which I saw many years ago in Griffiths textbook on Quantum mechanics. I cannot recognise it in the form that my lecturer is using.

We are studying a tight-binding model for a one-dimensional chain of atoms. This is what my lecturerer writes:

Consider a one-dimensional chain of (sodium) atoms, each one represented by a fix point charge of +1. In the search for the eigenstates for an electron in interaction with these charges we resort to a basis of one s-like orbital per atom, that we will call $X$ , indicating where it is located, $X = j a$, with $a$ the lattice parameter and $j$ an integer indicating the lattice site. Let us make an additional approximation: The Hamiltonian matrix elements will be considered non-zero only when involving neighbouring sites, the nearest-neighbour approximation, i.e., $\langle X | H | X \rangle = \varepsilon$ and $\langle X +a | H | X \rangle = -t$. Therefore the standard matrix of the hamiltonian in this basis is tridiagonal: $$ H = \begin{bmatrix} \ddots& \\ &\varepsilon & -t & 0 \\ &-t& \varepsilon & -t & \\ & 0 &-t&\varepsilon&\\ &&&& \ddots \end{bmatrix} $$ The solution is trivial if we construct Bloch states as linear combinations of the basis states: $$ |k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle $$ For the basis set we have, with one single basis state per unit cell, there is only one such Bloch state per k value. Our Hamiltonian will thus be a block diagonal matrix with blocks 1x1: It will be directly diagonal!

I have some questions about this:

• How can my lecturer assume that if $|k \rangle$ is a bloch function, that there exists a formula like this: $|k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle$? The bloch theorem says that there exists a fuction $u$ that can take the role of $|X \rangle$ in this sum, if I (falsely) assume that $|X\rangle$ is a basis. Does he arrive at this in some other way?
• My lecturer says that there is only one Bloch state per k value. Is this to say that there is exactly one band? If so, how can we prove this?

My lecturer is teaching the Bloch theorem, which I saw many years ago in Griffiths textbook on Quantum mechanics. I cannot recognise it in the form that my lecturer is using.

We are studying a tight-binding model for a one-dimensional chain of atoms. This is what my lecturerer writes:

Consider a one-dimensional chain of (sodium) atoms, each one represented by a fix point charge of +1. In the search for the eigenstates for an electron in interaction with these charges we resort to a basis of one s-like orbital per atom, that we will call $X$ , indicating where it is located, $X = j a$, with $a$ the lattice parameter and $j$ an integer indicating the lattice site. Let us make an additional approximation: The Hamiltonian matrix elements will be considered non-zero only when involving neighbouring sites, the nearest-neighbour approximation, i.e., $\langle X | H | X \rangle = \varepsilon$ and $\langle X +a | H | X \rangle = -t$. Therefore the standard matrix of the hamiltonian in this basis is tridiagonal: $$ H = \begin{bmatrix} \ddots& \\ &\varepsilon & -t & 0 \\ &-t& \varepsilon & -t & \\ & 0 &-t&\varepsilon&\\ &&&& \ddots \end{bmatrix} $$ The solution is trivial if we construct Bloch states as linear combinations of the basis states: $$ |k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle $$ For the basis set we have, with one single basis state per unit cell, there is only one such Bloch state per k value. Our Hamiltonian will thus be a block diagonal matrix with blocks 1x1: It will be directly diagonal!

I have some questions about this:

• What does my lecturer mean by $|k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle$ ? Does $|k\rangle$ denote a planewave? I.e. $\langle x | k \rangle = L^{-1/2}e^{ikx}$? Does he mean that $\langle x | X \rangle = 2 \alpha^{3/2} e^{-|x - ja|/\alpha}$ where $\alpha$ is the Bohr radius?
• My lecturer says that there is only one Bloch state per k value. Is this to say that there is exactly one band? If so, how can we prove this?
• My lecturer says the hamiltonian is diagonal. This is to say that $t = 0$. How can I prove this?

My lecturer is teaching the Bloch theorem, which I saw many years ago in Griffiths textbook on Quantum mechanics. I cannot recognise it in the form that my lecturer is using.

We are studying a tight-binding model for a one-dimensional chain of atoms. This is what my lecturerer writes:

Consider a one-dimensional chain of (sodium) atoms, each one represented by a fix point charge of +1. In the search for the eigenstates for an electron in interaction with these charges we resort to a basis of one s-like orbital per atom, that we will call $X$ , indicating where it is located, $X = j a$, with $a$ the lattice parameter and $j$ an integer indicating the lattice site. Let us make an additional approximation: The Hamiltonian matrix elements will be considered non-zero only when involving neighbouring sites, the nearest-neighbour approximation, i.e., $\langle X | H | X \rangle = \varepsilon$ and $\langle X +a | H | X \rangle = -t$. Therefore the standard matrix of the hamiltonian in this basis is tridiagonal: $$ H = \begin{bmatrix} \ddots& \\ &\varepsilon & -t & 0 \\ &-t& \varepsilon & -t & \\ & 0 &-t&\varepsilon&\\ &&&& \ddots \end{bmatrix} $$ The solution is trivial if we construct Bloch states as linear combinations of the basis states: $$ |k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle $$ For the basis set we have, with one single basis state per unit cell, there is only one such Bloch state per k value. Our Hamiltonian will thus be a block diagonal matrix with blocks 1x1: It will be directly diagonal!

I have some questions about this:

• How can my lecturer assume that if $|k \rangle$ is a bloch function, that there exists a formula like this: $|k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle$? The bloch theorem says that there exists a fuction $u$ that can take the role of $|X \rangle$ in this sum, if I (falsely) assume that $|X\rangle$ is a basis, that means (at best) that $ u = \sum_i \alpha_i |X \rangle$. I assume the lecturer means that $\langle x | X \rangle = 2 \alpha^{3/2} e^{-|x - ja|/\alpha}$ where $\alpha$ is the Bohr radius. Does he arrive at this in some other way?
• My lecturer says that there is only one Bloch state per k value. Is this to say that there is exactly one band? If so, how can we prove this?

My lecturer is teaching the Bloch theorem, which I saw many years ago in Griffiths textbook on Quantum mechanics. I cannot recognise it in the form that my lecturer is using.

We are studying a tight-binding model for a one-dimensional chain of atoms. This is what my lecturerer writes:

Consider a one-dimensional chain of (sodium) atoms, each one represented by a fix point charge of +1. In the search for the eigenstates for an electron in interaction with these charges we resort to a basis of one s-like orbital per atom, that we will call $X$ , indicating where it is located, $X = j a$, with $a$ the lattice parameter and $j$ an integer indicating the lattice site. Let us make an additional approximation: The Hamiltonian matrix elements will be considered non-zero only when involving neighbouring sites, the nearest-neighbour approximation, i.e., $\langle X | H | X \rangle = \varepsilon$ and $\langle X +a | H | X \rangle = -t$. Therefore the standard matrix of the hamiltonian in this basis is tridiagonal: $$ H = \begin{bmatrix} \ddots& \\ &\varepsilon & -t & 0 \\ &-t& \varepsilon & -t & \\ & 0 &-t&\varepsilon&\\ &&&& \ddots \end{bmatrix} $$ The solution is trivial if we construct Bloch states as linear combinations of the basis states: $$ |k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle $$ For the basis set we have, with one single basis state per unit cell, there is only one such Bloch state per k value. Our Hamiltonian will thus be a block diagonal matrix with blocks 1x1: It will be directly diagonal!

I have some questions about this:

• What does my lecturer mean by $|k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle$ ? Does $|k\rangle$ denote a planewave? I.e. $\langle x | k \rangle = L^{-1/2}e^{ikx}$? Does he mean that $\langle x | X \rangle$ is (the unknown) solution of the shrödinger equation for each $X$?
• My lecturer says that there is only one Bloch state per k value. Is this to say that there is exactly one band? If so, how can we prove this?
• My lecturer says the hamiltonian is diagonal. This is to say that $t = 0$. How can I prove this?

My lecturer is teaching the Bloch theorem, which I saw many years ago in Griffiths textbook on Quantum mechanics. I cannot recognise it in the form that my lecturer is using.

We are studying a tight-binding model for a one-dimensional chain of atoms. This is what my lecturerer writes:

Consider a one-dimensional chain of (sodium) atoms, each one represented by a fix point charge of +1. In the search for the eigenstates for an electron in interaction with these charges we resort to a basis of one s-like orbital per atom, that we will call $X$ , indicating where it is located, $X = j a$, with $a$ the lattice parameter and $j$ an integer indicating the lattice site. Let us make an additional approximation: The Hamiltonian matrix elements will be considered non-zero only when involving neighbouring sites, the nearest-neighbour approximation, i.e., $\langle X | H | X \rangle = \varepsilon$ and $\langle X +a | H | X \rangle = -t$. Therefore the standard matrix of the hamiltonian in this basis is tridiagonal: $$ H = \begin{bmatrix} \ddots& \\ &\varepsilon & -t & 0 \\ &-t& \varepsilon & -t & \\ & 0 &-t&\varepsilon&\\ &&&& \ddots \end{bmatrix} $$ The solution is trivial if we construct Bloch states as linear combinations of the basis states: $$ |k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle $$ For the basis set we have, with one single basis state per unit cell, there is only one such Bloch state per k value. Our Hamiltonian will thus be a block diagonal matrix with blocks 1x1: It will be directly diagonal!

I have some questions about this:

• What does my lecturer mean by $|k\rangle = N^{-1/2} \sum_{X}^N e^{ikX} |X \rangle$ ? Does $|k\rangle$ denote a planewave? I.e. $\langle x | k \rangle = L^{-1/2}e^{ikx}$? Does he mean that $\langle x | X \rangle = 2 \alpha^{3/2} e^{-|x - ja|/\alpha}$ where $\alpha$ is the Bohr radius?
• My lecturer says that there is only one Bloch state per k value. Is this to say that there is exactly one band? If so, how can we prove this?
• My lecturer says the hamiltonian is diagonal. This is to say that $t = 0$. How can I prove this?