User:Eg-T2g/Sandbox

Dimerization and phase transition
Now let's consider that half of the atoms shift position by an amount d and that there are two inter atomic matrix elements $$\Delta_1$$ and $$\Delta_2$$ for the short spacing a - d and the long spacing a + d respectively. The length of the unit cell doubles to 2 a, so the size of the reciprocal lattice cell now becomes $$-\frac{\pi}{2 a}\leqq k\leqq\frac{\pi}{2 a}$$. In each lattice cell we have two states, one on each atom:
 * $$|n,1\rangle = \begin{pmatrix}

1\\ 0 \end{pmatrix} ,\quad 0\\ 1 \end{pmatrix} $$
 * n,2\rangle = \begin{pmatrix}


 * $$ \langle n\pm 1,i|H|n,i\rangle= 0 \ ;$$ &ensp; $$\langle n,i|H|n,i\rangle=E_0 \ .$$
 * $$ \langle n,1|H|n,2\rangle= \langle n,2|H|n,1\rangle= -\Delta_1 \ .$$
 * $$ \langle n\pm 1,1|H|n,2\rangle= \langle n\pm 1,2|H|n,1\rangle=-\Delta_2$$

Lattice
Graphite consists of carbon atoms arranged in hexagonal layers which are held together by weak Van der Waals forces. The 2-dimensional layers of the crystal structure belong to the awesome hexagonal wallpaper group p6m. The surrounding of the carbon atoms has the symmetry of the prismatic point group D3h, with a three fold achiral dihedral symmetry. the same point group as the atoms in a HCP crystal structure. Although graphite has a hexagonal crystal structure it is very different from elements with a HCP crystal structure. The structure is very open and light while the HCP structure is close packed and dense. The primitive plane lattice vectors are


 * $$\vec{\mathbf{A}} = a \begin{pmatrix} 1 \\ 0 \end{pmatrix}

,\quad \vec{\mathbf{B}} = \frac{a}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} $$

There are two carbon atoms in a primitive unit cell of graphite which are denoted by the orange triangles in the figure on the right. It is good practice to place one atom in the origin and to place the other atoms in positions with the highest posible symmetry. The positions we choose here are


 * $$\vec{\mathbf{r}}_1 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

,\quad \vec{\mathbf{r}}_2 = \frac{a}{2} \begin{pmatrix}1\\ 1/\sqrt{3} \end{pmatrix} $$

Hybrid states
The carbon atoms are surrounded by three other carbon atoms. From the 2s and the three 2p orbitals a new basis set of have three sp2-hybrid orbitals and a pz orbital can be constructed. The sp2-hybrid orbitals all have an even symmetry with respect to reflection in the plane of the lattice
 * $$\psi_1 = \frac{1}{\sqrt{3}} \left(s + \sqrt{2}p_x\right)$$
 * $$\psi_{2,3} = \frac{1}{\sqrt{3}} s - \frac{1}{\sqrt{6}} p_x \pm \frac{1}{\sqrt{2}}p_y$$

and the pz orbital has an odd symmetry with respect to reflection in the plane of the lattice
 * $$\psi_4 = p_z$$

The states of $$\psi_1$$, $$\psi_2$$ and $$\psi_3$$ are orthonormal and belong to the same representation. They don't mix with $$\psi_4$$ because $$\psi_4$$ belongs to a different representation. We neglect interactions between layers. Orthonormal means that the integrals of the products of the states meet the condition:
 * $$\langle i|j\rangle = \int_{\Omega_r}\psi_i^*(\mathbf{r})\psi_j(\mathbf{r})d\mathbf{r} = \delta_{i,j}$$

The $$sp_2$$-hybrid orbitals on the atoms are rotated to an orientation in which the hybrid orbitals form $$\sigma$$ bond orbitals that have a high symmetry round the axes between the stoms. Atom 1 is rotated 30o counter clockwise and atom 2 is rotated 150o clockwise. Now $$\psi_1$$ on atom 1 points to $$\psi_1$$ on atom2, $$\psi_2$$ on atom 1 points to $$\psi_2$$ on atom2 and $$\psi_3$$ on atom 1 points to $$\psi_3$$ on atom2. The $$\psi_4$$, or $$p_z$$ orbitals form an independent system of $$\pi$$-orbitals. There are only bonds between atom 1 and atom 2 within or between primitive unit cells. There are no bonds between atom 1 with atom 1 in an adjacent cell and no bonds between atoms 2 in adjacent cells. There are only $$\pi$$-bonds between $$\psi_4$$-states and $$\sigma$$-bonds between the $$sp_2$$-states.
 * Bond orbitals

Atom 1 has three neighbors with positions $$\vec{\mathbf{p}}_i = \vec{\mathbf{r}}_i - \vec{\mathbf{r}}_1$$



\vec{\mathbf{p}}_1 = \frac{a}{2} \begin{pmatrix}1\\ 1/\sqrt{3} \end{pmatrix} ,\quad \vec{\mathbf{p}}_2 = \frac{a}{2} \begin{pmatrix}-1\\ 1/\sqrt{3} \end{pmatrix} ,\quad \vec{\mathbf{p}}_3 = a \begin{pmatrix} 0 \\ -1/\sqrt{3} \end{pmatrix} $$

Similarly atom 2 has three neighbors with positions $$\vec{\mathbf{q}}_i = \vec{\mathbf{r}}_i - \vec{\mathbf{r}}_2$$



\vec{\mathbf{q}}_1 = -\frac{a}{2} \begin{pmatrix}1\\ 1/\sqrt{3} \end{pmatrix} ,\quad \vec{\mathbf{q}}_2 = \frac{a}{2} \begin{pmatrix}1\\ -1/\sqrt{3} \end{pmatrix} ,\quad \vec{\mathbf{q}}_3 = a \begin{pmatrix} 0 \\ 1/\sqrt{3} \end{pmatrix} $$ or
 * $$ \vec{\mathbf{p}}_i = -\vec{\mathbf{q}}_i $$

The Hamiltonian
The tight binding Hamiltonian of graphite has two 4 dimensional diagonal $$3 sp_2 + p_z$$ atomic blocks and two 4 dimensional $$3 sp_2 + p_z$$ bond blocks. The matrix elementts of the Hamiltonian can be expresssed like
 * $$\langle i,n|H|j,m\rangle = H_{i,j}^{n,m}$$

in which $$i$$ and $$j$$ denote the atoms and $$n$$ and  $$m$$ denote the (hybrid) atomic orbitals $$\psi_n$$. The matrix representation of the tight binding Hamiltonian has four $$4 \times 4$$-blocks that look like



H_{i,j} = \begin{pmatrix} H_{i,j}^{1,1} & H_{i,j}^{1,2} & ...\\ H_{i,j}^{2,1} & H_{i,j}^{2,2} & ...\\ ... & ... & ... \end{pmatrix} $$

For a graphite layer the hamiltonian looks like:

H = \begin{pmatrix} H_{1,1} & H_{1,2} \\ H_{2,1} & H_{2,2} \end{pmatrix} $$

So the diagonal blocks of the Hamiltonian are

H_{i,i} = \begin{pmatrix} E_{sp_2} & 0 & 0 & 0 \\ 0 & E_{sp_2} & 0 & 0 \\ 0 & 0 & E_{sp_2} & 0 \\ 0 & 0 & 0 & E_{p_z} \end{pmatrix} $$


 * $$|\vec{\mathbf{k}},m\rangle =\frac{1}{\sqrt{N}}

\sum_{n=1}^N e^{i \vec{\mathbf{k}}\cdot\vec{\mathbf{R}}_n} |n,m\rangle $$


 * $$|\vec{\mathbf{k}},m\rangle =\frac{1}{\sqrt{N}}\sum_{n=1}^N

\cos(\vec{\mathbf{k}}\cdot\vec{\mathbf{R}}_n) |n,m\rangle $$



H_{i,i\pm 1} = \begin{pmatrix} V_{\sigma} \left[ e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_1} + e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{q}}_1}\right] & 0 & 0 & 0 \\ 0 & V_{\sigma} \left[ e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_2} + e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{q}}_2}\right] & 0 & 0 \\ 0 & 0 & V_{\sigma} \left[ e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_3} + e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{q}}_3}\right] & 0 \\ 0 & 0 & 0 & V_{\pi} \sum_{i=1}^3 \left[e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_1} + e^{\pm i\vec{\mathbf{k}}\cdot\vec{\mathbf{q}}_i} \right] \end{pmatrix} $$
 * $$\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_1 = a (k_x + k_y / \sqrt{3}) / 2$$
 * $$\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_2 = a (-k_x + k_y / \sqrt{3}) / 2$$
 * $$\vec{\mathbf{k}}\cdot\vec{\mathbf{p}}_3 = -a k_y / \sqrt{3}$$
 * $$V_{\sigma} = (V_{ss\sigma} + \sqrt{2}V_{sp\sigma} + 2 V_{pp\sigma})/3 $$
 * $$V_{\pi} = V_{pp\pi} $$



H_i = \begin{pmatrix} E_i & f_i(\mathbf{k}) \\ f_i^*(\mathbf{k}) & E_i \end{pmatrix} $$
 * $$|\epsilon - H_i| = 0$$

Relativistic corrections to the Schrödinger equation
The relativistic Hamiltonian for electrons in an atom has to be derived from the Dirac equation. The Dirac equation is derived from Einsteins famous classical relativistic expression for the energy
 * $$E = \sqrt{m^2c^4 + p^2c^2}$$

Solving the Dirac equation is a rather tedious job in comparison to solving the Schrödinger equation. By writing the expression for the energy like
 * $$E = m c^2 \sqrt{1 + \left(\frac{p}{mc}\right)^2}$$

and transforming this expression like the Taylor series expansion of
 * $$\sqrt{1 + x^2} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} - \frac{5x^6}{128} + ... $$

we get an expression like
 * $$E = m c^2 + \frac{p^2}{2m} - \frac{p^4}{8m^3c^2} + \frac{p^6}{16m^5c^4} - ... $$

in which we recognise the mass-energy equivalence in the first term and the well known classical kinetic energy expression $$p^2 / (2m)$$ in the second term. The higher order $$p^{2n}$$-terms are relativistic corrections. Realise that $$x \approx 0.5$$ at most in the Taylor series expansion and that this series is rapidly converging even for values of $$x$$ rather close to $$x = 1$$. The Hamiltonian expression, from which the Schrödinger equation is derived,
 * $$H = T + V$$

In most cases it suffices to replace the kinetic energy term
 * $$T = \frac{p^2}{2m}$$

with an additional $$p^4$$-term:
 * $$T = \frac{p^2}{2m} - \frac{p^4}{8m^3c^2}$$

Heat capacity and Fermi level of free electron gas

 * $$E = \frac{(\hbar k)^2}{2m} \ ,$$
 * $$D_1\left(E\right) = \frac {1}{2\hbar}\sqrt{\frac {2 m}{E}}$$
 * $$D_3\left(E\right) = \frac{4 \pi}{\hbar^3} \sqrt{2 m E}$$
 * $$n = \int f(E)g(E)\,dE$$
 * $$C = \frac{\partial}{\partial T}\int E f\left(E\right)g\left(E\right)\,dE$$
 * $$f(E) = \frac{1}{\exp\left(\frac{E-\mu}{k_BT}\right)+1}$$

Atomic orbital and electron configuration calculations
An atomic electron configuration is often expressed like [Ng]nsinpj(n-1)dk(n-2)fl to denote that the inner orbitals of the atom have a noble gas configuration and the outer s, p, d and f orbitals, if present, are occupied with (i + j + k + l) electrons, if present. These kind of representations of atomic electron configurations appear in electron configuration tables and they make up the structure of the Periodic Table. These tables suggest that electrons are neatly and independently arranged in independent orbitals.

Experts on the subject of the physical properties of atoms, like atomic physicists, chemical physicists, astrophysicists and physical chemists, have more detailed views on the subjects of atomic and molecular orbitals, wave functions and electron configurations. Though these subjects may seem difficult and incomprehensible to a layman or inexperienced student at first sight, it is not difficult to understand for somebody who understands the principles of atomic orbitals and electron configurations as generally presented to the layman or inexperienced student.
 * Ab Initio or First Principles calculations

An atom or molecule is described by a single wave function that incorporates all the particles that belong to the system, The one-electron wave functions, or spin-orbitals as they are called by physicists, of the particles in the system can't be separated from each other or from the system as a whole. This is not difficult to see if you take a look at the expression of such a many particle wave function. A wave function of a system in which all available states are occupied is described by a single Slater determinant.
 * Many-particle wave functions

A determinant is a mathematical shorthand notation for a special summation of products. If you expand a Slater-determinant you get a complete list of products of all the possible arrangements of one electron spin-orbitals with alternating signs. This means that the determinant, or the wave function of the whole system, changes sign if you exchange two particles in the determinant expression of the wave function. This is required by the Pauli principle. The Pauli principle states that a Fermion wave function should be anti-symmetric. This property is also reflected in the anti-symmetry of the Pauli spin matrix. So a Slater determinant is not only a compact way of expressing a complicated sum of products. It also meets the special symmetry requirements of the Pauli principle.
 * Determinants as wave function expression

The closed shell wave function of a noble gas atom can be described by a single Slater determinant. If you take a look at this expression you will see that it is not possible to "look" at one of the particles in the expression separately because the wave function of a single particle is mixed with all the other particles and wave functions that make up the system. The wave function that is calculated for this kind of electron configurations is the so called ground state of the system.
 * Noble gas and closed shell configurations

Atoms with an open shell electron configuration have more complicated wave functions which are expressed by sums and series of Slater determinants. In this case the average over the entire system doesn't even consist of states which are filled or empty. Some states are only partially filled.
 * Open shell configurations