Materials Science and Engineering/Doctoral review questions/Discussion List

Discussion List

• Definition of a mode (i.e., for optical fibers) (Melissa Smith)
• Quantum mechanical descriptions of electron (Megan)
  • Free electron
  • Infinite well
  • Finite well
  • Harmonic oscillator
  • Hydrogen (H)
  • Hydrogen (H₂)
    • Hartree Fock
• Fermi energy, occupation levels for electrons
• Photonic Bandgap and Periodic Mediums (George)

Wave equation

A pulse traveling through a string with fixed endpoints as modeled by the wave equation.

Electromagnetic waves as a general phenomenon were predicted by the classical laws of electricity and magnetism, known as Maxwell's equations. If you inspect Maxwell's equations without sources (charges or currents) then you will find that, along with the possibility of nothing happening, the theory will also admit nontrivial solutions of changing electric and magnetic fields. Beginning with Maxwell's equations for a vacuum:

$\mathrm{\nabla }\cdot 𝐄=0(1)$

$\mathrm{\nabla }\times 𝐄=-\frac{\partial }{\partial t}𝐁(2)$

$\mathrm{\nabla }\cdot 𝐁=0(3)$

$\mathrm{\nabla }\times 𝐁={\mu }_0{ϵ}_0\frac{\partial }{\partial t}𝐄(4)$

where

$\mathrm{\nabla }$ is a vector differential operator (see Del).

One solution,

$𝐄=𝐁=\mathrm{𝟎}$,

is trivial.

To see the more interesting one, we utilize vector identities, which work for any vector, as follows:

$\mathrm{\nabla }\times (\mathrm{\nabla }\times 𝐀)=\mathrm{\nabla }(\mathrm{\nabla }\cdot 𝐀)-{\mathrm{\nabla }}^2𝐀$

To see how we can use this take the curl of equation (2):

$\mathrm{\nabla }\times (\mathrm{\nabla }\times 𝐄)=\mathrm{\nabla }\times (-\frac{\partial 𝐁}{\partial t})(5)$

Evaluating the left hand side:

$\mathrm{\nabla }\times (\mathrm{\nabla }\times 𝐄)=\mathrm{\nabla }(\mathrm{\nabla }\cdot 𝐄)-{\mathrm{\nabla }}^2𝐄=-{\mathrm{\nabla }}^2𝐄(6)$

where we simplified the above by using equation (1).

Evaluate the right hand side:

$\mathrm{\nabla }\times (-\frac{\partial 𝐁}{\partial t})=-\frac{\partial }{\partial t}(\mathrm{\nabla }\times 𝐁)=-{\mu }_0{ϵ}_0\frac{{\partial }^2}{{\partial }^{2}t}𝐄(7)$

Equations (6) and (7) are equal, so this results in a vector-valued differential equation for the electric field, namely

${\mathrm{\nabla }}^2𝐄={\mu }_0{ϵ}_0\frac{{\partial }^2}{\partial t^2}𝐄$

Applying a similar pattern results in similar differential equation for the magnetic field:

${\mathrm{\nabla }}^2𝐁={\mu }_0{ϵ}_0\frac{{\partial }^2}{\partial t^2}𝐁$.

These differential equations are equivalent to the wave equation:

${\mathrm{\nabla }}^{2}f=\frac{1}{c^2}\frac{{\partial }^{2}f}{\partial t^2}$

where

c is the speed of the wave and

f describes a displacement

Or more simply:

${◻}^{2}f=0$

where ${◻}^2$ is d'Alembertian:

${◻}^2={\mathrm{\nabla }}^2-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}$

Notice that in the case of the electric and magnetic fields, the speed is:

$c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}$

Which, as it turns out, is the speed of light. Maxwell's equations have unified the permittivity of free space ${ϵ}_0$, the permeability of free space ${\mu }_0$, and the speed of light itself, c. Before this derivation it was not known that there was such a strong relationship between light and electricity and magnetism.

Snell's Law

Wavefronts due to a point source

In optics and physics, Snell's law (also known as Descartes' law or the law of refraction), is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves, passing through a boundary between two different isotropic media, such as air and glass. The law says that the ratio of the sines of the angles of incidence and of refraction is a constant that depends on the media.

Named after Dutch mathematician Willebrord Snellius, one of its discoverers, Snell's law states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of velocities in the two media, or equivalently to the inverse ratio of the indices of refraction:

$\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}$

Incident, Reflected, and Transmitted Wave

File:Angles of incidence, reflection, and transmission.png

Equations of wave resulting from movement initially from left to right:

Incident Wave: $\overrightarrow{E_i}{e}^{\omega t-i\overrightarrow{k_i}\cdot \overrightarrow{r}}$

Reflected Wave: $\overrightarrow{E_r}{e}^{\omega t-i\overrightarrow{k_r}\cdot \overrightarrow{r}}$

Transmitted Wave: $\overrightarrow{E_t}{e}^{\omega t-i\overrightarrow{k_t}\cdot \overrightarrow{r}}$

Boundary Equations

Dispersion relations with regard to a homogeneous medium provide the following equations:

$\left|\overrightarrow{k_i}\right|=\left|\overrightarrow{k_r}\right|=\frac{\omega n_1}{c}$

$\left|\overrightarrow{k_t}\right|=\frac{\omega n_2}{c}$

The phases of the field be equal in order that the boundary conditions are satisfied.

${(\overrightarrow{k_1}\cdot \overrightarrow{r})}_{x=0}={(\overrightarrow{{k_1}^{\prime }}\cdot \overrightarrow{r})}_{x=0}={(\overrightarrow{k_2}\cdot \overrightarrow{r})}_{x=0}$

$(k_{1_y}y+k_{1_z}z)=(k_{{1_y}^{\prime }}y+k_{{1_z}^{\prime }}z)=(k_{2_y}y+k_{2_z}z)$

$k_{1_y}=k_{{1_y}^{\prime }}=k_{2_y}$

$k_{1_z}=k_{{1_z}^{\prime }}=k_{2_z}$

Two Important and Related Consequences

The vector $\overrightarrow{r}$ is an arbitrary vector in the z-y plane.

$\overrightarrow{r}=(x=0,y,z)={\overrightarrow{r}}_{z-y}$

$(\overrightarrow{k_{1t}}\cdot \overrightarrow{r_t})=(\overrightarrow{k_{1^{\prime }t}}\cdot \overrightarrow{r_t})=(\overrightarrow{k_{2t}}\cdot \overrightarrow{r_t})$

The vectors $\overrightarrow{k_1},\overrightarrow{k_{1^{\prime }}},\overrightarrow{k_2}$ are all in the plane of incidence. The coordinate system is oriented such that the plane of incidence is the same as the x-z plane.

$\overrightarrow{E}=\overrightarrow{E}{e}^{i(\omega t-k_{x}x-k_{z}z)}$

The tangential components of the wavevector are the same regardless of the medium is lossless of absorbing

$k_{1z}=k_{1^{\prime }z}=k_{2z}=\beta$

$k_{iz}=k_{tz}\to |k_i|\sin {\theta }_1=|k_t|\sin {\theta }_2$

$\frac{\omega n_1}{c}\sin {\theta }_1=\frac{\omega n_2}{c}\sin {\theta }_2$

Conclusion of Derivation

Angle of reflection is equal to angle of incidence:

 ${\theta }_i={\theta }_r$

Snell's law:

 $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Density of States for 1D, 2D, 3D

System

$N$ electrons in box of side length $L$

Hamiltonian

$\hat{H}(\overrightarrow{r_1},\overrightarrow{r_2},...,\overrightarrow{r_N})=\sum _{i=1}^N\frac{{\hslash }^2}{2m}\underset{i}{\overset{2}{\mathrm{\nabla }}}$

$\hat{H}(\overrightarrow{r_1},\overrightarrow{r_2},...,\overrightarrow{r_N})=\sum _{i=1}^N\widehat{H_i}$

Eigenvalues and Eigenfunctions

${\hat{H}}_i{u}_k(x,y,z)=E_k{u}_k(x,y,z)$

$u_k(x,y,z)=u_k(\overrightarrow{r})$

$\frac{1}{V^{1/2}}{e}^{i\overrightarrow{k}\cdot \overrightarrow{r}}$

${ϵ}_k=\frac{{\hslash }^2(k_x^2+k_y^2+k_z^2)}{2m}$

Born Von-Karman Boundary Conditions

$u_k(x+L,y,z)=u_k(x,y,z)$

$u_k(x,y+L,z)=u_k(x,y,z)$

$u_k(x,y,z+L)=u_k(x,y,z)$

k vectors

• k vector: momentum
• Energy eigenfunctions: momentum eigenfunctions

$\hat{P}{u}_k(\overrightarrow{r})=\frac{\hslash }{i}\overrightarrow{\mathrm{\nabla }}{u}_k(\overrightarrow{r})$

$\hat{P}{u}_k(\overrightarrow{r})=\hslash \overrightarrow{k}{u}_k(\overrightarrow{r})$

• The k vector can be interpreted as a wave vector

$\left|\overrightarrow{k}\right|=\frac{2\pi }{\lambda }$

• $\lambda$: de Broglie wavelength

Boundary Conditions

• Apply boundary condition to develop identities

$e^{i{k}_{x}L}=e^{i{k}_{y}L}=e^{i{k}_{z}L}=1$

• The boundary conditions result in quantization of k vectors

$k_x=\frac{2\pi n_x}{L}$

$k_y=\frac{2\pi n_y}{L}$

$k_z=\frac{2\pi n_z}{L}$

${ϵ}_k=\frac{{\hslash }^2(2\pi {)}^3(n_x^2+n_y^2+n_z^2)}{2m{L}^2}$

Enumerate States

File:Ky - kx space.png

• $N_{spatialstates}=\frac{\mathrm{\Omega }}{{\left(\frac{2\pi }{L}\right)}^2}$

• $N_{spatialstates}={\left(\frac{L}{2\pi }\right)}^2$

• $N_{spatialstates}=DOS\times area$

${\left(\frac{L}{2\pi }\right)}^2$: density of spatial states in k space

Fermi Surface

After introducing electrons into the system, what states will they occupy? Lowest energy levels are filled, and the Fermi surface, which is a circle in 2D and a sphere in 3D, separates the filled from unfilled states.

• $N_{electrons}=2\frac{2\pi k_F^3}{3}\cdot \frac{1}{{\left(\frac{2\pi }{L}\right)}^3}$

• $N_{electrons}=2\frac{4\pi k_F^3}{3}\frac{V}{(2\pi {)}^3}$

The number of electrons per unit volume, $n$, is equal to the total number of electrons, $N_{electrons}$, divided by the volume:

• $n=\frac{N}{V}$

• $n=\frac{k_F^3}{3{\pi }^2}$

Fermi energy

The Fermi energy is typically between $1.5eV$ and $15eV$

• ${ϵ}_F=\frac{{\hslash }^2}{2m}{k}_F^2$

Ground State Energy

• $E=2\sum _{k\le k_F}\frac{{\hslash }^2}{2m}{k}^2$

• $E=\underset{k<k_F}{\int }{d}^{3}k\frac{V}{8{\pi }^3}\frac{{\hslash }^2}{2m}{k}^2$

• $E=\frac{V}{{\pi }^2}\frac{{\hslash }^2{k}_F^5}{10m}$

Energy per electron:

• $\frac{E}{N}=\frac{3}{5}{ϵ}_F$

Density of States

The density of states is used to determine how many states are in the interval $ϵ\to ϵ+dϵ$

The number of states provided in the free electron case is as expressed below:

• $N=2\frac{4\pi k^3}{3}\cdot \frac{1}{{\left(\frac{2\pi }{L}\right)}^3}$

• $N={\left(\frac{2mϵ}{{\hslash }^2}\right)}^{3/2}$

The density of states is the derivative of the number of electrons with respect to $ϵ$:

• $\frac{dN}{dϵ}=g(ϵ)$

• $\frac{dN}{dϵ}=\frac{m}{{\hslash }^2{\pi }^2}\sqrt{\frac{2mϵ}{{\hslash }^2}}$

The number of energy levels in the range $ϵ\to ϵ+dϵ$ is the density of states multiplied by $dϵ$:

• $g(ϵ)dϵ$

The density of states in a particular band is expressed below:

• $g_n(ϵ)=\frac{1}{4{\pi }^3}\underset{S_n(ϵ)}{\int }\frac{dS}{|\overrightarrow{{\mathrm{\nabla }}_k}ϵ(k)|}$

• $g(ϵ)=\sum _n{g}_n(ϵ)$

Bloch Theorem

Periodic Potential and Expression with Fourier Series

$V(r+T)=V(r)$

$T$: Translation vector of the lattice

 $V(r)=\sum _G{V}_G\exp (i\cdot Gr))$

$\sum _G$: sum over all reciprocal vectors

$V_G$: Fourier coefficients of the potential

$G$: Reciprocal lattice vector

Expression of Wave Function with Fourier Transformation

 $\mathrm{\Psi }(r)=\sum _k{C}_k\exp [ikr]$

$C_k$: Fourier coefficients of the wave function

$k$: Wave vectors

k vectors

 $k_x=\pm \frac{n_x\cdot 2\pi }{N\cdot a}$

$\frac{2\pi }{a}$: magnitude of reciprocal lattice vector

$\frac{n_{x}2\pi }{a}$: whole set of reciprocal lattice vectors

$\frac{1}{N}$: intersperse $N$ points between reciprocal lattice points

Connection between wave vectors and reciprocal vectors

Express any wave vector, $k$, as the sum of a reciprocal vector, $G$, and wave vector that is within the Brillouin zone.

 $k=G+k^{\prime }$

$k^{\prime }$: wavevector always confined to the Brillouin zone of the reciproval lattice

Substitute expressions into the Schrodinger equation

$[\frac{-{\hslash }^2{\mathrm{\nabla }}^2}{2m}+V(\overrightarrow{r})]\mathrm{\Psi }=E\mathrm{\Psi }$

 $\sum _k\frac{(\hslash k{)}^2}{2m}{C}_k\exp (ikr)+\sum _{k^{\prime }}\sum _G{C}_{k^{\prime }}{V}_G\exp (i[k^{\prime }+G]\cdot r=E\sum _k{C}_k\exp (ikr)$

Rename the summation indices and replace $k^{\prime }$

$k^{\prime }+G=k$

$k^{\prime }=k-G$

$V(r)\mathrm{\Psi }=\sum _{G,k}{V}_G{C}_{k-G}{e}^{ikr}$

$\sum _k{e}^{ikr}[(\frac{{\hslash }^2{k}^2}{2m}-E)C_k+\sum _G{V}_G{C}_{k-G}]=0$

The equation is true with regard to an space vector $r$

 $(\frac{(\hslash \cdot k{)}^2}{2m}-E)\cdot C_k+\sum _G(C_{k-G}\cdot V_G)=0$

The known Fourier coefficients of the periodic potential are coupled to the Fourier coefficients of the wave function.

$\left[\begin{array}{ccc}\frac{\hslash }{2m}(k-G{)}^2& V& 0\\ V& \frac{\hslash }{2m}{k}^2& V\\ 0& V& \frac{\hslash }{2m}(k+G{)}^2\end{array}\right]\left[\begin{array}{c}{C}_{k-G}\\ C_k\\ C_{k+G}\end{array}\right]=E_k\left[\begin{array}{c}{C}_{k-G}\\ C_k\\ C_{k+G}\end{array}\right]$

de Broglie Relation

 $E=hf=\hslash \omega$

$\hslash \omega =\frac{1}{2}m{v}_g^2$

$\hslash \frac{\partial \omega }{\partial k}=m{v}_g\frac{\partial v_g}{\partial k}$

$\hslash =m\frac{\partial v_g}{\partial k}$

$\hslash k=m{v}_g$

 $\lambda =\frac{h}{m{v}_g}$