Materials Science and Engineering/Equations/Quantum Mechanics

< Materials Science and Engineering < Equations

Relation between energy and frequency of a quanta of radiation

  $E=hf\;$ 
  $E=\hbar \omega$

  $\mathbf{p}=\hbar \mathbf{k}\;$

Energy:

E

Frequency:

f

Angular Frequency:

\omega = 2\pi f

Wavenumber:

k = 2\pi / \lambda

Plank's Constant:

h

De Broglie Hypothesis

  $p=h/\lambda\;$

Wavelength:

\lambda

Momentum:

p

Phase of a Plane Wave Expressed as a Complex Phase Factor

  $\psi \approx e^{i(\mathbf{k}\cdot\mathbf{x}- \omega t)}$

Time-Dependent Schrodinger Equation

  $i\hbar\frac{\partial}{\partial t}\Psi=-\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi\;$

  $\mathrm{i}\hbar \frac{d}{d t} \left| \psi \left(t\right) \right\rangle = H(t)\left|\psi\left(t\right)\right\rangle$

Ket:

|\psi(t)\rangle

Reduced Planck's Constant:

\hbar

Hamiltonian:

H(t)

The Hamiltonian describes the total energy of the system.

Partial Derivative:

\partial / \partial t

Mass:

m

Potential:

V

Derivation

Begin with a step from the time-independent derivation

  $\frac{1}{\Psi} \frac{d^2 \Psi}{dx^2} = \frac{1}{c^2 \zeta} \frac{d^2 \zeta}{dt^2}$

Set each side equal to a constant, $- \kappa^2$

$-\kappa^2 = \frac{1}{c^2 \zeta} \frac{d^2 \zeta}{dt^2}$

Multiply by $c^2$ to remove constants on the right side of the equation.

$-\beta^2 = \frac{1}{\zeta} \frac{d^2 \zeta}{dt^2}$

The solution is similar to what was found previously

$\zeta (t) = Ne^{\pm i \beta t}$

The amplitude at a point $t$ is equal to the amplitude at a point $t + \tau$

$N e^{\pm i \beta t} = N e^{\pm i \beta (t + \tau)}$

The following equation must be true:

$\beta \tau = 2 \pi \;$

Rewrite $\beta$ in terms of the frequency

$\beta = 2 \pi v \;$

Enter the equation into the expression of $\zeta$

$\zeta (t) = Ne^{\pm 2 \pi i v t}$

$\zeta (t) = N e^{-i E t / \hbar}$

The time-dependent Schrodinger equation is a product of two 'sub-functions'

$\Psi (x,t) = \psi (x) \zeta (t)\;$

$\Psi (x,t) = \psi e^{-iEt/\hbar}$

To extract $E$ , differentiate with respect to time:

$\frac{\partial \Psi}{\partial t} = \frac{-iE}{\hbar} \psi e^{-iEt/\hbar}$

$\frac{\partial \Psi}{\partial t} = \frac{E}{i \hbar} \psi e^{-iEt/\hbar}$

Rearrange:

  $i \hbar \frac{\partial \Psi}{\partial t} = E \Psi$ 
  $\hat H \Psi = E \Psi$

Time-Independent Schrodinger Equation

  $H\Psi = E\Psi\;$

-\frac{\hbar^2}{2 m} \frac{d^2 \psi (x)}{dx^2} + U(x) \psi (x) = E \psi (x)

\left[-\frac{\hbar^2}{2 m} \nabla^2 + U(\mathbf{r}) \right] \psi (\mathbf{r}) = E \psi (\mathbf{r})

  $-\frac{\hbar^2}{2 m} \nabla^2 \psi + (U - E) \psi = 0$

  $H \left|\psi_n\right\rang = E_n \left|\psi_n \right\rang.$

Del Operator:

\nabla

Derivation

The Schrodinger Equation is based on two formulas:

The classical wave function derived from the Newton's Second Law
The de Broglie wave expression

Formula of a classical wave:

  $\frac{d^2z}{dx^2} = \frac{1}{c^2} \frac{d^2 z}{dt^2}$

Separate the function into two variables:

$z(x,t) = \Psi (x) \zeta (t)\;$

Insert the function into the wave equation:

$\zeta \frac{d^2 \Psi}{dx^2} = \frac{\Psi}{c^2} \frac{d^2 \zeta}{dt^2}$

Rearrange to separate $\Psi$ and $\zeta$

$\frac{1}{\Psi} \frac{d^2 \Psi}{dx^2} = \frac{1}{c^2 \zeta}\frac{d^2 \zeta}{dt^2}$

Set each side equal to an arbitrary constant, $-\kappa^2$

$\frac{1}{\Psi} \frac{d^2 \Psi}{dx^2} = - \kappa^2$

$\frac{d^2 \Psi}{dx^2} = - \kappa^2 \Psi$

Solve this equation

$\Psi (x) = Ne^{\pm i \kappa x}$

The amplitude at one point needs to be equal to the amplitude at another point:

$N e^{\pm i \kappa x} = N e^{\pm i \kappa (x + \lambda)}$

The following condition must be true:

  $\kappa \lambda = 2 \pi\;$

Incorporate the de Broglie wave expression

  $\frac{h}{mv} = \lambda$

$\kappa = \frac{2 \pi m v}{h}$

Use the symbol $\hbar$

$\hbar = \frac{h}{2\pi}$

$\frac{d^2 \Psi}{dx^2} = - \frac{m^2 v^2}{\hbar^2} \Psi$

$\frac{- \hbar^2}{m^2 v^2} \frac{d^2 \Psi}{dx^2} = \Psi$

Use the expression of kinetic energy, $E_{kinetic} = \frac{1}{2} mv^2$

$\frac{-\hbar^2}{2m} \frac{d^2 \Psi}{dx^2} = E \Psi$

Modify the equation by adding a potential energy term and the Laplacian operator

  $\frac{-\hbar^2}{2m} \nabla^2 \Psi + V \Psi = E \Psi$ 
  $\hat H \Psi = E \Psi$

Non-Relativistic Schrodinger Wave Equation

In non-relativistic quantum mechanics, the Hamiltonian of a particle can be expressed as the sum of two operators, one corresponding to kinetic energy and the other to potential energy. The Hamiltonian of a particle with no electric charge and no spin in this case is:

  $H \psi\left(\mathbf{r}, t\right) = \left(T + V\right) \psi\left(\mathbf{r}, t\right)$

  $H \psi\left(\mathbf{r}, t\right) = \left[ - \frac{\hbar^2}{2m} \nabla^2 + V\left(\mathbf{r}\right) \right] \psi\left(\mathbf{r}, t\right)$

  $H \psi\left(\mathbf{r}, t\right) = \mathrm{i} \hbar \frac{\partial \psi}{\partial t} \left(\mathbf{r}, t\right)$

kinetic energy operator:

T = \frac{p^2}{2m}

mass of the particle:

m\;

momentum operator:

\mathbf{p} = -\mathrm{i}\hbar\nabla

potential energy operator:

V = V\left(\mathbf{r}\right)

real scalar function of the position operator

\mathbf{r}

V

Gradient operator:

\nabla

Laplace operator:

\nabla^2

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