Materials Science and Engineering/Equations/Quantum Mechanics

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: $$\nabla^2$$