Materials Science and Engineering/Equations/Quantum Mechanics

Relation between energy and frequency of a quanta of radiation

 $E=hf$
 $E=\hslash \omega$

 $𝐩=\hslash 𝐤$

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

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

Time-Dependent Schrodinger Equation

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

 $\mathrm{i}\hslash \frac{d}{dt}|\psi \left(t\right)⟩=H(t)|\psi \left(t\right)⟩$

Ket: $|\psi (t)⟩$

Reduced Planck's Constant: $\hslash$

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

 $\frac{1}{\mathrm{\Psi }}\frac{d^2\mathrm{\Psi }}{d{x}^2}=\frac{1}{c^2\zeta }\frac{d^2\zeta }{d{t}^2}$

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

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

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

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

The solution is similar to what was found previously

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

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 $\zeta$

$\zeta (t)=N{e}^{\pm 2\pi ivt}$

$\zeta (t)=N{e}^{-iEt/\hslash }$

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

$\mathrm{\Psi }(x,t)=\psi (x)\zeta (t)$

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

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

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

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

Rearrange:

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

Time-Independent Schrodinger Equation

 $H\mathrm{\Psi }=E\mathrm{\Psi }$

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

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

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

 $H|{\psi }_n⟩=E_n|{\psi }_n⟩.$

Del Operator: $\mathrm{\nabla }$

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:

 $\frac{d^{2}z}{d{x}^2}=\frac{1}{c^2}\frac{d^{2}z}{d{t}^2}$

Separate the function into two variables:

$z(x,t)=\mathrm{\Psi }(x)\zeta (t)$

Insert the function into the wave equation:

$\zeta \frac{d^2\mathrm{\Psi }}{d{x}^2}=\frac{\mathrm{\Psi }}{c^2}\frac{d^2\zeta }{d{t}^2}$

Rearrange to separate $\mathrm{\Psi }$ and $\zeta$

$\frac{1}{\mathrm{\Psi }}\frac{d^2\mathrm{\Psi }}{d{x}^2}=\frac{1}{c^2\zeta }\frac{d^2\zeta }{d{t}^2}$

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

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

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

Solve this equation

$\mathrm{\Psi }(x)=N{e}^{\pm i\kappa x}$

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

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

$\kappa =\frac{2\pi mv}{h}$

Use the symbol $\hslash$

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

$\frac{d^2\mathrm{\Psi }}{d{x}^2}=-\frac{m^2{v}^2}{{\hslash }^2}\mathrm{\Psi }$

$\frac{-{\hslash }^2}{m^2{v}^2}\frac{d^2\mathrm{\Psi }}{d{x}^2}=\mathrm{\Psi }$

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

$\frac{-{\hslash }^2}{2m}\frac{d^2\mathrm{\Psi }}{d{x}^2}=E\mathrm{\Psi }$

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

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

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:

 $H\psi (𝐫,t)=(T+V)\psi (𝐫,t)$

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

 $H\psi (𝐫,t)=\mathrm{i}\hslash \frac{\partial \psi }{\partial t}(𝐫,t)$

kinetic energy operator: $T=\frac{p^2}{2m}$

mass of the particle: $m$

momentum operator: $𝐩=-\mathrm{i}\hslash \mathrm{\nabla }$

potential energy operator: $V=V\left(𝐫\right)$

real scalar function of the position operator $𝐫$: $V$

Gradient operator: $\mathrm{\nabla }$

Laplace operator: ${\mathrm{\nabla }}^2$