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

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

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

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 $t+\tau$

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

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 $\zeta$

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

$\zeta (t)=Ne^{-iEt/\hbar }$

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

 $H\Psi =E\Psi \;$

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

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

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

 $H\left|\psi _{n}\right\rangle =E_{n}\left|\psi _{n}\right\rangle .$

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:

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

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:

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

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 $\hbar$

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

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

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

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: $\nabla ^{2}$