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: $\nabla ^{2}$ 