Schrodinger Equation

de Broglie Wavelength

For a photon, because of the photoelectric effect, we know that $$E = \hbar\omega = hf$$ where $h$ is the Planck's constant and $f$ is the frequency By special relativity, the energy of a photon is also known to be $$E = pc$$ where $p$ is the momentum and $c$ is the speed of the photon Thus, we have $$E = hf = h(\frac{c}{\lambda}) = pc$$ $$\bbox[5px,border:2px solid #666] { p = \frac{h}{\lambda} }$$ In 1924 a French physicist Louis de Broglie proposed that not just photon, but other particles at a scale where quantum effects are significant, also satisfy the relation $p = \frac{h}{\lambda}$, just the speed is not necessary to be $c$.

Momentum Operator

$$\begin{align} k &= \frac{2\pi}{\lambda} \\ &= (2\pi)\frac{p}{h} \\ &= \frac{p}{\hbar} \\ \end{align}$$ The equation of a plane wave is given by $$\begin{align} \phi &= e^{i(kx - \omega t)} \\ \frac{d}{dx} \phi &= ike^{i(kx-\omega t)} \\ &= i(\frac{p}{\hbar})e^{i(kx-\omega t)} \\ \frac{\hbar}{i}\frac{d}{dx} \phi &= pe^{i(kx-\omega t)} \\ \end{align}$$ As any wavefunction can be expressed as $$\psi = e^{i\omega t}\phi(x)$$ We concluded that the operator of momentum is given by $$\hat{p} = \frac{\hbar}{i}\frac{d}{dx}$$ with $$e^{ikx}$$ as its eigenfunction.

Time Independent Schrodinger Equation

Total energy of the particle is $KE + PE$, i.e. $$E = \frac{p^2}{2m} + V$$ where $V$ is the potential energy as we know that the momentum operator is $\hat{p} = \frac{\hbar}{i}\frac{d}{dx}$, we have the energy operator $$\hat{E} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V$$ $$\bbox[5px,border:2px solid #666] { \hat{E}\psi = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi + V\psi }$$ This is known as the time independent Schrodinger equation.

Time Dependent Schrodinger Equation

As any wavefunction can be expressed as $$\psi = e^{i\omega t}\phi(x)$$ Differentiate with respect to time, $$\begin{align} \frac{d}{dt}\psi &= -i\omega e^{i(kx-\omega t)} \\ i\hbar\frac{d}{dt}\psi &= -(i)(i\hbar)\omega e^{(i(kx - \omega t)} \\ &= \hbar \omega e^{i(kx - \omega t)} \\ \end{align}$$ As we know $E = \hbar \omega$, we can write $$\hat{E}\psi = i\hbar\frac{d}{dt}\psi$$ Substitute the energy operator, we have $$\bbox[5px,border:2px solid #666] { i\hbar\frac{d}{dt}\psi = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi + V\psi }$$ This is known as the time dependent Schrodinger equation.