QHO
#Physics
Professor Dave Explains Video
Quantum Physics
Topics
$\displaystyle U(x)=\frac{1}{2}kx^{2}$
- This is the potential of a quantum harmonic oscillator, and it applies to all the equations below
$\displaystyle \psi_{n}(x)=\left( \frac{m\omega}{\pi \hbar} \right)^{\frac{1}{4}}\frac{1}{\sqrt{ 2^{n}n! }} H_{n}(\xi)e^{-\frac{1}{2} \xi^{2}}$
- When the potential is that of a harmonic oscillator, the wave eigenfunctions of energy $\displaystyle E_{n}$ takes the above form
- You may also substitute $\displaystyle k=m\omega ^{2}$ to get $\displaystyle U(x)=\frac{1}{2}m\omega ^{2}x^{2}$
- $\displaystyle \xi\equiv \sqrt{ \frac{m\omega}{\hbar} }x$ for substitution
- $\displaystyle \left( \frac{m\omega}{\pi \hbar} \right)^{\frac{1}{4}}\frac{1}{\sqrt{ 2^{n}n! }}$ is the normalization factor
- $\displaystyle H_{n}(\xi)$ is a Hermite polynomial
- Desmos Demo of Different Standing Waves
$\displaystyle \psi_{n}(x)=\frac{1}{\sqrt{ n }}(\hat{a}{+})^{n}\psi{0}(x)$
- $\displaystyle \hat{a}_{+}$ is the raising operator
- $\displaystyle {\psi}_{0}(x)$ is the ground state and is equal to $\displaystyle\left( \frac{m\omega}{\pi \hbar} \right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^{2}}$
$\displaystyle \psi(\xi)=h(\xi)e^{-\xi ^{2}/2}$
- General wave function of a harmonic oscillator
- $\displaystyle h(\xi)=a_{0}+a_{1}\xi+a_{2}\xi ^{2}+\ldots=\sum_{j = 0}^{\infty}a_{n}\xi^{n},a_{n+2}=\frac{2n-k+1}{(n+1)(n+2)}a_{n}$
$\displaystyle E_{n}=\left( n+\frac{1}{2} \right)\hbar \omega=\frac{1}{2}{\left\langle{K}\right\rangle}=\frac{1}{2}{\left\langle{U}\right\rangle}$
- Shows how energy is quantized in a harmonic oscillator
- $\displaystyle E_{n}$ is the energy of $\displaystyle n$th energy state
- $\displaystyle \hbar$ is the reduced planck constant
- $\displaystyle \omega$ is the angular frequency of oscillation
- $\displaystyle {\left\langle{K}\right\rangle}$ is the average kinetic energy of the particle
- $\displaystyle {\left\langle{U}\right\rangle}$ is the average potential energy of the particle
Statistical Mechanics
Topics
$\displaystyle H=\frac{1}{8\pi}\int \left( \epsilon E^{2}+\frac{B^{2}}{\mu} \right) , \mathrm{d}V\rightarrow \varepsilon=\hbar {\omega}_{0}\left( n+\frac{1}{2} \right)$
- Delocalized harmonic oscillator
$\displaystyle \frac{ \mathrm{d}^2 \psi}{ \mathrm{d} x^{2}}=\frac{2m}{\hbar}[V(x)-E]\psi$