$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
FullPage
overview
significance
derivation
related experiments
associated concepts
This is kind of like a finite well, but instead we have the potential increasing in a quadratic function instead of a sudden jump. That is, $$\begin{align*} V(x)=\frac{1}{2}kx^2. \end{align*}$$ Classically, a particle would sit at the very bottom. Quantum mechanically, due to the uncertainty principle, the particle can't just sit at the bottom, having zero energy.

Significance

Coming soon

Derivation

Once again we want to solve the eigenvalue equation $H\psi=E\psi$, where $H$ is the Hamiltonian. The total energy of the system is the potential energy plus the kinetic energy. The kinetic energy, written in the position basis, is $$\begin{align*} -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}, \end{align*}$$ while the potential energy is $$\begin{align*} V(x)=\frac{1}{2}kx^2\psi(x). \end{align*}$$ This is a pain to solve. We use ladder operators. $\hat{H}$ can be written in terms of ladder operators, where $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega, \end{align*}$$ and $\hat{a}$ raises the energy level and $\hat{a}^{\dagger}$ lowers the energy. Thar means there's a state $\ket{E_0}$ for which the energy $E_0$ is the minimum possible energy for the system, and $\hat{a}|E_0\rangle=0$.

Associated Concepts

Coming soon
This is kind of like a finite well, but instead we have the potential increasing in a quadratic function instead of a sudden jump. That is, $$\begin{align*} V(x)=\frac{1}{2}kx^2. \end{align*}$$ Classically, a particle would sit at the very bottom. Quantum mechanically, due to the uncertainty principle, the particle can't just sit at the bottom, having zero energy.

Significance

Coming soon

Derivation

Once again we want to solve the eigenvalue equation $H\psi=E\psi$, where $H$ is the Hamiltonian. The total energy of the system is the potential energy plus the kinetic energy. The kinetic energy, written in the position basis, is $$\begin{align*} -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}, \end{align*}$$ while the potential energy is $$\begin{align*} V(x)=\frac{1}{2}kx^2\psi(x). \end{align*}$$ This is a pain to solve. We use ladder operators. $\hat{H}$ can be written in terms of ladder operators, where $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega, \end{align*}$$ and $\hat{a}$ raises the energy level and $\hat{a}^{\dagger}$ lowers the energy. Thar means there's a state $\ket{E_0}$ for which the energy $E_0$ is the minimum possible energy for the system, and $\hat{a}|E_0\rangle=0$.

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts
FullPage
overview
significance
derivation
related experiments
associated concepts
This is kind of like a finite well, but instead we have the potential increasing in a quadratic function instead of a sudden jump. That is, $$\begin{align*} V(x)=\frac{1}{2}kx^2. \end{align*}$$ Classically, a particle would sit at the very bottom. Quantum mechanically, due to the uncertainty principle, the particle can't just sit at the bottom, having zero energy.

Significance

Coming soon

Derivation

Once again we want to solve the eigenvalue equation $H\psi=E\psi$, where $H$ is the Hamiltonian. The total energy of the system is the potential energy plus the kinetic energy. The kinetic energy, written in the position basis, is $$\begin{align*} -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}, \end{align*}$$ while the potential energy is $$\begin{align*} V(x)=\frac{1}{2}kx^2\psi(x). \end{align*}$$ This is a pain to solve. We use ladder operators. $\hat{H}$ can be written in terms of ladder operators, where $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega, \end{align*}$$ and $\hat{a}$ raises the energy level and $\hat{a}^{\dagger}$ lowers the energy. Thar means there's a state $\ket{E_0}$ for which the energy $E_0$ is the minimum possible energy for the system, and $\hat{a}|E_0\rangle=0$.

Associated Concepts

Coming soon
This is kind of like a finite well, but instead we have the potential increasing in a quadratic function instead of a sudden jump. That is, $$\begin{align*} V(x)=\frac{1}{2}kx^2. \end{align*}$$ Classically, a particle would sit at the very bottom. Quantum mechanically, due to the uncertainty principle, the particle can't just sit at the bottom, having zero energy.

Significance

Coming soon

Derivation

Once again we want to solve the eigenvalue equation $H\psi=E\psi$, where $H$ is the Hamiltonian. The total energy of the system is the potential energy plus the kinetic energy. The kinetic energy, written in the position basis, is $$\begin{align*} -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}, \end{align*}$$ while the potential energy is $$\begin{align*} V(x)=\frac{1}{2}kx^2\psi(x). \end{align*}$$ This is a pain to solve. We use ladder operators. $\hat{H}$ can be written in terms of ladder operators, where $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega, \end{align*}$$ and $\hat{a}$ raises the energy level and $\hat{a}^{\dagger}$ lowers the energy. Thar means there's a state $\ket{E_0}$ for which the energy $E_0$ is the minimum possible energy for the system, and $\hat{a}|E_0\rangle=0$.

Associated Concepts

Coming soon
FullPage
overview
significance
derivation
related experiments
associated concepts