$ \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}} $
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Overview
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The raising operator (also known as the creation operator) is defined as $$\begin{align*} \hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}-\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The lowering operator (also known as the annhiliation operator) is defined as $$\begin{align*} \hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The two are Hermitian conjungates of each other. They are very useful for quantum harmonic oscillators.

Notations

The lowering/annihilation operator is denoted with $\hat{a}$ and the raising/creation operator is denoted with $\hat{a}^{\dagger}$

Concepts

We can find the commutator. It's 1, so the two do not commute. $[\hat{a}^{\dagger}\hat{a}]=\hat{a}^{\dagger}\hat{a}-\hat{a}\hat{a}^{\dagger}=1$
For a potential energy of $V(x)=kx^2$ for some $x$ and a kinetic energy of $\frac{\hat{p}^2}{2m}$, where the total energy is constant we can write the Hamiltonian in terms of the ladder operators. Reminder that $[\hat{x}, \hat{p}]=i\hbar$ $$\begin{align*} a^{\dagger}a &=\frac{m\omega}{2\hbar}(\hat{x}-\frac{i\hat{p}}{m\omega})(\hat{x}+\frac{i\hat{p}}{m\omega}) \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2+\frac{i}{m\omega}(\hat{x}\hat{p}-\hat{p}\hat{x})+\frac{\hat{p}^2}{m^2\omega^2}] \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2-\frac{\hbar}{m\omega}+\frac{\hat{p}^2}{m^2\omega^2}], \end{align*}$$ Dividing both sides by $\hbar\omega$ gives $$\begin{align*} \hbar\omega\hat{a}^{\dagger}\hat{a}&=\frac{1}{2}m\omega^2 \hat{x}^2 -\frac{1}{2}\hbar\omega + \frac{\hat{p}^2}{2m} \\ &=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}-\frac{1}{2}\hbar\omega \end{align*}$$ If we let $k=m\omega^2$, then $\hat{H}=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}$ and so $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega \end{align*}$$
The ladder operators raise and lower the energy eigenvalues.
Assume that $|E\rangle$ is an eigenvector of $\hat{H}$ with the associated eigenvalue $E$ (ie. $\hat{H}|E\rangle=E|E\rangle$). Let's use a ladder operator on $|E\rangle$, giving us $\hat{a}|E\rangle$ and measure the total energy of $\hat{a}|E\rangle$. $$\begin{align*} \hat{H}(\hat{a}|E\rangle)&=\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-1+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-\frac{1}{2})\hat{a}|E\rangle \end{align*}$$ Since $\frac{1}{2}\hbar\omega$ is a scalar, we can multiply it by $\hat{a}$ in front or behind. As well, $\hbar\omega\hat{a}\hat{a}^{\dagger}\hat{a}=\hat{a}\hbar\omega\hat{a}^{\dagger}\hat{a}$, which gives $$\begin{align*} &=\hat{a}(\hbar\omega\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\hbar\omega)|E\rangle \\ &=\hat{a}(E-\hbar\omega)|E\rangle \\ &=(E-\hbar\omega)\hat{a}|E\rangle \end{align*}$$ That means if $|E\rangle$ is an eigenvector of $\hat{H}$, then $\hat{a}|E\rangle$ is an eigenvector with the corresponding eigenvalue $E-\hbar\omega$. Thus, applying the operator $\hat{a}$ to a system with the state $|E\rangle$ lowers the energy. That is, we can model the lowering of energy by $\hbar\omega$ by applying $\hat{a}$.
To avoid the energy going to $-\infty$, we define a lowest energy state $E_0$ such that $\hat{a}|E_0\rangle=0$, where $0$ isn't an eigenvalue, it just means we're out of the Hilbert space.
Similarly, we can show that $$\begin{align*} H(\hat{a}^{\dagger}|E_0\rangle) = (E_0+\hbar\omega)\hat{a}^{\dagger}|E_0\rangle \end{align*}$$
What's the ground state? It's the associated eigenvalue for $\hat{H}|E_0\rangle$, where $\hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega$, which gives $$\begin{align*} \hat{H}|E_0\rangle &=(\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega)|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}\hat{a}|E_0\rangle+\frac{1}{2}\hbar\omega|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}0 + \frac{1}{2}\hbar\omega|E_0\rangle, \end{align*}$$ meaning the ground state is $\frac{1}{2}\hbar\omega|E_0\rangle$. Hence the energy eigenvalues are $E_n(n\hbar\omega+\frac{1}{2}\hbar\omega)$ with corresponding eigenvectors $(\hat{a}^{\dagger})^n|E_0\rangle$.
The raising operator (also known as the creation operator) is defined as $$\begin{align*} \hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}-\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The lowering operator (also known as the annhiliation operator) is defined as $$\begin{align*} \hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The two are Hermitian conjungates of each other. They are very useful for quantum harmonic oscillators.

Notations

The lowering/annihilation operator is denoted with $\hat{a}$ and the raising/creation operator is denoted with $\hat{a}^{\dagger}$

Concepts

We can find the commutator. It's 1, so the two do not commute. $[\hat{a}^{\dagger}\hat{a}]=\hat{a}^{\dagger}\hat{a}-\hat{a}\hat{a}^{\dagger}=1$
For a potential energy of $V(x)=kx^2$ for some $x$ and a kinetic energy of $\frac{\hat{p}^2}{2m}$, where the total energy is constant we can write the Hamiltonian in terms of the ladder operators. Reminder that $[\hat{x}, \hat{p}]=i\hbar$ $$\begin{align*} a^{\dagger}a &=\frac{m\omega}{2\hbar}(\hat{x}-\frac{i\hat{p}}{m\omega})(\hat{x}+\frac{i\hat{p}}{m\omega}) \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2+\frac{i}{m\omega}(\hat{x}\hat{p}-\hat{p}\hat{x})+\frac{\hat{p}^2}{m^2\omega^2}] \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2-\frac{\hbar}{m\omega}+\frac{\hat{p}^2}{m^2\omega^2}], \end{align*}$$ Dividing both sides by $\hbar\omega$ gives $$\begin{align*} \hbar\omega\hat{a}^{\dagger}\hat{a}&=\frac{1}{2}m\omega^2 \hat{x}^2 -\frac{1}{2}\hbar\omega + \frac{\hat{p}^2}{2m} \\ &=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}-\frac{1}{2}\hbar\omega \end{align*}$$ If we let $k=m\omega^2$, then $\hat{H}=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}$ and so $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega \end{align*}$$
The ladder operators raise and lower the energy eigenvalues.
Assume that $|E\rangle$ is an eigenvector of $\hat{H}$ with the associated eigenvalue $E$ (ie. $\hat{H}|E\rangle=E|E\rangle$). Let's use a ladder operator on $|E\rangle$, giving us $\hat{a}|E\rangle$ and measure the total energy of $\hat{a}|E\rangle$. $$\begin{align*} \hat{H}(\hat{a}|E\rangle)&=\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-1+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-\frac{1}{2})\hat{a}|E\rangle \end{align*}$$ Since $\frac{1}{2}\hbar\omega$ is a scalar, we can multiply it by $\hat{a}$ in front or behind. As well, $\hbar\omega\hat{a}\hat{a}^{\dagger}\hat{a}=\hat{a}\hbar\omega\hat{a}^{\dagger}\hat{a}$, which gives $$\begin{align*} &=\hat{a}(\hbar\omega\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\hbar\omega)|E\rangle \\ &=\hat{a}(E-\hbar\omega)|E\rangle \\ &=(E-\hbar\omega)\hat{a}|E\rangle \end{align*}$$ That means if $|E\rangle$ is an eigenvector of $\hat{H}$, then $\hat{a}|E\rangle$ is an eigenvector with the corresponding eigenvalue $E-\hbar\omega$. Thus, applying the operator $\hat{a}$ to a system with the state $|E\rangle$ lowers the energy. That is, we can model the lowering of energy by $\hbar\omega$ by applying $\hat{a}$.
To avoid the energy going to $-\infty$, we define a lowest energy state $E_0$ such that $\hat{a}|E_0\rangle=0$, where $0$ isn't an eigenvalue, it just means we're out of the Hilbert space.
Similarly, we can show that $$\begin{align*} H(\hat{a}^{\dagger}|E_0\rangle) = (E_0+\hbar\omega)\hat{a}^{\dagger}|E_0\rangle \end{align*}$$
What's the ground state? It's the associated eigenvalue for $\hat{H}|E_0\rangle$, where $\hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega$, which gives $$\begin{align*} \hat{H}|E_0\rangle &=(\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega)|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}\hat{a}|E_0\rangle+\frac{1}{2}\hbar\omega|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}0 + \frac{1}{2}\hbar\omega|E_0\rangle, \end{align*}$$ meaning the ground state is $\frac{1}{2}\hbar\omega|E_0\rangle$. Hence the energy eigenvalues are $E_n(n\hbar\omega+\frac{1}{2}\hbar\omega)$ with corresponding eigenvectors $(\hat{a}^{\dagger})^n|E_0\rangle$.
FullPage
Overview
Notations
Concepts
FullPage
Overview
Notations
Concepts
The raising operator (also known as the creation operator) is defined as $$\begin{align*} \hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}-\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The lowering operator (also known as the annhiliation operator) is defined as $$\begin{align*} \hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The two are Hermitian conjungates of each other. They are very useful for quantum harmonic oscillators.

Notations

The lowering/annihilation operator is denoted with $\hat{a}$ and the raising/creation operator is denoted with $\hat{a}^{\dagger}$

Concepts

We can find the commutator. It's 1, so the two do not commute. $[\hat{a}^{\dagger}\hat{a}]=\hat{a}^{\dagger}\hat{a}-\hat{a}\hat{a}^{\dagger}=1$
For a potential energy of $V(x)=kx^2$ for some $x$ and a kinetic energy of $\frac{\hat{p}^2}{2m}$, where the total energy is constant we can write the Hamiltonian in terms of the ladder operators. Reminder that $[\hat{x}, \hat{p}]=i\hbar$ $$\begin{align*} a^{\dagger}a &=\frac{m\omega}{2\hbar}(\hat{x}-\frac{i\hat{p}}{m\omega})(\hat{x}+\frac{i\hat{p}}{m\omega}) \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2+\frac{i}{m\omega}(\hat{x}\hat{p}-\hat{p}\hat{x})+\frac{\hat{p}^2}{m^2\omega^2}] \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2-\frac{\hbar}{m\omega}+\frac{\hat{p}^2}{m^2\omega^2}], \end{align*}$$ Dividing both sides by $\hbar\omega$ gives $$\begin{align*} \hbar\omega\hat{a}^{\dagger}\hat{a}&=\frac{1}{2}m\omega^2 \hat{x}^2 -\frac{1}{2}\hbar\omega + \frac{\hat{p}^2}{2m} \\ &=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}-\frac{1}{2}\hbar\omega \end{align*}$$ If we let $k=m\omega^2$, then $\hat{H}=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}$ and so $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega \end{align*}$$
The ladder operators raise and lower the energy eigenvalues.
Assume that $|E\rangle$ is an eigenvector of $\hat{H}$ with the associated eigenvalue $E$ (ie. $\hat{H}|E\rangle=E|E\rangle$). Let's use a ladder operator on $|E\rangle$, giving us $\hat{a}|E\rangle$ and measure the total energy of $\hat{a}|E\rangle$. $$\begin{align*} \hat{H}(\hat{a}|E\rangle)&=\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-1+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-\frac{1}{2})\hat{a}|E\rangle \end{align*}$$ Since $\frac{1}{2}\hbar\omega$ is a scalar, we can multiply it by $\hat{a}$ in front or behind. As well, $\hbar\omega\hat{a}\hat{a}^{\dagger}\hat{a}=\hat{a}\hbar\omega\hat{a}^{\dagger}\hat{a}$, which gives $$\begin{align*} &=\hat{a}(\hbar\omega\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\hbar\omega)|E\rangle \\ &=\hat{a}(E-\hbar\omega)|E\rangle \\ &=(E-\hbar\omega)\hat{a}|E\rangle \end{align*}$$ That means if $|E\rangle$ is an eigenvector of $\hat{H}$, then $\hat{a}|E\rangle$ is an eigenvector with the corresponding eigenvalue $E-\hbar\omega$. Thus, applying the operator $\hat{a}$ to a system with the state $|E\rangle$ lowers the energy. That is, we can model the lowering of energy by $\hbar\omega$ by applying $\hat{a}$.
To avoid the energy going to $-\infty$, we define a lowest energy state $E_0$ such that $\hat{a}|E_0\rangle=0$, where $0$ isn't an eigenvalue, it just means we're out of the Hilbert space.
Similarly, we can show that $$\begin{align*} H(\hat{a}^{\dagger}|E_0\rangle) = (E_0+\hbar\omega)\hat{a}^{\dagger}|E_0\rangle \end{align*}$$
What's the ground state? It's the associated eigenvalue for $\hat{H}|E_0\rangle$, where $\hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega$, which gives $$\begin{align*} \hat{H}|E_0\rangle &=(\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega)|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}\hat{a}|E_0\rangle+\frac{1}{2}\hbar\omega|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}0 + \frac{1}{2}\hbar\omega|E_0\rangle, \end{align*}$$ meaning the ground state is $\frac{1}{2}\hbar\omega|E_0\rangle$. Hence the energy eigenvalues are $E_n(n\hbar\omega+\frac{1}{2}\hbar\omega)$ with corresponding eigenvectors $(\hat{a}^{\dagger})^n|E_0\rangle$.
The raising operator (also known as the creation operator) is defined as $$\begin{align*} \hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}-\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The lowering operator (also known as the annhiliation operator) is defined as $$\begin{align*} \hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(\hat{x}+\frac{i\hat{p}}{m\omega}) \end{align*}$$.
The two are Hermitian conjungates of each other. They are very useful for quantum harmonic oscillators.

Notations

The lowering/annihilation operator is denoted with $\hat{a}$ and the raising/creation operator is denoted with $\hat{a}^{\dagger}$

Concepts

We can find the commutator. It's 1, so the two do not commute. $[\hat{a}^{\dagger}\hat{a}]=\hat{a}^{\dagger}\hat{a}-\hat{a}\hat{a}^{\dagger}=1$
For a potential energy of $V(x)=kx^2$ for some $x$ and a kinetic energy of $\frac{\hat{p}^2}{2m}$, where the total energy is constant we can write the Hamiltonian in terms of the ladder operators. Reminder that $[\hat{x}, \hat{p}]=i\hbar$ $$\begin{align*} a^{\dagger}a &=\frac{m\omega}{2\hbar}(\hat{x}-\frac{i\hat{p}}{m\omega})(\hat{x}+\frac{i\hat{p}}{m\omega}) \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2+\frac{i}{m\omega}(\hat{x}\hat{p}-\hat{p}\hat{x})+\frac{\hat{p}^2}{m^2\omega^2}] \\ &=\frac{m\omega}{2\hbar}[\hat{x}^2-\frac{\hbar}{m\omega}+\frac{\hat{p}^2}{m^2\omega^2}], \end{align*}$$ Dividing both sides by $\hbar\omega$ gives $$\begin{align*} \hbar\omega\hat{a}^{\dagger}\hat{a}&=\frac{1}{2}m\omega^2 \hat{x}^2 -\frac{1}{2}\hbar\omega + \frac{\hat{p}^2}{2m} \\ &=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}-\frac{1}{2}\hbar\omega \end{align*}$$ If we let $k=m\omega^2$, then $\hat{H}=\frac{1}{2}m\omega^2\hat{x}^2+\frac{\hat{p}^2}{2m}$ and so $$\begin{align*} \hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega \end{align*}$$
The ladder operators raise and lower the energy eigenvalues.
Assume that $|E\rangle$ is an eigenvector of $\hat{H}$ with the associated eigenvalue $E$ (ie. $\hat{H}|E\rangle=E|E\rangle$). Let's use a ladder operator on $|E\rangle$, giving us $\hat{a}|E\rangle$ and measure the total energy of $\hat{a}|E\rangle$. $$\begin{align*} \hat{H}(\hat{a}|E\rangle)&=\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-1+\frac{1}{2})\hat{a}|E\rangle \\ &=\hbar\omega(\hat{a}\hat{a}^{\dagger}-\frac{1}{2})\hat{a}|E\rangle \end{align*}$$ Since $\frac{1}{2}\hbar\omega$ is a scalar, we can multiply it by $\hat{a}$ in front or behind. As well, $\hbar\omega\hat{a}\hat{a}^{\dagger}\hat{a}=\hat{a}\hbar\omega\hat{a}^{\dagger}\hat{a}$, which gives $$\begin{align*} &=\hat{a}(\hbar\omega\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\hbar\omega)|E\rangle \\ &=\hat{a}(E-\hbar\omega)|E\rangle \\ &=(E-\hbar\omega)\hat{a}|E\rangle \end{align*}$$ That means if $|E\rangle$ is an eigenvector of $\hat{H}$, then $\hat{a}|E\rangle$ is an eigenvector with the corresponding eigenvalue $E-\hbar\omega$. Thus, applying the operator $\hat{a}$ to a system with the state $|E\rangle$ lowers the energy. That is, we can model the lowering of energy by $\hbar\omega$ by applying $\hat{a}$.
To avoid the energy going to $-\infty$, we define a lowest energy state $E_0$ such that $\hat{a}|E_0\rangle=0$, where $0$ isn't an eigenvalue, it just means we're out of the Hilbert space.
Similarly, we can show that $$\begin{align*} H(\hat{a}^{\dagger}|E_0\rangle) = (E_0+\hbar\omega)\hat{a}^{\dagger}|E_0\rangle \end{align*}$$
What's the ground state? It's the associated eigenvalue for $\hat{H}|E_0\rangle$, where $\hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega$, which gives $$\begin{align*} \hat{H}|E_0\rangle &=(\hbar\omega\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hbar\omega)|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}\hat{a}|E_0\rangle+\frac{1}{2}\hbar\omega|E_0\rangle \\ &=\hbar\omega\hat{a}^{\dagger}0 + \frac{1}{2}\hbar\omega|E_0\rangle, \end{align*}$$ meaning the ground state is $\frac{1}{2}\hbar\omega|E_0\rangle$. Hence the energy eigenvalues are $E_n(n\hbar\omega+\frac{1}{2}\hbar\omega)$ with corresponding eigenvectors $(\hat{a}^{\dagger})^n|E_0\rangle$.
FullPage
Overview
Notations
Concepts