양자 조화 진동자의 사다리 연산자
Definition
Let’s define two operators $a_{+}$ and $a_{-}$ as follows. Then $a_{\pm}$ is the ladder operator of the Hamiltonian in a quantum harmonic oscillator.
$$ \begin{align*} a_{+} &= \dfrac{1}{\sqrt{2\hbar m \omega}}(- \i P + m\omega X) \\ a_{-} &= \dfrac{1}{\sqrt{2\hbar m \omega}}(+ \i P + m\omega X) \end{align*} $$
Here, $P$ is the momentum operator, $X$ is the position operator, $\hbar$ is the Planck constant, $m$ is the mass of the particle, and $\omega$ is the angular frequency.
Explanation
$a_{\pm}$ is an operator that raises the state (eigenvalue) $\ket{\psi}$ of the Hamiltonian operator $H$ by $\pm \hbar \omega$. That is, the following holds. Let the eigenfunction of $H$ be $\ket{\psi}$, and the corresponding eigenvalue be $E$,
$$ Ha_{\pm} \ket{\psi} = (E \pm \hbar \omega)a_{\pm} \ket{\psi} $$
or
$$ [H, a_{\pm}] \ket{\psi} = \pm \hbar \omega a_{\pm} \ket{\psi} $$
It is useful when solving the harmonic oscillator algebraically.
Properties
$$ H = \hbar\omega\left( a_{\pm}a_{\mp} \pm \dfrac{1}{2} \right) \tag{1} $$
The commutator of the two ladder operators is as follows.
$$ \begin{equation} \begin{aligned} [a_{-}, a_{+}] &= 1 \\ [a_{+}, a_{-}] &= -1 \end{aligned}\tag{2} \end{equation} $$
The commutator with the Hamiltonian is as follows. That is, these are the ladder operators of $H$.
$$ \begin{equation} \begin{aligned} [H, a_{+}] = + \hbar \omega a_{+} \\ [H, a_{-}] = - \hbar \omega a_{-} \end{aligned}\tag{3} \end{equation} $$
Proof
$(1)$
The Hamiltonian of the quantum harmonic oscillator is as follows.
$$ H = \dfrac{\hbar^{2}}{2m} \dfrac{d^{2}}{dx^{2}} + \dfrac{1}{2} m^{2}\omega^{2}X^{2} = \dfrac{1}{2m}P^{2} + \dfrac{1}{2} m^{2}\omega^{2}X^{2} $$
The result can be shown through simple calculations.
$$ \begin{align*} \hbar \omega a_{+}a_{-} &= \dfrac{\hbar \omega}{\sqrt{2\hbar m \omega}}(- \i P + m\omega X)\dfrac{\hbar \omega}{\sqrt{2\hbar m \omega}}(+ \i P + m\omega X) \\ &= \dfrac{\hbar \omega}{2\hbar m \omega}(- \i P + m\omega X)(\i P + m\omega X) \\ &= \dfrac{1}{2m}\Big[ (-\i\i)P^{2} + \i m\omega XP - \i m\omega PX + m^{2}\omega^{2}X^{2} \Big]\\ &= \dfrac{1}{2m}\Big[ P^{2} + \i m\omega(XP - PX) + m^{2}\omega^{2}X^{2} \Big]\\ &= \dfrac{1}{2m}\Big[ P^{2} + \i m\omega[X, P] + m^{2}\omega^{2}X^{2} \Big]\\ &= \dfrac{1}{2m}\Big[ P^{2} + \i m\omega \i \hbar + m^{2}\omega^{2}X^{2} \Big]\\ &= \dfrac{1}{2m}\Big[ P^{2} + m^{2}\omega^{2}X^{2} \Big] - \dfrac{\hbar\omega}{2}\\ &= H - \dfrac{\hbar\omega}{2}\\ \end{align*} $$
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$(2)$
From $(1)$, the following is obtained.
$$ \begin{align*} [a_{-}, a_{+}] &= a_{-}a_{+} - a_{+}a_{-} \\ &\overset{\text{by } (1)}{=} \left( \frac{1}{\hbar\omega} H + \frac{1}{2} \right) - \left( \frac{1}{\hbar\omega} H - \frac{1}{2} \right) \\ &= 1 \end{align*} $$
By the property of commutators $[A, B] = - [B, A]$, the following is obtained.
$$ [a_{+}, a_{-}] = - [a_{-}, a_{+}] = -1 $$
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$(3)$
$$ \begin{align*} [A + B, C] &= [A,C] + [B,C] \tag{3-1} \\ [AB,C] &= A[B,C] + [A,C]B \tag{3-2} \\ [A, A] &= 0 \tag{3-3} \\ \end{align*} $$
It must be shown that it is $[H, a_{\pm}] = \pm\hbar\omega a_{\pm}$ by the definition of the ladder operator.
$$ \begin{align*} [H,a_{+}] &= \textstyle [\hbar\omega (a_{+}a_{-}+\frac{1}{2}) ,a_{+}] &\text{by (1)} \\ &= \textstyle [\hbar\omega a_{+}a_{-},a_{+}]+[\frac{1}{2}\hbar\omega, a_{+}] &\text{by (3-1)} \\ &= \textstyle \hbar \omega[a_{+}a_{-},a_{+}] \\ &= \hbar \omega(a_{+}[a_{-},a_{+}] + [a_{+},a_{+}]a_{-}) &\text{by (3-2)} \\ &= \hbar \omega a_{+} &\text{by (2) and (3-3)} \\ \end{align*} $$
And
$$ \begin{align*} [H,a_{-}] &= \textstyle [\hbar\omega (a_{-}a_{+}-\frac{1}{2}) ,a_{-}] &\text{by (1)} \\ &= \textstyle [\hbar\omega a_{-}a_{+},a_{-}]-[\frac{1}{2}\hbar\omega, a_{-}] &\text{by (3-1)} \\ &= \textstyle \hbar \omega[a_{-}a_{+},a_{-}] \\ &= \hbar \omega(a_{-}[a_{+},a_{-}] + [a_{-},a_{-}]a_{+}) &\text{by (3-2)} \\ &= - \hbar \omega a_{-} &\text{by (2) and (3-3)} \\ \end{align*} $$
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