📂Quantum Mechanics

What is a Ladder Operator in Quantum Mechanics?

Definition

For an arbitrary operator $N$, let $n$ be its eigenvalue and $\ket{n}$ be the corresponding eigenfunction.

$$N \ket{n} = n \ket{n}$$

An operator $A$ that satisfies the following conditions is called the ladder operator corresponding to $N$.

$$\left[ N, A \right] = cA \tag{1}$$

Here, $c$ is a constant, and $[N, A]$ is the commutator.

Explanation

The reason why $A$ is called the ladder operator is that $A$ can raise or lower the eigenvalue of $\ket{n}$. When $A$ raises the eigenvalue of $\ket{n}$, it is called the raising operator, and when it lowers the eigenvalue of $\ket{n}$, it is called the lowering operator.

Consider the case where $c \gt 0$. By solving $(1)$,

$$NA - AN = cA \implies NA = AN + cA$$

It can be shown that $A\ket{n}$ is an eigenfunction of $N$ with an eigenvalue greater by $c$ than $\ket{n}$.

$$\begin{align*} N(A\ket{n}) &= (AN + cA)\ket{n} \\ &= AN\ket{n} + cA\ket{n} \\ &= nA\ket{n} + cA\ket{n} \\ &= (n + c)A\ket{n} \end{align*}$$

So, it is $A\ket{n} = \ket{n+c}$. In this scenario, the raising operator is commonly denoted as $A_{+}$. If $N$ is a Hermitian operator, then the adjoint operator of $A{+}$ is, conversely, the lowering operator that lowers the eigenvalue of $\ket{n}$ by $c$, and it is denoted as $A_{-} = A^{\ast}$. Since $N$ is a Hermitian operator, $c$ is real,

$$\begin{align*} && NA - AN &= cA \\ \implies && (NA - AN)^{\dagger} &= (cA)^{\dagger} \\ \implies && A^{\dagger}N - NA^{\dagger} &= cA^{\dagger} \\ \implies && NA^{\dagger} - A^{\dagger}N &= -cA^{\dagger} \\ \implies && \left[ N, A^{\dagger} \right] &= -cA^{\dagger} \\ \end{align*}$$

Therefore, it is $A^{\dagger} = A_{-}$ and the lowering operator lowers the eigenvalue of $\ket{n}$ by $c$.

• Ladder operators of angular momentum $L_{\pm}$