# Similarity between unitary operators and ladder operators

I observed a similarity. Is this a co-incidence?:

$$(I+\epsilon P)|x\rangle =|x+\epsilon\rangle$$

And,

$$(X+iP)|n\rangle=A_n|n+1\rangle$$

Here, $$|x\rangle$$ is an eigenfunction of position. $$|n\rangle$$ is an eigenfunction of the Hamiltonian $$X^2+P^2$$

The similarity that I observe is that $$(I+\epsilon P)$$ works as an "infinitesmial ladder operator" for $$|x\rangle$$

Is the motivation for ladder operators rooted in this similarity? If so, how? Can ladder operators be systematically derived by exploiting this similarity? I've only seen ladder operators introduced abruptly as a "mathematical trick".

EDIT There's one more clue. The first equation works only because of the value of the commutator $$[X,P]$$. The second equation also works because of the commutator $$[a,H]$$

Given all these clues, can we systematically motivate ladder operators?

The defining equation for a ladder (or lowering) operator $$\hat{L}$$ with respect to some observable $$\hat{X}$$ is* $$[\hat{X},\hat{L}] = -\Delta \hat{L},$$ where $$\Delta$$ is some difference in eigenvalues of $$\hat{X}$$. It follows that if $$|\xi\rangle$$ is some eigenvector of $$\hat{X}$$ with eigenvalue $$\xi$$, then $$\hat{L}|\xi\rangle$$ is another eigenvector with eigenvalue $$\xi-\Delta$$, since $$\hat{X}\hat{L}|\xi\rangle = [\hat{X},\hat{L}]|\xi\rangle +\hat{L}\hat{X}|\xi\rangle = -\Delta \hat{L}|\xi\rangle + \hat{L}\xi |\xi\rangle = (\xi-\Delta)\hat{L} |\xi\rangle.$$
So far this is completely general, but there are numerous famous and familiar examples in quantum physics. For the simple harmonic oscillator, the relevant ladder operator is $$\hat{L} = \hat{a} = \hat{x}+{\rm i}\hat{p}$$ (in appropriate adimensional units), the observable is the Hamiltonian $$\hat{X} = \hat{H} = \omega\hat{a}^\dagger\hat{a},$$ and we have $$[\hat{H},\hat{a}] = -\omega\hat{a}$$ (taking $$\hbar=1$$). The OP raises another example: the unitary translation operator $$\hat{T}(y) = {\rm e}^{{\rm i}y \hat{p}}$$, whose action is usually written as $$\hat{T}^\dagger(y) \hat{x}\hat{T}(y) = \hat{x}+y,$$ but this is equivalent to a commutation relation $$[\hat{x},\hat{T}(y)] = y \hat{T}(y),$$ which defines $$\hat{T}(y)$$ as a "ladder operator" that "lowers" the observable $$\hat{x}$$ by an amount $$\Delta = -y$$. (The OP describes the example of an infinitesimal transformation $$\hat{T}(\varepsilon)$$ up to order $$O(\varepsilon)$$ but the relation actually holds in general).
*It is not always so easy to gain physical intuition from commutator relations but you can check out this nice question for some insights. Mathematically, the defining commutation relation of a ladder operator encapsulates the fact that $$\hat{L}$$ changes the (eigen)value of $$\hat{X}$$.
• 1. Can you also edit in a physical intuition for the general ladder operator requirement that you wrote at the top? 2. $X$ and $P$ are canonical conjugates. Are $H$ and the Ladder operator also somehow canonical conjugates or something similar? Feb 15 at 11:52
• Isn't $[I+i\epsilon P, X]=X+i\epsilon XP - X -i\epsilon PX= \epsilon i [X,P] =-\epsilon h\neq \epsilon (I+i\epsilon P)$ Feb 15 at 15:16
• @RyderRude 1. I added an edit regarding physical intuition. 2. No, $\hat{X}$ and $\hat{L}$ are not canonically conjugate. 3. $[I-{\rm i}\epsilon \hat{p},\hat{x}] = -\epsilon I = -\epsilon (I -{\rm i}\epsilon \hat{p})$ up to order $\epsilon$, which is the same order that your equations in the OP are valid to. That is, $I -{\rm i}\epsilon \hat{p}$ is only the translation operator at lowest order in $\epsilon$. Feb 16 at 17:34