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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?

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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 should be obvious from the above argument that any other continuous unitary symmetry operator is also a ladder/lowering operator with respect to the transformed observable, which accords intuitively with our notion of translations, rotations etc. as a shift of spatial, angular coordinate etc.


*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 $\hat{X}$.

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  • $\begingroup$ 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? $\endgroup$
    – Ryder Rude
    Feb 15 at 11:52
  • $\begingroup$ Isn't $[I+i\epsilon P, X]=X+i\epsilon XP - X -i\epsilon PX= \epsilon i [X,P] =-\epsilon h\neq \epsilon (I+i\epsilon P)$ $\endgroup$
    – Ryder Rude
    Feb 15 at 15:16
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    $\begingroup$ @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 $\epsilon$. $\endgroup$ Feb 16 at 17:34

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