Translation Operator and Position Basis

Question

In Modern Quantum Mechanics by Sakurai, at page 46 while deriving commutator of translator operator with position operator, he uses $$\left| x+dx\right\rangle \simeq \left| x \right\rangle.$$ But for every $\epsilon > 0$ $$\langle x+ \epsilon \left| x \right\rangle = 0.$$ Therefore this limiting process $$\lim_{\epsilon \rightarrow 0} \left| x+ \epsilon \right\rangle = \left| x \right\rangle$$ does not make sense for me. I couldn't derive commutator relation without using these.

Answer 1

Here's the most logical way to proceed if you ask me. Given any $a\in\mathbb R$, we define the translation operator $T_a$ by its action on position basis vectors $$ T_a|x\rangle = |x + a\rangle $$ One can prove the following properties:

1. $T_a$ is unitary for each $a\in\mathbb R$.

2. $T_aT_b = T_{a+b}$ for all $a,b\in\mathbb R$.

It follows (by Stone's theorem up to some mathematical details), that there exists a hermitian operator $K$ for which $$ T_a = e^{-iaK} $$ The operator $K$ whose existence is guaranteed in this way is called the infinitesimal generator of translations. Next, we want to show that $K$ and $X$ have certain commutation relations. To do so, we note the following fact (see here) $$ T_aXT_{-a} = e^{-iaK}Xe^{+iaK} = X - ia[K, X] + \mathcal O(a^2) $$ Now acting with both sides on a position eigenvector $|x\rangle$ gives $$ (x-a)|x\rangle = (x-ia[K,X])|x\rangle + \mathcal O(a^2) $$ and equating terms of the same order in $a$ gives $$ [X,K] = iI $$ as desired.

Answer 2

The derivation by Sakurai is by no means mathematically rigorous, so you should expect something like your argument about the scalar product. Indeed, we have everything more or less fine until $$ [x,\mathcal{T}(\epsilon)]|z\rangle=\epsilon|z+\epsilon\rangle $$ where we want to replace $|z+\epsilon\rangle$ by $|z\rangle$ and claim that it is ok in the first order in $\epsilon$. As soon as position eigenstates are non-normalizable, there is no measure of 'smallness' to use in our reasoning about orders. However, what makes sense is to deduce $[x,\mathcal{T}(\epsilon)]=\epsilon\mathcal{T}(\epsilon)$, which is true for any finite $\epsilon$. Here the reason why everything works nice is that $\mathcal{T}$ is a good bounded(=continuous) operator which is defined on the whole Hilbert space of states, and is easily understood even on the generalized vectors like $|x\rangle$. In fact, if you work in coordinate representation, you can deduce this commutator working only with normalizable wavefunctions, on which the action of $x$ is defined (they remain normalizable after this action), giving completely rigorous mathematical sense to your calculation.

What is different when you try to deal with Sakurai's $K$ (which you are trying to do every time when talking about infinitesimal translations) rigorously, is that it is a bad (unbounded, discontinuous) operator. Indeed, in a sense, $$ K=i\left.\frac{d}{d\epsilon}\mathcal{T}(\epsilon)\right|_{\epsilon=0}. $$ But the only way to give sense to this formula is to define the action of $K$ on states: $$ K|\psi\rangle=i\lim_{\epsilon\to0}\frac{\mathcal{T}(\epsilon)|\psi\rangle-|\psi\rangle}{\epsilon} $$ But this limit exists only for certain good states, which we say are in the domain of $K$. In fact, if you look at $K$ in the coordinate rep, it is just $-i\frac{d}{dx}$, which is defined on the (everywhere dense) subspace of differentiable functions of the space $L_2$ of square-integrable functions. When you deal with $K$ rigorously, you have to restrict yourself to the domain of $K$ (for example, if you consider joshphysics answer, where every formula with $K$ is restricted to the domain, it is almost a rigorous proof).

However, due to some reason, which is surely related to the fact that the domain $D(K)$ of $K$ is everywhere dense -- any state can be approximated by a state from $D(K)$ to any desired accuracy, a careless treatment like that of Sakurai works.