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In Quantum Mechanics, a translation by a distance $a$ is given by $\hat T(a)$-

$$\hat T(a) \psi(x)=\psi(x-a)$$ which shifts the graph towards the right by a distance $a$.

When we want to find the explicit form of the above transformation, we expand $\psi(x-a)$ in a Taylor series about the point $x$-

$$\psi(x-a)=\sum_{n=0}^\infty \dfrac{(-a)^n}{n!}\dfrac{d^n}{dx^n}\ \psi(x)$$

Using the fact that $\hat p\equiv-i\hbar\dfrac{d}{dx}$, we get-

$$\psi(x-a)=\sum_{n=0}^\infty \dfrac{(-ia/\hbar)^n}{n!} \hat p^n\ \psi(x)$$

We see that RHS is the definition of- $\exp \Big(\dfrac{-ia\hat p}{\hbar}\Big) \psi(x)$

and thus we conclude $\hat T(a) \equiv\exp \Big(\dfrac{-ia\hat p}{\hbar}\Big)$

My doubt is that: In general the radius of convergence of any function in a Taylor series expansion about a point is finite, so the above proof tells us that $|a|$ can not be arbitrarily large. Why then, do we keep using the above generator for any value of $a$. My guess is that for well behaved functions, which $\psi$s obviously are, the radius of convergence is almost always large enough to talk about a sensible taylor series approximation.

Any help is appreciated.

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  • $\begingroup$ See Stone's theorem $\endgroup$ Jul 7, 2021 at 18:52
  • $\begingroup$ Also, even considering only smooth wavefunctions is not enough for their Taylor series to have a non-zero convergence radius; $\endgroup$ Jul 7, 2021 at 18:58

2 Answers 2

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One could argue that you're going about this backwards, if you want an explicit definition of $T_a$. The definition $$\big(T_a\psi\big)(x)= \psi(x-a)$$ is about as explicit a definition of an operator as you could ever hope for. From there, we can define its infinitesimal generator $\hat p$ via $$\hat p\psi = \lim_{a\rightarrow 0} \left[i\hbar\left(\frac{T_a - \mathbb I}{a}\right) \right]\psi = -i\hbar \psi'$$ whose domain consists of all functions $\psi\in L^2(\mathbb R)$ such that the aforementioned limit exists.

In fact, the expression $T_a = e^{-ia\hat p/\hbar}$ is not defined via Taylor series, but rather via the spectral theorem. In broad strokes, the spectral theorem says that there is a unitary operator $U$ such that $U\hat p \psi = \hat \mu U \psi \iff \hat p = U^\dagger \hat \mu U$ where $\mu$ is a multiplication operator. From there, one defines $e^{-ia\hat p/\hbar}:= U^\dagger e^{-ia\hat \mu/\hbar} U$.


In concrete terms for $\hat p$, differentiation is the same as Fourier transforming $\psi(x) \mapsto \tilde \psi(k)$, multiplying by $ik$, and then transforming back:

$$ \frac{d}{dx} \psi(x) = \frac{d}{dx} \frac{1}{\sqrt{2\pi}}\int e^{ikx} \tilde \psi(k) dk = \frac{1}{\sqrt{2\pi}}\int e^{ikx} \bigg(ik \tilde \psi(x)\bigg) dk$$

so $U$ is the Fourier transform operator and $\mu: \tilde \psi(k) \mapsto \hbar k \tilde \psi(k)$ simply multiplies by $\hbar k$. As a result, the definition of $e^{-ia\hat p/\hbar}$ is given by Fourier transforming $\psi(x)\mapsto \tilde \psi(k)$, multiplying by $e^{-ia(\hbar k)/\hbar}$, and then transforming back:

$$\bigg(e^{-ia\hat p/\hbar} \psi\bigg)(x) = \frac{1}{\sqrt{2\pi}} \int e^{ikx} \bigg( e^{-iak}\tilde \psi(k)\bigg) dk = \frac{1}{\sqrt{2\pi}} \int e^{ik(x-a)} \tilde \psi(k) dk = \psi(x-a)$$

for all functions $\psi\in L^2(\mathbb R)$.

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  • $\begingroup$ Why did you define momentum as the translation generator? $\endgroup$ Jul 10, 2021 at 10:23
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    $\begingroup$ @IndischerPhysiker I didn't actually use the word momentum anywhere; $\hat p = -i\hbar \frac{d}{dx}$ is the infinitesimal generator of spatial translations by definition. The momentum operator is defined to be the generator of spatial translations because in Hamiltonian mechanics the momentum observable generates spatial translations (see e.g. here) and want to impose the same algebraic structure on our quantum theory. There are other ways to think about it [...] $\endgroup$
    – J. Murray
    Jul 10, 2021 at 14:17
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    $\begingroup$ [...] but this answer was written to explain how the right-hand side of $T_a := e^{-a \frac{d}{dx}}$ acts on non-analytic functions. $\endgroup$
    – J. Murray
    Jul 10, 2021 at 14:17
  • $\begingroup$ Okay, so does that prove that wave functions have an infinite radius of convergence? $\endgroup$ Jul 10, 2021 at 17:54
  • $\begingroup$ @IndischerPhysiker No. Almost no functions in $L^2(\mathbb R)$ are everywhere analytic. If you define $e^{-a\frac{d}{dx}}$ to be a shorthand for the Taylor series $\sum_{n=0}^\infty \frac{(-a)^n}{n!} \frac{d^n}{dx^n}$, then it's true that you'd need $\psi$ to be everywhere analytic in general; however, this is explicitly not how that expression is defined, as I mentioned in my answer. Defining it via the spectral theorem requires only that $\psi$ have a well-defined Fourier transform, which is true for all functions in $L^2(\mathbb R)$. $\endgroup$
    – J. Murray
    Jul 10, 2021 at 18:13
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You can get the result in another way which doesn't assume an infinite radius of convergence. All we need is that for infinitesimal $\delta a$, \begin{equation} \psi(x - \delta a) = \left [ 1 - \frac{i \delta a}{\hbar} \hat{p} \right ] \psi(x). \end{equation} In other words, the first two terms of the Taylor series. Now, a way to make $\delta a$ arbitrarily small is to set it equal to a finite value $a$ divided by a number $N$ which is going to infinity. \begin{equation} \delta a \equiv a / N \end{equation} Then we use the fact that translations form an abelian group so shifting by $a$ is the same as shifting $N$ times by $\delta a$. As such, \begin{align} \psi(x - a) &= \psi(x - N \delta a) \\ &= \left [ 1 - \frac{i \delta a}{\hbar} \hat{p} \right ]^N \psi(x) \\ &= \left [ 1 - \frac{i a}{N \hbar} \hat{p} \right ]^N \psi(x). \end{align} If we now take the limit as $N \to \infty$, the definition of the exponential function nicely turns this into $\exp \left ( \frac{-ia \hat{p}}{\hbar} \right )$.

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