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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.

Physiker
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2 Answers2

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

Albatross
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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 $\exp \left ( \frac{-ia \hat{p}}{\hbar} \right )$.

Connor Behan
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