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Let $T_\epsilon=e^{i \mathbf{\epsilon} P/ \hbar}$ an operator. Show that $T_\epsilon\Psi(\mathbf r)=\Psi(\mathbf r + \mathbf \epsilon)$.

Where $P=-i\hbar \nabla$.

Here's what I've gotten: $$T_\epsilon\Psi(\mathbf r)= e^{i \mathbf{\epsilon} P/ \hbar}\Psi(\mathbf r)=\sum^\infty_{n=0} \frac{(i\epsilon \cdot (-i\hbar \nabla)/\hbar)^n}{n!} \Psi(\mathbf r)=\sum^\infty_{n=0} \frac{(\mathbf \epsilon \cdot \nabla)^n}{n!}\Psi(\mathbf r)= \Psi(\mathbf r) + (\epsilon \cdot \nabla) \Psi(\mathbf r) + \frac{(\epsilon \cdot \nabla)^2 \Psi(\mathbf r)}{2} + \cdots$$

This looks somewhat like a Taylor expansion of $\Psi(\mathbf r)$, but it's different than I've seen before -- I've never seen it in terms of a directional derivative. Can you confirm if this is the Taylor expansion of $\Psi(\mathbf r + \mathbf \epsilon)$? Or if not, what I should be getting when expanding $e^{i \mathbf{\epsilon} P/ \hbar}\Psi(\mathbf r)$? Thanks.

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  • $\begingroup$ you can show that it is true in Fourier space. $\endgroup$ – mike Sep 13 '14 at 19:24
  • $\begingroup$ I have trouble with the series expansion too. Did you check that $P$ is bounded? If it is not, does the series definition of exponential still make sense? (You can still define arbitrary functions of a given operator by defining a function on its eigenpairs.) $\endgroup$ – Argyll Sep 16 '14 at 1:42
  • $\begingroup$ physics.SE cross-post $\endgroup$ – user137731 Sep 16 '14 at 2:07
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In short, yes.

One thing that you need to clarify is, $\epsilon$ is not a vector while the argument of $\Psi$ is. I'll assume that $\epsilon$ is multiplied by some constant vector $v$.

Consider $\Psi(\vec{r}+\epsilon\vec{v})$ as a function of both $\vec{r}$ and $\epsilon$. Name it $\psi$ for clarity.

$$ \frac{\partial\psi}{\partial\epsilon}|_{\epsilon=0}=\nabla\Psi(\vec{r}) $$

So after the expansion,

$$ \begin{gather} T_\epsilon\Psi(\vec{r})=&\sum^\infty_{n=0} \frac{(\mathbf \epsilon \cdot \nabla)^n}{n!}\Psi(\vec{r}) \\ =&\sum^\infty_{n=0} \frac{\epsilon^n \cdot \nabla^n}{n!}\Psi(\vec{r}) \\ =&\sum^\infty_{n=0} \frac{\epsilon^n}{n!}\frac{\partial^n\psi(\vec{r})}{\partial\epsilon^n} \\ =&\psi(\vec{r}+\epsilon\vec{v}) \end{gather} $$ as required.

However, I have problem with the validity of the Taylor expansion. $P$ isn't bounded right? Does the Taylor series of an unbounded operator converge? I don't know the math here.

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