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.
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.