The exponential of an operator I have a problem to get an intuitive idea about these operators:
$\hat{D_x}(x)=e^{-i\frac{x}{\hbar}\hat{p_{x}}}$: the spatial displacement operator, moves the wave function $\psi$ along the x coordinate, $\hat{p_{x}}$ is the momentum operator which generates the displacement.
$\hat{R_x}(\theta)=e^{-i\frac{\theta}{\hbar}\hat{L_{x}}}$: the rotation operator, $\hat{L_{x}}$ is the angular momentum operator who generates the rotation.
I believe that the term $e^{-i}$ translate a rotation on a 2D plan.
My questions are:
In the case of $\hat{D_x}$ operator, how the operator $\hat{p_{x}}$ can be responsible for the displacement, How this work? and how can I visualize the operation of the terms $e^{-i}$ and $\hat{p_{x}}$ at the same time to generate this displacement?
the same questions apply to the case of $\hat{R_x}$.
Correct me If I am wrong somewhere, and thank you in advance.
 A: Short answer: 
Intuitively, for the displacement operator, the exponential accumulates an infinite number of infinitesimal displacements, and this gives rise to an overall macroscopic finite displacement. The same principle holds for the rotation operator, i.e., accumulation of many small rotations. 
Since the momentum operator generates a displacement via exponentiation, the momentum operator is called the generator of displacement. 
Long Answer: 
The exponentiation of an operator can be understood from the first principle using the definition of differentiation $-$
$$\frac{d}{{dx}}\psi \left( x \right) = \lim_{h\to 0} \frac{{\psi \left( {x + h} \right) - \psi \left( x \right)}}{h}.$$
And then we juggle the terms around to find a linearized approximation of $\psi(x)$,
\begin{align}
      \lim_{h\to 0}\psi \left( {x + h} \right) & = \lim_{h\to 0} \left[\psi \left( x \right) + h\frac{d}{{dx}}\psi \left( x \right)\right], \\ \\ 
                & = \lim_{h\to 0} \left[1 + h\frac{d}{{dx}}\right]\psi \left( x \right), \\ \\
                & = {\mathcal{T}}_h\cdot \psi \left( x \right). 
\end{align} 
The last line of the above equation tells us that if we would like to march forward (or backward) by a small quantity $h$, we need to myltiply the function $\psi(x)$ by the operator $$\mathcal{T}_h := \lim_{h\to 0} \left[ 1 + h \frac{d}{dx}\right].$$ 
For small $h$, we can find the value of the function $\psi \left( x +2h \right)$ by applying $\mathcal{T}_h$ twice on the function $\psi \left( x \right)$, 
$$\psi \left( x +2h \right) = \mathcal{T}_h \cdot \mathcal{T}_h \cdot \psi \left( x \right) = \mathcal{T}_h^2 \cdot \psi \left( x \right).$$ 
We can continue further and evaluate the function $\psi$ at $x+Nh$, i.e., $\psi(x+Nh)$, by applying $\mathcal{T}_h$ operator $N$ times on $\psi(x)$:
\begin{align}
\psi(x+Nh) & = \mathcal{T}_h^N \psi(x). \\
           & = \lim_{h\to 0}\left[1+h \frac{d}{dx}\right]^N\psi(x). \tag{1}\label{marchF}
\end{align}
If we define the quantity $Nh$ as $a$, this reduces to the folloing substitution
$$h:=\frac{a}{N},$$
and plugging this into Eq. \eqref{marchF}, we arrive
\begin{align}
\psi(x+a) & = \lim_{N\to \infty}\left[1+ \frac{a}{N} \frac{d}{dx}\right]^N\psi(x) \\ \\ 
          & = \exp\left[a \frac{d}{dx}\right] \cdot \psi(x), \\ \\ 
          & = T_a \cdot \psi(x).
\end{align}
In the second line, we have used the definition of exponential function which is one of the key points towards an understanding of how all these generators come about. 
We can massage translation operator $T_a$ a little further to look like the standard form
\begin{align}
    T_a & := \exp\left[a\frac{d}{dx}\right], \\ \\
        & = \exp\left[-i \frac{a}{\hbar}\cdot \left(i \hbar \frac{d}{dx}\right)\right], \\ \\
        & = \exp\left[-i \frac{a}{\hbar}\cdot \hat{p}_x\right].
\end{align}
A: 
$\hat{D}_x(x)=e^{-i\frac{x}{\hbar}\hat{p}_x}$:
  the spatial displacement operator, moves the wave
  function $\psi$ along the x coordinate, $\hat{p_{x}}$
  is the momentum operator which generates the displacement.

First let me say that using $x$ in the definition
of $\hat{D}_x(x)$ is really confusing, because $x$ is also used
as the spatial coordinate of the wave function $\psi(x,y,z)$.
Therefore I prefer to write it as
$$\hat{D}_x(a)=e^{-i\frac{a}{\hbar}\hat{p}_x} \tag{1}$$
We need to show that this operator is displacing
a wave function $\psi(x,y,z)$ by a distance $a$ in $x$-direction.
Recall the definition of the momentum operator
$$\hat{p}_x=\frac{\hbar}{i}\frac{\partial}{\partial x} \tag{2}$$
Plugging (2) into (1) we get
$$\hat{D}_x(a)=e^{-a\frac{\partial}{\partial x}} \tag{3}$$
Now we use the well-known expansion of the exponential function
$$e^A=1+A+\frac{1}{2}A^2+\frac{1}{3!}A^3+\frac{1}{4!}A^4+... $$
with $A=-a\frac{\partial}{\partial x}$ to rewrite (3)
and get:
$$\hat{D}_x(a)=1
 -a\frac{\partial}{\partial x}
 +\frac{a^2}{2 }\frac{\partial^2}{\partial x^2}
 -\frac{a^3}{3!}\frac{\partial^3}{\partial x^3}
 +\frac{a^4}{4!}\frac{\partial^4}{\partial x^4}
 \pm... \tag{4}$$
This is still an operator equation.
We let the operators on the left and right side
operate on an arbitrary wave function $\psi(x,y,z)$
and get:
$$\begin{align}
\hat{D}_x(a)\psi(x,y,z)
&=\left(1
 -a\frac{\partial}{\partial x}
 +\frac{a^2}{2}\frac{\partial^2}{\partial x^2}
 -\frac{a^3}{3!}\frac{\partial^3}{\partial x^3}
 +\frac{a^4}{4!}\frac{\partial^4}{\partial x^4}
 \pm…
  \right)\psi(x,y,z) \\
&=\psi(x,y,z)
 -a\frac{\partial \psi(x,y,z)}{\partial x}
 +\frac{a^2}{2}\frac{\partial^2\psi(x,y,z)}{\partial x^2}
 -\frac{a^3}{3!}\frac{\partial^3\psi(x,y,z)}{\partial x^3}
 +\frac{a^4}{4!}\frac{\partial^4\psi(x,y,z)}{\partial x^4}
 \pm…
\end{align} \tag{5}$$
Here we recognize the right side as the Taylor series
expansion of $\psi(x-a,y,z)$. So we finally have:
$$\hat{D}_x(a)\psi(x,y,z)=\psi(x-a,y,z) \tag{6}$$
Now we have proven that the operator $\hat{D_x}(a)$ has displaced
the wave function $\psi$ by a distance $a$ in $x$-direction.

Showing that $\hat{R}_x(\theta)=e^{-i\frac{\theta}{\hbar}\hat{L}_x}$
rotates the wave function $\psi$ by an angle $\theta$
around the $x$-axis,
can be done in a similar way by using the definition of
the angular momentum operator $\hat{L}_x$.
A: Expressing the momentum operator as spatial derivative gives us:
$$
\hat D_z=\exp\left(-i\frac{z}{\hbar}(-i\hbar\partial_x)\right)=\exp(z\partial_x)
$$
The result of $\left.\hat D_z\psi\middle|x\right>$ can be expressed as:
$$
\left.\hat D_z\psi\middle|x\right> = \psi(x)+z\psi'(x)+\frac{z^2}2\psi''(x)+\frac{z^3}6\psi'''(x)\ldots
$$
which can be seen as a Taylor expansion of function $\psi(x+z)$ around the point $x$.
$$
\left.\hat D_z\psi\middle|x\right> = \left.\psi\middle|x+z\right>
$$
Rotational shift operator works similarly.
A: I'm going to add some more 'classical' intuition, without the added confusion of quantum operators. You know that $e^{i\theta}$ rotates a complex number by an angle $\theta$:
$$z'=e^{i\theta}z.$$
Let's expand the exponential:
$$e^{i\theta}=\left(\lim_{n\rightarrow\infty}\left(1+\frac 1 n\right)^n\right)^{i\theta}=\lim_{n\rightarrow\infty}\left(1+\frac 1 n\right)^{in\theta}$$
Now define $m=n/(i\theta)$. The exponential becomes
$$e^{i\theta}=\lim_{m\rightarrow\infty}\left(1+\frac{i\theta}m\right)^{m}$$
Inside the brackets you see an 'infinitessimal rotation'. If you multiply by this number you rotate by $\theta/m$ and since $m$ goes to infinity this rotation is infinitessimaly small. But this is not an actual rotation! The length of this number is $\sqrt{1+\theta^2/m^2}$ which is not one. But is very close to one. If you Taylor expand this length you get $\sqrt{1+\theta^2/m^2}\approx1+\frac 1 2\theta^2/m^2+\mathcal O(m^{-4})$. The angle also isn't exactly $\theta/m$, but it is still pretty close if $m$ is large. You can now interpret $e^{i\theta}$ as rotating by this 'almost a rotation' $m$ times. After each 'almost a rotation' the length gets multiplied by $1+\frac 1 2\theta^2/m^2$ and the angle also gets an error so this seems bad.
But now comes the important part: because the length only deviates up to second order the error goes to zero as $m\rightarrow\infty$; when $m$ becomes infinite the length becomes 1 and we have a true rotation.
$$|e^{i\theta}|=\lim_{m\rightarrow\infty}\left(1+\frac 1 2\theta^2/m^2+\mathcal O(m^{-4})\right)^m=1$$
Ok so why did I start this whole rant? This proces can be slightly generalised. Start with some transformation and make it infinitessimal. The infinitessimal version is way easier to work with and once you figured out its basic properties you can re-exponentiate to get the full transformation back. The infinitessimal versions form the Lie algebra of this transformation.
Consider for example an infinitessimal translation:
$$f(x+a)\approx \left(1+a\frac d{dx}\right)f(x)\implies f(x+a)=e^{a\frac d{dx}}f(x)$$
Or the rotation of a 2D vector:
define $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ noticing that $J^2=-I$. Then you have
$$v'\approx(I+J\theta)v\implies v'=\exp(J\theta)v$$
with $\exp$ the matrix exponential.
