Why is $\textbf{J}$ called angular momentum operator? Can anyone explain why the expectation value of J is angular momentum?
Here is how $J$ is defined: The rotation operator $$ U(\alpha)=\exp(-i {\bf \alpha} \cdot \mathbf{J}) $$ Rotates the system around the direction of $\alpha$, through an angle $|\alpha |$.
Also, How to typeset $\alpha$ is bold face?
A way to get bold greek letters is to use \boldsymbol: $$ \boldsymbol\alpha .$$
We can most easily see it is angular momentum by considering how an infinitesimal rotation operates on a spatial wave function. Consider an infinitesimal rotation of angle $\theta$ about the z axis. Remember a small rotation acts on coordinates like $$ \pmatrix{x\\y} \to \pmatrix{\cos\theta & \sin\theta\\-\sin\theta & \cos\theta}\pmatrix{x\\y} \approx \pmatrix{x+\theta y\\y-\theta x}.$$ So we have via Taylor expansion $$ \psi(x,y,z) \to \psi(x+\theta y,y-\theta x,z) \approx \psi(x,y,z) +\theta y\partial_x\psi -\theta x\partial_y\psi \\= [1-i\theta(x(-i\partial_y)-y(-i\partial_x))]\psi\\=[1-i\theta(xp_y-yp_x)]\psi \\ = (1-i\theta J_z)\psi $$ where we identified $J_z= xp_y-y p_x$ as the z component of angular momentum. This form for the infinitesimal rotations can be iterated to give a finite rotation as in $$ \lim_{n\to \infty} (1-\frac{1}{n}iJ_z\theta)^n\psi = e^{-iJ_z\theta}\psi$$ for rotations of size $\theta/n$ (where now $\theta$ is not restricted to be small.) So we see that the $z$ component of angular momentum generates rotations about the $z$ axis. This can of course be generalized to the other directions and general rotations.
One can also see it by looking at the abstract properties of the rotation, in particular, that for rotation matrices we have $[R_x(\theta),R_y(\theta)] \approx R_z(\theta^2)-I$ for small $\theta,$ where $R_i$ is the rotation about the $i$ axis. Taking $R_i(\theta)\to e^{-iJ_i \theta},$ and expanding for small $\theta,$ one will find that $[J_x,J_y] = iJ_z,$ i.e. the canonical commutation relations for angular momentum.
Note that inside the first method, we also see how we can identify linear momentum as the generator of spatial translations: for we have $$ \psi(x) \to \psi(x-a) = (1-a\partial_x)\psi = (1-ia p_x)\psi.$$