I am reading about the total, orbital and spin angular momentum, and I am not clear as to what these generators actually do after exponentiating.
Could you give me a physical picture of what happens to the quantum set, after being acted upon by the operator obtained after exponentiating these?
For example does the total angular momentum rotate the ket in a circle, about the normal? What about the other two?
Did you try the wikipedia article about angular momentum operators?
The arrows schematically represent the internal state of the particle (its spin state). The blue and red dots represent two particles at different locations. Or, if you prefer, it could be two parts of the wavefunction of a single particle, with blue meaning negative phase and red meaning positive phase.
When you exponentiate Jz (the total angular momentum operator), you get real-world rotation about the z axis -- part A in the figure. The whole system and everything in it is rotated.
When you exponentiate Lz (the orbital angular momentum operator), you get "spatial-only" rotation about the z axis -- part B in the figure. The positions of particles get rotated but their spin states stay exactly the same.
When you exponentiate Sz (the spin operator), you get "internal-only" rotation about the z axis -- part C in the figure. The internal state of the particle is rotated, but the particle itself stays in the same place.
By the way, you should avoid phrases like "ket changes direction" and "rotate the ket". You're thinking about rotation in real three-dimensional space, but the ket is not an object in real three-dimensional space. The phrase "rotate the ket" sort of works for a spin-1/2 particle at a single point, but anyway using that phrase is a very bad habit.
Angular momenta are sets of three operators $\{\hat{L}_{i}\}$ such that
$$[\hat{L}_i,\hat{L}_j]=i\hbar\epsilon_{ijk}\hat{L}_{k} $$
where $\epsilon_{ijk}$ is the Levi-Civita symbol. Orbital angular momentum and spin are angular momenta because they satisfy that commutation relation. (That commutator can be derived if you define $\vec{L}=\hat{\vec{r}}\times\hat{\vec{p}}$, a formal vector product between vector operators)
As in classical physics, angular momentum is useful when you have spherical symmetry. Given the case, instead of using Cartesian coordinates $x,y,z$ you may want to use $r,\theta,\phi$ to represent the wave function $\Psi(\vec{r})$.
The eigenstates of these operators $|l,m\rangle$ can be used to write your wave function as
$$\Psi(\vec{r})=R(r)\varphi(\theta,\phi) $$
with
$$ \varphi(\theta,\phi)=\displaystyle\sum_{l,m}c_{lm}|l,m\rangle$$
You can write $|l,m\rangle$ as the spherical harmonics.
Angular momentum is related to rotations, they are the infinitesimal generators of rotations. What does that mean? It means that you can write a rotation of angle $\alpha$ over an axis $\vec{n}$ exponential of the angular momentum about that axis $\hat{L}_{n}$
$$\hat{R}(\vec{n},\alpha)=\exp(-i\frac{\alpha}{\hbar}\vec{n}\cdot\vec{L}) $$
where the dot product is understood as $n_{1}\hat{L}_{1}+n_{2}\hat{L}_{2}+n_{3}\hat{L}_{3}$. As you said the rotation operator $\hat{R}(\vec{n},\alpha)$ does rotate the quantum state represented by $\Psi(\vec{r})$.
Spin case is a little more tricky because the introduction of spin (e.g. $1/2$) in the formalism requires the Hilbert space of states of the system, say $\mathcal{H}$, as well as a two dimensional complex vector space $\mathbb{C}^2$ to the variables of spin. And this lead us to the introduction of quantities called spinors, usually represented by column vectors. The idea of the spin is the same as discussed above, the spin-space rotation operator can be written as exponentials of the spin operators and they do rotate the spinor of the state.
An arbitrary rotation operator acting upon an arbitrary state vector may be a little hard to write, but you can expand the state as a series of the angular momentum operator eigenstates and the spin operator eigenstates to simplify the calculations. The answer of course depends on the particular rotation and state, but as I said, it is always a rotation (in $\mathcal{H}$, in $\mathbb{C}^2$ or both).
Reference: http://students.washington.edu/tkarin/rotations.pdf it a crash course but it may be helpful
If you rotate your quantum system (together with any external fields acting on it) with respect to your coordinate axes (along which you make measurements), nothing changes internally in the system, but the measurements along the coordinates axes will change because the system is at different angles to the measuring axes/devices.
Therefore, the quantum system must be described by a new wave function after the rotation.
For example, the spatial part of the new wave function is obtained by simply rotating the spatial part of the old function together with the system. This is effected by the exponent of the orbital angular momentum operator L, acting on the old wave function. You can visualize this as rotating a probabilistic cloud (the spatial part of the wave function).
The spin part of the new wave function also has to change to reflect the fact that your measurements on it will be different due to the different angles between the system and the measuring coordinate axes. For example, the expectation values of Sx, Sy, Sz will change as if you rotated the old vector with such components, together with the system. The exponent of the spin angular momentum operator S effects that change by acting on the spin part of the old wave function in the spin space. I found a detailed illustration of that here: Rotation Operators in Spin Space, Prof. Richard Fitzpatrick.
The exponent of the total angular momentum operator J = L + S, can be written as a product of the exponent of L and the exponent of S, since L and S commute (they act on different parts of the wave function). The result is that the exponent of J rotates the entire wave function - L rotates the spatial part, S rotates the spin part.
