How rotation of basis is possible in Heisenberg picture? Shouldn't it be fix? I was trying to understand Heisenberg picture. According to my understanding in this picture operators(position x(t) and momentum p(t)) change with time but basis (x,y,z coordinates) remain fix. I was trying to understand this concept by imaging a particle(p) in x,y plane. x and y are basis and vector that connects the origin of the plane with the particle p is state vector (R). Now according to Heisenberg picture operators change with time it means the the position of the particle P can vary with time but x and y remain fix.
But in Shankar's book author has mentioned that "in Heisenberg picture basis can rotate like state vectors. So in such basis,the vectors appear to be frozen."
So according to the author x,y are moving but the particle P remains at the same position. Which is opposite to the Heisenberg picture. I'm really confused after reading this concept from Shankar's book. Can some one please explain where I am wrong?
 A: You have to keep in mind that particles in QM do not have definite positions or momentum. The state vector can tell us the probability of measuring the particle with a specific position or momentum. This seems like where some of your confusion lies. You talk about the particle moving or being at rest at a particular position, and this view is not supported here. In other words, the "motion" of the state vector has nothing to do with the change in position or momentum of that particle itself, since these things are not defined for the particle in the first place.
Another related confusion I see is how the state vector relates to the position basis vector. In QM, $R\neq a\hat x+b\hat y$. In other words,$R$ is not the position vector of the particle in space. It is an abstract vector that we can choose to represent in different bases that tells us the probability of measuring various observables. Therefore, this also means that a "fixed" state vector as in the Heisenberg picture does not mean the particle is at rest, since "at rest" doesn't exist here.
With that being said, we can still gain some intuition thinking about the state vector as a spatial vector in our classical 3D space. In the Schrodinger picture, our operators and basis vectors (coordinate system) are constant in time (for time independent Hamiltonians). Therefore, it is analagous to us seeing a fixed coordinate system. However, we will see that the state vector ends up "rotating" in this picture due to a phase factor of the state vector $e^{-iEt/\hbar}$ (For example, with a time independent $H$ in the position basis we end up with $\langle x|\Psi\rangle=e^{-iEt/\hbar}\psi(x)$).
But what if we want to "rotate" along with the state vector so that it appears to be stationary to us? Well then to us we will see the coordinate system as rotating around us now. This is analagous to changing (or "rotating") operators, which were fixed before, and hence also having time dependent basis vectors. (This is a very hand-wavy argument, but it is for the sake of the analogy)$^*$. This is the Heisenberg picture.
Therefore, the difference between the two pictures can be analagous to having fixed vs. rotating coordinates. Of course you need to keep in mind that the state vector does not actually "live" in 3D space. We are dealing with more abstract vector spaces here. 
Keep in mind these two pictures are purely mathematical differences. They are just two different ways to work with the formalism of QM. At the end of the day, both pictures give the same physical results. It seems like you are giving the state vector more physical significance than it actually has. The state vector gives us probabilities of measuring values of certain observables. The state vector itself is not physical, and it is not the system or any observable of the system.

$^*$ To be more formal, let's look at the expectation value of an observable. In the Schrodinger picture, the state vectors have a time dependence given by 
$$|\psi(t)\rangle=U(t)|\psi(0)\rangle$$
where $U(t)$ is a unitary transformation (which means $U^\dagger (t)U(t)=I$ )
Therefore, the expectation value of some observable $A_S$ is given by
$$\begin{align} \langle A_S\rangle(t) & = \langle\psi(t)|A_S|\psi(t)\rangle \\ & = \langle\psi(0)|U^\dagger(t)A_SU(t)|\psi(0)\rangle \end{align}$$
So if we treat $U^\dagger(t)A_SU(t)$ as a time dependent operator $A_H(t)$, then we arrive at 
$$\langle A_S\rangle(t)=\langle A_H(t)\rangle$$
where now $A_H$ has the time dependence, and we perform inner products with state vectors $|\psi(0)\rangle=|\psi_0\rangle$ that do not change in time
Therefore, if $A_H$ has a time dependency, it must be that the eigenvectors of this operator also have a time dependency (although you can show, thanks to the nature of the unitary transformation, that the eigenvalues do not change, which is what we want). 
This is the difference between the two pictures. In the Schrodinger picture the operators are fixed (as well as the basis vectors) and our state vectors evolve in time. In the Heisenberg picture the operators evolve in time and the state vectors remain fixed. The two pictures are related by a unitary transformation, which is analogous to active (rotate the vector) vs. passive (rotate the coordinate system) classical transformations of vectors in 3D space.
