Why time evolution of state vector is governed by only perturbation part of Hamiltonian?Why we not consider unperturbed part of Hamiltonian Why in the interaction picture time evolution of state vector only considers time time dependent part of Hamiltonian and and while calculating expectation value in interaction picture unperturbed part of Hamiltonian is used? And while representing the state vector in interaction picture ,time propagator only considers time independent part of Hamiltonian. Can some one please explain the time evolution of state vectors and operators in the interaction picture with equation. 
$$
    |\psi_{I}(t)\rangle = U^{\dagger}|\psi_{s}(t)\rangle
$$
where time propagator operator is 
$$
    U(t-t_{0}) = e^{-iH_{0}(t-t_{0})/h}    
$$
In the first equation shouldn't they mention U rather than its complex conjugate? And why they have only used time independent part of Hamiltonian?where $H=H^{0}+H(t)$. 
Same is the case for the expectation value of interaction picture, it also considers unperturbed part. How to deal with perturbed part? 
 A: The state vector in the interaction picture $|\psi_I\rangle$ is defined to be $U_0^\dagger |\psi_S\rangle$, where $|\psi_S\rangle$ is the state vector in the Schrodinger picture, and $U_0=e^{-iH_0(t-t_0)/\hbar}$ with $H=H_0+H_1(t)$ as you say$^*$. If you don't like the complex conjugate in the equation you can instead write it as $|\psi_S\rangle=U_0|\psi_I\rangle$ if you want. But this is where you can start with the interaction picture and derive other useful relationships from there. In other words, this is a relationship between the state vectors within each picture. It is not a full representation of the interaction picture itself.
This also answers "why are we just using $H_0$ and not the full $H$? Once again, this is how we are defining the state vector in this picture in terms of the state vector in the Schrodinger picture. The time dependent part of the Hamiltonia does come into play in the interaction picture though. I will give some of this now.
You have to keep in mind that the Hamiltonian $H$ above is really the Hamiltonian in the Schrodinger picture. If we define $H_{0,I}=U_0^\dagger H_0 U_0=H_0$ (since $U_0$ and $H_0$ commute) and $H_{1,I}=U_0^\dagger H_1 U_0$, then we can get the time evolution of our state vector in the interaction picture.
If we use these definitions and start with the time dependent Schrodinger equation, you can show that
$$i\hbar\frac{\partial}{\partial t}|\psi_I\rangle=H_{1,I}|\psi_I\rangle$$
(If you don't like how we just defined these operators in this way, substitute the state vector into the TDSE first and see that defining the operators in this way ends up being very useful).
This is what should help with your confusion. $H_{1,I}=U_0^\dagger H_1 U_0=e^{iH_0(t-t_0)/\hbar}H_1e^{-iH_0(t-t_0)/\hbar}$ depends on both $H_0$ and $H_1(t)$, so we still have both parts of the original Hamiltonian.
This shows how really any operator transforms from the Schrodinger picture to the interaction picture:
$$A_I=U_0^\dagger A_S U_0$$
But, once again, just because we are not using $H_1$ here does not mean it is not important. This is just how we relate the two pictures (interaction and Schrodinger). This is then why the expectation values do not depend on $H_1$ as you point out. It is just an artifact of how we define the interaction picture to be "relative to" the Schrodinger picture.

$^*$ The Hamiltonian $H=H_0+H_1(t)$ is viewed in the "Schrodinger picture". Now in your comments and earlier questions we discussed that operators in the Schrodinger picture have no time dependence. This was somewhat loose speaking. What we should have said was that there is no time dependence due to unitary transformations. Operators in the Schrodinger picture can still have a time dependence if something is physically changing. And example of this is if we have a particle in a time dependent electric field. The problem is simpler to solve if we can break the Hamiltonian into a sum time-independent and time-dependent, although this does not have to be the case.
As for the usage of $U_0$ versus $U_0^\dagger$, this is purely a definition. If you want to put something "more physical" with this, you could say that $|\psi_S\rangle=U_0|\psi_I\rangle$ means that $|\psi_S\rangle$ is a time evolved state of $|\psi_I\rangle$ if the Hamiltonian was just $H_0$. But this is not the actual system in question, since $H=H_0+H_1(t)$
A: You should get comfortable converting between thinking about the unitary time evolution operator $\hat{U}(t)$ and the Hermitian Hamiltonian $\hat{H}(t)$. The two are related by ($\hbar = 1$)
$$
\frac{d}{dt}\hat{U}(t) = -i\hat{H}(t)\hat{U}(t)
$$
If $\hat{H}(t)$ is time independent we can integrate this equation without worrying about non-commutativity of operators and find
$$
\hat{U}(t) = \exp(-i\hat{H} t)
$$
We can take either $\hat{U}$ or $\hat{H}$ to be fundamental and derive one from the other. I personally think of $\hat{U}$ as being the fundamental thing so I'll talk about things that way.
Suppose we have a system whose Hamiltonian has two parts:
$$
\hat{H}_T = \hat{H}_X + \hat{H}_Y
$$
Suppose $\hat{H}_X$ is a well understood, probably time-independent, Hamiltonian while $H_Y$ is more complicated and possibly time-dependent.
There are unitary operators corresponding to $\hat{H}_X$ and $\hat{H}_Y$ which satisfy
\begin{align}
\frac{d}{dt}\hat{X} &= -i\hat{H}_X \hat{X}\\
\frac{d}{dt}\hat{Y} &= -i\hat{H}_Y \hat{Y}\\
\end{align}
These are the corresponding time evolution operators for $\hat{H}_X$ and $\hat{H}_Y$. The question is how does $\hat{U}$ (which gives rise to the total Hamiltonian $\hat{H}_T$) relate to $\hat{X}$ and $\hat{Y}$?
To investigate this we see what happens if we write $\hat{U}$ as a product of two unitary operators $\hat{A}$ and $\hat{B}$. We can also define Hamiltonian $\hat{H}_A$ and $\hat{H}_B$ which are related to the time derivatives of $\hat{A}$ and $\hat{B}$.
\begin{align}
\hat{U} &= \hat{A}\hat{B}\\
\frac{d}{dt}\hat{U} &= \left(\frac{d}{dt}\hat{A}\right)\hat{B} + \hat{A}\frac{d}{dt}\hat{B}\\
&=-i\left(\hat{H}_A\hat{A}\hat{B} + \hat{A}\hat{H}_B\hat{B} \right)\\
&= -i\left(\hat{H}_A + \hat{A}\hat{H}_B\hat{A}^{\dagger}\right)\hat{U}\\
&= -i\left(\hat{H}_X + \hat{H}_Y \right)\hat{U}
\end{align}
Note that above $\hat{A}$ and $\hat{B}$ were arbitrary unitary operators so we can choose them however we like. We see that if we choose $\hat{A} = \hat{X}$ then we have 
\begin{align}
\hat{H}_A &= \hat{H}_X\\
\hat{H}_B &= \hat{X}^{\dagger}\hat{H}_Y\hat{X} = \hat{H}_{Y,I}
\end{align}
We need a new unitary operator $\tilde{Y}$ which gives rise to Hamiltonian $\hat{H}_{Y,I}$
$$
\frac{d}{dt}\tilde{Y} = -i\hat{H}_{Y,I}\tilde{Y}
$$
Of course $\tilde{Y}=\hat{B}$
This means that if $\hat{H}_T = \hat{H}_X+\hat{H}_Y$ then we know that $\hat{U}$ can be written
$$
\hat{U} = \hat{X}\tilde{Y}
$$
Where $\hat{X}$ and $\tilde{Y}$ give rise to the Hamiltonians indicated above.
Now we require a few conditions to make the working in the interaction picture useful.


*

*$\hat{X}$ represents well understood "simple" time evolution which is previously solved.

*$\hat{H}_Y$ is in some sense complicated and not yet solved but the combination $\hat{H}_{Y,I} = \hat{X}^{\dagger}\hat{H}_Y\hat{X}$ is somehow more simple and solvable. For example, $\hat{H}_Y$ may be time dependent while, by luck and cleverness, $\hat{H}_{Y,I}$ is time-independent.


Suppose now that we are working in some complete orthogonal "calculation basis" given by the kets $\{|q_i\rangle\}$. To say that $\hat{X}$ is "solved" means that we know how to act it on any of these kets and get a time-dependent resultant ket.  This could be found by solving the Schrodinger equation with the Hamiltonian $\hat{H}_X$.
What the interaction picture says is that, since $\hat{U} = \hat{X}\tilde{Y}$, if we can figure out the action of $\tilde{Y}$ on a ket in the calculation basis, then we can figure out the action of $\hat{U}$ on a ket in the calculation basis.
Suppose
$$
\hat{X} |q_i\rangle = \sum_{j} c^X_{ij}(t)|q_j\rangle
$$
with $c^X_{ij}(t)$ known time-dependent coefficients capturing the time-evolution under $\hat{X}$ or equivalently under $\hat{H}_X$.
Suppse we also solve the Schrodinger equation with $\hat{H}_{Y,I}$. then we also know
$$
\tilde{Y} |q_i\rangle = \sum_j c^Y_{ij}(t)|q_j\rangle
$$
We "know" tihs formula in the sense that the time-dependent coefficients $c^Y_{ij}(t)$ are known.
Then we can put these two solution together to find
$$
\hat{U}|q_i\rangle = \hat{X}\tilde{Y}|q_i\rangle = \sum_j \hat{X} c^Y_{ij}(t)|q_j\rangle = \sum_{jk} C^X_{kj}(t)C^Y_{ij}(t)|q_k\rangle
$$
A few translations to make the abstract stuff I'm talking about here commensurate with what you'll usually see. In comparison with Aaron Stevens' answer
\begin{align}
\hat{H}_X &= H_0\\
\hat{H}_Y &= H_1\\
\hat{H}_{Y,I}&=H_{1,I}\\
|\psi_I\rangle &= \hat{X}^{\dagger}|\psi_S\rangle\\
\hat{X} &= U_0\\
\end{align}
One final note: Often when the interaction picture is used authors simply ignore $\hat{X}$ and act as if it doesn't exist and continue on with the problem. This is done for two reasons. 1) the dynamics under $\hat{X}$ are boring and well understood so authors assume that if the reader is worried about those dynamics they can carry out the last step of calculation themselves to return into the non-interaction picture. 2) One gets a lot of intuition about the solution to problems by looking at their solutions in the interaction picture alone. All of this ends up being a bit confusing because it seems like authors are ignoring something at least somewhat important.
Edit:
Another point I'd like to add. Suppose the initial state is $|\psi_0\rangle$. Note that this is a ket at a fixed moment in time. It will never change. We can define the Schrodinger picture ket as
$$
|\psi_S\rangle = \hat{U}|\psi_0\rangle = \hat{X}\tilde{Y}|\psi_0\rangle
$$
We can see by the above manipulations that
\begin{align}
|\psi_I\rangle &= \hat{X}^{\dagger}|\psi_S\rangle = \hat{X}^{\dagger}\hat{U}|\psi_0\rangle = \hat{X}^{\dagger}\hat{X}\tilde{Y}|\psi_0\rangle\\
&= \tilde{Y}|\psi_0\rangle
\end{align}
This explains the sense in which "moving into the interaction picture" does the job of "removing or rotating out the time evolution under $\hat{X}$" from the problem.
