Quantum mechanics stationary states I am taking an intro course to quantum mechanics and I have some trouble understanding what exactly is going on with the particle in stationary states as opposed to their linear combination which is the full solution to the Schrödinger equation. 
In Griffith's book, he says the Schrödinger equation is analogous to Newton's second law so the wave equation is analogous to the trajectory of the particle. So $Ψ(x,t=a)$ is analogous to classically saying where the particle is at x during time a. So far the particle isn't moving, it just might appear in certain places (defined by $|Ψ|^2$ if we were to measure it). 
In chapter 2 we talk about stationary states. It says that they are states of defined total energy $K+V=H$ and $Ψ(x,t)=Ψ(x)e^{-\mathrm{i}Et/h}$. Is the particle moving or what is it doing in these states? Is a stationary state analogous to classical closed system where total energy is conserved and no work is done on the system? If so what does $Ψ(x,t)=\sumΨ(x)e^{-\mathrm{i}Et/h}$ represent, work being done on the particle?   
 A: Solutions that can be written as $$\Psi (x,  t) = \Phi (x) e^{-iEt/\hbar} $$ are called stationary.
To prove this $$|\Psi (x,  t)|^2 = \Psi^* (x,  t)\Psi (x,  t)$$
$$=
(\Phi (x)e^{-iEt/\hbar})^*\Phi (x)e^{-iEt/\hbar}$$
\begin{align}
&=\Phi ^*(x)e^{iEt/\hbar}\Phi (x)e^{-iEt/\hbar}\\
&=\Phi ^*(x)\Phi(x)\\
&=|\Psi (x)|^2\\
\end{align}
The dependance on $t $  has disappeared.  The spatial part of the wave function satisfies the Time Independent Schrodinger Equation.  (T.I.S.E.)

In chapter 2 we talk about stationary states. I says that they are states of defined total energy K+V=Hamiltonian . Is the particle moving or what it doing in these states? Is a stationary state analogous to classical closed system where Total energy is conserved and no work is done on the system? If so what does the wavefunction represent, work being done on the particle?

In the extract from your  post, it is clear that you are thinking in classical terms,  it's actually the word stationary that has you confused.
The solution to the Schroedinger Equation is called stationary because the probability density does depend not on time. $V(x)$ has no time dependence. 
It has nothing to do with work, it's just that the word "stationary" now means that the potential term $V (x) $ in the T.I.S.E. has no time variable. You are getting caught up in classical ideas because it has not been made clear to you what stationary means in quantum mechanics, you are thinking of what it means in classical mechanics. 

But what is the particle doing? In classical I know we find position, path of the particle, velocity, energy ext. So far in quantum ive just found solutions to a deferential equation, but im not sure what the particle is doing in the states. Probabilities change with time in the superposition states, what does that mean for the particle what is happening to it. Is it moving when the probabilities change, what about velocity does that play a role in this while solution thing

But what is the particle doing? 
To be explicit, this question has no meaning when you consider a superposition of states as, if there are say, 5 energy levels the electron "could" be in (or actually is in) all or any of them  at the same time, the only energy level that is of any real importance, is the one we find it in when we measure the system.
Motion and velocity are concepts that don't carry well over to the quantum world. If you kick a soccer ball towards goal, you are abe to easily measure it's velocity and position (by looking at it) at any time.
But an electron say, does not have a definite  trajectory, if you don't measure it, it can be considered to be in an superposition of states. So there is no definite path, only a probability of finding  the electron where you expect it to be. The same reasoning applies to the  "velocity"  of the particle.  
When you solved  a 1 dimensional  S. E, you get a superposition of states, all evolving over time. But later you will need to solve a real 3 d equation, like the electron states around a H nucleus. 
Think energy instead of motion and velocity. In that energy level, motion and velocity are not important, only energy is, because the electron  needs to follow the rules of that energy level. 
You have a superposition of states, all with different energies and a certain probability of being in any particular energy state state. When you measure, you find 1 state with a certain energy and the superposition is gone. When you go away and then come back to the H atom, you get the whole thing starting again. As soon as you measure it, you "freeze" the system but then it goes back in a system of "maybe this energy level",  "maybe that energy level". 
This is what QM is like. 


*

*The energy levels around a H atom are at certain discrete values. 

*When an electron is in that energy level, it's "motion and velocity" are determined by its energy.

*The other  superposition states vanish, so you don't have to worry about them when you do a measurement. 

*As soon as you finish your  measurement, it all goes back to a superposition of states again. 

A picture of the states a given electron could be in, you don't know until you measure it.
