Definite energy states for a single non-relativistic particle with a time dependent potential Do definite energy states exist for a single particle when its potential itself changes with time?
I tried solving it and the equations seem to show that they do not exist. But then i am confused as to what energies will be observed when it is measured. What does the expectation value of energy mean in this case?
As a specific example, consider a 1D infinite potential well with $V(x,t)=t$. What energies are observed and with what probability when the system's energy is measured?
 A: In this case, the eigenstates of the Hamiltonian are not useful to solve the problem, and one has to work with the Schrödinger equation directly:
$$
i\hbar \, \partial_t \psi(x,t)=\frac{-\hbar^2}{2m}\partial_x^2\psi(x,t)+P\,t\,\psi(x,t)
$$
Using a Fourier transform in the variable $x$ you can show that the general solution is
$$
\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \hat{\psi}(k,0)\,\exp\left(\frac{-i\hbar\,k^2\,t}{2m} - \frac{i\,P\,t^2}{2\hbar}+ikx \right)\,dk
$$
where $\hat{\psi}(k,0)\in L^2(\mathbb{R})$ is the Fourier transform of the initial condition $\psi(x,0)$. Note that the kernel of the integral operator
$$
\exp\left(\frac{-i\hbar\,k^2\,t}{2m} - \frac{i\,P\,t^2}{2\hbar}+ikx \right)
$$
is a eigenfuction of the Hamiltonian $\hat{H}(t)$ for all $t$ with eigenvalue
$$
E(t) = \frac{\hbar^2k^2}{2m} + Pt
$$
This is because $[\hat{H}(t),\hat{H}(t')]=0$ for different times $t$ and $t'$. When you make a measurement at time $t$ your eigenfunction will collapse to the instantaneous eigenvector of $\hat{H}(t)$. Suppose now you have an eigenvector $\psi(x,t_0)$ of $\hat{H}(t_0)$ with eigenvalue $E(t_0)$, how will it evolve? Using the formula from before I get
$$
\psi(x,t) = \psi(x,t_0) \, \exp\left(\frac{-i\,E(t_0)\,(t-t_0)}{\hbar} -\frac{i\,P\,(t-t_0)^2}{2\hbar} \right)
$$
So $\psi$ will be an instantaneous eigenvector of $\hat{H}(t)$ for all times.  In this sense the state is "stationary", since you will always find the state of the system in the corresponding eigenvector, but the value of energy I measure will depend on time since the eigenvalues of the Hamiltonian depend on time too.
I'll like to add that for time dependent Hamiltonians with the property $[\hat{H}(t),\hat{H}(t')]=0$ the evolution operator can be written
$$
\hat{U}(t_1,t_0) = \exp\left(-\frac{i}{\hbar} \int_{t_0}^{t_1} \hat{H}(t) \, dt \right)
$$
A: For a particle in a time dependant external potential any Hermitian the Hamiltonian will still have a complete set of eigenstates, and the corresponding eigenvalues will still be the allowed energies of the system. These energies and eigenstates will however generally be time dependant and this means that they are of far less use for solving the TDSE. 
For example in the infinite-square-well-with-rising-bottom that you proposed it is pretty simple to show that, for $0< x < L$
\begin{align} \hat{H}\alpha \sin(kx) & = 
\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + Pt \right]\alpha\sin(kx)\\
& = \left[\frac{\hbar^2}{2m}k^2 + Pt\right]\alpha\sin(kx) \end{align}
So $\alpha\sin(kx)$ is an eigenstate of $\hat{H}$ with energy $E(t) = \frac{\hbar^2}{2m}k^2 +Pt$ and $k$ chosen to satisfy the conditions at the edge of the well. However if we try to plug this into the TDSE, writing $\psi(x) = \alpha \sin(kx)$, we find that \begin{align} \imath\hbar \frac{\partial}{\partial t} e^{-\imath \frac{E(t)}{\hbar}t}\psi(x) & = \left(E+\dot{E}t\right)e^{-\imath \frac{E}{\hbar}t}\psi(x) \\
& = \left[\hat{H}+Pt\right]e^{-\imath \frac{E}{\hbar}t}\psi(x) \end{align} So the states of definite energy are not stationary states of the system, as they must have a more complex time dependence than $e^{-\imath \frac{E}{\hbar}t}$. 
In general this type of problem cannot be solved analytically. There are results for the limiting behavior for very rapid or very slowly changing potentials but normally these problems are solved using numerical techniques or time dependant perturbation theory. 
Also it should be noted that a problem with a time varying potential should not be confused with the Heisenberg picture of QM, which is equivalent to the Schrodinger picture, but with the time dependence moved from the state vectors to the operators and is essentially a change of basis. 
