Consider a two-state system. Let $E_0$ and $E_1>E_0$ be the eigenenergies of this system and $|0\rangle$, $|1\rangle$ the corresponding eigenkets.
Consider a system that is initially in the state $$|\psi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle).$$ Given that the parities of $|0\rangle$, $|1\rangle$ are even and odd respectively, What is the time revolution of the expectation value of $x$, the position operator? (I want to express $x$ in terms of $\langle0|x|1\rangle$)
I failed because I cannot solve the $TISE$ (Time independent Schr\"{o}dinger equation) $$-\frac{\bar{h}^2}{2m}\frac{d^2\langle x|0\rangle}{dx^2}+V(x)\langle x|0\rangle=E_0\langle x|0\rangle$$ and another TISE for $|1\rangle$.
I would suggest you ask questions like this over at physics se, but I'll answer on here anyway.
First off, there are no two-state systems that are represented by wave functions on some space that I know of (though there are ones with two bound states.). Still, your question makes sense if you are considering the lowest two energy levels (or any pair of levels for that matter). To get the time evolution of the expected value, you need the time evolution of the state, which is given by $$ \vert \Psi(t)\rangle = \frac{1}{\sqrt{2}}(e^{-iE_0t}\vert 0\rangle+e^{-iE_1t}\vert 1\rangle)$$ (with $\hbar =1).$ Then you multiply out $$ \langle x\rangle (t) = \langle \Psi(t)\vert x\vert\Psi(t)\rangle = \frac{1}{2}(\langle 0\vert e^{iE_0t}+\langle 1 \vert e^{iE_1t})x(e^{-iE_0t}\vert 0\rangle+e^{-iE_1t}\vert 1\rangle) \\= \frac{1}{2}(\langle 0\vert x\vert 0 \rangle +\langle 1\vert x\vert 1 \rangle + e^{i(E_0-E_1)t}\langle 0\vert x\vert 1\rangle +e^{i(E_1-E_0)t}\langle 1\vert x\vert 0 \rangle).$$
You will need to know the eigenstates of the specific system to get the matrix elements $\langle i\vert x\vert j\rangle$. This is where solving the TISE comes into play. $\langle x \vert 0\rangle$ and $\langle x \vert 1\rangle$ are solutions to the TISE by definition. The TISE can be written abstractly (i.e. not in the $\vert x\rangle$ basis) as $$ H\vert \psi\rangle = E\vert\psi \rangle$$ just says that $\vert\psi \rangle$ is an eigenvector of the Hamiltonian with eigenvalue $E$ and $\vert 0\rangle$ and $\vert 1\rangle$ are eigenvectors of $H$ with eigenvalues $E_0$ and $E_1,$ resp.