# Schrodinger equation of linear combination of quantum states

We know that the solution for $$i\hbar \frac{\partial}{\partial t}|\psi (t) \rangle = H|\psi (t)\rangle$$ where $$H$$ is time-independent Hamiltonian, is $$|\psi(t)\rangle = e^{-iHt/\hbar}|\psi(t=0)\rangle$$.

Now, suppose $$|\psi\rangle = \frac{1}{\sqrt{N}}(|\psi_1\rangle + |\psi_2\rangle)$$ with some proper normalization factor $$N$$. A simple example would be $$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$, where $$|+\rangle$$ is $$+$$ eigenvector of Pauli-X operator and $$|0\rangle, |1\rangle$$ are $$\pm$$ eigenvectors of Pauli-Z operator. (I used the notation that is commonly used in quantum information area)

If $$|\psi\rangle = \frac{1}{\sqrt{N}}(|\psi_1\rangle + |\psi_2\rangle)$$, then does the solution of Schrodinger equation for $$|\psi(t)\rangle$$ is simply

$$|\psi(t)\rangle = \frac{1}{\sqrt{N}} e^{-iHt/\hbar}(|\psi_1(t=0)\rangle + |\psi_2(t=0)\rangle)$$?

Can we generalize this to time-dependent Hamiltonian case?

Yes, indeed. In fact, for a two-by-two Hamiltonian one could evaluate the operator $$e^{-iHt}$$ exactly and compare the result with the solution of the SE for the same Hamiltonian (when solving it as a system of differential equations).
A 2-by-2 Hamiltonian can be written as a sum of Pauli matrices: $$H=\mathbf{\omega}\cdot\mathbf{\sigma}=\omega_x\sigma_x+\omega_y\sigma_y+\omega_z\sigma_z=\begin{bmatrix}\omega_z&\omega_x-i\omega_y\\\omega_x+i\omega_y&-\omega_z\end{bmatrix}$$ One can than evaluate the exponential operator $$e^{-iHt}=e^{-i\mathbf{\omega}\cdot\mathbf{\sigma}t}= \sum_{n=0}^{+\infty}\frac{(-i\mathbf{\omega}\cdot\mathbf{\sigma}t)^n}{n!}$$ using the properties of Pauli matrices $$\sigma_x\sigma_y=i\sigma_z, \sigma_y\sigma_z=i\sigma_x, \sigma_z\sigma_x=i\sigma_y,$$ reducing it to a 2-by-2 matrix.
The Schrödinger equation is in this case: $$i\partial_t|\psi(t)\rangle = H|\psi(t)\rangle,\\ |\psi(t)\rangle=a(t)|+\rangle + b(t)|-\rangle\\ \Rightarrow \begin{bmatrix}\partial_t a(t)\\ \partial_t b(t)\end{bmatrix}= \begin{bmatrix}\omega_z&\omega_x-i\omega_y\\\omega_x+i\omega_y&-\omega_z\end{bmatrix} \begin{bmatrix} a(t)\\ b(t)\end{bmatrix}$$