Time evolution unclear Let $$H := \sigma_z \otimes \sum_{k=1}^{n}\mathrm{id} \otimes ...\otimes g_k \sigma_z \otimes...\otimes \mathrm{id}$$ be the Hamiltonian ($\sigma_z$ is the Pauli-matrix of couse)
and $$|\psi_0 \rangle:=(a|\!\uparrow\rangle+b |\!\downarrow\rangle ) \otimes_{k=1}^{n} (\alpha_k |\!\uparrow \rangle + \beta_k |\!\downarrow\rangle).$$
Then my script says that the time-evolution of this state is
\begin{align}
|\psi(t)\rangle
&=a|\!\uparrow\rangle \otimes_{k=1}^{n} (\alpha_k e^{i g_kt}|\!\uparrow \rangle + e^{-i g_k t} \beta_k |\!\downarrow\rangle)
\\ &\quad 
+b |\!\downarrow\rangle\otimes_{k=1}^{n} (\alpha_k e^{-i g_kt}|\!\uparrow \rangle + e^{i g_k t} \beta_k |\!\downarrow\rangle).
\end{align}
Does anybody have an explanation for this? 
My concern about this equation arises from the fact that for $n=0$, we would have $|\psi(t)\rangle= |\psi(0)\rangle$ which should not be case, but rather $|\psi(t)\rangle= e^{-it}a|\!\uparrow\rangle+ e^{it} b|\!\downarrow\rangle.$
What do you think?
If anything is unclear, please let me know.
 A: You seem to be making the assumption $e^{i(H_1 \otimes H_2)t} = e^{iH_1t}\otimes e^{iH_2t}.$ This is not true in general. Write down the spectral decompositions of the operators and then exponentiate them carefully. To be more specific $$e^{i (\sigma _z \otimes g\sigma _z) t} \neq e^{i \sigma _z t}\otimes e^{i g\sigma _zt}.$$ You can write  the matrices explicitly and check it yourself. 
A: From a simpler perspective, take a look at your hamiltonian, 
$$
H := \sigma_z \otimes \sum_{k=1}^{n}\mathrm{id} \otimes ...\otimes g_k \sigma_z \otimes...\otimes \mathrm{id},
$$
when $n=0$. Here the sum is over a null set, so the convention is that the empty sum is zero, i.e. $H=0$. Under that hamiltonian, the time evolution
$$|\psi(t)⟩\equiv |\psi(0)⟩$$
is perfectly reasonable.
In general, finding solutions of the Schrödinger equation - like all other ODEs - can be complicated, but verifying if a given function is a solution is in principle very simple. It is relatively simple (though it's a pain to typeset in LaTeX) to plug the solution you've been given into the Schrödinger equation and see that it's a solution.
