Time-dependent Schrodinger equation from variational principle In the paper,  "Density-functional theory for time-dependent systems" Physical Review Letters 52 (12): 997 the authors mentioned that the action
$$ A= \int_{t_0}^{t_1} \mathrm dt \langle \Phi(t) | i \hbar\;\partial / \partial t - \hat{H}(t) | \Phi(t)  \rangle \tag{1} $$
provides the solution of time-dependent Schrödinger equation at its stationary point.  Wikipedia  called (1) as the Dirac action without further reference. 
If I do a variation, indeed the stationary point of action (1) gives
$$  i \hbar\;\partial / \partial t | \Phi(t)  \rangle = \hat{H}(t) | \Phi(t)  \rangle $$
However, from path-integral point of view, the least action principle is only a limiting case when $\hbar \rightarrow 0$. In general, there is no least action principle in quantum mechanics.  
My question is, how to reconcile these two aspects? What does vary of action (1) mean?
 A: There definitely is a  least action principle in quantum mechanics, indeed, the path-integral method is based on it. Feynman's doctoral thesis is titled:" the least action principle in quantum mechanics". Please see, e.g., http://cds.cern.ch/record/101498/files/?ln=en
A: If you are interested in an path integral with the action:
$$
\mathcal{S}= \int_{t_0}^{t_1} dt \langle \Phi(t) | i \hbar\partial / \partial t - \hat{H}(t) | \Phi(t)  \rangle \tag{1}
$$
then $\Phi(k, t)=\langle k| \Phi(t)  \rangle$ is now an operator or a mude variable inside the path integral. The bridge between the operator and path integral linguage is:
$$
\langle \alpha|\mathcal{T}\left(...\hat{\Phi}(k,t)...\right)|\beta\rangle=\int \mathcal{D}\phi(k,t)\mathcal{D}\bar{\phi}(k,t)\left(...\phi(k,t)...\right)e^{\frac{i}{\hbar}\mathcal{S}}
$$
And the Action is now written as:
$$
\mathcal{S}= \int_{t_0}^{t_1} dt \int dk\, \bar{\phi}(k,t) ( i \hbar\partial / \partial t - \hat{H}_k(t) ) \phi(k, t)  
$$
with $\hat{H}_k(t)$ being an linear operator acting on functions $\phi(k,t)$.
This is the second quantization. Now we have an complex quantum field theory. Taking the canonical momentum of the fields and using the Dirac rule of quantization:
$$
\left[\hat{\Phi}(k,t),\hat{\Phi}^{\dagger}(k',t')\right]_\pm=\delta(k'-k)\delta(t'-t)
$$
This is the algebra of annihilation and creation operators. Because the theory are linear ($\mathcal{L}$ bilinear in $\Phi$) the number operator 
$$
N=\sum_{k}\Phi(k, t)^{\dagger}\Phi(k, t)
$$
commutes with $\mathcal{H}$, the hamiltonian related to the action $\mathcal{S}$. This imply that $N$ is a constant of motion. The schrodinger equation (equation of motion) is obeyed by a field operator:
$$
i \hbar\frac{\partial}{\partial t} \hat{\Phi}(k,t)= \hat{H}_k(t)\hat{\Phi}(k,t)
$$
and if you find eigenfunctions of $u_n(k,t)$ for the $\hat{H}_k(t)$ we have:
$$
\hat{\Phi}(k,t)=\sum_{n}u_n(k,t)\hat{c}_n
$$ 
where $\hat{c}_n$ is the annihilation operator of a particle. You see that your prediction would be the same.
