Why is there no current between two capacitors connected in series? I want to ask a question about the figure below.

Why the current does not flow from $-Q$ term to $+Q$ term or from a to b since there is a voltage?
 A: The charges will remain as they are even after disconnecting the battery(provided you don't connect the other two terminals together). You can imagine the other two plates(say C and D) as producing a uniform E-field and the plates A and B as the opposite faces of a metal block(as they are connected together). The charges are induced on the faces but no current flows as these charges are attracted towards the plates C and D due to Coulomb force of attraction.

capacitor fully charged and then disconnected
Now if you look at the electric field lines, you can see that the electric field lines from C to D(not labelled in the image) kinda cancel the electric field lines present in the region between plate A and plate B. Hence you end up with electric field in the region between plate A and plate C (or D)(depends on your consideration) and plate B and plate D(or C)
and no field between A and B and hence the charges within the block (A and B) won't be tempted to move towards each other.
What happens is something like this:

(This animation is from Wikipedia)
And as we know,
$$E=-\frac{\text dV}{\text dr}=0$$
So the potential difference among the faces $A$ and $B$ inside the block will be zero as the potential is constant.
A: The voltage drop is the same over both capacitors. The voltage level is not. For instance, if there is a total voltage of 2 V across the whole circuit, and there is nothing in the circuit other than the capacitors and the voltage source, then both capacitors will have a voltage drop of 1 V. This does not mean that they are "at" one volt, it means that for each capacitor, the difference in voltage between their two ends is one volt. So if you start at the negative side of the voltage source and consider this to be "zero" volts (remember, all voltage levels are relative to some reference voltage), and go around the circuit, then when you get to the negative side of the first capacitor, you will still be at 0 V (where "zero volts" means "zero volts relative to the negative terminal of the voltage source"). After traversing this capacitor, you will be at 1 V (where "one volt" means "one volt relative to the negative terminal of the voltage source"). When you get to the negative side of the second capacitor, you will still be at 1 V. After you traverse the second capacitor, you will be at 2 V (where "two volts" means "two volts relative to the negative terminal of the voltage source").
Each capacitor has their positive side at one volt higher than their negative side, so for each capacitor, their positive side is at the same voltage relative to the negative side of that capacitor. Their positive sides are not both the same voltage relative to the voltage source.
A: Short answer: Potential is defined depending on the choice of a origin (e.g. ground). The positive plate of a capacitor has potential $Q/C$ greater than negative plate of the same capacitor. We do not know its potential compared to anything else, unless we know how they are connected in a circuit. Here the points a and b are connected by an ideal conducting wire, hence the potential difference between them must be zero, so is the current.
Detailed answer: If you connect two uncharged capacitors in series to a battery, there will be a current in the circuit until equilibrium is reached. As current flows, the capacitors will start charging, and there will be a voltage drop along each capacitor.
In equilibrium, the net voltage drop in the two capacitors will be equal to the voltage in the battery.

Reason: Suppose the instantaneous charge in each capacitor at time $t$ is $Q(t)$ (The charges in the two capacitors must be equal - can be proved from law of conservation of charge), and net electrical resistance in the circuit is $IR = dq/dt R$. Then $V = Q(t)/C_1 + Q(t)/C_2 + IR$. Solve this differential equation and take the limit of very large time. You will find that $V = Q(t = \infty)/C_1 + Q(t = \infty)/C_2$.
Hence, the voltage drop of two capacitors is same as the voltage of the battery, $V = V_{C_1} + V_{C_2}$ (As $V_{C_1} = Q/C_1$). But applying Kirchoff's loop law, $V - V_{C_1} - V_{\textrm{between }a\textrm{ and }b} - V_{C_2} = 0$. Comparing these two equations, $V_{\textrm{between }a\textrm{ and }b} = 0$. As there is no voltage, there will be no current.
However, if initially $Q_1(t=0)/C_1 + Q_2(t=0)/C_2 \neq V$, there will be a net current flow until the voltage drops along the capacitors balance the voltage of the battery, and then there will be no current.
Consider this analogous situation, where there are batteries instead of capacitors. Will there be a current flow, now that a and b have different polarities? There won't be, as the net voltage between a and b is zero.

A: I assume that the question is about the DC case. There is no current between a and b, as there is no voltage difference. The voltage difference between the plates opposite to a and b, say a' and b', is canceled out by the charges on a and b. The plates a and b are connected so form, a single conductor. The charges on a and b are an example of how a conductor cancels out any voltage across it by building up surface charge. 
