To solve the problem, let's start by considering the energy stored in each capacitor before they are connected together.

The energy stored in a capacitor is given by the formula:

$$E = \frac{1}{2} C V^2$$

Where $E$ is the energy, $C$ is the capacitance, and $V$ is the potential.

For the first capacitor charged to the potential $V$, the energy stored is:

$$E_1 = \frac{1}{2} C V^2$$

For the second capacitor charged to the potential $2V$, the energy stored is:

$$E_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \cdot 4V^2 = 2 CV^2$$

The total initial energy stored in the system is the sum of $E_1$ and $E_2$:

$$E_{\text{initial}} = E_1 + E_2 = \frac{1}{2} CV^2 + 2 CV^2 = \frac{5}{2} CV^2$$

When the two capacitors are connected together, their potentials will become equal because they are identical capacitors. Let's denote this final potential as $V_f$. The total charge before and after the connection remains constant because charge is conserved. Therefore, we can write:

$$C \cdot V + C \cdot 2V = 2C \cdot V_f$$

Simplifying this equation gives us the final potential:

$$V + 2V = 2 V_f$$

$$3V = 2 V_f$$

$$V_f = \frac{3}{2} V$$

The final energy stored in the system when the capacitors are connected is now the total energy stored across both capacitors at the final potential $V_f$:

$$E_{\text{final}} = 2 \cdot \frac{1}{2} C V_f^2 = 2 \cdot \frac{1}{2} C \left(\frac{3}{2} V\right)^2 = C \cdot \frac{9}{4} V^2$$

The decrease in energy $ \Delta E $ of the combined system is the initial energy minus the final energy:

$$\Delta E = E_{\text{initial}} - E_{\text{final}}$$

$$\Delta E = \frac{5}{2} CV^2 - \frac{9}{4} CV^2$$

$$\Delta E = \frac{10}{4} CV^2 - \frac{9}{4} CV^2$$

$$\Delta E = \frac{1}{4} CV^2$$

The correct answer is:

Option A: $\frac{1}{4} CV^2$