What is the potential difference between $a$ and $b$ in this basic circuit, when the switch is open? My professor says the potential difference between $a$ and $b$ is the difference between the voltages on the top left capacitor and the bottom right capacitor.
My friend and I however both think it's zero. 
Why does the professor think that the voltages on the top right capacitor and the bottom left capacitor don't contribute any potential and should be excluded?
Take C to equal 1 Farad.
Assume the loose ends of the circuit shown connect back to a battery source of 10 V.

 A: The potential difference across the top two capacitors must be the same as the difference across the bottom two. I will number the capacitors $C_{11}$ for top left, $C_{12}$ for top right etc.
If we assume the charge on each capacitor is the same, then the voltage difference must be zero. But if we can assume that each capacitor may have a different charge on it, we are left with the constraint that the voltage across $C_{11}$ plus $C_{12}$ is the same as the voltage across $C_{21}$ plus $C_{22}$.
This means that the potential difference $V_{ab}$ can be written as $-V_{11}+V_{21}$; there is no reason why this should equal the voltage your teacher specified unless the two wires sticking out are connected - so there is no net voltage from left to right. But you stated that they "connect back to a 10 V battery".
If the capacitors were uncharged when the circuit was first connected to the battery, there will be no net voltage (the capacitors are forming the capacitor-equivalent of a Wheatstone bridge). And if they did have a charge, then the net potential difference is definitely not the difference due to $C_{11}$ and $C_{22}$ - for one thing, you would have to account for the 10 V if that was the calculation you were using...
A: First, let's assume the left-most terminal is connected to the positive terminal of the battery and the capacitor voltage reference direction is left-most terminal positive.
Now, consider a KVL loop clockwise through the top 2C capacitor, the switch, the bottom C capacitor and the battery:
$$10 \mathrm V = V_{2C_{top}} + V_{ab} + V_{C_{bot}}$$
So, the statement 

"the potential difference between a and b is the difference between
  the voltages on the top left capacitor and the bottom right
  capacitor"

is false.
Now, an equally valid KVL equation is
$$V_{2C_{top}} + V_{ab} - V_{2C_{bot}} = 0$$
as well as
$$V_{C_{top}} - V_{C_{bot}} - V_{ab} = 0$$
Thus
$$V_{ab}= V_{2C_{bot}} - V_{2C_{top}} = V_{C_{top}} - V_{C_{bot}}$$
A: Without even doing any circuit calculations, you can conclude the voltage between a and b is zero by symmetry. Proof:
Assume there's a voltage between the two points. If you close the switch, a current would flow. If you take the mirror image of the circuit, you'd expect the same current to flow, but in the opposite direction. Except the circuit is left unchanged by taking the mirror image, so the two currents should be equal. Mathematically, this can be written:
$I_{ab} = -I_{ab}$
Evidently, the only way for this to be true is if the current is zero, and hence, the voltage zero. This proof works for any circuit that shows this kind of symmetry.
Maybe your professor drew the diagram incorrectly, and meant to switch the capacitances on one side?
