Capacitor Stacking Charge Pump
If you took 2 capacitors and held each to the terminals of a battery, each would contain a charge with a voltage across each equal to the battery voltage. Assuming these capacitors don't leak, you could then connect them in series (just like 2 batteries), and the voltage across the pair would be twice battery voltage - and would light a LED if connected to it.

Redraw the circuit without R3, R4, Q2. Replace Q1 with a push-button switch. Over a period of time both capacitors charge fully through the resistors R1, R2, R5. The resistors have high values, so over a short time (when you press the button) will draw so little current as to have no effect on the capacitor charges. Then C1 (2V) is in series with C2 (2V) across the load.

You can minimise the circuit even further : remove R2 (connect the 2V supply directly to the switch), and C1. 
Now when you press the button the battery and the charged C2 are in series across the load, which sees (briefly) 4V.

That is the basis of charge-pump circuits used to produce an output that is higher in voltage than the DC supply.