"Enough of that...what mathematical formula are you using to determine the amount of energy required to heat a 2km by 1000 km section of atmosphere to the point that it would dissipate a hurricane?"
http://www.universetoday.com/8361/taking-the-temperature-of-a-hurricanes-eye/Ok from that article, you can see that the temperature difference from the hottest to coolest point is around 21 fahrenheit. Hurricane functions on temperature differences, so heating up the top of the hurricane to eliminate that will wreck the convection currents.
21 fahrenheit=around 10 Celsius, a round number to work with
largest hurricane, igor or ivan or whatever was 1,200 km diameter. i'll assume 1,000 km diameter in these calculations. With a cylinder of 500km radius and 2km height i got approximately 1,500,000 km^3 volume (wolfram alpha). The specific heat of air water vapor is about 1.8j/kg which i rounded to 2. The density of water vapor is 0.6kg/m^3 which meant 9*10^17 grams (rounded to 10^18) and 2*10^18 joules to heat 1 degree. So 2*10^19 joules to heat 10 degrees. I took into account that bombs or any method wouldn't heat the atmosphere evenly, so i multiplied that by 10 to 2*10^20 joules and doubled it because only 30-50% of an atomic bomb goes to thermal energy. so 5*10^20 joules.
What was your question about antimatter. It doesn't really matter. a 3,000 km radius frensel lens and 3 hours of sunlight should do the same thing quite nicely.