I'll approach the problem purely as a physics/engineering problem and see where we end up...
In my garden the energy delivered from the sun to the surface of the earth is about:
1800 Kwh/M2 per year in the Great Basin Desert where days are often cloudless and haze free. It would be less in more cloudy areas. That is equal to 167 KWH per square foot per year. Since there are 860 food Calories per KWH that is equivalent to 140,000 Calories per square foot per year
Then we need to consider what portion of that light can be used.
1% Plant efficiency at converting sunlight to food.
50% Of the year the garden is not covered with snow.
50% Plants only use half of available light due to starting small.
75% Percentage of growing season or day when it's warm enough for photosynthesis.
Which results in 270 Calories per square foot potentially being converted into food.
That number is right in line with the data discussed at:
http://www.waldeneffect.org/blog/Calories_per_acre_for_various_foods/
which if converted to Calories per square foot works out to:
Potatoes 390
Corn 280
Wheat 150
Soybeans 48
In my garden I harvest approximately 130 Calories per square foot when growing dry beans, and approximately 120 for corn. I don't keep good enough records to calculate calories for other crops.
How much food does a person actually need?
Let's say 2000 Calories per Day which equals 730,000 Calories per year which works out to
2700 square feet per person per year. So an acre of ground could feed 16 people.