As was suggested twice already you do not want to power this directly from the line.
You should have an isolated low voltage supply.
I would use a current-sink to provide a regulated current. An easy way
to make a current-sink is to use an op-amp, resistor and FET. I have been working on a prototype that is four channels at 20A per channel. For an LED application the practical limits would be around 2A per channel (due to cooling). A picture of my prototype that is powering 5 1W LEDs is at http://tinyurl.com/yzg9kd7
I have mounted the LEDs on a heatsink. I use thermal grease between the LEDs and
heatsink. I have a schematic of a current-sink (a.k.a. load cell) at
http://tinyurl.com/6cbn6h (scroll down to the "Electronic Load" section.
The most important thing in powering LEDs is to keep the V/A under limits, otherwise they'll easily burn. That's why it's not advisable to use conventional power sources for them. When the voltage rises, an LED resistance will drop, so there needs to be a mechanism to limit the current in such instances that is not present in those sources.
An LED driver is specially made to fit those requirements. For the power you mentioned, you could buy the finished device at eBay, make a search for "power LED driver".
If you want to build your own, start searching a component supplier like Digi-key for a suitable driver; take a look at its datasheet, usually it has sample schematics for applications.
If you want to power the LEDs from the AC outlet, consider that usually those drivers work at DC and lower voltages, so you'll also need to make (or use) an AC-DC converter. For those specs you mentioned, I'd say it's recommended to use a switching power supply, which is another entire topic.
Even then, I'd advise to build an independent supply (power source+driver) for each set of LEDs (in fact, for the LEDs the better is to have one driver for each of them). Consider that you'll have 4x0.7+1+0.5=4.3A - that's a somewhat high current, but OK. Now, 9x4.3=38A - that's a lot. Not only it's much more difficult to make that, also is more dangerous to handle, and more expensive. And if that one is somehow damaged, all is lost - possibly including the LEDs themselves.
Best Answer
As I see, there are two problems:
Your LED is getting 11.5V @ 0.45A. This means it's consuming ~5W on energy. Part of this energy is converted to light, part of it to heat. What part of the consumed energy goes to light depends on the LED, it's usually 15% ... 25%. The rest of the consumed energy is converted to thermal energy. This thermal energy has to go somewhere. If the LED is mounted on a PCB, the PCB draws the heat away. If it's not the LED melts itself into selfdestruction.
High power LEDs have special pads on the case itself designed to transfer heat away. The datasheet of you LED states: thermal resistance: 1.7°C/W. This is the thermal resistance between the PN junction inside the LED and the thermal pad.This means that per every Watt of energy going through the pad , the temperature difference between the junction and the pad is going to be 1.7°C But the thermal resistance between a pad not connected to anything and the air surrounding it is very high, so the LED and pad quickly heat up, not being able to transfer the heat away. For this pad to be able to draw heat away it needs to be thermally connected to something that has a low thermal resistance between it and the surrounding air. The simplest way to thermally connect a pad is to solder it (since metal conducts heat very well) and the "something" that draws the heat away is usually a PCB. FR4 is not a good heat conductor, so usually the thermal pad is soldered to a polygon, and if that's not enough special aluminium-core PCBs are used.
Here are some articles that go into more detail on this subject: a b c. This wikipedia page also has some good links.
A LED needs to be powered by a current source, not a voltage source. Each LED is unique, its precise forward voltage changes from device to device, varies with temperature (it usually decreases as the temperature rices, this is why connecting diodes in parallel to increase forward current is useless), and for one LED it may be 11.5V, for another one (of the same type) it may be 11.3V. Again, a LED should be driven with a current source, NOT a voltage source.