Whitefork,
PixyMisa, couple questions:
Does the decay of a radioactive atom require some form of activation energy from somewhere else?
No. Here's an example: Carbon 14. This element has 8 protons and 10 neutrons. Neutrons have a higher rest mass than protons, but the repulsive electrostatic potential energy between protons also has mass. In this case, if one of those neutrons spontaneously decays into a proton, electron, and neutrino, then the total combined mass/energy of all of these particles, plus the extra potential energy, will still add up to less than the current rest mass of the nucleus. There is a principle in QM, anything that can happen eventually will. Since this decay can happen, without violating any conservation laws, it does. The half-life is basically just a function of the strengths of the interactions between the relative particles.
If you can completely isolate a given atom from the outside world (which you can't), than, in theory, would it remain in its current quantum state indefinitely?
I guess it depends on what you mean by "outside world". Take a solitary neutron, for example. A neutron all by itself is unstable, because as mentioned above, its rest mass is higher than the combined rest mass of a proton, electron, and neutrino. It will therefore decay into such a triplet, with a half-life of about 15 minutes.
But how does this decay happen? One way to look at it is as an interaction with the vacuum. This example is too complicated, but you can see the basic principle by looking at simple electron excitation states in atoms.
Imagine a Hydrogen atom in its ground state. A photon with just the right energy necessary to kick it into its first excited state comes along, and it absorbs the photon and jumps into its excited state. What happens then? The excited state is unstable. After a very short period of time, the atom emits a photon of the right energy, and drops back to its ground state. This is called spontaneous emission.
But why does this unstable state decay? To understand that, you need to know about stimulated emission. Let's say you've got an atom in its excited state, and you hit it with a photon corresponding to the energy difference between the excited state and the ground state. In this case, the atom drops to its ground state, and emits a second photon of the same energy. That photon will also, incidentally, be in phase with the first one. This is how lasers work.
So, why does spontaneous decay occur? Because what actually happens is a virtual photon appears from the vacuum with the right energy. This virtual photon causes stimulated emission, and then disappears back into the vacuum.
This is the basis of all "spontaneous" decay in QM. Spontaneous decay is essentially a particle interacting with the constantly fluctuating vacuum. So in principle, an unstable particle will still decay, even if cut off from the rest of the World, as long as it is still in the vacuum. Of course, nobody knows of any way to even meaningfully talk about particles somehow not being "in the vacuum". In fact, many of the properties we think of as intrinsic characteristics of particles, such as mass, appear to simply be manifestations of the interactions between the particle and the vacuum.
hammegk,
Math. Physics. TLOP. You also remain clueless. Have fun with your non-deterministic QM effect computer.
If you really believe that computers are not subject to quantum indeterminacy, then you are the one who is clueless. You also have no excuse for this willful ignorance, since several people here have already pointed out specific concrete examples of the indeterminacy inherent in computer hardware.
Dr. Stupid
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