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As the title says, I am thinking about the question that whether a quantum phase transition has latent heat. If so, at 0 temperature, we can drive the system by some parameter from disorder phase to ordered phase, where does the heat goes since it is already 0 temperature.

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heat and temperature aren't the same thing. a first order finite-temperature phase transition also has latent heat; the heat goes into reconfiguring the degrees of freedom, but the temperature doesn't change upon adding more heat until after the phase transition is complete. –  wsc Jan 4 '13 at 16:15
    
Shouldn't this depend on the quantum phase transition? There isn't just one type of these. Is there a particular quantum phase transition you're thinking of? –  Peter Shor Jan 4 '13 at 22:29
    
Some do, some don't. Just like classical phase transitions, quantum ones can be first order (with latent heat) or second order (without) -- or something else. Most of the traditional examples that leap to my mind are second order. –  emarti Jan 12 '13 at 6:00
    
Now, I think changing the external parameter to drive a quantum phase transition is going to add heat into the system. But this is not latent heat. If I input external work to change the parameter to drive one ground state A to the other B. If there is no heat out, it means that B has a higher energy. And using the same argument, A also has a higher energy than B. So here is contradiction. Thus it seems that some of the work using to change the external parameter must become heat, then the system is heated. –  Xiao-Qi Sun Feb 1 '13 at 8:47
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See wikipedia.

It says that a quantum phase transition is second-order, which means there is no latent heat.

There is apparently some controversy about this. From the abstract of this article:

"It is frequently argued that only second order phase transitions at T = 0 deserve to be called quantum phase transitions, while first order quantum phase transitions are a contradiction in terms."

I will try to explain in some more detail. From the Wikipedia article: "Contrary to classical phase transitions, quantum phase transitions can only be accessed by varying a physical parameter—such as magnetic field or pressure—at absolute zero temperature." Let's assume that the parameter we change is pressure (magnetic field works the same way, but it will be convenient to have a specific name for this parameter). The quantum phase transition takes place at temperature T = 0. That is, at any given pressure, the system will be in its lowest energy state. However, it is not required that the energy of the material be the same at all pressures. When we vary the pressure, the energy may change. If this happens, then changing the pressure will either require work or extract work from the system. For example, increasing the pressure may make the ground state of the material more dense; in this case, it will take energy to increase the pressure, as we are doing work to compress the material.

The difference between a first order quantum phase transition and a second order quantum phase transition is whether the slope of the energy/pressure curve changes at the phase transition. For how a first order quantum phase transition works, consider a material with two different states. For $P < P_c$, the first state has lower energy. For $P > P_c$, the second state has lower energy. However, the energy/pressure curve for each individual state is smooth as it passes through the transition. At $P_c$, then, the material will ideally change from one state to the other at $P_c$. I would suspect that for most substances, this will only happen smoothly if you change the pressure to a point slightly beyond $P_c$ and wait a long time at this point.

Latent heat is energy that you must put into a system during the phase transition. For ordinary phase transitions, there is an entropy difference between the two phases, and the added energy is necessary because the entropy of the material changes when you go through the phase transition; this is related to the Second Law of Thermodynamics. At T = 0, there is no entropy in either phase. At T = 0, if two phases can coexist, they have to have the same energy. This means that even first order phase transitions at T = 0 cannot have any latent heat.

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Let us think of two ground states. If I input external work to change the parameter to drive one ground state A to the other B. If there is no heat out, it means that B has a higher energy. And using the same argument, A also has a higher energy than B. So here is contradiction. Thus it seems that some of the work using to change the external parameter must become heat, then the system is heated. –  Xiao-Qi Sun Feb 1 '13 at 8:44
    
But it seems to be not the same as the latent heat. Just there is a heat capacity related to the parameter driving the QPT. –  Xiao-Qi Sun Feb 1 '13 at 8:53
    
If you input external work to change the system from A to B, then changing the system from B to A will perform work. –  Peter Shor Feb 1 '13 at 11:56
    
If changing the system from B to A as you say will perform work, then the system is not stable. If the system is stable, it means that change in either direction cost energy. –  Xiao-Qi Sun Feb 1 '13 at 16:33
    
@Xiao-Qi Sun: I tried to address your comments in my revised answer. It cannot be the case that a change in either direction costs energy, because energy is conserved. –  Peter Shor Feb 3 '13 at 14:22
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