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Temperature and quantum physics

  1. Jun 8, 2008 #1
    Hi everyone, first of all let me state I'm no expert in quantum physics. But I'd be glad if any of you would just introduce a little to how temperature is understood from the quantum model stand point. From a classical point of view I believe temperature is a measure of the internal energy of an object, which is given by the kinethic energy of the particles that compose the object, the vibrations of these particles. I might be wrong here, please correct me if so. In quantum physics, the behavior of sumatomic particles is caotic, so how can one understand the vibrations at this level, which ultimately give the temperature of the object?
    I hope I was clear here. I just want to know if my understanding of temperature is correct, and what the concept of temperature means in the world of quantum physics? WHAT is temperature, as scientists understand it today?
     
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  3. Jun 9, 2008 #2

    malawi_glenn

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    average kinetic energy of an ensamble of particles is still valid, temperature is a macroscopic entity, just as colour is etc. Kinetic energy is just not only vibrational energy.
     
  4. Jun 9, 2008 #3
    Temperature is a statistical concept, so in order to talk about the temperature of a quantum system, you need an ensemble of things that behave quantum mechanically. If they are in different energy eigenstates, then you can appeal to the usual methods of statistical mechanics to find out what the temperature is. Another way to go about it is to look at the thermal density matrix. However, in order to have a well defined temperature, like any system, it has to be in equilibrium.
     
  5. Jun 9, 2008 #4
    Thanks for your responses guys, that helped.
     
  6. Jun 10, 2008 #5

    Fredrik

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    It isn't. It's a measure of how much the energy changes when you change the entropy of the system, or equivalently 1 divided by how much the entropy changes when you change the energy. It's usually defined by taking entropy S to be a function of energy E and defining the temperature T(E)=1/S'(E).

    The point of the concept of temperature is that it tells you in which direction energy will flow when you put two systems in thermal contact. If the total energy of the two systems is constant (it will be unless they interact with a third system), energy will flow from the system with the higher temperature to the one with the lower temperature. It's actually not very difficult to show that this must happen if the combined system goes towards a state of higher entropy.
     
  7. Jun 10, 2008 #6

    Fredrik

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    I forgot about how easy it is to motivate this definition...

    Consider two systems that are put in thermal contact with each other. If we keep them isolated from other systems, the total energy is [itex]E=E_1+E_2[/itex] and the total entropy is [itex]S(E)=S(E_1)+S(E_2)[/itex]. Energy will flow from one system to the other if that increases the total entropy of the combined system. Let's say that the total entropy increases when energy flows from system 1 to system 2. That means that

    [tex]0<\frac{d}{dE_2}(S_1(E-E_2)+S_2(E_2))=-S_1'(E_1)+S_2'(E_2)[/tex]​

    [tex]\frac{1}{S_1'(E_1)}>\frac{1}{S_2'(E_2)}[/tex]​

    This shows that the energy flows from the system with the higher value of [itex]1/S_i'(E_i)[/itex] to the one with the lower value. We therefore define the temperature at energy E of a system with entropy S(E) as

    [tex]T(E)=\frac{1}{S'(E)}[/tex]​
     
  8. Jun 10, 2008 #7
    And to add to Fredrik's post the Entropy of a system (S) in a quantum sense is a measure of how many states are available to the system (S=k ln \omega) where \omega is the number of states a system has (1 single coin has two states H or T (heads or tails) a system of two coins has 4 states HH,HT,TH,TT (neglecting the complications due to indistinguishability)). Therefore, entropy is a quantum concept just as much as it is a thermodynamic one.
     
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