I do not know if this is a stupid question as I am not an expert in thermodynamics and certainly no expert in QM.

So, we know that the laws of thermodynamics are laws based on statistics. They therefore require more than one body (more than one particle) in order to have meaning. In QM we have the smallest particles currently known and thus we have problems that have to with one or two particles (in exercises for example).

So, referring to the 2nd law, must entropy always increase in QM? If not, how does it translate into always occurring in the macro world? What are the differences in the approach and why do they exist?

Note: Again, I am a newbie in these subjects, so do not give an answer that has definitions that only someone with expertise in those subjects will know about. Also,if you need any clarification, just say it and I will edit the question.

  • $\begingroup$ Hi no offence, could you just edit it slightly, with paragraphs. My guess is, as a newbie to q. m. that the laws are different, but I will be very interested in the answers you get. Great question, imo regards $\endgroup$ – user74893 Mar 22 '15 at 22:38
  • $\begingroup$ I actually do not know how to put paragraphs.I press enter and it creates them but when i am finished,the paragraphs are gone! :D $\endgroup$ – TheQuantumMan Mar 22 '15 at 22:40
  • $\begingroup$ Found it!double space! $\endgroup$ – TheQuantumMan Mar 22 '15 at 22:42
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    $\begingroup$ Yeah, that happens on my comments too, and I end up sending a half finished comment on this tablet when I press the return key. Also the maths symbols are not rendered on my tablet, John Rennie told me it was a bug in the system. $\endgroup$ – user74893 Mar 22 '15 at 22:46
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    $\begingroup$ Since thermodynamics is a subset of classical statistical physics, its laws will not all be preserved when we pass to quantum statistical physics. It would help if you focused on a particular law you have in mind and elaborate on it. $\endgroup$ – ACuriousMind Mar 22 '15 at 22:47

To answer your question first we need to know why do we need quantum mechanics in thermodynamics: In Quantum mechanics you can attribute a wave function(to be precise wave-packet) to a particle. Single Particle Wave Function.

At high temperatures particles can be pictured as billiard balls because their size is much smaller compared to interparticle distance. But as the gas cools down the particles get closer and closer and their wave packets overlap with each other Two Particles Wave Functions

"De Broglie Wavelength" is the wavelength of these particle waves. In the limit "Thermal Wavelength"(a constant multiplied by De Broglie Wavelength) is smaller than the interparticle distance we are in the classical regime but if the Thermal wavelength is comparable or longer than interparticle distance then we need to consider quantum mechanical effects.

Now that we know that at lower temperatures we need to use quantum mechanics instead of classical mechanics we get back to your question. The fundamental difference that leads to different and interesting phenomenon in quantum statistics is that particles in quantum mechanical system are indistinguishable. Meaning if you swap two particles the Hamiltonian of the system doesn't change. which yields the wave function describing the system has to be either symmetric or antisymmetric under interchanging particles.(for a proof look at P.104 of Introduction to Statistical Physics by Kerson Huang)

If the wave function is symmetric we call the system "Bose Gas" and if it's antisymmetric we call it's a "Fermi Gas". In Bose gas two or more particles can occupy same state, While in Fermi gas each state is either empty or occupied by one particle. This leads to so different behaviour between Bose gas and Fermi gas. For example electrons in solids can be thought of as Fermi gas and Photons as Bose gas.

The quantum mechanical behaviours are so different with what you expect from classical thermodynamics but in the high temperature limit one can show that both Fermi and Bose gases approach to classical statistics and you get the same classical thermodynamic.

One last interesting example is that in classical thermodynamics we can find the entropy of gas up to a constant but in quantum statistics we can find the exact constants in entropy formula so in that sense it gives more information than classical thermodynamics.

  • $\begingroup$ @Landos Adam hi is it too simplistic to say that in the macro world, for example, you place all the different coloured billiard balls at one end of the table , low entropy here, then you shake the table around, balls move all over the table, all distinct so easy to see high entropy here. But try the same thing with say, electrons, then no distinction so entropy is harder (impossible?) to quantify? I will put this as my own question later, so yes/no is fine by me for now. Thanks and regards $\endgroup$ – user74893 Mar 22 '15 at 23:55
  • $\begingroup$ @Mohammadreza,good analysis,but in practice what does a bose gas and a fermi gas do?You explained what happens with their wave function but you did not explain what thos gasses do. $\endgroup$ – TheQuantumMan Mar 23 '15 at 0:16
  • $\begingroup$ @irishphysics it is a great example!Couldn't say it better! $\endgroup$ – TheQuantumMan Mar 23 '15 at 0:17
  • $\begingroup$ This is just me, but I interpreted the question as being about whether thermodynamics is applicable at the microscopic level, not whether there are quantum effects in macroscopic thermodynamics. But maybe OP will correct me. $\endgroup$ – Javier Mar 23 '15 at 1:23
  • $\begingroup$ @LandosAdam There are several behaviors for example you can have a Bose-Einstein condensate state of Bose gas by cooling it down close to absolute zero and it'll change to a new state of matter;Liquid Helium-4 becomes superfluid at temperatures lower than ~2K. A superfluid has weird properties like it has no viscosity. In the case of Fermi gas for example you can explain semiconductor behaviours using fermi statistics. Also "Black Body Radiation" can be explained(which is basically a box full of photons) by Bose statistics. So you have to see their physical importance case by case. $\endgroup$ – Zakeri Mar 23 '15 at 6:06

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