The usage of temperature in quantum mechanics

Question

This is a follow up question for my previous question, because now I am more confused than I was before.

• In a regular quantum mechanics course, temperature plays no role as it does not appear in the formulas. This is discussed e.g. here, and roughly speaking the explanation is that you need a system of many particles to talk about temperature.
• Suppose now we consider a system of many particles, for example bosons. Then one can either do "regular" quantum mechanics, i.e. study the Schrödinger equation, Green function etc or you can do quantum statistical mechanics, so that the system is at positive temperature and the equilibrium state is given by a density matrix.
• The first case is usually referred to as the zero temperature setting, whilst the second one is the positive temperature.

Many things are not clear to me. I wonder if you could help me clarify the following questions.

1. I always thought that the choice of describing a system via "regular" quantum mechanics or quantum statistical mechanics was connected to whether the temperature affected the distribution of energies of the system or not. In other words, if the temperature does not play a role in the description of the system, then quantum effects dominate and you use "regular" quantum mechanics. But this is what they call zero temperature, so I am confused whether I can use regular quantum mechanics to study systems with positive but somehow negligible temperature or not.
2. The relationship between temperature and energy in a quantum many-body system is not straightforward. At absolute zero ($T=0$), one might expect the de Broglie wavelength to be infinite, implying that the average energy and momentum of the particles would be zero and that all particles would be in the ground state. However, this assumption is apparently incorrect on several levels. Firstly, a system at $T=0$ does not necessarily have to be in its ground state, otherwise all many-body quantum mechanics books at $T=0$ would not exist. Secondly, the energy of the ground state is not necessarily zero. However, even in PSE the mention of the zero temperature being synonym of being at the ground state is found e.g. here.

I understand my question overlaps with my previous one, but now I am bringing up some points that have arisen from the other question

Answer 1

Many-body quantum mechanics is also often referred to as Quantum statistical physics, which is a more telling name in the sense that it points out that we are considering systems where one has to take into account both the quantum mechanics and the statistical mechanics/physics.

Most statistical physics texts answer this question at length, so I limit myself to just a few remarks:

• Pure and mixed states in QM A statistical system of quantum nature cannot be considered to be in a pure state, i.e., a state described by a wave function. Instead it is in a mixed state, described by a density matrix, which obeys von Neumann equation (rather than the time-dependent Schrödinger equation for the wave function.) In thermodynamic equilibrium the density matrix is just the quantum Boltzmann distribution. In this respect I recommend Fetter&Walecka's book on quantum many-body systems - it may be rather pedestrian and outdated, but it does a review of necessary statistical physics and provides some statistical derivations before jumping into diagrammatics.
• The eigenstates and energies are still there Note that the particle eigenstates are still obtained from the time-independent Schrödinger equation, even though it is usually happily forgotten when we jump into the second quantization. The ground state formalism deals with only one state and the problem of calculating this state and its energy. It is essentially dealing with solving the time-independent Schrödinger equation, so there is no contradiction here.
• Quantum coherence at low temperatures Perhaps beyond what the OP had in mind is the zero temperature limit when considering dynamics of quantum systems. This usually means neglecting the relaxation processes that are either slow or inactive at small energies (i.e., at low temperatures) energy - like optical phonons in solids. This resulted in the running joke that Zero temperature is higher than the Kondo temperature, referring to the studies where the temperature/relaxation effects are slow enough to consider coherent quantum effects, but still high enough to neglect finer many-body effects, like Kondo effect, supercodnuctivity, etc.

Answer 2

Statistical mechanics can be derived from suitable models in the decoherent limit of quantum theory.

In many quantum experiments, such as single particle interference, the square amplitudes of states don't obey the rules of probability, see Section 2 of this paper for an example:

https://arxiv.org/abs/math/9911150

If information is copied out of a quantum system interference is suppressed: this effect is called decoherence:

https://arxiv.org/abs/0903.5082

There are measures of the extent of decoherence such as quantum discord that can be used to quantify the extent to which you can take advantage of quantum effects in any situation including in a heat engine:

https://arxiv.org/abs/quant-ph/0301127

It is possible to derive statistical mechanics in the decoherent limit where quantum correlations can be neglected:

https://arxiv.org/abs/2104.11223

https://arxiv.org/abs/1806.03532

For the kinds of systems considered in statistical mechanics decoherence gives us a mixed state in which the temperature is a parameter describing how the probabilities of states change with energy. In standard statistical mechanics the different ensembles express different constraints on how a system interacts with its environment, see Chapter 4-6 of "Modern Classical Physics" by Thorne and Blandford. Quantum physics doesn't change this picture except that it explains how probabilities arise rather than just postulating them and explains the limits of the applicability of the rules of probability.