For an ensemble of quantum systems, the temperature can be defined easily. But for a single quantum system, we know that the system can be in a quantum state with a large energy fluctuation. Then, if the temperature T of the quantum system can be defined by
KB T \approx Variance(E) ?
Is that right? Or if there is a more suitable definition of temperature for a quantum system in a specific quantum state?
Thank you for joining the discussion!
Temperature is not a concept of Quantum Mechanics itself.
You have to look into the statistical mechanics books, specifically
Quantum Statistical Mechanics; Fermi statistics, Bose statistics , these distributions contain temperature.
Well, zero temperature -> ground energy E_0. Therefore, E ~ T. The thermal energy of i.e. an electron can be looked up easily: https://en.wikipedia.org/wiki/Electron_temperature
The problem is that QM does not necessarily follow statics. Construct a finite well, and play around with the S.E. Maybe it helps you finding the solution for which I don't understand the question yet.
Formally, a quantum thermal state (i.e. a density matrix) is given by a matrix exponential of a Hamiltonian \rho(\beta, H)=Z^-1 exp(-\beta H) where Z is the partition function and \beta^-1=k_B T defines the temperature. Given such a state you can get the temperature by looking at it's mean energy (not fluctuations which is connected to the specific heat) and taking an appropriate derivative of the partition function.
Note, that mathematically any mixed state that has full rank can be written like this (simply take the matrix log of the density matrix) and for a pure state one would need to take a zero temperature limit \beta->\infinity.
It is another question, whether a given physical system is well-described by a thermal state of the Hamiltonian that is coincidentally also the generator of the dynamics of that given system. If it is not, then you have a non-thermal or a non-equilibrium state and clearly the temperature need not always be fixed e.g. due to heat transport. See the review in the link for a modern viewpoint on how systems reach thermal equilibrium.
Article Quantum many-body systems out of equilibrium
If your question is about characterising *pure* quantum states, and their relation to thermal equilibrium, this is a rather vast subject, which you will get acquainted with by googling ``typicality''. There is more to say about that than I can conveniently state in one post, besides which it is not my specialty.
If a system is in a pure state, its temperature is not defined unless it is a ground state, in which case its temperature is zero. One may nonetheless talk about effective temperature, that is, the temperature at which the energy of the system equals that of the quantum state. Whether effective temperature is a useful concept depends on the situation.
Thank you Juan Weisz. Yes, as you said: "Temperature is not a concept of Quantum Mechanics itself". But, for one single quantum system without any statistical (related to different distributions) consideration, how to define temperature? Quantum mechanics has a core of probability, which is exactly the statistical property. Therefore, there must be temperature in the quantum mechanics. The questions is how to define?
Thank you, Alex Risos. The temperature of ground state is Zero? I don't think so! In the thermodynamics, T for one freedom is =K_B T/2 when the system is in equilibrium state. For one single quantum system, for example an atom (electrons with its nuclei), the temperature of the electrons in its ground state must has temperature, but is not absolutely zero. The ground state has its statistical properties which can be used to define effective temperature to estimate the energy transfer of kinetic energy from electrons to other coupled system. Right?
Thank you Marek Gluza. You provided a very good paper and I am reading it now. If we can talk about partition function then we have introduced temperature. The effective temperature assumes that E \approx K_B T, where E is the total energy of a quantum state and the Boltzmann constant K_B is just derived from typical thermodynamic processes. However for one quantum state, the relation E \approx K_B T is unacceptable because the energy can not be the free energy to become heat. The fluctuation is really the random property of heat, so it will be a proper quantity to estimate the thermal effect of a quantum state.
Thank you L. I. Plimak, you make me think over the temperature of the ground state. I recognize that if you get an electron, only one electron without anything around, you can't give the temperature or estimate the thermal effect of the electron. That means you can't separate the quantum system or the quantum state from its background to define T or to estimate thermal effect. Therefore, at least, a quantum system should have an equilibrium temperature of cosmic background radiation. But it is not the case! For a ground quantum state, we can keep it from any background in the Lab. In this case, T is zero because the energy can not be released at all! So the temperature of a quantum state is actually a non-equilibrium property!
Dear Lin Zhang,
You suggestion to identify temperature with energy fluctuation contradicts conventional thermodynamics, where temperature = energy. On the other hand, it makes perfect physical sense to me. Let's call it fluctuation temperature, and nobody is the wiser.
Best
Lev
Lin
Quantum theory is posed for a situation where there is
energy conservation. Fluctuations in energy will limit
Quantum properties. Quantum Statistical Mechanics also
allows you to calculate fluctuations. You cannot do anything
with a single quantum problem alone. You need a heat bath or
something to give you temperature.
A paper is recommended:
M. Gring, M. Kuhnert,1 T. Langen, T. Kitagawa, B.
Rauer,1 M. Schreitl, I. Mazets, D. Adu Smith, E. Demler,
J. Schmiedmayer, Relaxation and Prethermalization in
an Isolated Quantum System, Science 337, 1318 (2012).
This article is also enlightening:
Dunkel J.,Consistent thermostatistics forbids negative absolute temperatures, Nature Physics, 10 (1) 67-72 (2013). DOI: http://dx.doi.org/10.1038/nphys2815
i think that there are derivatives to this question. how would one measure this temperature? could one hold this single quantum system in a heat bath? for example suppose one performs a two slits interference experiment on one particle at a time, does it interact with the heat bath or is the time scale too short for that? i think that time and temperature are two parameters that crawl together into this question. one can not speak about thermodynamics without a (large enough) time scale. on the other hand uncertainty in energy is over short periods of time.
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Materials Science
- Materials
- Inorganic Materials
- Ceramics
- Glass