My understanding is that temperature is not a quantum mechanical operator. Hence, thereby it should not have any uncertainty. However, any instrument that tries to take the measurement of temperature will have some uncertainty in it for example the mercury postion of the thermometer! Further, we know temperature is $$ T = \frac{\partial U}{\partial S}_{(N,V)} $$

But we cannot measure $U$ with arbitrary accuracy so there must be some uncertainty in the slope of $U$ as well!


Does temperaure have uncertainty? If yes, how is this possible when temperature is not a quantum mechanical observable? If no, how is this possible when it is the derivative of energy with respect to entropy and energy has uncertainty?

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    $\begingroup$ Ask yourself: Is temperature a quantum mechanical operator? $\endgroup$ – ACuriousMind Aug 13 '15 at 14:23
  • $\begingroup$ But the position of mercury or any thermometer instrument is a quantum mechanical operator ... right? $\endgroup$ – drewdles Aug 13 '15 at 14:24
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    $\begingroup$ Okay, are you asking about the experimental uncertainty of a temperature measurement, or about the fundamental uncertainty of a quantum mechanical observable? $\endgroup$ – ACuriousMind Aug 13 '15 at 14:33
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    $\begingroup$ The temperature of a system is just a measure of its energy, so it's subject to the same uncertainty as measuring the energy. $\endgroup$ – John Rennie Aug 13 '15 at 15:37
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    $\begingroup$ @JohnRennie, temperature is not a measure of internal energy - there is no simple general relationship between the two. Internal energy also cannot be directly measured; but temperature can - there are no "energometers", but there are thermometers. $\endgroup$ – Ján Lalinský Sep 16 '15 at 19:39

The Heisenberg uncertainty principle applies to dimensions commensurate to $\hbar$, i.e. at the level of particles ( atoms, molecules, elementary particles). Temperature is a classical observable appearing in thermodynamics as a variable, but when analyzed from the emergent statistical ensemble it is not a variable but a statistical average of the kinetic energy of the sample macroscopically displaying temperature $T$.

$$E_k=\frac12 m v^2_\mathrm{rms}=\frac32 kT$$

The formula above is for an ideal gas, but similar formulas exist for all bulk matter. Temperature is called an intensive thermodynamic variable.

As it is not a variable or a function/differential of variables at the quantum mechanical level, it cannot enter into the quantum mechanical equations as an operator. It thus cannot participate in the commutator relations that define whether an observable is limited, when measured, by the uncertainty principle.

  • $\begingroup$ No, that is not what temperature is. For example, the temperature of the electron gas in a metal is nowhere close to $kT$. $\endgroup$ – Pieter Feb 19 '18 at 16:45
  • $\begingroup$ @Pieter the fomula is for an ideal gas. Of course it will not hold for solids or whatever, but still temperature is an average statistical thermodynamic measure, emergent from the underlying quantum mechanical operator formulation .en.wikipedia.org/wiki/Intensive_and_extensive_properties it is an intensive property of bulk matter. $\endgroup$ – anna v Feb 19 '18 at 17:13
  • $\begingroup$ Actually, the formula also holds for solids (classical systems without zero point motion). But one can define temperature for a pure quantum system. like a spin system, counting states, determine the change with temperature. One can then order these systems in warmer and cooler, depending on the likely direction of energy transfer when they are in contact. Of course this is a bit of a toy system, but it is the same temperature as in the weather forecast. $\endgroup$ – Pieter Feb 19 '18 at 17:23
  • $\begingroup$ @Pieter I understood the question to be about the uncertainty principle, i.e. if there is an operator for the temperature with commutators defining uncertainty. There is another answer here where a link is given with the claim of temperature being a quantum mechanical operator, but commutators would be further down the line. this is the link to the paper arxiv.org/abs/1708.09374 Temperature as a quantum observable . Rob deleted the answer $\endgroup$ – anna v Feb 19 '18 at 17:49

There are extensions of temperature, I believe, for microscopical objets, but usual temperature is an intensive property of macroscopic objects. The number of degrees of freedom is at least Avogadro Number, and quantum effects tend to cancel.

You could perhaps as the question for mesoscopic objects.


protected by Qmechanic Feb 19 '18 at 19:46

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