Quantum thermodynamics of single particle

[Konte]

Hello everybody,

I have two questions:

1) Is it possible to define a temperature for single particle (or atom or molecule)? If so, how?

2) How to model with quantum Hamiltonian an exchange of energy between a single atom (or molecule) and a reservoir at given temperature ##T## ?

Thank you everybody.

Konte

 

[Bystander]

1) Is it possible to define a temperature for single particle (or atom or molecule)? If so, how?
2) How to model with quantum Hamiltonian an exchange of energy between a single atom (or molecule) and a reservoir at given temperature TTT ?

What are your thoughts?

 

[Konte]

What are your thoughts?

I don't understand your answer.

 

[Demystifier]

1) Is it possible to define a temperature for single particle (or atom or molecule)? If so, how?

Yes, by taking
$$\rho=e^{-H/kT}$$
where ##H## is the single-particle Hamiltonian.

 

[Konte]

Yes, by taking
$$\rho=e^{-H/kT}$$
where ##H## is the single-particle Hamiltonian.

Thank you for your answer.
I still have question:
- in this post https://www.physicsforums.com/threa...echanics-and-temperature.426455/#post-2869963, the forumer xerma mentioned a ##\rho = \frac{e^{-H/kT}}{Z}##. What is the difference between those two definitions of ##\rho## (that I suppose both density matrix operator) ?

Thank you very much.

 

[vanhees71]

The latter formula is correct. The statistical operator must be of trace ##1##. Thus the statistical operator of the canonical ensemble is
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}), \quad \beta=\frac{1}{k_{\text{B}} T}.$$

 

[Konte]

The latter formula is correct. The statistical operator must be of trace ##1##. Thus the statistical operator of the canonical ensemble is

$$\hat{\rho}=\frac{1}{Z} exp(−\beta H),\,\,Z=Tr \,\, exp(−\beta \hat{H}),\,\,\beta=\frac{1}{k_BT}$$​

Thank you for your answer.
How to demonstrate this expression of ##\hat{\rho}##?
Because, after searching, I always have another alternative form ##\hat{\rho}= \sum_i p_i | \psi_i \rangle \langle \psi_i|##

Thanks.

Konte

 

[vanhees71]

This expression you get from the maximum entropy principle. If you look for all statistical operators that lead to a given expectation value ##U=\mathrm{Tr} (\hat \rho \hat{H})## of the energy and minimize the entropy,
$$S=-k_{\text{B}} \mathrm{Tr} \hat{\rho} \ln \hat{\rho},$$
you get to the canonical statistical operator (of course you need the normalization ##\mathrm{Tr} \hat{\rho}=1## as another constraint).

The "alternative form" is just the expansion of the statistical operator in terms of its eigenvectors. Note that there can be also generalized eigenvectors if the operator has a continuous spectrum.

 

[Konte]

@vanhees71
Ok, thank you for this interesting answer. Could you indicate me some lectures that can help to understand and make it clearer for a novice as me, please?
I suppose, all of those concepts are valid and applicable for the case of a single system (like single atom or single molecule) ?

Thanks.
Konte

 

[vanhees71]

Well, these are very general concepts. My favorite book, using the information-theoretical approach, is

A. Katz, Principles of Statistical Mechanics, W. H. Freeman and Company, San Francisco and London, 1967.

Very good is also Landau&Lifshitz, vol. 5 or Reif.

 

[Konte]

Well, these are very general concepts. My favorite book, using the information-theoretical approach, is

A. Katz, Principles of Statistical Mechanics, W. H. Freeman and Company, San Francisco and London, 1967.

Very good is also Landau&Lifshitz, vol. 5 or Reif.

Thanks a lot.

Konte

 

[Konte]

Hello,

I'm back, just because a little doubt persist on my understanding:
Is the operator ##\hat{\rho} = \frac{e^{-\beta \hat{H}}}{Z}## still meaningful even for describing a pure state?
Thanks

Konte

 

[vanhees71]

Of course not. An equilibrium state is only a pure state for ##T \rightarrow 0##. Note that using the equilibrium (canonical) distribution means that you look at a single particle within a heat bath at temperature ##T##!

 

[Konte]

Note that using the equilibrium (canonical) distribution means that you look at a single particle within a heat bath at temperature ##T##!

Thank you,

So even for a single particle within a heat bath at ##T##°, the concept of mixed states is meaningful?

In other words, single particle is describable as a canonical ensemble once it is "surrounded" by a heat bath at fixed ##T##° (equilibrium)?

Konte.