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I Stationary states -- Boltzmann distribution

  1. Oct 13, 2016 #1
    Hello everybody,

    - In quantum mechanics, the state ## | \psi \rangle ## of a system that is in thermodynamic equilibrium can be expressed as a linear combination of its stationary states ## | \phi _n \rangle ## : $$ | \psi \rangle = \sum_n c_n | \phi _n \rangle $$
    It permit us to express the mean value of energy as:
    $$ \langle E \rangle _{\psi}= \sum_n E_n | c_n |^2 $$

    - In other approach, one way to express the mean value of energy is by using the Boltzmann distribution. So my question is:
    As the system is in thermodynamic equilibrium, is it allowed to think that Boltzmann distribution ## \frac{N_i}{N} = \frac{g_i e^{-\frac{E_i}{k_BT}}}{Z(T)} ## are equivalent to the ## | c_n |^2 ## ?

    Thank you everybody.

    Konte
     
  2. jcsd
  3. Oct 13, 2016 #2

    kith

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    That's not correct, a system in thermodynamic equilibrium can't be described by a ket vector. Where did you get this idea? What we need is the density matrix, see wikipedia.

    If you use the density matrix, a similar relation is essentially correct. Note however that for a system of bosons resp. fermions, the Boltzmann distribution gets replaced by the Bose-Einstein distribution resp. the Fermi-Dirac distribution.
     
  4. Oct 14, 2016 #3
    Thank you for the answer,

    Because I have a lack of knowledge about matrix density, I would like to be sure:
    Let say, I want to describe and to study a single ammonia molecule at room temperature. So, even in this case, the correct way is using density matrix?

    Thanks.
    Konte
     
  5. Oct 14, 2016 #4

    DrClaude

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    If you have a single molecule, then you can't really talk about its temperature. If it is exchanging energy with a reservoir at given T, then yes, the density matrix formalism is necessary.
     
  6. Oct 14, 2016 #5

    Demystifier

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    In QM, the state describes you knowledge about the system. Since ##c_n=| c_n |e^{i\varphi_n}##, knowledge of ##c_n## involves more knowledge than knowledge of ##| c_n |##. If you know ##c_n## then you can write the expansion
    $$ | \psi \rangle = \sum_n c_n | \phi _n \rangle $$
    in which case you don't necessarily need to work with density matrix. But often you know only ##| c_n |## and not ##c_n##, in which case you cannot write the expansion above. In this case, it is necessary to use the density matrix. Thermodynamic equilibrium is a special example of the case in which you know only ##| c_n |## and not ##c_n##.
     
  7. Oct 14, 2016 #6

    rubi

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    That doesn't seem right. Of course, you can write the energy expectation value of a thermal state ##\rho=\frac{e^{-\beta \hat H}}{Z}## as a sum ##\sum |c_n|^2 E_n##, but there is no pure state ##\left|\psi\right>=\sum c_n \left|E_n\right>## that reproduces the statistics of ##\rho## exactly, no matter what phase information you supply. Of course, there is a state that reproduces the energy expectation value, but the density matrix allows you to compute more than just the expectation value of the energy observable. If you want to reproduce the statistics of all observables, then no pure state can be found that matches the thermal state.
     
    Last edited: Oct 14, 2016
  8. Oct 14, 2016 #7
    Thanks for your answer. So, for the case of this single molecule, how can I model this exchange of energy with a reservoir at given T?
     
  9. Oct 14, 2016 #8
    How do we construct that state? May be in the same way as I mention in my first post?

    It is really interesting. Is there any mathematical way or formal demonstration that can prove it to us once and for all?

    Thanks a lot.
     
  10. Oct 14, 2016 #9

    rubi

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    One example would be ##\rho_\text{pure}=\left|\Psi\right>\left<\Psi\right|## with ##\left|\Psi\right>=\frac{1}{\sqrt{Z}}\sum_n e^{-\frac{1}{2}\beta E_n}\left|E_n\right>##, but you can also add arbitrary phases as Demystifier has mentioned. However, this state is not statistically equivalent to the density matrix ##\rho_\text{mixed}=\frac{1}{Z}e^{-\beta \hat H}##. It only reproduces the energy expectation value.

    Take the projector ##P=\left|\theta\right>\left<\theta\right|## onto the state ##\left|\theta\right>=\frac{1}{\sqrt{2}}\left(\left|E_1\right>+\left|E_2\right>\right)##. You get ##\left<P\right>_\text{pure}=\frac{1}{2 Z}\left|e^{-\frac{1}{2}\beta E_1}+e^{-\frac{1}{2}\beta E_2}\right|^2## but ##\left<P\right>_\text{mixed}=\frac{1}{2 Z}\left(e^{-\beta E_1}+e^{-\beta E_2}\right)##. I did the calculation in my head, so potentially the factors aren't exactly right. o0) Anyway, this means that the states predict different probabilities for finding the state ##\left|\theta\right>## in a measurement. If you give me a different state ##\left|\Psi\right>## with different phases to fix the problem for this particular ##P##, I will just add a relative phase factor in ##\left|\theta\right>## to make the problem reappear.
     
    Last edited: Oct 14, 2016
  11. Oct 17, 2016 #10

    Demystifier

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    What exactly "doesn't seem right"? I don't see any contradiction between my statements and your statements.
     
  12. Oct 17, 2016 #11

    rubi

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    Maybe I have misunderstood you, but it seemed to me as if you were claiming that in a thermal state we know ##\left|c_n\right|## and once you supply the phase information, i.e. a list of numbers ##\varphi_n## such that ##c_n=\left|c_n\right|e^{\mathrm i\varphi_n}##, we know the "real" state of the system. But that can't be right, because no matter what list of numbers ##\varphi_n## you give me, I will always find an observable such that its statistics in the thermal state doesn't match its statistics in the state with the additional phase information incorporated. Hence, it can't be true that there is additional phase information and we just don't know it. The ##c_n## can't be the coefficients of a pure state whose phase information we have dropped.
     
  13. Oct 17, 2016 #12

    Demystifier

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    Well, one should distinguish what one can (in principle) know about some physical system from what one actually knows about that system. It is certainly possible that someone does not know something which in principle can be known. For example, there can be a system in which ##\varphi_n## can be known, but someone does not know them. In this case, that person with incomplete information will describe the system with a mixed (e.g. thermal) state, despite the fact a more complete information is also possible. Of course, that person may perform additional measurements and attain additional information about phases, after which he can describe the system with a pure state and not with a mixed state. But before that, the best he can is to describe the system with a mixed state.
     
  14. Oct 17, 2016 #13

    rubi

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    But that new state would be statistically inconsistent with the thermal state and as far as we know, the statistics predicted by the thermal state is consistent with experiments, so collecting additional information should not require us to modify the state.
     
  15. Oct 17, 2016 #14

    Demystifier

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    i) Suppose that I have a machine with two buttons; button A and button B. When I press A, the machine prepares a particle in the state ##|A\rangle##. When I press B, the machine prepares state ##|B\rangle##. So I press one of these buttons, but I don't tell you which one I pressed. What can you say about the state before you make any measurement? Can your knowledge at that time be expressed as
    $$\rho=\frac{1}{2}|A\rangle \langle A| + \frac{1}{2}|B\rangle \langle B| ?$$

    ii) Now consider the variation of the experiment in which I have a 1000 machines of the kind above. For each of them I press either A or B. (I am not obliged to press the same button each time.) In this way I prepare 1000 particles, each of which is either in the state ##|A\rangle## or ##|B\rangle##. You pick one of these particles. What can you say about the state of that particle before you make any measurement? Can your knowledge on that particle at that time be expressed as
    $$\rho=\frac{1}{2}|A\rangle \langle A| + \frac{1}{2}|B\rangle \langle B| ?$$
     
  16. Oct 17, 2016 #15

    rubi

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    None of these states is likely to describe the experiments correctly. In the first case, I can only make one experiment and in the second case I don't know the probabilities for pressing A or B. The states are just best guesses based on the maximum entropy method. However, this method only makes sense if you can improve your knowledge by Bayesian inference. In physics, this makes no sense, since the microscopic dynamics of a system is governed by the certain laws of physics and the approach to equilibrium is a physical process that doesn't depend on our knowledge. The laws of physics are supposed to single out a definite state. Only after the laws of physics have made the system approach some equilibrium state, we can measure it, compute its entropy and find that it happens to correspond to a maximum entropy distribution. However, the maximum entropy method doesn't explain the state. A MaxEnt advocate would expect that the prior state would have to be adjusted using Bayesian inference, but that would require one to modify it. However, experimentally, the thermal states don't seem to require any modification. The only explanation can be that the theory generically predicts an approach to equilibrium.
     
  17. Oct 17, 2016 #16

    Demystifier

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    If I understood you correctly, you claim that quantum state is objective thing that does not depend on our knowledge. I would agree that this is so for pure states, but I don't agree that this is so for mixed states. In addition, many physicists would even disagree that pure states are objective. So this is a somewhat controversial topic, which depends on the interpretation of QM. So perhaps it's better to stop further discussion.
     
  18. Oct 17, 2016 #17

    rubi

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    No, I'm not making any interpretational statements. The situation is analogous to classical statistical mechanics. In principle, you could also use the MaxEnt method to arrive at the classical canonical ensemble, but it's not physically correct to do so. The statistial ensembles must arise from a relaxation into equilibrium, since the laws of physics are defined by the microscopic theory and no additional input should be required. That's why people study ergodic theory or the Fokker-Planck equation and so on. The same reasoning applies to quantum statistical mechanics.

    In other words: Do you think the MaxEnt formalism is a valid way to derive the canonical ensemble in classical statistical mechanics? If not, why would it be valid in quantum mechanics? If yes, how do you explain that it describes the correct statistics even though no Bayesian updating is needed?
     
    Last edited: Oct 17, 2016
  19. Oct 18, 2016 #18

    vanhees71

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    Well, according to the H theorem the equibrium state is the state of maximum entropy under the constraints set by conservation laws, and this leads to the (grand-)canonical statistical operators
    $$\hat{\rho}=\frac{1}{Z} \exp[-\beta (\hat{H}-\sum_j \mu_j \hat{Q}_j)], \quad Z=\mathrm{Tr} \exp(\ldots).$$
     
  20. Oct 18, 2016 #19

    Demystifier

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    Yes I do.

    In addition, let me also note that I think that approaches based on (quasi)ergodicity are not a valid way to derive the canonical ensemble in classical statistical mechanics. If you want, I can explain why do I think so.

    Sometimes it is needed. For instance, if you measure the velocity of a single molecule, then the probability distribution for this molecule is no longer given by Maxwell-Boltzmann distribution.

    I am glad that you translated the problem into a discussion of classical statistical mechanics, where the concepts are much clearer.
     
    Last edited: Oct 18, 2016
  21. Oct 18, 2016 #20

    Demystifier

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    Yes but H-theorem assumes that some information about the system is ignored. The ignored information is some information which can be known in principle, but not in practice. It usually corresponds to some fine details which are ignored by coarse-graining. If nothing is ignored, i.e. if all microscopic degrees are taken into account, then unitarity of QM implies that pure state evolves into a pure state and von-Neumann entropy does not change. The ignorance of information is a subjective thing, so in this sense increase of entropy by H-theorem can be considered subjective.
     
    Last edited: Oct 18, 2016
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