3
$\begingroup$

This is a basic question but I struggle to grasp it really. In the Wikipedia page for Ground state, it's written

According to the third law of thermodynamics, a system at absolute zero temperature exists in its ground state; thus, its entropy is determined by the degeneracy of the ground state. [...] It is also possible for the highest excited state to have absolute zero temperature for systems that exhibit negative temperature.

I don't see how this happen. To illustrate what my question is, for example, consider this very simple model like the Ising model in 1D under 3 sites.

$$H = -\sigma_1 \sigma_2 -\sigma_2 \sigma_3 = \begin{bmatrix} -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2\\ \end{bmatrix}$$

From what I understand, if we start from any eigenstate, they are all stationary states and if the system were to be driven by this Hamiltonian $e^{-iHt}$, it would just stay in that eigenstate until the end of time. Moreover, any arbitrary state is just a linear combination of eigenstates, so the time evolution is nothing but some local phase change to each of the eigenstates. Which means that the energy observable is basically identical for the initial and final states. For example, let's pick some random state

$$|\Psi_0\rangle = \frac{1}{2\sqrt{2}} \begin{bmatrix} 1\\1\\1\\1\\1\\1\\1\\1\end{bmatrix} \rightarrow e^{-iHt}|\Psi_0\rangle =\frac{1}{2\sqrt{2}}\begin{bmatrix} e^{i2t}\\e^{i1t}\\1\\e^{i1t}\\e^{i1t}\\1\\e^{i1t}\\e^{i2t}\end{bmatrix} $$

Clearly, nothing changes here, and there's no sight of the system somehow fall into the "ground state".

I can see, technically, how the ground and the highest excited states here both have $S = k \ln 2$ while the middle excited states have $S = k \ln 4$, thus making them the lowest entropy states and be where the system "falls" into at $T=0$, but how does this process happen quantum mechanically? And what exactly is the effect of "temperature" in shifting the most probable state?

Thank you!

$\endgroup$
4
  • 5
    $\begingroup$ The quote from Wikipedia is (as so often) highly misleading (or incomplete). Of course, if you prepare your system in some state and the system (then) is isolated, it evolves according to the Schrödinger equation. In particular, if you prepare it in some energy eigenstate (not necessarily the ground state), it will stay in this state forever. But you are here interested in systems which are coupled to some environment. To start, you can for example read about spontaneous emission. $\endgroup$ Commented Mar 18 at 16:25
  • $\begingroup$ I think the connection between spontaneous emission and the grand canonical ensemble arises because both involve the concept of energy exchange with a reservoir. In the case of spontaneous emission, the reservoir is the quantized electromagnetic field, and the energy exchange results in the emission of a photon. In the grand canonical ensemble, the reservoir is a thermal reservoir, and energy exchange occurs between the system and the reservoir. Both frameworks exhibit the concept of "falling to the ground state". But mathematically, can they be made to be equivalent? $\endgroup$
    – Kim Dong
    Commented Mar 18 at 16:47
  • 1
    $\begingroup$ I don't really understand what you mean, sorry. I am not an expert in open systems either. It is a very complicated topic. Here however, it might suffice if you consider the canonical density matrix and check the zero temperature limit; this should give you a density matrix only in the ground subspace. So if you accept that the thermal/canonical density matrix is a good description of a system coupled to a heat bath, then, as I said above, the zero temperature limit of that state is the ground state (or any density matrix on that subspace). $\endgroup$ Commented Mar 18 at 17:04
  • $\begingroup$ The question "How does quantum system get into ground state" is a fine question in quantum theory (see e.g. spontaneous emission) or in statistical physics (evolution towards thermodynamic equilibrium), but it has nothing to do directly with the quote or 3rd law of thermodynamics. 3rd law of thermodynamics is a somewhat obscure statement about behaviour of macroscopic systems in the limit where temperature "goes to zero". $\endgroup$ Commented Mar 18 at 17:50

2 Answers 2

6
$\begingroup$

As others have pointed out, that part of the wikipedia article is written about what the quantum mechanical system does when it is coupled to a thermal environment. In general, that's what thermodynamics is - you have some exchange of energy between the system (your Ising chain) and some huge environment, which you don't know exactly what it is made up of, all you know is its statistical properties. The third law is then a statement about the statistics of the energy states that you find yourself in for the Ising chain, when you are in thermal equilibrium, in the limit that the temperature $T \to 0$.

You can check this works out exactly as stated. If your Ising chain has Hamiltonian $H = \Omega (- \sigma_1 \sigma_2 - \sigma_2 \sigma_3)$ (let $\Omega > 0$ be a characteristic energy scale, which is missing in your Hamiltonian). If your Ising chain is in thermal equilibrium with a reservoir, then it has a density matrix $$ \rho:=\frac{e^{-H /T}}{\mathrm{Tr}[e^{-H /T}]} = \mathrm{diag}\left\{ \frac{e^{\frac{2 \Omega }{T}}}{2 \left(e^{\Omega /T}+1\right)^2},\frac{1}{4 \cosh \left(\frac{\Omega }{T}\right)+4},\frac{1}{2 \left(e^{\Omega /T}+1\right)^2},\frac{1}{4 \cosh \left(\frac{\Omega }{T}\right)+4},\frac{1}{4 \cosh \left(\frac{\Omega }{T}\right)+4},\frac{1}{2 \left(e^{\Omega /T}+1\right)^2},\frac{1}{4 \cosh \left(\frac{\Omega }{T}\right)+4},\frac{e^{\frac{2 \Omega }{T}}}{2 \left(e^{\Omega /T}+1\right)^2} \right\} $$ You can explicitly calculate the (von Neumann) entropy associated with this thermal state: $$ \mathrm{entropy} = - \mathrm{Tr}[ \rho \log \rho ] = \frac{\frac{\Omega }{T}\left(1-e^{{2 \Omega }/{T}}\right) +\left(e^{\Omega /T}+1\right)^2 \log \left(4 \cosh \left({\Omega }/{T}\right)+4\right)}{\left(e^{\Omega /T}+1\right)^2} $$ In the limit $T \to 0$, so expanding the above for $ T \ll \Omega$, one can check that this becomes $\simeq \log(2)$, which is exactly what wikipedia says it will be, since $2$ is the degeneracy of the ground state in your Hamiltonian

$\endgroup$
3
  • $\begingroup$ Hmm, but this doesn't seem to answer the question of what gives rise to this canonical density matrix? Like what kind of interactions/processes onto the sites, in say an open quantum system, that leads to the various energy states being distributed exponentially through the Boltzmann distribution $\sim e^{-H/T}$? I am aware of the principle of maximum entropy which gives rise to this distribution, but it seems axiomatic even, for something that I think is reasonably describable through a physical picture. $\endgroup$
    – Kim Dong
    Commented Mar 19 at 11:12
  • 1
    $\begingroup$ @KimDong You have to postulate something. The MaxEnt principle is, IMHO, more than a good motivation for "ordinary" stat. mech. situations. $\endgroup$ Commented Mar 19 at 13:08
  • 1
    $\begingroup$ @KimDong That is actually a very difficult and nontrivial question to answer which is actively being researched - "Eigenstate Thermalization Hypothesis" (see en.wikipedia.org/wiki/Eigenstate_thermalization_hypothesis) is an attempt to answer how quantum mechanical systems end up in a thermal state. This is very complicated. Alternatively you can also consider open systems coupled to thermal baths (assuming they are in a thermal state to start with), and you can see how systems like your Ising chain time evolve towards a thermal state like the one above. $\endgroup$ Commented Mar 19 at 13:08
3
$\begingroup$

Well, this quip is also true for the pure Coulomb hamiltonian of the hydrogen atom. Since electromagnetic coupling is weak, we still use $|n, l, m\rangle$ as approximate states and then consider some interaction or background field that couples states.

Since:

$$ \langle n, l, m|n',l',m'\rangle = \delta_{nn'}\delta_{ll'}\delta_{mm'} $$

if your perturbing potential is $V(\vec r) = constant$, nothing ever happens.

Throw in a dipole term, such as a background field, $\vec E = E\hat z$:

$$ V'(\vec r) = qEz $$

and now "approximate" eigenstates are coupled: interactions.

I'm not too versed on the Ising model, but perhaps a similar prescription applies.

Addendum: when you go to QED, the perturbing field is dynamic, so it must be included in the definition of ground-state, so again, we kind of factor the problem in to "absorbed/emitted" photons and the atom, for which we still use the Coulomb eigenstate language.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.