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Consider the following statement:

If we know that the system is in the ground state, then the temperature is zero.

How does this follow from the statistical definition of temperature?

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Mad props for a cool question. I'm going to justify essentially the converse of the statement because it doesn't make much sense to talk about the temperature of a system that is in a pure state.

Let's assume that we're talking about a quantum system with discrete energy spectrum (with no accumulation points) in thermal equilibrium. Let $\beta = 1/kT$ be the inverse temperature. Then recall that the Boltzmann distribution tells us that the population fraction of systems in the ensemble corresponding to energy $E_i$ is given by $$ p_i = \frac{g_ie^{-\beta E_i}}{Z} $$ where $g_i$ is the degeneracy of the energy level. In particular, note that the relative frequency with which energies $E_i$ and $E_j$ will be found in the ensemble is $$ p_{ij}(\beta) = \frac{g_ie^{-\beta E_i}}{g_je^{-\beta E_j}} = \frac{g_i}{g_j}e^{-\beta(E_i - E_j)} $$ In particular, let $i=0$ correspond to the ground level, then the frequency of any level relative to the ground level is $$ p_{i0}(\beta) = \frac{g_i}{g_0}e^{-\beta(E_i - E_0)} $$ Notice that since the ground level has the lowest energy by definition, we have $E_i - E_0 \geq 0$, but zero temperature corresponds to the limit $\beta \to \infty$, and we have $$ \lim_{\beta \to \infty } p_{i0}(\beta) = \delta_{i0} $$ In other words, at zero temperature, every member of the ensemble must be in the ground energy level; the probability that a system in the ensemble will have any other energy becomes vanishingly small compared to the probability that a member of the ensemble has the lowest energy.

Addendum - September 18, 2017

In response to a question in the comments about whether or not at zero temperature the system is in a pure state:

Recall that a quantum state (density matrix) $\rho$ is said to be pure if and only if $\rho^2 = \rho$. We now show that as $T\to 0$, or equivalently as $\beta\to+\infty$, the thermal density matrix $\rho$ approaches a density matrix $\rho_*$ that is pure if the ground level is non-degenerate and not pure otherwise. We will rely on an argument quite similar in character to the one given above in which we compared the probabilities of finding a system in a given energy level when we approach zero temperature.

For any positive integer $d$, let $I_d$ denote the $d\times d$ identity matrix. As above, we consider a system with discrete energy levels $E_0<E_1<\dots$ and with corresponding degeneracies $g_0, g_1, \dots$. Recall that the thermal density matrix, namely the density matrix for a system in thermal equilibrium with a heat bath at inverse temperature $\beta$, is given in the energy eigenbasis by:

\begin{align} \rho = \frac{1}{Z} \begin{pmatrix} e^{-\beta E_0}I_{g_0} & & & \\ & e^{-\beta E_1} I_{g_1} & & \\ && e^{-\beta E_2} I_{g_2} \\ & & &\ddots \end{pmatrix}, \qquad Z = \sum_j g_j e^{-\beta E_j}. \end{align}

Let $i\geq 0$ be given, and let us concentrate on the scalar factor in front of the identity matrix $I_{g_i}$ in the $i^\mathrm{th}$ block of the density matrix:

\begin{align} \frac{e^{-\beta E_i}}{Z} = \frac{e^{-\beta E_i}}{\sum_j g_je^{-\beta E_j}} = \frac{e^{-\beta(E_i -E_0)}}{\sum_jg_j e^{-\beta(E_j - E_0)}} = \frac{e^{-\beta(E_i -E_0)}}{g_0 + \sum_{j>0}g_j e^{-\beta(E_j - E_0)}}. \end{align}

Therefore, we have

\begin{align} \lim_{\beta\to +\infty} \frac{e^{-\beta E_i}}{Z} = \frac{\lim_{\beta\to +\infty}e^{-\beta(E_i -E_0)}}{g_0 + \lim_{\beta\to +\infty}\sum_{j>0}g_j e^{-\beta(E_j - E_0)}} = \frac{\delta_{i0}}{g_0 + 0} = \frac{\delta_{i0}}{g_0}. \end{align} Care may need to be taken in the case of an infinite-dimensional Hilbert space in asserting that the limit of the sum in the denominator approaches zero since it's an infinite series. It follows that \begin{align} \rho_* = \lim_{\beta\to+\infty} \rho = \frac{1}{g_0}\begin{pmatrix} I_{g_0} & & & \\ & 0 & & \\ && 0 \\ & & &\ddots \end{pmatrix} \end{align}

and therefore in particular $\rho_*^2 = \rho_*$ if and only if $g_0 = 1$.

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  • $\begingroup$ Suppose there are multiple ground states. Does this change the answer? $\endgroup$ May 19 '13 at 22:31
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    $\begingroup$ @AlecS I went over the answer to make sure that "ground state" is replaced by "ground level" to make certain that their is no confusion even when the ground level is degenerate. If the ground level is degenerate, then the best one can say is that the density operator becomes the projector onto the lowest energy eigenspace. $\endgroup$ May 19 '13 at 22:41
  • $\begingroup$ Aha thank you, that is a nice solution. I think that it is also generalizable for an isolated system (micro-canonical ensemble) by placing this system (in ground state) in a larger system in ground state and by making the wall diathermal. Energy levels and temperature won't change by doing this, and I can use your reasoning to conclude that $T=0$. Do you think that this reasoning makes sense? $\endgroup$
    – yarnamc
    May 19 '13 at 22:50
  • $\begingroup$ @yarnamc Well by making the wall diathermal, I imagine what you have in mind is that the two systems will come to equilibrium with one another and have the same temperature. But as soon as this happens, the small system will now be in the canonical ensemble, not the microcanonical ensemble, so it's not clear to me what your argument is. $\endgroup$ May 19 '13 at 23:00
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    $\begingroup$ @SRS I just looked back at my math, and I made a crucial error -- in the case of a degenerate ground state, there is a $1/g_0$ factor in the expression for the limiting density matrix that makes it non-pure when the ground level is degenerate. I therefore revise my claim: the thermal density matrix is pure in the zero temperature limit if and only if the ground level is non-degenerate. I encourage you to check my math again seeing as how I made a bad error earlier. My apologies. $\endgroup$ Sep 19 '17 at 22:28
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Here I will provide an answer from the point of view of equilibrium density matrix:

  • The eigenequation of Hamiltonian $\hat H$: $$\hat{H}|\phi_n\rangle = E_n|\phi_n\rangle \qquad (n=0,1,2,\cdots) \tag{1}$$
  • The spectrum decomposition of Hamiltonian: $$\hat{H}=\sum_{n\in\mathbb N_0}E_n|\phi_n\rangle\langle\phi_n| \tag{2} $$
  • Equilibrium density matrix: $$ \hat{\rho} \equiv \dfrac{e^{-\beta \hat{H}}}{Z}=\sum_{n\in\mathbb N_0} \dfrac{e^{-\beta E_n}}{Z}|\phi_n\rangle\langle \phi_n| = \dfrac{\sum_{n\in\mathbb N_0}e^{-\beta E_n}|\phi_n\rangle\langle\phi_n|}{\sum_{n\in\mathbb N_0}e^{-\beta E_n}} \tag{3}$$ where $\beta=\dfrac{1}{k_B T}$ is the inverse temperature and we have used the fact $\hat{H}$ commutes with $\hat{\rho}$ in equilibrium.
  • The zero temperature limit $T \rightarrow 0$ or $\beta\rightarrow\infty$: $$\lim_{\beta\rightarrow\infty}\hat{\rho} = \lim_{\beta\rightarrow\infty} \dfrac{\sum_{n\in\mathbb N_0}e^{-\beta E_n}|\phi_n\rangle\langle\phi_n|}{\sum_{n\in\mathbb N_0}e^{-\beta E_n}} = \lim_{\beta\rightarrow\infty} \dfrac{\sum_{n\in\mathbb N_0}e^{-\beta (E_n-E_0)}|\phi_n\rangle\langle\phi_n|}{\sum_{n\in\mathbb N_0}e^{-\beta (E_n-E_0)}} = |\phi_0\rangle\langle\phi_0| \tag{4} $$ For Hamiltonian $\hat{H}$ with degenerate ground states $(4)$ reduces instead to an equally weighted ensemble of degenerate ground states.
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A simpler and probably more naive explanation is to note that, by the equipartition theorem, $E \propto T$. So in the ground-state, where $E = 0$, it must be that $T = 0$ as well. However, this only works in the classical case (for example, the ground state energy of a quantum harmonic oscillator is not equal to 0).

The answer by joshphysics is great but I think there is small typo in his post. Shouldn't the inequality read $E_i - E_0 \geq 0$. I tried to edit the post, but "edits must be at least 6 characters". And I'm unable to comment on the post.

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  • $\begingroup$ The energy of the ground-state is not necessarily zero. Consider for example the harmonic oscillator. Important physical effects, such as the Casimir force, are a result of the non-zero value of the ground state energy. $\endgroup$
    – zap
    Jan 4 '17 at 16:39
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What is zero temperature?
Zero temperature in this context means the temperature that is lower than the lowest excitation energy of the system, which means that a system with a continuous excitation spectrum might actually never be in the ground state.

When do we speak about zero temperature?
Nevertheless, zero temperature is often used as a convenient approximation for various calculations, where its smallness is determined by the characteristic terms included in the Hamiltonian, rather than true phsyics. Thus, it is not uncommon to hear statements like

My zero temperature is much higher than the Kondo temperature. (In this case it says that we neglect the Kondo effect, assuming $T\gg T_K$, but the temperature is low enough to, e.g., approximate Fermi functions by step function, consider that the level broadening is homogeneous, etc.)

Zero temperature is not attainable
Actually, zero temperature can never be reached. See, e.g., this question.

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