I am trying to help myself clarify some concepts. The purpose of this post is twofold: 1) to get some comments/feedback, and 2) to help me organize my thoughts. I have been approaching these concepts from a more mathematical point of view, but quickly realize that this seems to be not enough.

Assumption: I am considering a closed system.

Here are the definitions that I am using:

Mixed states by definition are operators of the form \begin{align} \rho = \sum_{n} p_n|\psi_n\rangle\langle\psi_n| \end{align} where $p_n\ge 0$ and $\sum_n p_n = 1$. For convenience, I assume the sequence $\{\psi_n\}$ is an orthonormal family of vectors in some Hilbert space $\mathcal{H}$ (infinite dimensional, say $\mathcal{H}= L^2(\mathbb{R}^n)$). Note that this definition also contains pure states.

A Gibbs State by definition is a mixed state of the form \begin{align} \rho_G = \frac{e^{-\beta H}}{Z}= Z^{-1}\sum_n e^{-\beta E_n}|E_n\rangle\langle E_n| \end{align} where $H=\sum_n E_n |E_n\rangle\langle E_n|$ and $\{|E_n\rangle\}$ (clearly an abuse of notation) is the family of eigenfunctions of $H$. Let us assume $\beta\ge 0$ and $Z:=\operatorname{Tr}(e^{-\beta H})<\infty$.

Lastly, we define the temperature of $\rho_G$ by $T:= 1/(k\beta)$ where $k$ is some constant (for simplicity, let us take $k=1$). Note that this means that temperature is dependent on the Hamiltonian.


It is clear to me that the set of mixed states is clearly larger than the set of Gibbs states. The claim comes from the observation that $\rho = \frac{1}{2}\left(|\psi_1\rangle\langle \psi_1|+|\psi_2\rangle\langle \psi_2|\right)$ cannot be written as a Gibbs state since the spectrum of \begin{align} H= -\Delta+V \ \ \ (1) \end{align} is usually countably infinite at the very least (am I wrong?) This means that infinitely many $p_n$ are zero so that $e^{-\beta E_n} = 0$ implies $E_n=0$ or $T = \infty$. However, no one is demanding me to assume $H=-\Delta+V$. In fact, in theory, we could consider $H= E_1|\psi_1\rangle\langle\psi_1|+ E_2|\psi_2\rangle\langle\psi_2|$ (a 2-level system). Then, we have the Gibbs states \begin{align} \rho_G(\beta) = Z^{-1}(e^{-\beta E_1}|\psi_1\rangle\langle\psi_1|+e^{-\beta E_2}|\psi_2\rangle\langle\psi_2|). \end{align}


  1. Given a mixed state, can we view it as a Gibbs state of some Hamiltonian? (From a physics perspective, this question doesn't make too much sense.)
  2. Given a mixed state, can we associate a temperature to it intrinsically? (From my first question, it is clear that you can't since, there, the temperature depends on the Hamiltonian.)

In these questions, I have completely ignored the physics and just view the objects mathematically (computationally).

  • 2
    $\begingroup$ You can put a Hamiltonian on any mixed state to make it a Gibbs state. But most of the time it would be a totally unphysical, non-local, impossible-to-construct Hamiltonian. $\endgroup$ Commented Jul 9, 2021 at 12:31

1 Answer 1


Gibbs states are defined by the KMS condition not by the alegbraic expression. In a Hilbert space $ \mathcal{H}_d$ of dimension $d < \infty$, any state $\rho$ with rank $k\leq d$ can be interpreted as a Gibbs state with an arbitrary temperature $\beta$ and Hamiltonian $H$ like, $$ \rho = \frac{e^{-\beta H}}{ Z_\beta^H}$$

For states with less than full rank, one must define the Hamiltonian $H$ (and the corresponding dynamical flow) appropriately on a subspace. So, the set of all mixed states is identical to the set of Gibbs states.

The answer to your second question (2.) is a bit more delicate, the state $\rho$ does not describe the system but a only a preparation procedure of the thermal ensemble. Therefore, one cannot define a temperature of the system by specifying just a density matrix $\rho$.

Also, please note that the $\beta$ occuring in the expression is the temperature of the canonical bath and is not a feature of the system.

However, given a Hamiltonian of the system, one may infer the temperature by performing projective measurements in the energy eigenbasis. Alternatively, one may also perform tomography on a dynamical sequence of states to infer the temperature using the microcanonical expression, $$ \frac{\partial S}{\partial E} = \beta,$$

where $S$ is the von Neumann entropy and $E$ is the internal energy of the ensemble defined by $$E = \mbox{Tr}[\rho H]$$


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