# Mixed States, Gibbs States, and Temperature

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.

Observations:

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}

Questions:

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).

• 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. Jul 9, 2021 at 12:31

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 $$E = \mbox{Tr}[\rho H]$$