# Two definitions of the density matrix?

There seems to be two different definitions of definitions of density matrices in Physics.

In Quantum Information we define a the density matrix associated with a wave function $| \psi \rangle$ as $| \psi \rangle \langle \psi |$. These represent pure states. Mixed states can be defined by taking convex combinations of these. In any case the density matrix depends on the state of the system and can evolve in time

In Quantum statistical mechanics the density matrix is defined as $e^{ -\beta H}$ . Where $\beta$ is the inverse temperature. By this definition the density matrix depends on the Hamiltonian ($H$) and not on what state the system is in.

I have two questions,

1. Are these definitions the same? If not what connects them?
2. What is the significance of $\beta$ in the second definition?
• Actually there is only the first definition. The 2nd one defines the thermal equilibrium state $\rho_{eq} = e^{-\beta H}/Z$ for inverse temperature $\beta = 1/kT$ and is a particular case of the first. So no, the 2nd one is not independent of the state, it is the state. If the system's density matrix is some $\rho \neq \rho_{eq}$ then the system is away from equilibrium.
– udrv
Apr 28 '16 at 6:28
• Thanks. If I understand you correctly then the second definition defines the state of the system that is in equilibrium with its surrounding. (Which I assume is at a temperature $1/k \beta$). Apr 28 '16 at 6:34
• Technically, since $\rho_eq$ commutes with the Hamiltonian $H$, it is a time-independent solution of the Liouville-von Neumann eq. in the absence of interactions, that is, ${\dot \rho_eq} = 0$. Thermal equilibrium with the surroundings corresponds to the limit of vanishing interactions with the surroundings, so yes, you are correct. But it applies to an isolated system too.
– udrv
Apr 28 '16 at 18:52

The Gibbs form $\rho\sim \mathrm{e}^{-\beta H}$ is just a fancy way of writing the standard Boltzmann distribution. A quantum (mixed) state is written in general as $$\rho = \sum_n p_n \lvert \phi_n\rangle \langle \phi_n\rvert, \qquad (1)$$ where $p_n$ is the probability to find the system in the pure state $\lvert \phi_n\rangle$. The thermal equilibrium state (a.k.a. Gibbs state) is often written as $\mathrm{e}^{-\beta H}/\mathcal{Z}$, where $\mathcal{Z} = \mathrm{Tr}(\mathrm{e}^{-\beta H} )$ is the partition function, and $\beta = 1/k_B T$ is the inverse temperature. To see what this means, simply expand $H$ in its basis of eigenvectors $H \lvert \psi_n\rangle = E_n \lvert \psi_n\rangle$, where $E_n$ is the energy of the eigenstate $\lvert \psi_n\rangle$, whereby
$$\frac{\mathrm{e}^{-\beta H}}{\mathcal{Z}} = \frac{1}{\mathcal{Z}}\sum_n \mathrm{e}^{-\beta E_n} \lvert \psi_n\rangle \langle \psi_n\rvert, \qquad \mathcal{Z} = \sum_n \mathrm{e}^{-\beta E_n}$$ But this is just the same as (1), with $p_n$ given by the Boltzmann distribution.