As stated in UT's QML course for a QPU that is interacting with the environment:

The environment is defined by a temperature $T$, and if we let the system equilibrate, the QPU will become thermalized at temperature $T$. As we saw in the notebook on evolution in open and closed systems, the energy of the states will follow a Boltzmann distribution: $\rho_0=\frac{1}{Z} e^{-H/T}$ where $Z=tr (e^{-H/T})$ is a normalization factor (called the partition function), ensuring that $tr(\rho_0)=1$. The inverse temperature $1/T$ is often denoted by $\beta$, so the state can also be written as $\rho_0=\frac{1}{Z} > e^{-\beta H}$. If $H$ has a discrete basis of orthonormal eigenstates $\{|n\rangle\}$ with eigenvalues $\{E_n\}$, we can write $H=\sum_n E_n > |n\rangle \langle n|$ and $\rho_0=\frac{1}{Z} \sum_n e^{-E_n/T} > |n\rangle \langle n|$ (since exponentiating a diagonal operator consists in exponentiating the elements of the diagonal). Hence, the thermal density matrix is a mixed state where each eigenstate of $H$ with energy $E$ has a classical probability $P(E)=\frac{1}{Z} > e^{-E/T}$, a Boltzmann distribution. We can see that the minimum energy eigenstate will have the highest probability. When $T > \rightarrow 0$, the minimum energy eigenstate will have a probability close to $1$. When $T \rightarrow \infty$, all the eigenstates tend to have equal probability

Does this mean that after running an algorithm in a system interacting with the environment, instead of getting an output pure state, we get a mixed state described by $\rho_0$?


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