# understanding final state of a QPU interacting with the environment

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 $$\rho_0$$?