Quantum Physics Corner

The density operator of a quantum system in thermal equilibrium at temperature \(T\) is given by \[\rho_T = \frac{e^{-H / k_{\text{B}} T}}{Z} \,,\] where \(H\) is the system Hamiltonian, \(k_{\text{B}}\) is the Boltzmann constant, and \[Z = \operatorname{Tr} e^{-H / k_{\text{B}} T}\] is the partition function. This expression for the density matrix can be obtained by extremizing the von Neumann entropy \[S = -k_{\text{B}} \operatorname{Tr} \rho \ln \rho\] over all states \(\rho\) of the same mean energy \[E = \operatorname{Tr} \rho H \,.\] Proof: We want to extremize \(S\) subject to (i) the normalization constraint, \(\operatorname{Tr} \rho = 1\), and (ii) the mean energy constraint, \(\operatorname{Tr} \rho H = E\). To this end, we introduce two Lagrange multipliers, \(\alpha k_{\text{B}}\) and \(\beta k_{\text{B}}\), and perform unconstrained extremization of \[\mathcal{F} = -k_{\text{B}} \operatorname{Tr} \rho \ln \rho - \alpha k_{\text{B}} \left( \operatorname{Tr}

Let us consider a closed quantum system \(S\) consisting of two subsystems \(S_1\) and \(S_2\). The Hilbert space \(\mathcal{H}\) of system \(S\) is a tensor product of Hilbert spaces \(\mathcal{H}_1\) and \(\mathcal{H}_2\) of subsystems \(S_1\) and \(S_2\), respectively, i.e., \[\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2 \,.\] Let \(\{ | a \rangle \}\) and \(\{ | b \rangle \}\) be complete orthonormal bases in \(\mathcal{H}_1\) and \(\mathcal{H}_2\), respectively, so that \[\langle a | a' \rangle = \delta_{aa'} \,, \qquad \langle b | b' \rangle = \delta_{bb'} \,,\] and any state \(| \Psi \rangle\) of \(S\) can be written as \[| \Psi \rangle = \sum_{a,b} \Psi_{ab} | a b \rangle \,, \qquad | a b \rangle \equiv | a \rangle \otimes | b \rangle \,.\] We take \(| \Psi \rangle\) to be normalized to one: \[\langle \Psi | \Psi \rangle = \sum_{a,b} |\Psi_{ab}|^2 = 1 \,.\] For notational simplicity, we treat sets \(\{ | a \rangle \}\) and \(\{ | b \rangle \}\)