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Density operator at thermal equilibrium

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}

Motion under a spatially uniform time-dependent force

Consider a quantum particle of mass \(m\) moving along a line, taken to be the \(x\) axis, under the action of a spatially uniform time-dependent force \(F(t)\). The time-evolution of the particle's wave function \(\Psi(x,t)\) is governed by the Schrödinger equation, \[i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} - F(t) x \Psi(x,t) \,.\] The solution to this equation can be written as \[\Psi(x,t) = \exp \left[ i \frac{p(t) x - s(t)}{\hbar} \right] \Psi_0 \big( x-q(t), t \big) \,,\] where \(\Psi_0\) is the free-particle wave function that initially coincides with \(\Psi\), \[\Psi_0(x,0) = \Psi(x,0) \,,\] and the functions \(p(t)\), \(q(t)\), and \(s(t)\) are defined as \[p(t) = \int_0^t d\tau \, F(\tau) \,,\] \[q(t) = \frac{1}{m} \int_0^{t} d\tau \, p(\tau) = \frac{1}{m} \int_0^{t} d\tau \int_0^{\tau} d\tau' \, F(\tau') \,,\] \[s(t) = \frac{1}{2 m} \int_0^t d\tau \, p^2(\tau) = \frac{1}{2 m} \int_

Density operator of a subsystem

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 \(\{ | b \rangle \}\)

Chebyshev polynomial expansion of the time-evolution operator

Here we outline an efficient numerical method of solving the time-dependent Schrödinger equation, \[i \hbar \frac{\partial | \psi(t) \rangle}{\partial t} = H | \psi(t) \rangle \,.\] The method is based on the Chebyshev polynomial expansion of the time-evolution operator. Only the basic idea is presented in these notes. Further details on the method can be found, for instance, in the following papers: An accurate and efficient scheme for propagating the time dependent Schrödinger equation Unified framework for numerical methods to solve the time-dependent Maxwell equations (in particular, section 3.3) The Hamiltonian \(H\) is assumed to have no explicit dependence on time. For instance, if the system under consideration is a particle of mass \(m\) moving in the presence of a static potential \(V(x)\), then \[\psi(x,t) = \langle x | \psi(t) \rangle\] is the particle's wave function, and \[H = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x) \,.\] In what

Robertson-Schrödinger uncertainty relation

Consider a quantum system in a state \(| \psi \rangle\), and let \(A\) and \(B\) be Hermitian operators representing a pair of observables. One can choose to perform a measurement of \(A\) or a measurement of \(B\) on the system. Generally, the outcomes of these measurements cannot be predicted with certainty. The outcomes are statistical in nature and characterized by the respective expectation values \(\langle A \rangle\) and \(\langle B \rangle\). Hereinafter, \[\langle \cdot \rangle = \langle \psi | \cdot | \psi \rangle \,.\] The corresponding uncertainties are defined as \[\sigma_A = \sqrt{ \langle \big( A - \langle A \rangle \big)^2 \rangle } = \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \,,\] \[\sigma_B = \sqrt{ \langle \big( B - \langle B \rangle \big)^2 \rangle } = \sqrt{\langle B^2 \rangle - \langle B \rangle^2} \,.\] The Robertson-Schrödinger uncertainty relation sates that \[\sigma_A^2 \sigma_B^2 \ge \langle \tfrac{i}{2} [A,B] \rangle^2 + \Big( \langle

Cauchy-Schwarz inequality

Let \(A\) be a positive-definite Hermitian operator. The following inequality holds for any pair of states \(| u \rangle\) and \(| v \rangle\): \[\big| \langle u | A | v \rangle \big|^2 \le \langle u | A | u \rangle \langle v | A | v \rangle \,.\] This is the Cauchy-Schwarz inequality (in a generalized form). In the case when \(A\) is the identity operator, the inequality reads \[\big| \langle u | v \rangle \big|^2 \le \langle u | u \rangle \langle v | v \rangle \,.\] Proof: Consider the state \[| \psi \rangle = z | u \rangle + | v \rangle \,,\] where \(z\) is a complex number. Since \(A\) is positive-definite, we have \[\langle \psi | A | \psi \rangle \ge 0 \,.\] This can be rewritten as \[\Big( \langle u | z^* + \langle v | \Big) A \Big( z | u \rangle + | v \rangle \Big) \ge 0 \,,\] or \[\langle u | A | u \rangle |z|^2 + \langle u | A | v \rangle z^* + \langle v | A | u \rangle z + \langle v | A | v \rangle \ge 0 \,.\] Since \(A^{\dagger} = A\), the last inequality becom