Chapter 1: Fundamental Concepts

1.1 Fundamental Assumptions

Five postulates of quantum mechanics:

  1. a state can be described by a vector in a Hilbert space: \(\mathcal{H} \to \ket{\psi}\); also \(\ket{\psi} \equiv e^{i \delta} \ket{\psi}\)

  2. Observables are represented by Hermitian operators: \(\hat{A} = \hat{A}^\dagger\)

  3. \([\hat{x}, \hat{p}] = i \hbar\); also \([\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A}\)

    Operator is a kind of Operation.

  4. Schrödinger equation: \(i \hbar \frac{\partial}{\partial t} \ket{\psi(t)} = \hat{H} \ket{\psi(t)}\)

  5. Identical particles are indistinguishable. \(\psi(x_1, x_2) = \pm \psi(x_2, x_1)\)

Similarities between Hirbert space and Euclidean space:

Hirbert Space Euclidean Space
Basis \(\ket{\phi_n}, \bra{\phi_n}\ket{\phi_m} = \delta_{nm}\) \(\hat{e}_n, \hat{e}_n \cdot \hat{e}_m = \delta_{nm}\)
State \(\ket{\psi} = \sum \alpha_n \ket{\phi_n}\) \(\vec{v} = \sum a_n \hat{e}_n\)
Operator \(\ket{\phi'} = \hat{O} \ket{\phi}, \hat{O} = \sum \alpha_{mn} \ket{\phi_m} \bra{\phi_n}\) \(\vec{v'} = \hat{M} \vec{v}, \hat{M} = \sum m_{ij} \hat{e}_i \hat{e}_j\)
Matrix Representation \(\hat{O} \to O_{mn} = \bra{\phi_m} \hat{O} \ket{\phi_n}\) \(\hat{M} \to M_{ij} = \hat{e}_i \cdot \hat{M} \cdot \hat{e}_j\)

1.2 Quantum Bomb Thought Experiment

Beam Splitter:

\[ \begin{pmatrix} \ket{1} \\ \ket{2} \end{pmatrix} \to \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \ket{1} \\ \ket{2} \end{pmatrix} \]

Mach-Zehnder Interferometer:

A Mach-Zehnder interferometer consists of two beam splitters (BS1 and BS2) and two mirrors (M1 and M2). A single photon enters from path \(\ket{1}\).

The mirror simply reflects the photon without changing its path label, contributing a phase factor. For simplicity we absorb mirror phases into the path definition, so mirrors act as identity on the two-path space.

Denote the beam splitter matrix as:

\[ \hat{B} = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} \]

After BS1, the photon entering path \(\ket{1}\) becomes:

\[ \ket{1} \xrightarrow{BS1} \frac{1}{\sqrt{2}} (-\ket{1} + \ket{2}) \]

After the mirrors redirect both paths and BS2 acts again:

\[ \frac{1}{\sqrt{2}} (-\ket{1} + \ket{2}) \xrightarrow{BS2} \frac{1}{\sqrt{2}} \left[ -\frac{1}{\sqrt{2}}(-\ket{1}+\ket{2}) + \frac{1}{\sqrt{2}}(\ket{1}+\ket{2}) \right] = \frac{1}{2}(2\ket{1}) = \ket{1} \]

The total transformation is \(\hat{B}^2\):

\[ \hat{B}^2 = \frac{1}{2}\begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Due to destructive interference, the photon always exits through path \(\ket{1}\) and never through path \(\ket{2}\).

Elitzur–Vaidman Quantum Bomb Test:

Now place a “bomb” on path \(\ket{2}\) between the two beam splitters. The bomb is a detector: if the photon takes path \(\ket{2}\), the bomb explodes (measurement collapses the state); if the photon takes path \(\ket{1}\), nothing happens.

After BS1, the state is \(\frac{1}{\sqrt{2}}(-\ket{1} + \ket{2})\). Three outcomes are possible:

Probability Event Detector \(\ket{2}\) outcome
\(1/2\) Photon takes path \(\ket{2}\) \(\to\) bomb explodes Bomb triggered
\(1/4\) Photon takes path \(\ket{1}\) \(\to\) after BS2, exits as \(\ket{1}\) Inconclusive
\(1/4\) Photon takes path \(\ket{1}\) \(\to\) after BS2, exits as \(\ket{2}\) Bomb detected without triggering

The last case is remarkable: the photon exits through path \(\ket{2}\), which is impossible in the normal (no-bomb) interferometer. Detecting the photon at path \(\ket{2}\) certifies that a bomb is present, even though the photon never interacted with it. This is an example of interaction-free measurement.

The success probability of non-destructive detection is \(1/4\) per trial. Using the quantum Zeno effect (many weak beam splitters), this can be improved arbitrarily close to \(1\).

1.3 Hermitian Operators

  1. Linear operator: \(\hat{A} (\alpha \ket{\phi} + \beta \ket{\psi}) = \alpha \hat{A} \ket{\phi} + \beta \hat{A} \ket{\psi}\)

    Antilinear operator: \(\hat{A} (\alpha \ket{\phi} + \beta \ket{\psi}) = \alpha^* \hat{A} \ket{\phi} + \beta^* \hat{A} \ket{\psi}\)

    Antilinear operator reveals the time-reversal symmetry.

  2. Hermitian conjugate: \(\bra{\phi} \hat{A}^\dagger \ket{\psi} = (\bra{\psi} \hat{A} \ket{\phi})^*\)

    \(\bra{\psi} \hat{A} \ket{\psi}\) is valid only if \(\hat{A}\) is Hermitian, because we don’t know the sequence of the operations.

1.4 Non-commuting Observables and Uncertainty Relations

\([\hat{x}, \hat{p}] = i \hbar\), \(\left\langle(\Delta x)^2\right\rangle \left\langle(\Delta p)^2\right\rangle \geq \frac{\hbar^2}{4}\)

Proof of x-p Uncertainty Relation

Assume \(\hat{A}, \hat{B}, \hat{C}\) are Hermitian operators, and \([\hat{A}, \hat{B}] = i \hat{C}\). Let \(\Delta \hat{A} = \hat{A} - \left\langle\hat{A}\right\rangle\), \(\Delta \hat{B} = \hat{B} - \left\langle\hat{B}\right\rangle\). Then

\[ [\Delta \hat{A}, \Delta \hat{B}] = [\hat{A}, \hat{B}] = i \hat{C} \]

Construct a inequality:

\[ \left\langle(\Delta \hat{A} + i \lambda \Delta \hat{B})^\dagger (\Delta \hat{A} + i \lambda \Delta \hat{B})\right\rangle \geq 0 \]

so that

\[ \left\langle(\Delta \hat{A})^2\right\rangle + \lambda^2 \left\langle(\Delta \hat{B})^2\right\rangle + i \lambda \left\langle[\Delta \hat{A}, \Delta \hat{B}]\right\rangle = \left\langle(\Delta \hat{A})^2\right\rangle + \lambda^2 \left\langle(\Delta \hat{B})^2\right\rangle + \lambda \left\langle\hat{C}\right\rangle \geq 0 \]

we can get a quadratic inequality of \(\lambda\), and we have

\[ \left\langle (\Delta \hat{B})^2 \right\rangle \lambda^2 + \left\langle\hat{C}\right\rangle \lambda + \left\langle(\Delta \hat{A})^2\right\rangle \geq 0 \]

so we conclude the discriminant must be non-positive:

\[ \left\langle\hat{C}\right\rangle^2 - 4 \left\langle(\Delta \hat{A})^2\right\rangle \left\langle(\Delta \hat{B})^2\right\rangle \leq 0 \]

also the uncertainty relation:

\[ \left\langle(\Delta \hat{A})^2\right\rangle \left\langle(\Delta \hat{B})^2\right\rangle \geq \frac{\left\langle\hat{C}\right\rangle^2}{4} \]

Proof of E-t Uncertainty Relation

Lemma: \(\frac{d}{dt} \left\langle\hat{A}\right\rangle = \frac{i}{\hbar} \left\langle[\hat{H}, \hat{A}]\right\rangle + \left\langle\frac{\partial \hat{A}}{\partial t}\right\rangle\)

Hensenberg Picture: \(\hat{A}_H(t) = e^{i \hat{H} t / \hbar} \hat{A} e^{-i \hat{H} t / \hbar}\). So the Lemma is equivalent to \(\frac{d}{dt} \hat{A}_H(t) = \frac{i}{\hbar} [\hat{H}, \hat{A}_H(t)] + (\frac{\partial \hat{A}}{\partial t})_H\).

Proof:

so we have \(i \hbar \dot{\hat{f}} = [\hat{f}, \hat{H}]\),

\[ \left\langle(\Delta \hat{f})^2\right\rangle \left\langle(\Delta \hat{H})^2\right\rangle \geq \frac{\left\langle[\hat{f}, \hat{H}]\right\rangle^2}{4} = \frac{\hbar^2}{4} \left\langle\dot{\hat{f}}\right\rangle^2 \]

1.5 Schodinger Equation and Generator

consider a \(\ket{\psi (x,t)}\), and define a time evolution operator:

\[ \hat{U}(\Delta t) \ket{\psi (x, t)} = \ket{\psi (x, t + \Delta t)} \]

which has the following properties:

  • \(U^\dagger U = 1\)
  • \(U(\Delta t \to 0) \to 1\)

\(U\) can be expressed as \(\hat{U}(\Delta t) = e^{- \frac{i}{\hbar} \hat{H} \Delta t} \approx 1 - \frac{i}{\hbar} \hat{H} \Delta t\), where \(\hat{H}\) is Hermitian. So we have the Schrödinger equation:

\[ i \hbar \frac{\partial}{\partial t} \ket{\psi (t)} = \hat{H} \ket{\psi (t)} \]

Translation operator: \(\hat{T}(\Delta x) = e^{- \frac{i}{\hbar} \hat{p} \Delta x} \approx 1 - \frac{i}{\hbar} \hat{p} \Delta x\).

Angular momentum operator: \(\hat{R}(\Delta \theta) = e^{- \frac{i}{\hbar} \hat{L} \Delta \theta} \approx 1 - \frac{i}{\hbar} \hat{L} \Delta \theta\).

Active and Passive Transformations

  • Active transformation: \(\ket{\psi} \to \ket{\psi'} = \hat{U} \ket{\psi}\)
  • Passive transformation: \(\ket{\psi} \to \ket{\psi'} = \hat{U}^\dagger \ket{\psi}\)

Example: Translation operator. Contider the original state: \(\psi(x) = \bra{x} \ket{\psi}\), we have two ways to get the translated state \(\psi'(x)\):

  • Operating on the state (active): \(\psi'(x) = \bra{x} \hat{T}(\Delta x) \ket{\psi} = \psi(x - \Delta x)\)
  • Operating on the basis (passive): \(\psi'(x) = \bra{x} \hat{T}^\dagger(\Delta x) \ket{\psi} = \psi(x + \Delta x)\)

Heisenberg Equation of Motion

Define \(\dot{\left\langle \hat{f} \right\rangle} \equiv \frac{d}{d t} \left\langle f \right\rangle\), we have

\[ \begin{aligned} \dot{\left\langle \hat{f} \right\rangle} & \equiv \frac{d}{d t} \left\langle f \right\rangle = \frac{d}{d t} \bra{\psi(t)} \hat{f} \ket{\psi(t)} \\ & = \left( \frac{d}{d t} \bra{\psi(t)} \right) \hat{f} \ket{\psi(t)} + \bra{\psi(t)} \hat{f} \left( \frac{d}{d t} \ket{\psi(t)} \right) + \bra{\psi(t)} \frac{\partial \hat{f}}{\partial t} \ket{\psi(t)} \\ & = -\frac{i}{\hbar} \bra{\psi(t)} \hat{H} \hat{f} \ket{\psi(t)} + \frac{i}{\hbar} \bra{\psi(t)} \hat{f} \hat{H} \ket{\psi(t)} + \bra{\psi(t)} \frac{\partial \hat{f}}{\partial t} \ket{\psi(t)} \\ & = \frac{\partial}{\partial t} \left\langle \hat{f} \right\rangle + \frac{i}{\hbar} \left\langle [\hat{f}, \hat{H}] \right\rangle \end{aligned} \]

1.6 Density Matrix

We introduce the consept in a two-level system:

  • Pure state: \(\ket{\psi} = \alpha \ket{0} + \beta \ket{1}\)
  • Mixed state: \(\{(p_i, \ket{\psi_i})\}\), where \(p_i\) is the probability of the state \(\ket{\psi_i}\)

So the mixed state can be seen as an ensemble of pure states. The density matrix is defined as:

\[ \hat{\rho} = \sum_i p_i \ket{\psi_i} \bra{\psi_i} \]

Density Matrix of Pure/Mixed State:

  • Pure state: \(\hat{\rho} = \ket{\psi} \bra{\psi}\), \(\hat{\rho}^2 = \hat{\rho}\), \(\text{Tr}(\hat{\rho}^2) = 1\)
  • Mixed state: \(\hat{\rho} = \sum_i p_i \ket{\psi_i} \bra{\psi_i}\), \(\hat{\rho}^2 \neq \hat{\rho}\), \(\text{Tr}(\hat{\rho}^2) < 1\)

Evolution of Density Matrix

The evolution of the density matrix is given by the von Neumann equation:

\[ i \hbar \frac{\partial \hat{\rho}}{\partial t} = [\hat{H}, \hat{\rho}] \]

Proof: Consider time-dependent density matrix \(\hat{\rho}(t) = \sum_i p_i \ket{\psi_i(t)} \bra{\psi_i(t)}\), we have

\[ \begin{aligned} \frac{d}{d t} \hat{\rho} (t) & = \sum_i p_i \left( \frac{d}{d t} \ket{\psi_i(t)} \bra{\psi_i(t)} + \ket{\psi_i(t)} \frac{d}{d t} \bra{\psi_i(t)} \right) \\ & = \sum_i p_i \left( -\frac{i}{\hbar} \hat{H} \ket{\psi_i(t)} \bra{\psi_i(t)} + \ket{\psi_i(t)} \frac{i}{\hbar} \bra{\psi_i(t)} \hat{H} \right) \\ & = -\frac{i}{\hbar} \hat{H} \hat{\rho}(t) + \frac{i}{\hbar} \hat{\rho}(t) \hat{H} = -\frac{i}{\hbar} [\hat{H}, \hat{\rho}(t)] \end{aligned} \]

Common Properties of Density Matrix

  1. The expectation value of an observable \(\hat{A}\) in density matrix formalism:

    \[ \left\langle \hat{A} \right\rangle = \text{Tr}(\hat{\rho} \hat{A}) \]

    Proof: \[ \begin{aligned} \text{Tr}(\hat{\rho} \hat{A}) & = \sum_n \bra{n} \hat{\rho} \hat{A} \ket{n} = \sum_n \bra{n} \left( \sum_i p_i \ket{\psi_i} \bra{\psi_i} \right) \hat{A} \ket{n} \\ & = \sum_i p_i \sum_n \bra{n} \ket{\psi_i} \bra{\psi_i} \hat{A} \ket{n} = \sum_i p_i \bra{\psi_i} \hat{A} \ket{\psi_i} = \left\langle \hat{A} \right\rangle \end{aligned} \]

  2. \(\text{Tr}(\hat{\rho}) = 1, \,\hat{\rho} = \hat{\rho}^\dagger, \, \text{Tr}(\hat{\rho}^2) \leq 1\) (only equal for pure state)

    Proof: Obvious.

Reduced Density Matrix

Consider a bipartite system \(\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B\), the reduced density matrix is defined as:

\[ \hat{\rho}_A = \text{Tr}_B(\hat{\rho}) = \sum_j \left( I_A \otimes \bra{j}_B \right) \hat{\rho} \left( I_A \otimes \ket{j}_B \right) \]

Example: Double Spin-1/2 System. In this system, a common pure state can be expressed as:

\[ \ket{\psi} = c_1 \ket{\uparrow \uparrow} + c_2 \ket{\uparrow \downarrow} + c_3 \ket{\downarrow \uparrow} + c_4 \ket{\downarrow \downarrow}, \quad \sum_i c_i^2 = 1 \]

where the density matrix is \(\hat{\rho} = \ket{\psi} \bra{\psi}\). Operate partial trace on the second spin, we can get the reduced density matrix of the first spin:

\[ \begin{aligned} \hat{\rho}_1 & = \text{Tr}_2(\hat{\rho}) = \sum_{s=\uparrow, \downarrow} \left( I \otimes \bra{s} \right) \hat{\rho} \left( I \otimes \ket{s} \right) = \left( I \otimes \bra{\uparrow} \right) \hat{\rho} \left( I \otimes \ket{\uparrow} \right) + \left( I \otimes \bra{\downarrow} \right) \hat{\rho} \left( I \otimes \ket{\downarrow} \right) \\ & = (|c_1|^2 + |c_2|^2) \ket{\uparrow} \bra{\uparrow} + (c_1 c_3^* + c_2 c_4^*) \ket{\uparrow} \bra{\downarrow} + (c_3 c_1^* + c_4 c_2^*) \ket{\downarrow} \bra{\uparrow} + (|c_3|^2 + |c_4|^2) \ket{\downarrow} \bra{\downarrow} \end{aligned} \]

for example, if \(\ket{\psi} = \frac{1}{\sqrt{2}} (\ket{\uparrow \downarrow} - \ket{\downarrow \uparrow})\), we have

\[ \hat{\rho}_1 = \frac{1}{2} \left( \ket{\uparrow} \bra{\uparrow} + \ket{\downarrow} \bra{\downarrow} \right) \]

1.7 Quantum Entropy and Ensemble

Let’s review the classical entropy first (Gibbs-Shannon entropy):

\[ S(X) \equiv - \sum_{x=1}^n P_x \log P_x \]

In quantum system, the von Neumann entropy is defined as:

\[ S = -k_B \text{Tr}(\hat{\rho} \ln \hat{\rho}) \]

Here we discuss three common ensembles in statistical mechanics.

Micro-canonical Ensemble

Consider an isolated quantum system:

  • No interaction with the external environment;
  • Total energy is conserved;
  • Hamiltonian is \(\hat{H}\), satisfying \(\hat{H}\lvert n\rangle = E_n \lvert n\rangle\).

We focus on the eigenstates whose energies fall within a certain “energy shell interval” \((E, E+\Delta E)\). These eigenstates span a subspace \[ \mathcal{H}_{\text{shell}} = \mathrm{span}\{{\lvert n\rangle \mid E < E_n < E+\Delta E}\} \]

The dimension of this subspace is denoted as \[ \Omega(E,\Delta E) \equiv \dim \mathcal{H}_{\text{shell}}, \] and we take an orthonormal basis \[ \lvert 1\rangle,\lvert 2\rangle,\dots,\lvert\Omega\rangle. \]

In the microcanonical ensemble, our fundamental assumption is: the system can only appear in states within this energy shell subspace, and we do not consider eigenstates outside the energy shell.

Maximum Entropy Principle and Microcanonical Distribution

In microcanonical ensemble, the only constraint is normalization: \[ \sum_{i=1}^{\Omega} P_i = 1. \]

We want to maximize under this constraint \[ S = -\sum_{i=1}^{\Omega} P_i\log P_i. \]

Construct the Lagrangian function \[ \tilde S_L = -\sum_{i=1}^{\Omega} P_i\log P_i - \alpha\left(\sum_{i=1}^{\Omega} P_i - 1\right), \] where \(\alpha\) is the Lagrange multiplier.

Taking the partial derivative with respect to each \(P_i\) and setting it to zero: \[ \frac{\partial \tilde S_L}{\partial P_i} = -(\log P_i + 1) - \alpha = 0, \] we obtain \[ \log P_i = -1 - \alpha \quad\Rightarrow\quad P_i = e^{-(1+\alpha)}. \]

Note that the right-hand side is independent of \(i\), which means all states must have equal probability. Using the normalization condition \[ \sum_{i=1}^{\Omega} P_i = \Omega P_i = 1 \] we solve for \[ P_i = \frac{1}{\Omega},\quad i=1,\dots,\Omega. \]

Thus, the maximum entropy density matrix is \[ \hat{\rho}_{\text{MC}} = \sum_{i=1}^{\Omega} \frac{1}{\Omega}\lvert i\rangle\langle i\rvert = \frac{1}{\Omega}\hat{I}_{\text{shell}}, \] where \(\hat{I}_{\text{shell}}\) is the identity operator on the energy shell subspace.

This is the quantum microcanonical ensemble:

Within a given energy shell, the system is equally distributed over all accessible energy eigenstates.

The entropy is \[ S_{\text{MC}} = \log \Omega(E,\Delta E), \] which is consistent with the structure in classical statistical physics \[ S = k_B \log\Omega, \] differing only by a factor of \(k_B\).

Canonical Ensemble

Consider a system interacting with a large heat bath:

  • The system can exchange energy with the heat bath, but cannot exchange particles;
  • The total system (system + heat bath) is isolated;
  • At equilibrium, the system has a definite temperature \(T\) (or \(\beta = 1/k_B T\));
  • The average energy of the system is \[ \langle H\rangle = \mathrm{Tr}(\hat{\rho}\hat{H}) = E. \]

For the system, the constraints are: \[ \mathrm{Tr}(\hat{\rho}) = 1,\qquad \mathrm{Tr}(\hat{\rho}\hat{H}) = E. \]

The goal is still to maximize the von Neumann entropy under these constraints \[ S(\hat{\rho}) = -\mathrm{Tr}(\hat{\rho}\log\hat{\rho}). \]

Lagrange Variation and Gibbs State

Construct the Lagrangian functional \[ \tilde S[\hat{\rho}] = -\mathrm{Tr}(\hat{\rho}\log\hat{\rho}) - \alpha\big(\mathrm{Tr}(\hat{\rho}) - 1\big) - \beta \big(\mathrm{Tr}(\hat{\rho}\hat{H}) - E\big), \] where \((\alpha,\beta)\) are Lagrange multipliers.

Perform variation with respect to the operator \((\hat{\rho})\). Note the following operator differential formula: \[ \delta \mathrm{Tr}(\hat{\rho}\log\hat{\rho}) = \mathrm{Tr}\big[\delta\hat{\rho}(\log\hat{\rho} + \mathbb{I})\big]. \]

Therefore \[ \delta\tilde S = -\mathrm{Tr}\big[\delta\hat{\rho}(\log\hat{\rho} + \mathbb{I})\big] - \alpha\mathrm{Tr}(\delta\hat{\rho}) - \beta\mathrm{Tr}(\delta\hat{\rho}\hat{H}). \]

Using the cyclic property of the trace, we extract \((\delta\hat{\rho})\): \[ \delta\tilde S = -\mathrm{Tr}\big[\delta\hat{\rho}(\log\hat{\rho} + \mathbb{I} + \alpha\mathbb{I} + \beta\hat{H})\big] = 0. \]

Requiring \((\delta\tilde S = 0)\) for arbitrary \((\delta\hat{\rho})\), we obtain the operator equation \[ \log\hat{\rho} + (1+\alpha)\mathbb{I} + \beta\hat{H} = 0, \] i.e., \[ \log\hat{\rho} = -(1+\alpha)\mathbb{I} - \beta\hat{H}. \]

Exponentiating: \[ \hat{\rho} = \exp\big(-(1+\alpha)\mathbb{I} - \beta\hat{H}\big) = e^{-(1+\alpha)}e^{-\beta\hat{H}}. \]

Using the normalization condition \((\mathrm{Tr}(\hat{\rho})=1)\), we define the partition function \[ Z(\beta) \equiv \mathrm{Tr}\big(e^{-\beta\hat{H}}\big), \] then \[ \hat{\rho}_{\text{C}} = \frac{1}{Z(\beta)}e^{-\beta\hat{H}}. \]

This is the density matrix of the canonical ensemble (Gibbs state).

Energy Eigenbasis and Classical Gibbs Distribution

In the eigenbasis of \((\hat{H})\), \[ \hat{H}\lvert n\rangle = E_n\lvert n\rangle, \] we have \[ \hat{\rho}_{\text{C}} = \sum_n P_n \lvert n\rangle\langle n\rvert, \quad P_n = \frac{e^{-\beta E_n}}{Z}, \quad Z = \sum_n e^{-\beta E_n}. \]

In this case, the von Neumann entropy becomes \[ S(\hat{\rho}_{\text{C}}) = -\sum_n P_n\log P_n, \] which is formally identical to the classical Gibbs–Shannon entropy.

Partition Function, Free Energy and Thermodynamic Relations

Partition Function: \[ Z(\beta) = \mathrm{Tr}\big(e^{-\beta\hat{H}}\big) = \sum_n e^{-\beta E_n}. \]

Define the Helmholtz free energy: \[ F(T,V,N) = -k_B T \log Z. \]

Below are several key relations:

  1. Internal energy: \[ E = \langle H\rangle = \mathrm{Tr}(\hat{\rho}_{\text{C}}\hat{H}) = -\frac{\partial}{\partial\beta}\log Z. \]

  2. Entropy: \[ S = -\mathrm{Tr}(\hat{\rho}_{\text{C}}\log\hat{\rho}_{\text{C}}). \] Substituting \((\hat{\rho}_{\text{C}} = e^{-\beta H}/Z)\): \[ S = -\mathrm{Tr}\left(\frac{e^{-\beta H}}{Z} \big(-\beta H - \log Z\big)\right) = \beta E + \log Z. \] Using \((F = -k_B T\log Z)\) to eliminate \((\log Z)\), we obtain \[ S = -\left(\frac{\partial F}{\partial T}\right)_{V,N}. \]

  3. Pressure: If \((\hat{H} = \hat{H}(V))\), then \[ \left(\frac{\partial F}{\partial V}\right)_{T,N} = -P. \]

  4. Total differential of free energy: \[ dF = -SdT - PdV. \]

These relations completely connect quantum statistics with thermodynamics: through the quantum partition function \((Z)\), all equilibrium thermodynamic quantities can be calculated.

Grand Canonical Ensemble

Now consider a system that exchanges not only energy but also particles with a reservoir:

  • The system can exchange both energy and particles with the environment;
  • In equilibrium, the system has temperature \((T)\) and chemical potential \((\mu)\);
  • The average energy and average particle number are respectively \[ \langle H\rangle = \mathrm{Tr}(\hat{\rho}\hat{H}) = E, \quad \langle N\rangle = \mathrm{Tr}(\hat{\rho}\hat{N}) = N. \]

The constraints are \[ \begin{cases} \mathrm{Tr}(\hat{\rho}) = 1, \\ \mathrm{Tr}(\hat{\rho}\hat{H}) = E, \\ \mathrm{Tr}(\hat{\rho}\hat{N}) = N. \end{cases} \]

Again, we seek to maximize the von Neumann entropy \((S(\hat{\rho}))\) under these constraints.

Lagrange Variation and Grand Canonical Density Matrix

Construct the Lagrangian functional \[ \tilde S[\hat{\rho}] = -\mathrm{Tr}(\hat{\rho}\log\hat{\rho}) - \alpha\big(\mathrm{Tr}\hat{\rho} - 1\big) - \beta\big(\mathrm{Tr}(\hat{\rho}\hat{H}) - E\big) - \gamma\big(\mathrm{Tr}(\hat{\rho}\hat{N}) - N\big). \]

Following the same procedure as for the canonical ensemble, variation with respect to \((\hat{\rho})\) yields \[ \log\hat{\rho} = -(1+\alpha)\mathbb{I} - \beta\hat{H} - \gamma\hat{N}, \] thus \[ \hat{\rho} = \exp\big(-(1+\alpha)\mathbb{I} - \beta\hat{H} - \gamma\hat{N}\big) = e^{-(1+\alpha)}\exp(-\beta\hat{H} - \gamma\hat{N}). \]

Relating \((\gamma)\) to the chemical potential: let \[ \gamma = -\beta\mu, \] then \[ \hat{\rho}_{\text{GC}} = \frac{1}{Z_{\text{GC}}}\exp\big[-\beta(\hat{H}-\mu\hat{N})\big], \] where \[ Z_{\text{GC}} = \mathrm{Tr}\exp\big[-\beta(\hat{H}-\mu\hat{N})\big] \] is called the grand partition function.

In the common eigenbasis \((\lvert n,N\rangle)\) of \((\hat{H},\hat{N})\), \[ P_{n,N} = \frac{e^{-\beta(E_{n,N} - \mu N)}}{Z_{\text{GC}}} \] is the probability distribution of the grand canonical ensemble.

Grand Potential and Thermodynamic Relations

In the grand canonical ensemble, the commonly used thermodynamic potential is the grand potential: \[ \Omega(T,V,\mu) = -k_B T\log Z_{\text{GC}}. \]

Its relation to other thermodynamic quantities is \[ \Omega = F - \mu N = E - TS - \mu N, \] and it satisfies \[ \begin{aligned} S &= -\left(\frac{\partial \Omega}{\partial T}\right)_{V,\mu}\\ P &= -\left(\frac{\partial \Omega}{\partial V}\right)_{T,\mu}\\ N &= -\left(\frac{\partial \Omega}{\partial \mu}\right)_{T,V}. \end{aligned} \]

Therefore, the total differential of the grand potential is \[ d\Omega = -SdT - PdV - Nd\mu. \]