Principles of quantum statistical mechanics

The problem of quantum statistical mechanics is the quantum mechanical treatment of an $N$-particle system. Suppose the corresponding $N$-particle classical system has Cartesian coordinates

$$q_1,...,q_{3N}$$

and momenta

$$p_1,...,p_{3N}$$

and Hamiltonian

$$H = \sum_{i=1}^{3N} {p_i^2 \over 2m_i} + U(q_1,...,q_{3N})$$

Then, as we have seen, the quantum mechanical problem consists of determining the state vector $\vert\Psi(t)\rangle$ from the Schrödinger equation

$$H\vert\Psi(t)\rangle = i\hbar{\partial \over \partial t}\vert\Psi(t)\rangle$$

Denoting the corresponding operators, $Q_1,...,Q_{3N}$ and $P_1,...,P_{3N}$, we note that these operators satisfy the commutation relations:

$$\left[Q_i,Q_j\right] = \left[P_i,P_j\right] = 0$$

$$\left[Q_i,P_j\right] = i\hbar I \delta_{ij}$$

and the many-particle coordinate eigenstate $\vert q_1...q_{3N}\rangle$ is a tensor product of the individual eigenstate $\vert q_1\rangle ,...,\vert q_{3N}\rangle$:

$$\vert q_1...q_{3N}\rangle = \vert q_1\rangle \cdots \vert q_{3N}\rangle$$

The Schrödinger equation can be cast as a partial differential equation by multiplying both sides by $\langle q_1...q_{3N}\vert$:

$$\langle q_1...q_{3N}\vert H\vert\Psi(t)\rangle = i\hbar {\partial \over \partial t}\langle q_1...q_{3N}\vert\Psi(t)\rangle$$

$$\left[-\sum_{i=1}^{3N}{\hbar^2 \over 2m_i}{\partial^2 \over \partial q_i^2} + U(q_1,...,q_{3N})\right]\Psi(q_1,...,q_{3N},t) = i\hbar {\partial \over \partial t} \Psi(q_1,...,q_{3N},t)$$

where the many-particle wave function is $\Psi(q_1,....,q_{3N},t) = \langle q_1...q_{3N}\vert\Psi(t)\rangle$. Similarly, the expectation value of an operator $A=A(Q_1,...,Q_{3N},P_1,...,P_{3N})$ is given by

$$\langle A \rangle = \int dq_1\cdots dq_{3N} \Psi^*(q_1,...,... ... i}{\partial \over \partial q_{3N}}\right)\Psi(q_1,...,q_{3N})$$