Quantum-mechanical ensemble theory - the density matrix

1. What's the situation?

We don't look at one quantum system, but an ensemble: a huge number \(\mathcal{N}\) of identical systems, all described by the same Hamiltonian (same physics), but not necessarily in the same quantum state.

For the k-th system, the wavefunction is \(\psi^k(t)\).

We choose some convenient orthonormal basis of states \(\{\phi_n\}\) (for example, energy eigenstates) and expand each wavefunction in that basis:

\[ \psi^k(t) = \sum_n a_n^{(k)}(t)\,\phi_n \]

The numbers \(a_n^{(k)}(t)\) are the probability amplitudes:

\(|a_n^{(k)}(t)|^2\) = probability that system k is in state \(\phi_n\) at time \(t\),
and for each system \(k\),

\[ \sum_n |a_n^{(k)}(t)|^2 = 1. \]

So: each system has its own list of probabilities for being in each state.

2. From amplitudes to the density matrix

Now we don't want to track every single system separately. We want a statistical description of the whole ensemble.

We introduce the density operator \(\hat{\rho}(t)\). In the basis \(\{\phi_n\}\) its matrix elements are

\[ \rho_{mn}(t) = \frac{1}{\mathcal{N}}\sum_{k=1}^{\mathcal{N}} a_m^{(k)}(t)\,a_n^{(k)*}(t). \]

Interpretation:

Diagonal elements \(\rho_{nn}(t) =\) average of \(|a_n^{(k)}(t)|^2\) over all systems. → Probability that a randomly chosen system is in state \(\phi_n\) at time \(t\).
Off-diagonal elements \(\rho_{mn}\) with \(m\neq n\) contain information about coherence / quantum superposition between states \(m\) and \(n\).

Normalization:

\[ \sum_n \rho_{nn}(t) = 1 \quad\Longleftrightarrow\quad \mathrm{Tr}(\hat{\rho})=1. \]

So \(\rho\) is just a fancy way of saying: "here are the probabilities (and coherences) for different quantum states in the ensemble."

3. How does \(\rho\) evolve in time?

Each wavefunction obeys the Schrödinger equation

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

and from that you can derive the equation of motion for \(\rho\):

\[ i\hbar\,\frac{d\hat{\rho}}{dt} = [\hat{H},\hat{\rho}] \]

This is equation (10): the Liouville–von Neumann equation. It's the quantum analog of the classical Liouville equation.

4. Equilibrium (stationary ensemble)

If the ensemble is in equilibrium, its statistical properties don't change in time:

\[ \frac{d\hat{\rho}}{dt} = 0. \]

From the equation of motion, that means

\[ [\hat{H},\hat{\rho}] = 0. \]

So:

\(\hat{\rho}\) must be a function of \(\hat{H}\) (they commute).
They can be diagonalized in the same basis: take \(\phi_n\) as energy eigenstates

\[ \hat{H}\phi_n = E_n \phi_n. \]

In that basis,

\[ H_{mn} = E_n\delta_{mn},\qquad \rho_{mn} = \rho_n\,\delta_{mn}. \]

So in equilibrium the density matrix is diagonal in the energy basis. The diagonal entries \(\rho_n\) are simply:

\[ \rho_n = \text{probability the system has energy }E_n. \]

5. Expectation value of an observable

Let \(\hat{G}\) be any physical quantity (operator). For the k-th system,

\[ G^{(k)} = \int \psi^{k*}\,\hat{G}\,\psi^k\,d\tau. \]

The ensemble average is

\[ \langle G \rangle = \frac{1}{\mathcal{N}}\sum_{k=1}^{\mathcal{N}} G^{(k)}. \]

If you rewrite this using the expansion coefficients and \(\rho\), you get the key formula

\[ \boxed{\langle G \rangle = \mathrm{Tr}(\hat{\rho}\,\hat{G})}. \]

Special case: \(\hat{G} = \hat{I}\) (identity operator):

\[ \mathrm{Tr}(\hat{\rho}) = 1, \]

which is exactly our normalization of probabilities.

Also, because the trace is basis-independent, the expectation value does not depend on which basis \(\{\phi_n\}\) you use. That's why (17) and (19) show the same result in any representation.

6. Microcanonical ensemble (section 5.2)

Now they specialise to a specific ensemble: microcanonical.

Assume:

fixed number of particles \(N\),
fixed volume \(V\),
energy is between \(E-\tfrac{1}{2}\Delta\) and \(E+\tfrac{1}{2}\Delta\) (a small window).

Let \(\Gamma(N,V,E;\Delta)\) ≡ number of accessible microstates in that energy window.

Key postulate: equal a priori probabilities – all accessible microstates are equally likely.

In the energy eigenbasis, the density matrix is diagonal:

\[ \rho_{mn} = \rho_n \delta_{mn}, \]

with

\[ \rho_n = \begin{cases} 1/\Gamma & \text{if state } n \text{ is in the allowed energy window},\\[4pt] 0 & \text{otherwise}. \end{cases} \]

You can check normalization:

\[ \sum_n \rho_n = \Gamma\cdot(1/\Gamma) = 1. \]

Then the entropy is

\[ S = k \ln \Gamma. \]

If there is only one microstate (\(\Gamma=1\)), the system is in a pure state, entropy \(S=0\) — this links to the third law of thermodynamics.

Summary

All of this is just a careful way to say:

"An ensemble of quantum systems is described by a density matrix \(\rho\), whose diagonal entries are the probabilities of different energy states; \(\rho\) evolves by \(i\hbar \dot{\rho} = [H,\rho]\), and expectation values are given by \(\langle G\rangle = \mathrm{Tr}(\rho G)\). For the microcanonical ensemble, all allowed energy states are equally probable, so \(\rho\) is diagonal with entries \(1/\Gamma\)."