Quantum statistical mechanics

From Academic Kids

Quantum statistical mechanics is the study of statistical ensembles of quantum mechanical systems. A statistical ensemble is described by a density operator S, which is a non-negative, self-adjoint, trace-class operator of trace 1 on the Hilbert space H describing the quantum system. This can be shown under various mathematical formalisms for quantum mechanics. One such formalism is provided by quantum logic.



From classical probability theory we know that the expectation of a random variable X is completely determined by its distribution DX by

<math> \operatorname{Exp}(X) = \int_\mathbb{R} \lambda \, d \, \operatorname{D}_X(\lambda) <math>

assuming, of course that the random variable is integrable or the random variable is non-negative. Similarly, let A be an observable of a quantum mechanical system. A is given by a densely defined self-adjoint operator on H. The spectral measure of A defined by

<math> \operatorname{E}_A(U) = \int_U \lambda d \operatorname{E}(\lambda), <math>

uniquely determines A and conversely, is uniquely determined by A. EA is a boolean homomorphism from the Borel subsets of R into the lattice Q of self-adjoint projections of H. In analogy with probability theory, given a state S, we introduce the distribution of A under S which is the probability measure defined on the Borel subsets of R by

<math> \operatorname{D}_A(U) = \operatorname{Tr}(\operatorname{E}_A(U) S). <math>

Similarly, the expected value of A is defined in terms of the probability distribution DA by

<math> \operatorname{Exp}(A) = \int_\mathbb{R} \lambda \, d \, \operatorname{D}_A(\lambda).<math>

Note that this expectation is relative to the mixed state S which is used in the definition of DA.

Remark. For technical reasons, one needs to consider separately the positive and negative parts of A defined by the Borel functional calculus for unbounded operators.

One can easily show:

<math> \operatorname{Exp}(A) = \operatorname{Tr}(A S) = \operatorname{Tr}(S A). <math>

Note that if S is a pure state corresponding to the vector ψ,

<math> \operatorname{Exp}(A) = \langle \psi | A | \psi \rangle. <math>

Von Neumann entropy

Of particular significance for describing randomness of a state is the von Neumann entropy of S formally defined by

<math> \operatorname{H}(S) = -\operatorname{Tr}(S \log_2 S) <math>.

Actually the operator S log2 S is not necessarily trace-class. However, if S is a non-negative self-adjoint operator not of trace class we define Tr(S) = +∞. Also note that any density operator S can be diagonalized, that it can be represented in some orthonormal basis by a (possibly infinite) matrix of the form

<math> \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 & \cdots \\ 0 & \lambda_2 & \cdots & 0 & \cdots\\ & & \cdots & \\ 0 & 0 & \cdots & \lambda_n & \cdots \\ & & \cdots & \cdots \end{bmatrix} <math>

and we define

<math> \operatorname{H}(S) = - \sum_i \lambda_i \log_2 \lambda_i. <math>

This value is an extended real number (that is in [0, ∞]) and this is clearly a unitary invariant of S.

Remark. It is indeed possible that H(S) = +∞ for some density operator S. In fact T be the diagonal matrix

<math> T = \begin{bmatrix} \frac{1}{2 (\log_2 2)^2 }& 0 & \cdots & 0 & \cdots \\ 0 & \frac{1}{3 (\log_2 3)^2 } & \cdots & 0 & \cdots\\ & & \cdots & \\ 0 & 0 & \cdots & \frac{1}{n (\log_2 n)^2 } & \cdots \\ & & \cdots & \cdots \end{bmatrix} <math>

T is non-negative trace class and one can show T log2 T is not trace-class.

Theorem. Entropy is a unitary invariant.

In analogy with classical entropy, H(S) measures the amount of randomness in the state S. The more disperse the eigenvalues are, the larger the system entropy. For a system in which the space H is finite-dimensional, entropy is maximized for the states S which in diagonal form have the representation

<math> \begin{bmatrix} \frac{1}{n} & 0 & \cdots & 0 \\ 0 & \frac{1}{n} & \dots & 0 \\ & & \cdots & \\ 0 & 0 & \cdots & \frac{1}{n} \end{bmatrix} <math>

For such an S, H(S) = log2 n.

Recall that a pure state is one the form

<math> S = | \psi \rangle \langle \psi |, <math>

for ψ a vector of norm 1.

Theorem. H(S) = 0 iff S is a pure state.

For S is a pure state if and only if its diagonal form has exactly one non-zero entry which is a 1.

This incidentally is one justification for the use of entropy as a measure of quantum entanglement.

Gibbs canonical ensemble

Consider an ensemble of systems described by a Hamiltonian H with average energy E. If H has pure-point spectrum and the eigenvalues of H go to + ∞ sufficiently fast, e-r H will be a non-negative trace-class operator for ever positive r.

The Gibbs canonical ensemble is the state

<math> S= \frac{e^{- \beta H}}{\operatorname{Tr}(e^{- \beta H})} <math>

where β is such that the ensemble average of energy satisfies

<math> \operatorname{Tr}(S H) = E. <math>

Under certain conditions the Gibbs canonical ensemble maximizes the von Neumann entropy of the state subject to the energy conservation requirement.


  • J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
  • F. Reif, Statistical and Thermal Physics, McGraw-Hill, 1985.

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