# Quantum statistical mechanics

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.

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## Expectation

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

[itex] \operatorname{Exp}(X) = \int_\mathbb{R} \lambda \, d \, \operatorname{D}_X(\lambda) [itex]

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

[itex] \operatorname{E}_A(U) = \int_U \lambda d \operatorname{E}(\lambda), [itex]

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

[itex] \operatorname{D}_A(U) = \operatorname{Tr}(\operatorname{E}_A(U) S). [itex]

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

[itex] \operatorname{Exp}(A) = \int_\mathbb{R} \lambda \, d \, \operatorname{D}_A(\lambda).[itex]

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:

[itex] \operatorname{Exp}(A) = \operatorname{Tr}(A S) = \operatorname{Tr}(S A). [itex]

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

[itex] \operatorname{Exp}(A) = \langle \psi | A | \psi \rangle. [itex]

## Von Neumann entropy

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

[itex] \operatorname{H}(S) = -\operatorname{Tr}(S \log_2 S) [itex].

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

[itex] \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} [itex]

and we define

[itex] \operatorname{H}(S) = - \sum_i \lambda_i \log_2 \lambda_i. [itex]

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

[itex] 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} [itex]

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

[itex] \begin{bmatrix} \frac{1}{n} & 0 & \cdots & 0 \\ 0 & \frac{1}{n} & \dots & 0 \\ & & \cdots & \\ 0 & 0 & \cdots & \frac{1}{n} \end{bmatrix} [itex]

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

Recall that a pure state is one the form

[itex] S = | \psi \rangle \langle \psi |, [itex]

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

[itex] S= \frac{e^{- \beta H}}{\operatorname{Tr}(e^{- \beta H})} [itex]

where β is such that the ensemble average of energy satisfies

[itex] \operatorname{Tr}(S H) = E. [itex]

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

## References

• 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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