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Quantum statistical mechanics : ウィキペディア英語版
Quantum statistical mechanics

Quantum statistical mechanics is statistical mechanics applied to quantum mechanical systems. In quantum mechanics a statistical ensemble (probability distribution over possible quantum states) 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.
== Expectation ==

From classical probability theory, we know that the expectation of a random variable ''X'' is completely determined by its distribution D''X'' by
: \mathbb(X) = \int_\mathbb \lambda \, d \, \operatorname_X(\lambda)
assuming, of course, that the random variable is integrable or that 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
: \operatorname_A(U) = \int_U \lambda d \operatorname(\lambda),
uniquely determines ''A'' and conversely, is uniquely determined by ''A''. E''A'' 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
: \operatorname_A(U) = \operatorname(\operatorname_A(U) S).
Similarly, the expected value of ''A'' is defined in terms of the probability distribution D''A'' by
: \mathbb(A) = \int_\mathbb \lambda \, d \, \operatorname_A(\lambda).
Note that this expectation is relative to the mixed state ''S'' which is used in the definition of D''A''.
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:
: \mathbb(A) = \operatorname(A S) = \operatorname(S A).
Note that if ''S'' is a pure state corresponding to the vector ψ, then:
: \mathbb(A) = \langle \psi | A | \psi \rangle.
The trace of an operator A is written as follows:
: \operatorname(A) = \sum_ \langle m | A | m \rangle .
== Von Neumann entropy ==
(詳細は)) 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
: T = \begin \frac& 0 & \cdots & 0 & \cdots \\ 0 & \frac & \cdots & 0 & \cdots\\ & & \cdots & \\ 0 & 0 & \cdots & \frac & \cdots \\ & & \cdots & \cdots \end
''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 (notice the similarity in the definitions), H(''S'') measures the amount of randomness in the state ''S''. The more dispersed 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
: \begin \frac & 0 & \cdots & 0 \\ 0 & \frac & \dots & 0 \\ & & \cdots & \\ 0 & 0 & \cdots & \frac \end
For such an ''S'', H(''S'') = log2 ''n''. The state ''S'' is called the maximally mixed state.
Recall that a pure state is one of the form
: S = | \psi \rangle \langle \psi |,
for ψ a vector of norm 1.
Theorem. H(''S'') = 0 if and only if ''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.
Entropy can be used as a measure of quantum entanglement.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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