Editing Quantum statistical mechanics (section)

== Von Neumann entropy ==<!-- This section is linked from [[Physical information]] -->
{{main|Von Neumann entropy}}

Of particular significance for describing randomness of a state is the von Neumann entropy of ''S'' ''formally'' defined by
<math display="block"> \operatorname{H}(S) = -\operatorname{Tr}(S \log_2 S). </math>  
Actually, the operator  ''S'' log<sub>2</sub> ''S'' is not necessarily trace-class. However, if ''S'' is a non-negative self-adjoint operator not of trace class we define Tr(''S'') = +&infin;.  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 display="block"> \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 & \cdots \\ 0 & \lambda_2 & \cdots & 0 & \cdots\\ \vdots & \vdots & \ddots & \\ 0 & 0 & &  \lambda_n & \\ \vdots & \vdots & & & \ddots \end{bmatrix} </math>
and we define
<math display="block"> \operatorname{H}(S) = - \sum_i \lambda_i \log_2 \lambda_i. </math>
The convention is that <math> \; 0 \log_2 0 = 0</math>, since an event with probability zero should not contribute to the entropy. This value is an extended real number (that is in [0, &infin;]) and this is clearly a unitary invariant of ''S''.

'''Remark'''. It is indeed possible that H(''S'') = +&infin; for some density operator ''S''. In fact ''T'' be the diagonal matrix
<math display="block"> 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\\ \vdots & \vdots & \ddots &  \\ 0 & 0 & &  \frac{1}{n (\log_2  n)^2 } & \\ \vdots & \vdots & & & \ddots \end{bmatrix} </math>
''T'' is non-negative trace class and one can show ''T'' log<sub>2</sub> ''T'' is not trace-class.

{{math theorem | Entropy is a unitary invariant.}}

In analogy with [[Shannon entropy#Formal definitions|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
<math display="block"> \begin{bmatrix} \frac{1}{n} & 0 & \cdots & 0 \\ 0 & \frac{1}{n} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots &  \frac{1}{n} \end{bmatrix} </math>
For such an ''S'', H(''S'') = log<sub>2</sub> ''n''. The state ''S'' is called the maximally mixed state.

Recall that a [[pure state]] is one of the form 
<math display="block"> S = | \psi \rangle \langle \psi |, </math>
for &psi; a vector of norm 1.

{{math theorem | math_statement =  {{math|1=H(''S'') = 0}} if and only if {{mvar|S}} is a pure state.}}

For {{mvar|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]].