Trace of an Operator - Imported Knowlpedia Content

Let $H$ be a finite-dimensional complex Hilbert space and let $A:H\to H$ be a linear operator. The trace of $A$ , denoted $\operatorname{Tr}(A)$ , is defined by choosing any orthonormal basis $(e_1,\dots,e_n)$ and setting

\operatorname{Tr}(A) := \sum_{k=1}^n \langle e_k, A e_k\rangle.

This value is independent of the chosen orthonormal basis.

This notion agrees with the usual matrix trace (see Trace) after identifying $A$ with its matrix in an orthonormal basis.

Equivalent descriptions (finite dimension)

If $A$ is represented by a matrix $(A_{ij})$ in any basis, then $\operatorname{Tr}(A)=\sum_i A_{ii}$ .
$\operatorname{Tr}(A)$ equals the sum of eigenvalues of $A$ , counted with algebraic multiplicity.

Properties

For operators $A,B$ on $H$ and scalars $\alpha,\beta$ :

Linearity: $\operatorname{Tr}(\alpha A+\beta B)=\alpha\operatorname{Tr}(A)+\beta\operatorname{Tr}(B)$ .
Cyclic property: $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ .
Unitary invariance: if $U$ is unitary, then $\operatorname{Tr}(U^\ast A U)=\operatorname{Tr}(A)$ .
Positivity: if $A$ is positive semidefinite, then $\operatorname{Tr}(A)\ge 0$ .

Trace and quantum expectation values

If $\rho$ is a Density Operator and $A$ is an observable (self-adjoint operator), the expected value of $A$ in state $\rho$ is

\mathbb{E}_\rho[A] = \operatorname{Tr}(\rho A).

In particular, density operators are normalized by the condition $\operatorname{Tr}(\rho)=1$ .

Note on infinite dimension

On an infinite-dimensional Hilbert space, not every bounded operator has a trace. One typically restricts to trace-class operators to extend $\operatorname{Tr}$ with similar properties.