In mathematics, specifically functional analysis, a trace-class operator is a linear operator for which a trace may be defined, such that the trace is a finite number independent of the choice of basis used to compute the trace. This trace of trace-class operators generalizes the trace of matrices studied in linear algebra. All trace-class operators are compact operators.

In quantum mechanics, mixed states are described by density matrices, which are certain trace class operators.

Trace-class operators are essentially the same as nuclear operators, though many authors reserve the term "trace-class operator" for the special case of nuclear operators on Hilbert spaces and use the term "nuclear operator" in more general topological vector spaces (such as Banach spaces).

Note that the trace operator studied in partial differential equations is an unrelated concept.

Definition

Suppose is a separable Hilbert space and a bounded linear operator on which is non-negative (I.e., positive semidefinite) and self-adjoint. The trace of , denoted by is the sum of the series[1]

where is an orthonormal basis of . The trace is a sum on non-negative reals and is therefore a non-negative real or infinity. It can be shown that the trace does not depend on the choice of orthonormal basis.

For an arbitrary bounded linear operator on we define its absolute value, denoted by to be the positive square root of that is, is the unique bounded positive operator on such that The operator is said to be in the trace class if We denote the space of all trace class linear operators on H by (One can show that this is indeed a vector space.)

If is in the trace class, we define the trace of by

where is an arbitrary orthonormal basis of . It can be shown that this is an absolutely convergent series of complex numbers whose sum does not depend on the choice of orthonormal basis.

When H is finite-dimensional, every operator is trace class and this definition of trace of T coincides with the definition of the trace of a matrix.

Equivalent formulations

Given a bounded linear operator , each of the following statements is equivalent to being in the trace class:

  • [1]
  • For some orthonormal basis of H, the sum of positive terms is finite.
  • For every orthonormal basis of H, the sum of positive terms is finite.
  • T is a compact operator and where are the eigenvalues of (also known as the singular values of T) with each eigenvalue repeated as often as its multiplicity.[1]
  • There exist two orthogonal sequences and in and a sequence in such that for all [2] Here, the infinite sum means that the sequence of partial sums converges to in H.
  • T is a nuclear operator.
  • T is equal to the composition of two Hilbert-Schmidt operators.[1]
  • is a Hilbert-Schmidt operator.[1]
  • T is an integral operator.[3]
  • There exist weakly closed and equicontinuous (and thus weakly compact) subsets and of and respectively, and some positive Radon measure on of total mass such that for all and :

Trace-norm

We define the trace-norm of a trace class operator T to be the value

One can show that the trace-norm is a norm on the space of all trace class operators and that , with the trace-norm, becomes a Banach space.

If T is trace class then[4]

Examples

Every bounded linear operator that has a finite-dimensional range (i.e. operators of finite-rank) is trace class;[1] furthermore, the space of all finite-rank operators is a dense subspace of (when endowed with the norm).[4] The composition of two Hilbert-Schmidt operators is a trace class operator.[1]

Given any define the operator by Then is a continuous linear operator of rank 1 and is thus trace class; moreover, for any bounded linear operator A on H (and into H), [4]

Properties

  1. If is a non-negative self-adjoint operator, then is trace-class if and only if Therefore, a self-adjoint operator is trace-class if and only if its positive part and negative part are both trace-class. (The positive and negative parts of a self-adjoint operator are obtained by the continuous functional calculus.)
  2. The trace is a linear functional over the space of trace-class operators, that is,
    The bilinear map
    is an inner product on the trace class; the corresponding norm is called the Hilbert–Schmidt norm. The completion of the trace-class operators in the Hilbert–Schmidt norm are called the Hilbert–Schmidt operators.
  3. is a positive linear functional such that if is a trace class operator satisfying then [1]
  4. If is trace-class then so is and [1]
  5. If is bounded, and is trace-class, then and are also trace-class (i.e. the space of trace-class operators on H is an ideal in the algebra of bounded linear operators on H), and[1] [5][1]
    Furthermore, under the same hypothesis,[1]
    and The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
  6. If and are two orthonormal bases of H and if T is trace class then [4]
  7. If A is trace-class, then one can define the Fredholm determinant of :
    where is the spectrum of The trace class condition on guarantees that the infinite product is finite: indeed,
    It also implies that if and only if is invertible.
  8. If is trace class then for any orthonormal basis of the sum of positive terms is finite.[1]
  9. If for some Hilbert-Schmidt operators and then for any normal vector holds.[1]

Lidskii's theorem

Let be a trace-class operator in a separable Hilbert space and let be the eigenvalues of Let us assume that are enumerated with algebraic multiplicities taken into account (that is, if the algebraic multiplicity of is then is repeated times in the list ). Lidskii's theorem (named after Victor Borisovich Lidskii) states that

Note that the series on the right converges absolutely due to Weyl's inequality

between the eigenvalues and the singular values of the compact operator [6]

Relationship between common classes of operators

One can view certain classes of bounded operators as noncommutative analogue of classical sequence spaces, with trace-class operators as the noncommutative analogue of the sequence space

Indeed, it is possible to apply the spectral theorem to show that every normal trace-class operator on a separable Hilbert space can be realized in a certain way as an sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of the compact operators that of (the sequences convergent to 0), Hilbert–Schmidt operators correspond to and finite-rank operators to (the sequences that have only finitely many non-zero terms). To some extent, the relationships between these classes of operators are similar to the relationships between their commutative counterparts.

Recall that every compact operator on a Hilbert space takes the following canonical form: there exist orthonormal bases and and a sequence of non-negative numbers with such that

Making the above heuristic comments more precise, we have that is trace-class iff the series is convergent, is Hilbert–Schmidt iff is convergent, and is finite-rank iff the sequence has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when is infinite-dimensional:

The trace-class operators are given the trace norm The norm corresponding to the Hilbert–Schmidt inner product is

Also, the usual operator norm is By classical inequalities regarding sequences,

for appropriate

It is also clear that finite-rank operators are dense in both trace-class and Hilbert–Schmidt in their respective norms.

Trace class as the dual of compact operators

The dual space of is Similarly, we have that the dual of compact operators, denoted by is the trace-class operators, denoted by The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let we identify with the operator defined by

where is the rank-one operator given by

This identification works because the finite-rank operators are norm-dense in In the event that is a positive operator, for any orthonormal basis one has

where is the identity operator:

But this means that is trace-class. An appeal to polar decomposition extend this to the general case, where need not be positive.

A limiting argument using finite-rank operators shows that Thus is isometrically isomorphic to

As the predual of bounded operators

Recall that the dual of is In the present context, the dual of trace-class operators is the bounded operators More precisely, the set is a two-sided ideal in So given any operator we may define a continuous linear functional on by This correspondence between bounded linear operators and elements of the dual space of is an isometric isomorphism. It follows that is the dual space of This can be used to define the weak-* topology on

See also

References

  1. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Conway 1990, p. 267.
  2. Trèves 2006, p. 494.
  3. Trèves 2006, pp. 502–508.
  4. 1 2 3 4 Conway 1990, p. 268.
  5. M. Reed and B. Simon, Functional Analysis, Exercises 27, 28, page 218.
  6. Simon, B. (2005) Trace ideals and their applications, Second Edition, American Mathematical Society.

Bibliography

  • Conway, John B. (1990). A course in functional analysis. New York: Springer-Verlag. ISBN 978-0-387-97245-9. OCLC 21195908.
  • Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.
  • Schaefer, Helmut H. (1999). Topological Vector Spaces. GTM. Vol. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
  • Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.