Partial trace

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In linear algebra and functional analysis, the partial trace is a generalization of the trace. Whereas the trace is a scalar valued function on operators, the partial trace is an operator-valued function. The partial trace has applications in quantum information and decoherence which is relevant for quantum measurement and thereby to the decoherent approaches to interpretations of quantum mechanics, including consistent histories and the relative state interpretation..

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[edit] Details

Suppose V, W are finite-dimensional vector spaces over a field of dimension m, n respectively. The partial trace TrV is a mapping

T \in \operatorname{L}(V \otimes W) \mapsto \operatorname{Tr}_V(T) \in \operatorname{L}(V)

It is defined as follows: let

e_1, \ldots, e_m

and

f_1, \ldots, f_n

be bases for V and W respectively; then T has a matrix representation

\{a_{k \ell, i j}\} \quad 1 \leq k, i \leq m, \quad 1 \leq \ell,j \leq n

relative to the basis

e_k \otimes f_\ell

of

V \otimes W.

Now for indices k, i in the range 1, ..., m, consider the sum

b_{k, i} = \sum_{j=1}^n a_{k j, i j}.

This gives a matrix bk, i. The associated linear operator on V is independent of the choice of bases and is by definition the partial trace.

[edit] Invariant definition

The partial trace operator can be defined invariantly (that is, without reference to a basis) as follows: It is the unique linear operator

\operatorname{Tr}_V: \operatorname{L}(V \otimes W) \rightarrow \operatorname{L}(V)

such that

\operatorname{Tr}_V(R \otimes S) = R \, \operatorname{Tr}(S) \quad \forall R \in \operatorname{L}(V) \quad \forall S \in \operatorname{L}(W).

From this abstract definition, the following properties follow:

\operatorname{Tr}_V (1_{V \otimes W}) = \dim W \ 1_{V}
\operatorname{Tr}_V (T (1_V \otimes S)) = \operatorname{Tr}_V ((1_V \otimes S) T) \quad \forall S \in \operatorname{L}(W) \quad \forall T \in \operatorname{L}(V \otimes W).

[edit] Partial trace for operators on Hilbert spaces

The partial trace generalizes to operators on infinite dimensional Hilbert spaces. Suppose V, W are Hilbert spaces, and let

\{f_i\}_{i \in I}

be an orthonormal basis for W. Now there is an isometric isomorphism

\bigoplus_{\ell \in I} (V \otimes \mathbb{C} f_\ell) \rightarrow V \otimes W

Under this decomposition, any operator T \in \operatorname{L}(V \otimes W) can be regarded as an infinite matrix of operators on V

\begin{bmatrix} T_{11} & T_{12} & \ldots & T_{1 j} & \ldots \\ T_{21} & T_{22} & \ldots & T_{2 j} & \ldots \\ \vdots & \vdots & & \vdots \\ T_{k1}& T_{k2} & \ldots & T_{k j} & \ldots \\ \vdots & \vdots & & \vdots \end{bmatrix}

First suppose T is a non-negative operator. In this case, all the diagonal entries of the above matrix are non-negative operators on V. If the sum

\sum_{\ell} T_{\ell \ell}

converges in the strong operator topology of L(V), it is independent of the chosen basis of W. The partial trace TrV(T) is defined to be this operator. The partial trace of a self-adjoint operator is defined if and only if the partial traces of the positive and negative parts are defined.

[edit] Computing the partial trace

Suppose W has an orthonormal basis, which we denote by ket vector notation as \{| \ell \rangle\}_\ell. Then

\operatorname{Tr}_V\left(\sum_{k,\ell} T_{k \ell} \, \otimes \, | k \rangle \langle \ell |\right) = \sum_j T_{j j}

[edit] Partial trace and invariant integration

In the case of finite dimensional Hilbert spaces, there is a useful way of looking at partial trace involving integration with respect to a suitably normalized Haar measure μ over the unitary group U(W) of W. Suitably normalized means that μ is taken to be a measure with total mass dim(W).

Theorem. Suppose V, W are finite dimensional Hilbert spaces. Then

\int_{\operatorname{U}(W)} (1_V \otimes U^*) T (1_V \otimes U) \ d \mu(U)

commutes with all operators of the form 1_V \otimes S and hence is uniquely of the form R \otimes 1_W. The operator R is the partial trace of T.

[edit] Partial trace as a quantum operation

The partial trace can be viewed as a quantum operation. Consider a quantum mechanical system whose state space is the tensor product H_A \otimes H_B of Hilbert spaces. A mixed state is described by a density matrix ρ, that is a non-negative trace-class operator of trace 1 on the tensor product H_A \otimes H_B . The partial trace of ρ with respect to the system B, denoted by ρA, is called the reduced state of ρ on system A. In symbols,

\rho^A = \operatorname{Tr}_B \rho.

To show that this is indeed a sensible way to assign a state on the A subsystem to ρ, we offer the following justification. Let M be an observable on the subsystem A, then the corresponding observable on the composite system is M \otimes I. However one chooses to define a reduced state ρA, there should be consistency of measurement statistics. The expectation value of M after the subsystem A is prepared in ρA and that of M \otimes I when the composite system is prepared in ρ should be the same, i.e. the following equality should hold:

\operatorname{Tr} ( M \cdot \rho^A) = \operatorname{Tr} ( M \otimes I \cdot \rho).

We see that this is satisfied if ρA is as defined above via the partial trace. Furthermore it is the unique such operation.

Let T(H) be the Banach space of trace-class operators on the Hilbert space H. It can be easily checked that the partial trace, viewed as a map

\operatorname{Tr}_B : T(H_A \otimes H_B) \rightarrow T(H_A)

is completely positive and trace-preserving.

The partial trace map as given above is induces a dual map \operatorname{Tr}_B ^* between the C*-algebras of bounded operators on \; H_A and H_A \otimes H_B given by

\operatorname{Tr}_B ^* (A) = A \otimes I.

\operatorname{Tr}_B ^* maps observables to observables and is the Heisenberg picture representation of \operatorname{Tr}_B.

[edit] Comparison with classical case

Suppose instead of quantum mechanical systems, the two systems A and B are classical. The space of observables for each system are then abelian C*-algebras. These are of the form C(X) and C(Y) respectively for compact spaces X, Y. The state space of the composite system is simply

C(X) \otimes C(Y) = C(X \times Y).

A state on the composite system is a positive element ρ of the dual of C(X × Y), which by the Riesz-Markov theorem corresponds to a regular Borel measure on X × Y. The corresponding reduced state is obtained by projecting the measure ρ to X. Thus the partial trace is the quantum mechanical equivalent of this operation.

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