- Partial trace
In

linear algebra andfunctional analysis , the**partial trace**is a generalization of the trace. Whereas the trace is ascalar valued function on operators, the partial trace is an operator-valued function. The partial trace has applications inquantum information anddecoherence which is relevant forquantum measurement and thereby to the decoherent approaches tointerpretations of quantum mechanics , includingconsistent histories and therelative state interpretation .**Details**Suppose "V", "W" are finite-dimensional vector spaces over a field of

dimension "m", "n" respectively. The partial trace Tr_{"V"}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 "b"

_{"k", "i"}. The associated linear operator on "V" is independent of the choice of bases and is by definition the**partial trace**.**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 :$operatorname\{Tr\}\_V:\; operatorname\{L\}(V\; otimes\; W)\; ightarrow\; operatorname\{L\}(V)$ 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:

:$operatorname\{Tr\}\_V\; (1\_\{V\; otimes\; W\})\; =\; dim\; W\; 1\_\{V\}$

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

**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:$igoplus\_\{ell\; in\; I\}\; (V\; otimes\; mathbb\{C\}\; f\_ell)\; ightarrow\; V\; otimes\; W$

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

:$egin\{bmatrix\}\; T\_\{11\}\; T\_\{12\}\; ldots\; T\_\{1\; j\}\; ldots\; \backslash \; T\_\{21\}\; T\_\{22\}\; ldots\; T\_\{2\; j\}\; ldots\; \backslash \; vdots\; vdots\; vdots\; \backslash \; T\_\{k1\}\; T\_\{k2\}\; ldots\; T\_\{k\; j\}\; ldots\; \backslash \; vdots\; vdots\; vdots\; end\{bmatrix\}$

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 Tr_{"V"}("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.**Computing the partial trace**Suppose "W" has an orthonormal basis, which we denote by ket vector notation as $\{|\; ell\; angle\}\_ell$. Then

:$operatorname\{Tr\}\_Vleft(sum\_\{k,ell\}\; T\_\{k\; ell\}\; ,\; otimes\; ,\; |\; k\; angle\; langle\; ell\; |\; ight)\; =\; sum\_j\; T\_\{j\; j\}$

**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".

**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 adensity 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 $ho\; ^A$, is called the reduced state of ρ on system "A". In symbols,:$ho^A\; =\; operatorname\{Tr\}\_B\; ho.$

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 $ho^A$, there should be consistency of measurement statistics. The expectation value of "M" after the subsystem "A" is prepared in $ho\; ^A$ 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:

:$operatorname\{Tr\}\; (\; M\; cdot\; ho^A)\; =\; operatorname\{Tr\}\; (\; M\; otimes\; I\; cdot\; ho).$

We see that this is satisfied if $ho\; ^A$ 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

$operatorname\{Tr\}\_B\; :\; T(H\_A\; otimes\; H\_B)\; ightarrow\; T(H\_A)$is completely positive and trace-preserving.The partial trace map as given above is induces a dual map $operatorname\{Tr\}\_B\; ^*$ between the

C*-algebra s 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 $operatorname\{Tr\}\_B$.**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

:$C(X)\; otimes\; C(Y)\; =\; C(X\; imes\; Y).$

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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