How to define a quantum channel for the partial trace?

Question

I understand that the partial trace is a linear map, a completely positive map and a trace-preserving map. However, I have no idea how to define a quantum channel with the partial trace because partial trace depends on an index $i$(i.e. $i$ of $\mathrm{Tr}_i(\rho)$). Intuitively, the index $i$ is random, but then I don't know how to define a map that represents the partial trace. Could anyone teach me the definition of a quantum channel using the partial trace?

Answer 1

The first and foremost thing to realize is that the partial trace over a density matrix is indeed a linear CPTP map $\Lambda$, but it is not a map from any $\mathcal{C}^{n\times n} \rightarrow \mathcal{C}^{n\times n}$ (i.e. to `itself' - the same dimension), but rather to a density operator space with a lower dimension: $\mathcal{C}^{n\times n} \rightarrow \mathcal{C}^{m\times m}$ with $m < n$.

For this reason, some of the common representations are not really usable; representations of a map $\Lambda$ like the Choi matrix or the $\chi$-matrix are often only used for maps that preserve the dimension.

However, if you are familiar with Kraus operators, they do offer a very straightforward representation. A partial trace over some index $i$ can be seen as the following operation:

\begin{equation} tr_{i} \big[\rho\big] = \sum_{j} \Big(I^{\otimes i-1}\otimes \langle{j}|\otimes I^{\otimes n-i}\Big) \rho \Big(I^{\otimes i-1}\otimes |{j}\rangle\otimes I^{\otimes n-i}\Big) \end{equation} That is, if I want to trace out the $i$-th qubit of some system of $n$ qubits described by $\rho$, I sum over an entire basis $\{|j\rangle \}$ on that qubit, and I do nothing (i.e. $I$, the identity operator) on all other $n-1$ qubits.

This representation allows us to obtain a straightforward Kraus representation of the map:

\begin{equation} A_{j} = I^{\otimes i-1}\otimes \langle{j}|\otimes I^{\otimes n-i}, \end{equation}

which is an operation that acts trivially on the first $i-1$ qubits and last $n-i$ qubits, and only acts nontrivially on the $i$-th qubit. Note that the Kraus operator is not square, which reflects the fact that the linear map is from a space of higher dimension to a space of lower dimension.

Furthermore, note that we are free to choose what basis we use for $\{|j\rangle \}$; the computational basis is often used. This reflects the fact that a set of Kraus operators for a map is not unique.

Generalization to tracing out more qubits is easily done by either applying the single-qubit trace in succession, or by replacing the $I$ operator of the specific qubit by a ket state as well - an important thing to notice here is that one needs to sum over the entire basis of the combined qubits that are being traced out. (That means that to trace out $k$ qubits there are $2^{k}$ Kraus operators associated with that map.)

Answer 2

$\newcommand{\on}[1]{{\operatorname{#1}}}$The basic answer is to just write, for a given bipartite state $X$, $\Phi(X)\equiv\operatorname{Tr}_B(X)$. This defines a quantum channel which acts on the input as a partial trace. You can check that this does indeed qualify as a quantum channel etc.

A misconception that might have led you to ask this question can be found in your saying that "I have no idea how to define a quantum channel with the partial trace because partial trace depends on an index $i$ (i.e. $i$ of $\operatorname{Tr}_i(\rho)$). Intuitively, the index $i$ is random, but then I don't know how to define a map that represents the partial trace. This is wrong. The index used to denote the partial trace represents which space is being traced out. It is not "random", but rather is something that has to be decided when the operation is defined. For example, for a bipartite system, you can always define the partial trace with respect to either of the two subspace, thus getting two different, if related, maps.

Another way to interpret the question is: "What does the partial trace map look like in one of the common quantum channel representations?".

^{A note on notation: I'll use here Einstein's summation notation (though using only lower indices) for repeated indices, as well as numbers instead of letters to represent indices. This is because I believe it makes it easier to see the structure of some expressions. Furthermore, I will sometimes use primed indices when an index represents an element in a dual vector space (i.e. it corresponds to a bra). For example, I'll write the matrix elements of a density matrix $\rho$ as $\rho_{11'}$. For a bipartite state $\sigma$, I will write $\sigma=\sigma_{12,1'2'}|1\rangle\!\langle 1'|\otimes|2\rangle\!\langle2'|$, where "$1$" and "$2$" correspond to kets in first and second space, while "$1',2'$" correspond to bras in the same spaces. Whenever suitable, I will include both the expression in operatorial form and the explicit base-dependent one with indices.}

Natural representation

(Natural representation in general) Given an arbitrary channel $\Phi$, denote with $K(\Phi)$ its natural representation. You can think of this as the linear map that acts on states via their vectorisations. Fixing a canonical basis, we can write $K(\Phi)$ explicitly through its matrix components as $$K(\Phi)\on{vec}(X) = \on{vec}(\Phi(X)), \qquad [\Phi(X)]_{11'}=K(\Phi)_{11',22'}X_{22'}.$$ One thing to notice at this point is that what the "$1$" and "$2$" indices represent does not really matter at this point. If $X$ is bipartite, then "$2$" is an index denoting an element of a bipartite space, and so it might be written as something like $2=(2_1 2_2)$, or directly $2\sim 23$, as I will do in the following. However, the natural representation is still possible, regardless of what the input and output spaces actually are.

(Natural representation of partial trace) In the case of $\Phi=\on{Tr}_B$ (where $B$ denotes here the second space), the input $X$ has a bipartite structure, and so we can write its matrix elements as $X_{12,1'2'}$, while the output does not have a tensor product structure (or at least we are not considering this structure here). We can then write the action of the partial trace in the natural representation as $$[\operatorname{Tr}_B(X)]_{11'}=K(\operatorname{Tr}_B)_{11',232'3'}X_{232'3'}.$$ From the definition of partial trace, we know that $[\operatorname{Tr}_B(X)]_{11'}= \sum_\alpha X_{1\alpha1' \alpha}\equiv X_{1\alpha 1'\alpha}.$ For the two expressions to be consistent, you then need the partial trace to have the following natural representation: $$ K(\on{Tr}_B)_{11',232'3'} = \delta_{12}\delta_{1'2'}\delta_{33'}. \tag A $$

(Alternative expression) We can get an expression with less floating indices by reasoning directly in terms of the operatorial relations: as pointed out before, we must have $K(\on{Tr}_B)\on{vec}(X) = \on{vec}(\on{Tr}_B(X))$. But we can also observe that $$\on{vec}(\on{Tr}_B(X)) = \sum_{i,j,\alpha} |i,j\rangle X_{i\alpha,j\alpha} = \left(\sum_\alpha I\otimes\langle \alpha|\otimes I\otimes\langle \alpha|\right)\on{vec}(X),$$ and thus $$K(\on{Tr}_B) = \sum_\alpha (I\otimes\langle \alpha|\otimes I\otimes\langle \alpha|)$$

Kraus representation

(Natural vs Kraus representation) Write the Kraus decomposition of $\Phi$ as $\Phi(X)=\sum_a A_a X A_a^\dagger \equiv A_a X A_a^\dagger.$ I will use "$a$" to index the Kraus operators, to distinguish the associated "auxiliary space" from indices denoting Hilbert space components. I will also not give any meaning to whether an index is raised or not (i.e. $A_a\equiv A^a$), as the meaning of the indices will be obvious from the context. The relation between this and the natural representation (again using non-bipartite notation for now) is $$ K(\Phi) = \sum_a A_a\otimes \bar A_a, \qquad K(\Phi)_{11',22'}=A^a_{12} \bar A^a_{1'2'}.$$ In the case of the partial trace, the input space (the index "$2$" here) has a bipartite structure while the output one doesn't, so the components of the $A_a$ operators are written as $A^a_{1,23}$ etc.

(Kraus repr of partial trace) Using the previous result about the natural representation of the partial trace, we can write $$\delta_{12}\delta_{1'2'}\delta_{33'}=A^a_{1,23}\bar A^a_{1',2'3'},$$ which means that $A^a_{1,23}=\delta_{12}B^a_3$ for some object $B$ such that $$B^a_3 \bar B^a_{3'}=\delta_{33'}.\tag{Y}$$ There are many equivalent choices for such an object. What we can observe is that (Y) can be understood as the orthonormality relation between a set of vectors, where "$3$" indexes the vectors and "$a$" is used for the components. This means that for (Y) to be possible, the auxiliary space in which the "$a$" indices live must have dimension greater than or equal to the dimension of the space of the index "$3$" (because you cannot have $n$ orthogonal vectors in a space of dimension $< n$). We can then take the dimensions to be equal, which makes $B$ a unitary operator with respect to its two indices. Because unitary matrices are all and only those whose rows (columns) form an orthonormal system for the space, and can freely choose which basis to use, the easier choice is to use the computational basis, which means that $B^a_3=\delta_{a3}$. It is trivial to check that this choice satisfies (Y), and therefore produces the correct Kraus operators. We conclude that the most natural (or a natural choice) for the Kraus operators of the partial trace is $$ A_a = I\otimes\langle a|, \qquad A^a_{1,23} = \delta_{12}\delta_{a3}.$$

Choi representation

This is straightforward from our analysis of the natural representation. Indeed, the Choi operator $J(\Phi)$ as the same matrix components as the natural representation $K(\Phi)$, up to some rearrangement. More specifically, $$J(\Phi)_{12,1'2'} = K(\Phi)_{11',22'}.$$ It then follows from (A) that $$J(\Phi)_{123,1'2'3'} = \delta_{12}\delta_{1'2'}\delta_{33'}.$$ Equivalently, as an operator, this reads $$J(\Phi) = |m\rangle\!\langle m|\otimes I,$$ where $|m\rangle\equiv \sum_i|i,i\rangle\!\langle i,i|$ is the maximally entangled state (modulo normalisation).

If you want to then represent this operator as a matrix, in the simplest case of all spaces being two-dimensional, you find a matrix that is the same as the density matrix of the Bell state $|00\rangle+|11\rangle$, upon replacing $1$ with the two-dimensional identity, and $0$ with the $2\times2$ zero matrix. More explicitly: $$ \left( \begin{array}{cccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right). $$

The fact that the $J(\Phi)$ is a square matrix while $K(\Phi)$ isn't is not a coincidence. The Choi is a square matrix for any map $\Phi$, while the natural representation is only square when input and output spaces have the same dimensions.

^{The Mathematica code to generate the above matrix is the following:}

SparseArray[{{i_, j_} :> 1 /; With[
       {iList = IntegerDigits[i - 1, 2, 3], 
        jList = IntegerDigits[j - 1, 2, 3]},
       And[
        iList[[1]] == iList[[2]],
        jList[[1]] == jList[[2]],
        iList[[3]] == jList[[3]]
        ]]}, {8, 8}] // Normal // TeXForm