If we have a quantum channel mapping from a $d$-dimensional state to a $d$-dimensional state, it can be described by at most $d^2$ Kraus operators. Suppose our channel maps instead from a $d_1$-dimensional state to a $d_2$-dimensional state, with $d_1>d_2$, e.g. with the quantum operation of taking the partial trace over a mode. What is the maximum required number of Kraus operators to characterise the channel? Is it $d_1d_2$, analogous to the case where $d_1=d_2$?


The answer is yes, you need $d_1 d_2$ operators, as already pointed out in the other answer. Here I'll show explicitly why this is the case.

Let $\Phi\in\mathrm{T}(\mathcal X,\mathcal Y)$ be a CPTP operator sending maps in $\mathcal X$ into maps in $\mathcal Y$. The spaces $\mathcal X$ and $\mathcal Y$ have dimensions $d_1$ and $d_2$, respectively.

Short proof

The Kraus decomposition is essentially the spectral decomposition of the Choi operator $J(\Phi)$ of $\Phi$ (or rather, one can always choose Kraus operators that are eigenvectors, up to "unvectorisation", of the Choi), as shown for example in this question. The Choi operator $J(\Phi)$ is a linear operator in $\mathcal Y\otimes\mathcal X$, which is a space of dimension $d_1 d_2$. Therefore, the maximum rank of $J(\Phi)$ is $d_1d_2$.

The rank of a positive operator corresponds to the number of its non-zero eigenvalues, which in this case is the number of Kraus operators in the decomposition of the channel. You can thus see why the conclusion follows.

Longer proof

This is a longer proof that doesn't need to introduce the notion of Choi operator (although we are of course using it under the hood).

Complete positivity means that any extension $\Phi\otimes I_n\in\mathrm{T}(\mathcal X\otimes\mathcal Z,\mathcal Y\otimes \mathcal Z)$ is a positive map.

Let us write the action of $\Phi$ on a generic operator $X$ in terms of its matrix elements as $[\Phi(X)]_{ij}=\Phi_{ijk\ell}X_{k\ell}$ (this essentially amounts to be using the natural representation of the map). The action of an extension $\Phi\otimes I_n$ onto a generic map $Y\in\mathrm{Lin}(\mathcal X\otimes\mathcal Z)$ then reads $$[(\Phi\otimes I_n)Y]_{ij}^{\alpha\beta}=\sum_{k\ell}\Phi_{ijk\ell} Y_{k\ell}^{\alpha\beta},\tag A$$ where I use $\alpha,\beta$ to denote the indices of $\mathcal Z$. This must hold true for any such operator $Y$.

The complete positivity implies that $(\Phi\otimes I_n)Y$ must be a positive operator. Remembering that an operator $A$ is positive iff it can be written as $A_{ij}=\sum_k u^k_i\bar u^k_j$ with $\sum_i u^k_i \bar u^\ell_i=\lambda_k\delta_{k\ell}$, upon some reflection we can see that this means that we can "separate" the matrix elements of our operator $(\Phi\otimes I_n)Y$ as follows: $$[(\Phi\otimes I_n)Y]_{ij}^{\alpha\beta}=\sum_a A^a_{i\alpha} \bar A^a_{j\beta}\tag B$$ for some objects $A$. This must hold true for any choice of positive operator $Y$.

In particular, it must hold true for the positive operator with components $Y_{k\ell}^{\alpha\beta}=\delta_{\alpha k}\delta_{\beta \ell}$ (which corresponds to the/a maximally entangled state). With this choice, (B) becomes $$\Phi_{ij\alpha\beta}=\sum_a A^a_{i\alpha} \bar A^a_{j\beta},$$ which you should recognise as the Kraus decomposition of the channel written in terms of the matrix elements. The number of elements in this sum equals the rank of the operator sending second and fourth indices into first and third (which is nothing but the Choi operator $J(\Phi)$ of $\Phi$). This is a linear operator in $\mathrm{Lin}(\mathcal Y\otimes \mathcal X)$, and as such can have maximum rank equal to $\operatorname{dim}(\mathcal X)\operatorname{dim}(\mathcal Y)=d_1 d_2$.


Yes. Choi's Theorem a priori uses different Hilbert spaces of potentially different dimensions $d_1$ and $d_2$. Then $d_1=d_2$ is a corollary. The proof is included there.


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