definition of the partial trace and its complete positivity

Question

I know the definition of the partial trace by its Kraus decomposition (which converges in trace norm to some trace class operator) $$ \mathrm{tr}_2(A) = \sum_k ( \mathbb{1} \otimes \langle f_k \rvert) A ( \mathbb{1} \otimes \rvert f_k \rangle),$$ where $\{f_k\}$ are any ONB of the separable Hilbert space that gets traced out and $(\mathbb{1} \otimes \rvert f_k\rangle)e = e \otimes f_k, (\mathbb{1} \otimes \langle f_k \rvert)(e \otimes g) = e\langle f_k,g \rangle$.

For the infinite dimensional case I have two questions:

1. In $\langle \phi , \mathrm{tr}_2(A)\psi\rangle = \sum_k \langle \phi \otimes f_k, A(\psi \otimes f_k)\rangle$, how is the "pointwise" evaluation of $\mathrm{tr}_2(A)\psi$ and pulling out the limit of the inner product justified?

2. How does one justify complete positivity? In finite dimensions one simply has for a positive operator $B$ on $\mathbb{C}^n \otimes H$, $(\mathrm{id}_{\mathbb{C}^n} \otimes \mathrm{tr}_2)B = \sum_k ( \mathrm{id}_{\mathbb{C}^n} \otimes \mathbb{1} \otimes \langle f_k \rvert) B ( \mathrm{id}_{\mathbb{C}^n} \otimes \mathbb{1} \otimes \rvert f_k \rangle)$, which is easily seen to be a sum of positive operators, thus positive. I cannot justify this in infinite dimensions.

I would appreciate any references to rigorous sources.

Answer 1

$\bullet$ For the first question, the computation goes as follows \begin{align*} \langle \phi,\operatorname{tr}_2(A)\psi\rangle&= \langle \phi,\big(\sum_k ({\bf 1}\otimes\langle f_k|)A({\bf 1}\otimes |f_k\rangle)\big)\psi\rangle\\ &=\sum_k \langle \phi,({\bf 1}\otimes\langle f_k|)A({\bf 1}\otimes |f_k\rangle)\psi\rangle\\ &=\sum_k \langle ({\bf 1}\otimes\langle f_k|)^\dagger\phi,A({\bf 1}\otimes |f_k\rangle)\psi\rangle\\ &=\sum_k \langle ({\bf 1}\otimes|f_k\rangle)\phi,A({\bf 1}\otimes |f_k\rangle)\psi\rangle=\sum_k \langle \phi\otimes f_k,A(\psi\otimes f_k)\rangle\,. \end{align*} I only wrote out this computation as explicitely because I wasn't sure what you meant by "[justifying] pointwise evaluation". For pulling out the sum, you use that the inner product is continuous (w.r.t. the norm by Cauchy-Schwarz, so pulling out limits such as a convergent infinite sum is allowed).

$\bullet$ For the second question, your original idea still works: the partial trace is a positive (i.e. positivity-preserving) map as is readily verified, and $\operatorname{tr}_2\otimes\operatorname{id}_n$ is still a partial trace (although now on the trace class over the larger space $H_1\otimes H_2\otimes\mathbb C^n$ to $H_1\otimes\mathbb C^n$).

A different but unnecessarily advanced argument would go via the dual channel: the operator $\operatorname{tr}_2(A)$ is the unique trace-class operator which satisfies $$ \operatorname{tr}(\operatorname{tr}_2(A)B)=\operatorname{tr}(A(B\otimes{\bf 1}))\qquad\text{ for all }B\in\mathcal B(H_1)\,. $$ thus its so called dual channel $\operatorname{tr}_2^*:\mathcal B(H_1)\to\mathcal B(H_1\otimes H_2)$ is given by $\operatorname{tr}_2^*(B)=B\otimes{\bf 1}$. One can generally show that this dual channel given via this trace duality is well-defined for all positive linear maps $T$ acting on the trace-class and, more importantly, that $T$ is completely positive if and only if its dual channel $T^*$ is. But completely positivity of $B\mapsto B\otimes{\bf 1}$ is straightforward.

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If you want to read up further on this topic (partial traces, dual channels, ...), I can recommend

• Chapter 2.3 and Chapter 6.2.2 in the so far unpublished book "Lectures in Quantum Noise Theory" of S. Attal or
• Chapter 4 & 8 in the "Quantum Effects" lecture notes by M. Wolf.

Answer 2

This is the way I answered my first question in the meantime: Denote $ E_k := ( \mathbb{1} \otimes \langle f_k \rvert)$. Since I know the definition $\sum_k E_k B E_k^\ast$ converges in trace norm to $\mathrm{tr}_2(B) $, we have for any bounded operator $A$ by Hölder $$ \lvert \mathrm{tr}[A \mathrm{tr}_2(B)] - \sum^n\mathrm{tr}[AE_k B E_k^\ast]\rvert = \lvert \mathrm{tr}[A(\mathrm{tr}_2(B) - \sum^n E_k B E_k^\ast)] \rvert \le \lVert A \rVert \lVert \mathrm{tr}_2(B) - \sum^n E_k B E_k^\ast \rVert_1 \to 0 $$ where $\lVert \cdot \rVert_1$ is trace norm. Using this and the definition of the trace one shows $\mathrm{tr}[A \mathrm{tr}_2(B)] = \mathrm{tr}[(A \otimes \mathbb{1})B].$ Then $$ \langle \phi, \mathrm{tr}_2(B) \psi \rangle = \mathrm{tr}[\lvert\psi\rangle\langle\phi\rvert\mathrm{tr}_2(B)] = \mathrm{tr}[(\lvert\psi\rangle\langle \phi \rvert \otimes \mathbb{1}) B] = \sum_{ik} \langle e_i \otimes f_k,(\lvert\psi\rangle\langle \phi \rvert \otimes \mathbb{1}) B (e_i \otimes f_k) \rangle \\=\sum_{ik} \langle(\lvert\phi\rangle\langle\psi\rvert \otimes \mathbb{1})(e_i\otimes f_k),B(e_i \otimes f_k)\rangle = \sum_{ik} \langle\langle\psi,e_i\rangle \phi \otimes f_k, B(e_i \otimes f_k)\rangle = \sum_{ik} \langle \phi \otimes f_k, B(\langle e_i,\psi\rangle e_i \otimes f_k\rangle = \sum_k \langle \phi\otimes f_k, B (\psi \otimes f_k)\rangle$$ where I used continuity of the inner product and $B$ and norm convergence of $\sum_i \langle e_i,\psi\rangle e_i$ to $\psi$.