# definition of the partial trace and its complete positivity

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.

$$\bullet$$ For the first question, the computation goes as follows $$\langle \phi,\operatorname{tr}_2(A)\psi\rangle= \langle \phi,\big(\sum\nolimits_k (1\otimes\langle f_k|)A(1\otimes |f_k\rangle)\big)\psi\rangle=\sum\nolimits_k \langle \phi,(1\otimes\langle f_k|)A(1\otimes |f_k\rangle)\psi\rangle\\ =\sum\nolimits_k \langle (1\otimes\langle f_k|)^\dagger\phi,A(1\otimes |f_k\rangle)\psi\rangle=\sum\nolimits_k \langle (1\otimes|f_k\rangle)\phi,A(1\otimes |f_k\rangle)\psi\rangle=\sum\nolimits_k \langle \phi\otimes f_k,A(\psi\otimes f_k)\rangle\,.$$ 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 $$\operatorname{tr}(\operatorname{tr}_2(A)B)=\operatorname{tr}(A(B\otimes\operatorname{id}))\qquad\text{ for all }B\in\mathcal B(H_1)\,.$$ thus its so called dual channel $$\operatorname{tr}_2^*:\mathcal B(H_1)\to\mathcal B(H_1\otimes H_2)$$ is given by $$\operatorname{tr}_2^*(B)=B\otimes\operatorname{id}$$. 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 $$B\mapsto B\otimes\operatorname{id}$$ is relatively easy to see.

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

• First of all thank you very much for your detailed answer. For the first question, it is exactly your second equality for which I was searching justification. Of course the inner product is continuous, but then you somehow need strong convergence? I don't see how that works. In the meantime I found an alternative justification, which I will provide in an answer below.
– T'x
Jun 29 '19 at 7:59
• Ah, of course now I see why your second equality holds. Trace norm convergence implies weak-* convergence which in turn implies weak convergence. But Cauchy-Schwarz is nowhere appearing?
– T'x
Jun 29 '19 at 14:07
• Hmm... good point. Cauchy-Schwarz doesn't seem to be necessary here, "comparing topologies" suffices. Anyways, I hope that clears up all you asked for in your question! Jun 29 '19 at 14:22
• Yes, it does, thank you again! :)
– T'x
Jun 29 '19 at 14:32

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 $$\psi$$.

• Yep that works just fine! Remark: what you essentially just showed for this explicit oeprator is that the weak operator topology (here: convergence in trace) is weaker than the strong operator topology (convergence evaluated on a trace class element, in trace norm) Jun 29 '19 at 10:52