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In A theoretical framework for quantum networks is proven that a linear map $\mathcal{M} \in \mathcal{L}(\mathcal{H_0},\mathcal{H_1})$ is CP (completely positive) iff its Choi operator $M$ is semi definite positive. Something confuses me in this derivation.

First, some definition reminders.

Let $X \in \mathcal{L}(H_0,H_1)$, let $\{|i \rangle \}_i$ be an orthonormal basis of $H_0$, we have:

$$ | \mathcal{I} \rangle \rangle \equiv \sum_i |ii \rangle$$ $$|X \rangle \rangle \equiv (X \otimes \mathcal{I}) | \mathcal{I} \rangle \rangle$$

The Choi operator is defined as:

$$ M = \mathcal{M} \otimes \mathcal{I}_{H_0} | \mathcal{I} \rangle \rangle \langle \langle \mathcal{I} |$$

In his proof, he assumes $M \geq 0$ the goal is to show that it implies $\mathcal{M}$ is CP.

$M$ is semi definite positive which implies it is hermitian with positive eigenvalues. It can thus be diagonalized. With $\lambda_i \geq 0$, we have:

$$ M = \sum_i \lambda_i |u_i \rangle \langle u_i |=\sum_i | K_i \rangle \langle K_i |$$

With $|K_i \rangle = \sqrt{\lambda_i} |u_i \rangle$

But he seems to "automatically" consider that $|K_i \rangle = |K_i \rangle \rangle$. I don't understand that. Why would we necesseraly have $|K_i \rangle = (K_i \otimes \mathcal{I}) | \mathcal{I} \rangle \rangle$. It is a very particular case. Why can the state be written as a local operation acting on a maximally entangled state ?

I have a super vague memory that any quantum state can be written as $(K \otimes \mathbb{I}) | \mathcal{I} \rangle \rangle$. Said differently, there always exist a linear operation $K$ (not necesseraly unitary of course) such that any vector in $H_1 \otimes H_0$ can be written as $K \otimes \mathcal{I} | \mathcal{I} \rangle \rangle$ I guess it would solve the problem. But I cannot find the source of that and I may be totally wrong.

In the end, why can we write: $|K_i \rangle = |K_i \rangle \rangle$. I would like a proof of that (and if the property I just talked about holds I would like a link to a reference expressing it or a proof of that as well in the answer)

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  • $\begingroup$ so is the question specifically about $\lvert K\rangle=(K\otimes I)\lvert I\rangle$? If so, I don't think you need to mention the Choi in the question's title $\endgroup$ – glS Aug 19 at 10:03
  • $\begingroup$ @glS well I had the feeling such thing was necessary but I wasn't sure about it. Originally it was really about the derivation. But answering the question under this angle was for me the best way to understand it. $\endgroup$ – StarBucK Aug 19 at 10:05
  • $\begingroup$ Thus I wouldn't say my question was about this property. But this property being true or not solves my question the way I understand it better. $\endgroup$ – StarBucK Aug 19 at 10:06
  • $\begingroup$ to be clear, I don't think you should remove the context about the Choi from the body of the question, that is absolutely fine. However, the answers to the question are about that specific fact, so to make those answers easier to find in the future it would be much better to have a title reflecting that. Otherwise one would have no idea what the answers are going to be by simply reading the generic title "A question about proving this implication". Remember, questions are mostly useful when they can be used by other people having similar problems in the future $\endgroup$ – glS Aug 19 at 10:08
  • $\begingroup$ @glS I see. I will try to edit the title accordingly to what you suggest $\endgroup$ – StarBucK Aug 19 at 10:09
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Let $K$ be a vector $$ |K\rangle=\sum_{ij}K_{ij}|i,j\rangle. $$ We could rewrite this ias $$ |K\rangle=\left(\left(\sum_{ij}K_{ij}|i\rangle\langle j|\right)|j\rangle\right)\otimes|j\rangle, $$ and this is just the same as $$ |K\rangle=K\otimes 1\sum_j|j,j\rangle=|K\rangle\rangle $$ if we define the matrix $K$ to be $K=\sum_{ij}K_{ij}|i\rangle\langle j|$.

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  • $\begingroup$ Ooooh great it was actually that simple. Thanks ! $\endgroup$ – StarBucK Aug 19 at 9:26
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You've already defined the Choi matrix as $M = \rho_{\mathrm{Choi}} = \left(\mathcal{M}\otimes I\right)(|\mathcal{I}\rangle\rangle\langle\langle\mathcal{I}||)$. I'm gonna write the maximally entangled state as $|\mathcal{\Omega}\rangle$ because it's better readable to me and I'm more accustomed to it.

You've already pointed out that $M$ being positive-semidefinite means that we can perform a real-valued spectral decomposition:

$$ M = \sum_{i}\lambda_{i}|u_{i}\rangle\langle u_{i}| = \sum_{i}\sqrt{\lambda_{i}}|u_{i}\rangle\langle u_{i}| \sqrt{\lambda_{i}}. $$ We can decompose these $\sqrt{\lambda_{i}}|u_{i}\rangle$'s into a tensor product of a basis for both of the copies of the Hilbert spaces: $$ \sqrt{\lambda_{i}}|u_{i}\rangle = \sum_{l}|a^{i}_{l}\rangle \otimes |b^{i}_{l}\rangle, $$

which means that we can write: \begin{equation} \begin{split} M =& \sum_{i}\lambda_{i}|u_{i}\rangle\langle u_{i}| = \sum_{i}\sum_{l}\sum_{m} |a^{i}_{l}\rangle \otimes |b^{i}_{l}\rangle \langle a^{i}_{m}| \otimes \langle b^{i}_{m}| \\ =& \sum_{i,l,m} |a^{i}_{l}\rangle \langle a^{i}_{m}| \otimes |b^{i}_{l}\rangle \langle b^{i}_{m}|. \end{split} \end{equation}

As you may be well aware, we can write the 'output' of the map $\mathcal{M}$ on 'input' $\rho_{\mathrm{in}}$, which thus is $\rho_{\mathrm{out}} = \mathcal{M}\left(\rho_{\mathrm{in}}\right)$, in terms of the Choi matrix $M$:

$$ \mathcal{M}\left(\rho_{\mathrm{in}}\right) = d \mathrm{tr}_{2}\big[M\left(I \otimes \rho_{\mathrm{in}}^{T}\right)\big], $$ where the trace is the partial trace over the second subsystem, and the $T$ superscript means the transpose.

Now, we plug in our above decomposition for $M$: \begin{equation} \begin{split} \mathcal{M}\left(\rho_{\mathrm{in}}\right) &= d\mathrm{tr}_{2}\big[M\left(I \otimes \rho_{\mathrm{in}}^{T}\right)\big] \\ &= d\mathrm{tr}_{2}\big[\sum_{i,l,m} |a^{i}_{l}\rangle \langle a^{i}_{m}| \otimes |b^{i}_{l}\rangle \langle b^{i}_{m}| \left(I \otimes \rho_{\mathrm{in}}^{T}\right)\big] \\ &= d\sum_{i,l,m}\mathrm{tr}_{2}\big[ |a^{i}_{l}\rangle \langle a^{i}_{m}| \otimes |b^{i}_{l}\rangle \langle b^{i}_{m}| \rho_{\mathrm{in}}^{T}\big] \\ &= d\sum_{i,l,m}|a^{i}_{l}\rangle \langle a^{i}_{m}| \langle b^{i}_{m}| \rho_{\mathrm{in}}^{T}|b^{i}_{l}\rangle \\ &= d\sum_{i,l,m}|a^{i}_{l}\rangle \langle a^{i}_{m}| \langle b^{*i}_{l}| \rho_{\mathrm{in}}|b^{*i}_{m}\rangle \\ &= d\sum_{i,l,m}|a^{i}_{l}\rangle \langle b^{*i}_{l}| \rho_{\mathrm{in}}|b^{*i}_{m}\rangle \langle a^{i}_{m}| \\ &= \sum_{i} A_{i} \rho_{\mathrm{in}} A_{i}^{\dagger}, \end{split} \end{equation} with $A_{i} = \sum_{l}\sqrt{d} |a^{i}_{l}\rangle \langle b^{*i}_{l}|$. This is just the Kraus decomposition, which is enough for $\mathcal{M}$ being CP.

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  • $\begingroup$ Thank you for your answer. Two questions: why do you have a $d$ that multiplies $tr_2$ in your definition of the isomorphism ? Is it because in your convention $| \Omega \rangle$ is normalized (whereas in mine it is not) ? Second question: does that mean in an indirect way that indeed any vector belonging in $H_1 \otimes H_0$ can be written under the form $A \otimes \mathcal{I} | \Omega \rangle$ ? $\endgroup$ – StarBucK Aug 18 at 18:54
  • $\begingroup$ You're welcome! The $d$ is indeed because of normalization - I've been very sloppy with normalization in the above derivation (I also swept under the rug that the $a$ and $b$ vectors aren't properly normalized Hilbert-space vectors, but you can check that it works out in the end (if you still don't agree, I would be interested in a discussion :) )). I'm not $100 \%$ sure about the second question though - the partial trace is a vital part of the derivation so I don't think that the mapping works always. I'll have to think about it more. $\endgroup$ – JSdJ Aug 18 at 19:05
  • $\begingroup$ You might be able to use the first identity on this wikipedia page about the vectorization operator. (Coincidentally, this is also identity you can use to arrive at a 'natural representation' of a channel as a big matrix, which you asked about in one of your other questions today) $\endgroup$ – JSdJ Aug 18 at 19:06
  • $\begingroup$ Thanks for your answer/comment. It solved a part of my problem but DaftWullie gave me the exact step I was looking for so I accepted his answer instead. Thanks anyway for your instructive answer ! $\endgroup$ – StarBucK Aug 19 at 9:27
  • $\begingroup$ You're welcome! Daftwullie's answer is indeed really nice :) $\endgroup$ – JSdJ Aug 19 at 9:31
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Let $\newcommand{\kett}[1]{\lvert #1\rangle\!\rangle}\newcommand{\ket}[1]{\lvert#1\rangle}\ket m\equiv \sum_k \ket{k,k}$ denote the (unnormalised) maximally entangled state.

The relation $\kett X=(X\otimes I)\ket m$ amounts to some simple index juggling. By this, I mean that you are considering the same object, i.e. the same set of numbers, but interpreting it in different ways (as an operator rather than as a vector).

To see this, let $X\in\mathcal L(H_0,H_1)$ be your operator, whose matrix elements (in some choice of basis) we write as $X_{ij}$. Note that you can understand $X_{ij}$ as an operator ("sending the index $j$ to the index $i$"), or as a vector in $H_0\otimes H_1$. More formally, if we write with $\kett X$ the "vector interpretation" of $X$, we have $$\langle i,j\kett X = X_{ij} =\langle i|X|j\rangle = \langle i,j|(X\otimes I)\ket m,$$ where we used $\langle i,j|X\otimes I|k,\ell\rangle = X_{ik}\delta_{j\ell},$ and thus $\kett X=(X\otimes I)\ket m.$ This is also often written as $\kett X=\operatorname{vec}(X)$, with $\operatorname{vec}:\mathrm{Lin}(\mathcal X,\mathcal Y)\to\mathcal Y\otimes\mathcal X$ the "vectorisation" operation.

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