Why can every $|X\rangle\in H_1\otimes H_0$ be written as $|X\rangle=(X\otimes I_{H_0})|\Omega \rangle$ for some $X\in\mathcal L(H_0,H_1)$?

Question

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)

Answer 1

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 $K=\sum_{ij}K_{ij}|i\rangle\langle j|$.

Answer 2

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.

Answer 3

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.