# Homework Help: Quantum Information Theory 1

• maomao
maomao
Homework Statement
Given a vector space ##X = \mathbb{C}^\Sigma##, with a standard basis ##\left\{ |a\rangle, a \in \Sigma \right\}##. Given two operators ##V, W \in \mathcal{L}(\mathcal{X})##, the Schur product of ##V## and ##W##, denoted by ##V \odot W \in \mathcal{L}(\mathcal{X})##, is defined as the operator whose components respect to the standard basis are:
##\langle a| V \odot W |b \rangle = \langle a| V |b \rangle \langle a| W |b\rangle , \forall a, b \in \Sigma##.
Given a positive semidefinite operator ##P \in \text{Pos}(\mathcal{X}) ##, the associated Schur superoperator to ##P##, denoted ##\Lambda_P \in \text{T}(\mathcal{X})##, is defined as ##\Lambda_P (X) = P \odot X, \forall X \in \mathcal{L}(\mathcal{X})##.
Prove that ##\Lambda_P (X) \in \text{CP}(\mathcal{X})##.
Relevant Equations
The relevant points I have to prove that a superoperator is completely positive (CP) ##\Phi \in \text{T}(\mathcal{X},\mathcal{Y})##, where ##\mathcal{X}## and ##\mathcal{Y}## are complex Euclidean spaces, are the following statements:
1. ##\Phi## is CP, e.g., ##\Phi \in \text{CP}(\mathcal{X}, \mathcal{Y})##.

2. ##\Phi \otimes \mathbb{I}_{\mathcal{L}(\mathcal{X})}## is positive.

3. ##J(\Phi) \in \text{Pos}(\mathcal{Y} \otimes \mathcal{X}) ##.

4. ##A_1, \cdots , A_N \in \mathcal{L}(\mathcal{X},\mathcal{Y})## exist so that,
## \Phi (X) = \sum_{k=1}^N A_k X A^{\dagger}_k , \forall X \in \mathcal{L}(\mathcal{X})##

5. Point 4. holds for ##N = \text{rank} [J( \Phi)]##.

6. There exists a complex Euclidean space ##\mathcal{Z}## and an operator ##A \in \mathcal{L}(\mathcal{X}, \mathcal{Y} \otimes \mathcal{Z})## such that:
## \Phi (X) = \text{Tr}_{\mathcal{Z}} (AXA^\dagger) , \forall X \in \mathcal{L}(\mathcal{X})##

7. Point 6. holds for ##\text{dim} \mathcal{Z} = \text{rank} [J(\Phi)]##.
I considered an operator ##X \in \mathcal{L}(\mathcal{X} \otimes \mathcal{K})##, that is positive, ##X \geq 0##. And I defined it as it follows:
##X = \sum_{i,j} a_{ij} ∣x_i \rangle \langle x_i ∣ \otimes ∣k_j \rangle \langle k_j∣ ##​
Where ##x_i## are basis for ##\mathcal{X}## and ##k_j## basis for ##\mathcal{Y}##. Then, we remember the Schur product for the operator: ##\Lambda_P (X) = P \odot X##.
Another important point is that a superoperator ##\Phi## preserves the trace: ##\text{Tr} [\Phi (X)] = \text{Tr} (X), \forall X \in \mathcal{L}(\mathcal{X})##, and since the trace is ##\text{Tr}(A) = \sum_i \langle i ∣ A ∣i \rangle##, then,

##\text{Tr} [\Phi (X)] = \text{Tr} [ P \odot X ] = \sum_k \langle k∣ P \odot X ∣k \rangle = \langle k∣ P ∣k \rangle \hspace{1mm} \sum_k \langle k∣ X ∣k \rangle##​

The sum in ##k## represents the Trace of ##X##, ##\text{Tr}(X)##, and this is why ##\langle k∣ P ∣k \rangle = 1##, but I am not sure how could this prove that ##\Lambda_P (X)## is CP.
I understand that, since X is semi-definite positive, ##\langle k∣ X ∣k \rangle \geq 0##. Then, if ##\langle k∣ P ∣k \rangle##, which I already said that it has to be equal to ##1## to hold the equality, is also non-negative, the product ##\langle k∣ P ∣k \rangle \langle k∣ X ∣k \rangle## will also be non-negative. Is this enough to prove that ##\Lambda_P (X) \geq 0##?

Last edited:

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