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Can't find any info on Stinespring dilation so I thought I could post here. If I have a qubit complete positive map $\Lambda$, that maps all inputs to the output $|0\rangle$, $\Lambda(\rho)=|0\rangle\langle0|$ how can I derive a Stinespring dilation for $\Lambda$ and the corresponding Kraus operators?

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    $\begingroup$ This is a trap that students often fall into: the actual map is not $\Lambda(\rho)=|0\rangle\langle0|$, but $\Lambda(\rho)=|0\rangle\langle0|\operatorname{tr}(\rho)$. This looks like pedantism as $\operatorname{tr}(\rho)=1$, but it makes a difference; without $\operatorname{tr}(\rho)$ the map is not even linear, and you get the wrong result if you try to calculate its Choi-Jamiołkowski representation. $\endgroup$ – Mateus Araújo Mar 11 at 12:07
  • $\begingroup$ Thanks for the clarification $\endgroup$ – user14766 Mar 11 at 13:24
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Stinespring dilation can be thought of as a way of representing an arbitrary completely positive trace preserving map $\Lambda$ on a system $A$ as a composition of two simpler maps: a unitary evolution $U_{AE}$ in a Hilbert space obtained by adjoining an auxiliary system $E$ followed the partial trace over $E$

$$ \Lambda(\rho_A) = \mathrm{tr}_E\left(U_{AE} (\rho_A \otimes |0\rangle_E\langle 0|) U_{AE}^\dagger\right) $$

where the environment is initially in a pure state, for example $|0\rangle$. Thus, in order for $\Lambda$ to return $|0\rangle_A\langle 0|$ we need $U_{AE}$ to swap the states of $A$ and $E$, i.e.

$$ U_{AE} = \text{SWAP}_{AE} = \sum_{ij} |i_A\rangle|j_E\rangle\langle j_A|\langle i_E|. $$

Let us try it

$$ \begin{align} \Lambda(\rho_A) &= \mathrm{tr}_E\left(\sum_{iji'j'} |i_A\rangle|j_E\rangle\langle j_A|\langle i_E| \left(\rho_A \otimes |0_E\rangle\langle 0_E|\right)|i'_A\rangle|j'_E\rangle\langle j'_A|\langle i'_E|\right) \\ &= \mathrm{tr}_E\left(\sum_{iji'j'} |i_A\rangle\langle j_A|\rho_A|i'_A\rangle \langle j'_A| \otimes|j_E\rangle\langle i_E|0_E\rangle\langle 0_E|j'_E\rangle\langle i'_E|\right) \\ &= \mathrm{tr}_E\left(\sum_{ji'} |0_A\rangle\langle j_A|\rho_A|i'_A\rangle \langle 0_A| \otimes|j_E\rangle\langle i'_E|\right) \\ &= \sum_{ji'} |0_A\rangle\langle j_A|\rho_A|i'_A\rangle \langle 0_A| \, \langle i'_E|j_E\rangle \\ &= \sum_j |0_A\rangle\langle j_A|\rho_A|j_A\rangle \langle 0_A| \\ &= |0_A\rangle\langle 0_A| \, \mathrm{tr}\rho_A \\ &= |0_A\rangle\langle 0_A|. \end{align} $$

Having found $U_{AE}$, we can compute Kraus operators from the relation $E_k = \langle k_E|U_{AE}|0_E\rangle$

$$ \begin{align} E_k &= \langle k_E|U_{AE}|0_E\rangle \\ &= \langle k_E| \left(\sum_{ij} |i_A\rangle|j_E\rangle\langle j_A|\langle i_E|\right)|0_E\rangle \\ &= \sum_{ij} |i_A\rangle|\langle j_A|\langle k_E|j_E\rangle\langle i_E|0\rangle \\ &= |0_A\rangle|\langle k_A| \end{align} $$

and since we are working with qubits

$$ E_0 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad E_1 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$

Incidentally, the Kraus operators above look like those for the amplitude damping channel with decay probability $1$ which with hindsight was to be expected.

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  • $\begingroup$ That's wild! Thanks. Can you explain why the trE vanishes and only the sum is left on the above steps? $\endgroup$ – user14766 Mar 11 at 19:43
  • $\begingroup$ Trace has the property that $\mathrm{tr}(|a\rangle\langle b|) = \langle b|a\rangle$, similarly partial trace has the property that $\mathrm{tr_E}(V_A\otimes |a_E\rangle\langle b_E|) = V_A \langle b_E|a_E\rangle$. It's easy to show this directly from definition. You can also think about this as the special case of the cyclic property. $\endgroup$ – Adam Zalcman Mar 11 at 19:49
  • $\begingroup$ The step you're asking about uses the above together with linearity (to bring the sum out in front before applying the cyclic property to each term). $\endgroup$ – Adam Zalcman Mar 11 at 19:50
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(Stinespring) Given a basis $\{|u_k\rangle\}$ for the input space, you want an isometry $V$ such that $V |u_k\rangle=|0\rangle\otimes |u_k\rangle$ for some orthonormal set $\{|u_k\rangle\}$. Then you have $$\mathrm{Tr}_2\Big(V|u_k\rangle\!\langle u_k|V^\dagger\Big) = |0\rangle\!\langle0|,$$ and thus the same hold for any pure state $|\psi\rangle$ (because you can decompose in terms of $\{|u_k\rangle\}_k$), and thus for any input state $\rho$.

More generally, you don't need input and output bases to be the same, so any isometry of the form $V|u_k\rangle=|0\rangle\otimes|v_k\rangle$ for arbitrary bases $\{|u_k\rangle\}_k,\{|v_k\rangle\}_k$ for the respective bases will work the same. This is the most general form of such an isometry though. To see this you can notice that $V$ must be such that $V|\psi\rangle$ is a product state for all $|\psi\rangle$.

(Kraus) Denote with $A_a$ the Kraus operators. These are related to the Stinespring isometry by $A_a=(I\otimes \langle a|)V$, thus $A_a=|0\rangle\!\langle a|$.

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