# How do I derive Stinespring and Kraus representations of a map such that $\Lambda(\rho)=|0\rangle\langle0|$ for all $\rho$?

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?

• 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. Mar 11 at 12:07
• Thanks for the clarification
– user14766
Mar 11 at 13:24

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.

• That's wild! Thanks. Can you explain why the trE vanishes and only the sum is left on the above steps?
– user14766
Mar 11 at 19:43
• 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. Mar 11 at 19:49
• 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). Mar 11 at 19:50

(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 $$A_a=|0\rangle\!\langle a|$$.