# In Stinespring dilation, can we always use a mixed state as the ancilla?

The Stinespring dilation theorem states that any CPTP map $$\Lambda$$ on a system with Hilbert space $$\mathcal{H}$$ can be represented as $$\Lambda[\rho]=tr_\mathcal{A}(U^\dagger (\rho\otimes |\phi\rangle\langle \phi|)U)$$ where $$\mathcal{A}$$ is an ancilla, $$|\phi\rangle\in\mathcal{A}$$ is an arbitrary pure state, and $$U$$ is a unitary on the joint system $$\mathcal{H}\otimes\mathcal{A}$$. Importantly, this works for any choice of the state $$|\phi\rangle\in\mathcal{A}$$ -- that is, neither does $$|\phi\rangle\in\mathcal{A}$$ have to depend on $$\Lambda$$, nor is a specific choice of $$|\phi\rangle\in\mathcal{A}$$ required.

Would this representation still be valid if, instead of attaching an ancilla in the pure state $$|\phi\rangle\langle\phi|$$, we instead attached an ancilla in some arbitrary mixed state $$\sigma$$? That is, given a state $$\sigma$$ of $$\mathcal{A}$$, I would like to know whether for every CPTP channel $$\Lambda$$, there exists a unitary $$U$$ on $$\mathcal{H}\otimes\mathcal{A}$$ such that $$\Lambda[\rho]=tr_\mathcal{A}(U^\dagger (\rho\otimes \sigma)U).$$

• To be clear: are you asking whether, for any map you specify, there exists a choice of $U$ and $\sigma$ that implements the map, or whether for any map and $\sigma$ that you specify, there exists a $U$ that implements the map? Mar 1 '21 at 11:15
• Clearly, the mixed state cannot be arbitrary, see Mateus' answer and quantumcomputing.stackexchange.com/questions/12951/… . What makes sense is to ask whether, given a mixed-state dilation, you can find a pure one on the same ancilla system. I think this is generally false, as the Stinespring dilation may need a maximal ancilla system while the mixed-state version can get away with a lower-dimensional one using a higher-rank mixed state. Mar 1 '21 at 12:28
• Note that if you have found a mixed-state dilation, the mixed state is arbitrary, as long as the rank is fixed (same argument as for pure states!). Mar 1 '21 at 12:30
• Please mark cross-posts to qc.se or other se sites, for the very least. Otherwise your creating double work, which is rather inconsiderate. Mar 1 '21 at 21:22
• Mar 1 '21 at 21:23

No, that doesn't work. It's fine to use an arbitrary pure state because the unitary $$U$$ can always be used to take it to any pure state you want. This argument doesn't work for a mixed state, as unitaries cannot take mixed states to pure states.
As a concrete example, consider the CPTP map $$\Lambda(\rho) = |0\rangle\langle 0| \operatorname{tr}(\rho),$$ and let $$\sigma = I/d_{\mathcal A}$$, the maximally mixed state on $$\mathcal A$$. If you now apply this CPTP map to $$I/d_{\mathcal H}$$, the maximally mixed state on $$\mathcal H$$, you should get $$|0\rangle \langle 0|$$, but $$\operatorname{tr}_\mathcal{A}(U^\dagger (\rho\otimes \sigma)U) = \operatorname{tr}_\mathcal{A}(U^\dagger (I/d_{\mathcal H}\otimes I/d_{\mathcal A})U) = I/d_{\mathcal H}$$ for any unitary $$U$$.
• Why not? If you could take $\sigma$ to a pure state, you would be back at the usual Stinespring dilation with an arbitrary pure ancilla. Mar 2 '21 at 9:41
No. If you take $$\sigma=\tfrac1{d_A}\mathbb I$$, you will have that for $$\rho=\tfrac1{d_S}\mathbb I$$, $$\mathrm{tr}_A(U^\dagger(\rho\otimes\sigma)U)=\tfrac1{d_S}\mathbb I\ ,$$ and thus, it will not be possible to implement any channel for which $$\Lambda(\tfrac1{d_S}\mathbb I)\ne \tfrac1{d_S}\mathbb I\ ,$$ such as any channel mapping all inputs to a constant output other than the identity.