# How does the Kraus decomposition imply the Stinespring representation?

To show that the Kraus decomposition $$\Phi(\rho)=\sum_{k=1}^D M_k\rho_S M_k^\dagger$$ implies the Stinespring form $$\Phi(\rho)=\text{tr}_E[U_{SE}(\rho_S\otimes|0\rangle\langle 0|_E)U_{SE}^\dagger]$$ we first introduce an orthonormal basis $$\{|k\rangle_E\}$$ for $$\mathcal H_E$$ and choose a ket $$|0\rangle_E$$ (for simplicity, this can be done by increasing the dimension of $$\mathcal H_E$$ to $$D+1$$ in such a way that $$|0\rangle_E$$ is actually orthogonal to all $$|k\rangle_E$$), then we define the operator $$U_{SE}$$ by the action $$U_{SE}(|\psi\rangle_S\otimes |0\rangle_E)=\sum_{k=1}^D M_k|\psi\rangle_S\otimes |k\rangle_E.$$ Two things are not very clear to me: first, how to go from Kraus to Stinespring using this definition, and secondly how to show that $$U_{SE}$$ is unitary on all kets in $$\mathcal H_{SE}$$ when only the action on the hyperplane $$|\psi\rangle_S\otimes |0\rangle_E$$ is defined. I'd appreciate some pointers as to how to do both.

1. There's no need to increase the dimension of $$\newcommand{\calH}{{\mathcal H}}\calH_E$$ to $$D+1$$. All you care about is that $$U$$ must send orthonormal vectors to orthonormal vectors. What the state $$|0\rangle_E$$ actually represents is purely conventional. You could put an arbitrary (pure) state there, and get the same channel (as long as you also change the unitary $$U$$ accordingly, of course).

2. How to show that $$U_{SE}$$ is unitary on all kets? You don't really need to. Or to be more precise, if you want to say that $$U_{SE}$$ is unitary, then yes, you need to define its action on a full basis of $$\calH_{SE}$$. But the way it acts on states orthogonal to $$|0\rangle_E$$ is completely irrelevant from the point of view of the channel $$\Phi$$, which is, presumably, what you really care about. If for some reason you really want $$U_{SE}$$ to be unitary, you can pick an arbitrary unitary operator, as long as its action on $$|0\rangle_E$$ is what it should be.

3. This might be purely conventional, but I wouldn't call the representation you write as the "Stinespring representation". The "Stinespring representation" would be when you write the channel as $$\Phi(\rho)=\operatorname{Tr}_E[V\rho V^\dagger]$$ for some isometry $$V:\calH_S\to\calH_{SE}$$. Then again, the "unitary representation" (how I would refer to what you are using) and the Stinespring representation are de facto equivalent, so this does not matter that much.

The main advantage of the Stinespring representation with the isometry is that it describes the underlying channel/map in a less redundant way: the freedom in the choice of initial state $$|0\rangle_E$$ and action of $$U_{SE}$$ on the auxiliary space disappears when we only write it with $$V$$.

4. To go from the representation with the unitary to the Kraus operators, a simple way is to decompose in components the various objects, and realise you get a Kraus representation. A starting point would be to write $$\operatorname{Tr}_E[U(\rho\otimes |0\rangle\!\langle0|)U^\dagger] = \sum_k (I\otimes \langle k|)U(\rho\otimes|0\rangle\!\langle0|)U^\dagger (I\otimes|k\rangle) = \sum_k \underbrace{(I\otimes \langle k|)U(I\otimes |0\rangle)}_{\equiv A_{k}} \rho \underbrace{(I\otimes\langle 0|)U^\dagger (I\otimes|k\rangle)}_{= A_k^\dagger} \equiv \sum_k A_k \rho A_k^\dagger.$$

An immediate way to see the relation between Kraus operators and Stinespring isometry $$V$$, is to notice that $$V$$, as a matrix, is made up by stacking the operators $$A_k$$ one on top of the other. Equivalently, the relation between the unitary $$U_{SE}$$ and the Kraus is that $$U_{SE}$$ is some (any) unitary such that a subset of its columns is built by stacking the Kraus operators $$A_k$$ one on top of the other. This is the pictorial representation I also mention in this other answer on physics.SE.

5. To go from Kraus to Stinespring, reasoning is the same as above. Given $$\Phi(X)=\sum_a A_a X A_a^\dagger$$ with, note that the linear operator $$V\equiv \sum_a (A_a X) \otimes |a\rangle$$ is an isometry, if $$\sum_a A_a^\dagger A_a = I$$, and $$\Phi(X)=\operatorname{Tr}_E(V X V^\dagger)$$.