What is the "Stinespring Dilation"?

Question

I've consulted Nielsen and Chuang to understand the Stinespring Dilation, but wasn't able to find anything useful. How does this operation relate to partial trace, Kraus operators, and purification?

Answer 1

Not exactly sure what you find confusing, but the ultimate need for Stinespring dilation theorem is that in quantum mechanics the dynamics is in general defined by a completely positive trace preserving map (CPTP) $\rho \mapsto \Lambda(\rho)$.

Now, we have a belief (rightly or wrongly) that all there is is a unitary evolution governed by Schrodinger's equation. Stinespring dilation theorem is a way of stating this desire from a mathematical point of view. It guarantees that as long as I see the evolution of my density matrix as governed by a positive map (in physics we require trace preserving), this CPTP map can ultimately be lifted to a unitary operation on a higher dimensional Hilbert, where we can simply think of unitary evolution (purification).

Given that this is the case, how do we go back to the lower dimensional space? The prescription is that we trace over the system we appended to get unitary dynamics and then the dynamics in the lower Hilbert space is governed by a set of Kraus operators. This procedure is guaranteed by Stinespring dilation theorem.

Answer 2

I'll stick to the finite-dimensional case (although it's worth noting that the Stinespring dilation theorem works more generally to represent completely positive maps on C*-algebras).

The basic idea is: given a quantum map $\Phi$, how "far" is it from being unitary/isometric? We know that an isometric channel takes the form $\Phi_V(X)=VXV^\dagger$ for some isometry $V$. But can a generic channel be written in this form? Stinespring's dilation theorem tells us that the answer is, essentially, affirmative. Any channel $\Phi\in\mathrm{C}(\mathcal H_1,\mathcal H_2)$, sending states in $\mathcal H_1$ into states in $\mathcal H_2$, can be written as $\operatorname{Tr}_3(VXV^\dagger)$ for some isometry $V:\mathcal H_1\to\mathcal H_2\otimes\mathcal H_3$.

Note that the isometry $V$, to achieve this, enlarges the space. In other words, physically speaking, it exploits an ancillary degree of freedom, here $\mathcal H_3$, to achieve an isometric representation for the (generally non-isometric) channel. It's also interesting to note that the standard construction in Stinespring's dilation theorem doesn't give you the optimal such expansion. There are often ways to "isometrise" a channel using fewer additional degrees of freedom.

An alternative perspective is the following: we know that not all quantum maps describe physical evolutions. A notable example can be the transpose map, $\Phi(X)\equiv X^T$, which is known to not correspond to any physically implementable operations (because it's not completely positive). On the other hand, we know that all unitary/isometric evolutions are valid physical operations. A natural question then becomes: could it be that any physical evolution is such an isometric evolution? Stinespring's dilation tells us that the answer is, essentially, yes. Any physically valid operation (meaning, any quantum channel) can be understood as an isometric evolution in a possibly enlarged space, followed by ignoring some degrees of freedom. Vice versa, all such operations are obviously quantum channels, hence we conclude that any possible physically valid operation on quantum states can be described by some isometry possibly followed by ignoring (partial tracing) some output degrees of freedom.

---

More precisely, given a generic quantum map $\Phi:\operatorname{Lin}(\mathcal H_1)\to\operatorname{Lin}(\mathcal H_2)$ (which I'm thus not assuming to be positive or trace-preserving), you can find operators $A,B\in\operatorname{Lin}(\mathcal H_1,\mathcal H_2\otimes\mathcal H_3)$ such that $\Phi(X)=\operatorname{Tr}_3[AXB^\dagger]$ for all $X\in\operatorname{Lin}(\mathcal H_1)$. To prove this, one can for example observe that performing the SVD on the Choi of $\Phi$ amounts to finding operators $A_a,B_a$ such that $$\Phi(X) = \sum_a A_a X B_a^\dagger.$$ The "generalised Stinespring dilation" then immediately follows defining the operators $$A = \sum_a A_a\otimes e_a, \qquad B\equiv \sum_a B_a\otimes e_a,$$ which to be more precise, are the operators with actions $Av=\sum_a (A_a v)\otimes e_a$, and similarly for $B$, and $\{e_a\}_a$ is an orthonormal basis for $\mathcal H_3$.

In the particular case where $\Phi$ is not just a quantum map, but more specifically a quantum channel (i.e. a trace-preserving completely-positive quantum map), then the above reasoning will give a decomposition with $B_a=A_a$ and $\sum_a A_a^\dagger A_a=I$, $\Phi(X)=\sum_a A_a X A_a^\dagger$, which you'll recognise is nothing but the standard Kraus decomposition of $\Phi$, and then also $A=B$, and $A$ is isometric, and thus we get the Stinespring dilation.

For more details on this, see chapter 2 of Watrous' book.

Answer 3

This doesn't add anything beyond what has already been said, but I find the following to be helpful in that it is simple and straightforward, so I thought I'd share it in case it might also help others.

As already noted in the answer by @glS, a quantum channel is a map from $\cal{H}_1$ to $\cal{H}_2$ of the form $\Phi(X)=\sum_jK_jXK_j^\dagger$, which according to Stinespring, can also be written as $\Phi(X)=\textrm{Tr}_3\left(VXV^\dagger\right)$ with $V$ an isometry, $V^\dagger V=I_{1}$ (I'm using the notation given in the previous answer, so $V$ maps from $\cal{H}_1$ to $\cal{H}_2\otimes\cal{H}_3$).

How does one find $V$ given the Kraus operators $K_j$, and vice-versa? It's actually very simple. If one knows the Kraus operators, then $V=\sum_jK_j\otimes\vert j\rangle$, where states $\vert j\rangle$ constitute an orthonormal basis of $\cal{H}_3$, which may be thought of as the Hilbert space describing the "environment" of the channel (that is, it represents the physical system that the input to the channel interacts with, which is what makes the channel "noisy", because without access to the environment that interaction is, in some sense, uncontrolled). Note that with this relationship, we have $V^\dagger V=\sum_{ij}K_i^\dagger K_j\langle i\vert j\rangle=\sum_{ij}K_i^\dagger K_j\delta_{ij}=\sum_jK_j^\dagger K_j=I_1$, as expected. To go the other way, one finds the Kraus operators as $K_j=\left(I_2\otimes\langle j\vert\right) V$. One also sees why the set of Kraus operators is generally not unique. Just use a different orthonormal basis of $\cal{H}_3$ and you get a different set of Kraus operators, the new set being a linear (isometric, actually) combination of the ones you started with.

The nice thing about this is that it also gives you the Kraus operators for the complementary channel, $K_m^{(c)}$, which instead maps the input to the environment, which we mentioned above. That is, with $\vert m\rangle$ an orthonormal basis of $\cal{H}_2$, we have that $K_m^{(c)}=\left(\langle m\vert \otimes I_3\right) V$. Actually, what I really like about this is that we directly get a relationship between the Kraus operators for the direct and complementary channels: $\langle j\vert K_m^{(c)} = \langle m\vert K_j$.