In the Choi of the state preparation map, what is the maximally entangled state with the trivial space $\mathbb{C}$?

Question

I am following Section 4.6.1 of Mark Wilde's book. The preparation map goes from a trivial input Hilbert space $\mathbb{C}$ to some output Hilbert space $\mathcal{H}_A$. Let it prepare the state $\vert 0\rangle\langle 0\vert$. Then, it has a single Kraus operator $\{ \vert 0\rangle\}$.

The Choi state of the channel is obtained by acting the the channel on one half of a maximally entangled state. What is the maximally entangled state for the trivial Hilbert space $\mathbb{C}$ which I should use as input to the channel in order to construct its Choi state?

I also tried thinking in terms of vectorization but I'm not sure how to vectorize the Kraus operator here (since it is already a vector!).

Answer 1

The dimension of the input space is 1. Two copies of the input space are thus $\mathbb C\otimes \mathbb C \cong \mathbb C$, which is also one-dimensional and thus only contains a single (normalized) state, namely the number $1$ (up to a phase). This is therefore the state which you have to consider as an input.

In particular, this means that the "input system part" of the Choi state is one-dimensional. In other words, the Choi state is simply the output state (tensored with a one-dimensional Hilbert space).

Answer 2

The choi of a map $\Phi:\operatorname{Lin}(\mathbb{C}^n)\to\operatorname{Lin}(\mathbb{C}^m)$ is an operator of the form $J(\Phi)\in\operatorname{Lin}(\mathbb{C}^m\otimes \mathbb{C}^n)$, using the definition $J(\Phi)=(\Phi\otimes \operatorname{Id}_n)(\sum_{i,j=1}^n|i,i\rangle\!\langle j,j|)$.

Therefore if the input space is one-dimensional, $\Phi:\operatorname{Lin}(\mathbb{C})\to\operatorname{Lin}(\mathbb{C}^n)$, then $J(\Phi)\in\operatorname{Lin}(\mathbb{C}^n\otimes \mathbb{C})$. But one-dimensional vector spaces are trivial, being spanned by a single element. So in practice you have the isomorphism $\operatorname{Lin}(\mathbb{C})\simeq \mathbb{C}$, sending $A\in\operatorname{Lin}(\mathbb{C})$ into $A(1)\in\mathbb{C}$. Furthermore, $\mathbb{C}^n\otimes\mathbb{C}\simeq \mathbb{C}^n$ via $$\mathbb{C}^n\otimes\mathbb{C}\ni \mathbf v\otimes (\lambda 1)\mapsto \lambda\mathbf v\in\mathbb{C}^n.$$ Therefore via these isomorphism, you can safely consider $J(\Phi)\in\operatorname{Lin}(\mathbb{C}^n)$.

For example, say the map is defined by $\Phi_\rho(|1\rangle\!\langle 1|)=\rho$ for some $|1\rangle\!\langle 1|\in\operatorname{Lin}(\mathbb{C})$, for some normalised $|1\rangle\in\mathbb{C}$, and some $\rho\in\operatorname{Lin}(\mathbb{C}^n)$. The reason this notation might appear confusing, especially the $|1\rangle\in\mathbb{C}$, is that the symbol "$\mathbb{C}$" is actually a bit overloaded here: when writing $|1\rangle\in\mathbb{C}$ I mean that $|1\rangle$ is an element in $\mathbb{C}$ thought of as a one-dimensional vector space over itself. The Choi of this map is $$J(\Phi_\rho) = (\Phi_\rho\otimes \operatorname{Id}_1)(|1,1\rangle\!\langle 1,1|) = \rho\otimes|1\rangle\!\langle 1| \simeq \rho.$$ So in synthesis, the Choi is effectively just the image of the map at $|1\rangle\!\langle 1|$.

For an example of the same ideas applied to a map whose image is one-dimensional, consider the trace: $\Phi(X)\equiv \operatorname{Tr}(X)$. This is a map $\operatorname{Tr}:\operatorname{Lin}(\mathbb{C}^n)\to\operatorname{Lin}(\mathbb{C})\simeq\mathbb{C}$, though we usually directly think of it as simply having $\mathbb{C}$ as image, rather than $\operatorname{Lin}(\mathbb{C})$. Otherwise (if we wanted to be really pedantic) we'd have to write something like $\Phi(X)=\operatorname{Tr}(X)|1\rangle\!\langle1|$ for some vector $|1\rangle$ spanning the one-dimensional vector space $\mathbb{C}$. Its Choi is then $$J(\operatorname{Tr}) = \sum_{i,j=1}^n(\operatorname{Tr}\otimes \operatorname{Id}_n)(|i,i\rangle\!\langle j,j|) = \sum_{i=1}^n |i\rangle\!\langle i| = I_n.$$ So again, $J(\Phi)\in\operatorname{Lin}(\mathbb{C}\otimes\mathbb{C}^n)\simeq\operatorname{Lin}(\mathbb{C}^n)$.