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

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!).

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).

• I see, so $\mathbb{C} = \mathbb{C}\otimes\mathbb{C}$ and both have only one state? Since the maximally entangled state we consider lives on two copies of the input space Commented Aug 4 at 11:35
• Exactly. ______ Commented Aug 4 at 12:26

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, $$J(\Phi)\in\operatorname{Lin}(\mathbb{C}\otimes\mathbb{C}^n)\simeq\operatorname{Lin}(\mathbb{C}^n)$$.