# What is the maximally entangled state in Choi matrix?

For a quantum channel $$\mathcal{E}$$, Choi matrix is defined as follows:

$$C(\mathcal{E}) = (\mathcal{E} \otimes I) (\sum_{i}\lvert ii\rangle \sum_j \langle jj\lvert)$$, where $$\sum_{i}\lvert ii\rangle$$ is the maximally entangled state.

It is clear that for 2 qubit case, this maximally entangled state is just the Bell state, i.e. $$\sum_i \lvert ii \rangle = \frac{1}{\sqrt{2}}(\lvert 00 \rangle + \lvert 11 \rangle)$$. But I thought the generalized Bell state, which is generalized GHZ-state, is the maximally entangled state so that $$\sum_i \lvert ii \rangle =$$ generalized GHZ state for 3+ qubits. But the notation from many other references gives $$\sum_{i=1}^d \lvert ii \rangle$$ where $$d$$ is the dimension of Hilbert space that this maximally entangled state lives in. This confuses me since the generalized GHZ state will always have 2 terms. What am I missing here?

• what do you mean with "what is" the state? You just wrote it, it's $\sum_{ij} |ii\rangle\!\langle jj|$. Or if you prefer, the projection onto the state $\sum_i |ii\rangle$. It's a maximally entangled state between two systems of dimension $d$. GHZ with more than two qubits is a multipartite state, it's a bit of a different context.
– glS
Jun 5, 2022 at 9:06

For the Choi-Jamiołkowski isomorphism you need the bipartite maximally entangled state of dimension $$d$$, while the GHZ state is a multipartite entangled state of dimension 2.
The normalization is also relevant. You can use either $$\sum_{i=0}^{d-1} |ii\rangle$$ or $$\frac1{\sqrt d}\sum_{i=0}^{d-1} |ii\rangle,$$ both work, but need different formulas. The standard formulas use the first alternative, the non-normalized one.
Definition of Choi–Jamiołkowski isomorphism states that $$d$$ is the dimension of the hilbert space while your generalized GHZ states is still with hilbert space dimension $$2$$ for each subsystem. The definition in the link is very clear, the maximally entangled state should be $$|\Phi ^{+}\rangle ={\frac {1}{\sqrt {d}}}\sum _{i=0}^{d-1}|i\rangle \otimes |i\rangle ={\frac {1}{\sqrt {d}}}(|0\rangle \otimes |0\rangle +\cdots +|d-1\rangle \otimes |d-1\rangle )}$$