Implication of SWAP being not positive in terms of quantum channel

Question

I am going over chapter 3 of Preskill's lecture notes regarding complete positivity. Specifically, on page 19, it is mentioned that since SWAP has eigenstates with eigenvalue -1, it is not positive, thus not a physically realizable operation.

This is quite strange to me since SWAP is quite common in quantum circuits. Also, I tried more general form of SWAP, e.g., adding a global phase, or adding a phase to $|01\rangle$ and $|10\rangle$ like iSWAP, but it seems like there will always be some eigenstates with eigenvalues not positive, or not even real.

So, from the point of quantum channels, why are SWAP gates possible? Is it because we have to involve a third qubit into consideration? But I don't think on those superconducting chips, we need to take over a third qubit when performing a SWAP gate.

Answer 1

Quoting from the linked source: "thus SWAP has negative eigenvalues, which means that $T\otimes I$ is not positive and therefore $T$ is not completely positive", where $T$ is the transpose. So they are not saying that the SWAP is not a realisable operation; they are saying that $T$ isn't.

As you note, the SWAP is a perfectly fine unitary gate. That there are negative eigenvalues is not a problem for a unitary operation (which has in general complex eigenvalues in the unit circle).

The argument made in the linked textbook is that if a quantum map $\Phi$ is such that $\Phi\otimes I$ is not positive, i.e. sends some positive semidefinite operator to some non-positive-semidefinite operator, then $\Phi$ is not completely positive, and thus does not describe a physical quantum operation.

Just to try to clarify a few things:

1. Let $V$ be a (complex) finite-dimensional vector spaces. Let $U\in\mathrm{Lin}(V)$ be a linear operator. We say that $U$ is positive semidefinite if it is Hermitian and has non-negative eigenvalues, i.e. $U^\dagger=U$ and $\langle v,Uv\rangle\ge0$ for all $v\in V$.

2. Let $\Phi$ be a (quantum) map on $V$. This is, by definition, a linear operator acting on linear operators, that is, $\Phi\in\mathrm{Lin}(\mathrm{Lin}(V),\mathrm{Lin}(V))$. You can think of it as an object sending density matrices (which are linear operators in $V$) to other density matrices.

3. A map $\Phi$ is said to be positive if it sends positive semidefinite operators to positive semidefinite operators, that is, $\Phi(X)\ge0$ whenever $\mathrm{Lin}(V)\ni X\ge0$. It is said to be completely positive if $\Phi\otimes I$ is positive for any possible extension.

4. So, they want to show that the transpose map $T$ is positive but not completely positive. This is the map sending $|i\rangle\!\langle j|$ to $|j\rangle\!\langle i|$ for all $i,j$. Note that here $|i\rangle\!\langle j|\in\mathrm{Lin}(V)$. To show that this is the case, they argue that, if $|\Psi\rangle\in V\otimes V$ denotes the maximally entangled state (note the enlarged space), then $(T\otimes I_V)\in\mathrm{Lin}(\mathrm{Lin}(V\otimes V),\mathrm{Lin}(V\otimes V))$ is a map sending $|\Psi\rangle\!\langle\Psi|$ (note that we use the density matrix corresponding to the ket state here) to the SWAP operator. Again, here $\text{SWAP}\in\mathrm{Lin}(V\otimes V)$. Now, this SWAP is not positive semidefinite (again, as a matrix/operator), hence $T\otimes I$ is not positive as a map, hence $T$ is not completely positive (again, as a map).