Let's take an example of an entanglement witness of the form $W = | \phi \rangle \langle \phi | ^{T_2}$ where $ | \phi \rangle $ is some pure entangled state.

If I wanted to test some state $\rho$, I would have to perform $\mathrm{Tr}(W \rho)$. I assume this is done by measuring $\rho$ multiple times in the eigenbasis of $W$ and finding the expected eigenvalue, and that would be the solution to $\mathrm{Tr}(W \rho)$.

  1. Is this the way it is done?
  2. Therefore, specifically in the above case, it is very much possible to physically apply the witness? (Even though there is a mathematical partial transpose present.)

1 Answer 1


This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as.

Just as an example of what I mean, let $$ W=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right). $$ As it happens, this is just a swap gate, but ignore this for now. You might find the eigenvectors $|00\rangle,|11\rangle,(|01\rangle\pm|10\rangle)/\sqrt{2}$ and measure those expectation values directly. Or, you might write $$ W=(\mathbb{I}+Z\otimes Z+X\otimes X+Y\otimes Y)/2, $$ and them you might go off and measure the 3 separate observables $ZZ$, $XX$ and $YY$, those being particularly natural, accessible, things.

Note that there's no problem using $W$ to define a measurement. The partial transpose is irrelevant; it's still a Hermitian matrix. The partial transpose just means it might not be a valid state, but being a valid state is irrelevant as a measurement: if we say "do a Z measurement", the Z matrix certainly has nothing to do with being a state. It's just a Hermitian matrix.


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