I stumbled across a new paper on gate set tomography. Can gate set tomography be applied to a quantum channel or multiple quantum channels? Will the same advantages still apply of not having to 'rely on pre-calibrated state preparations and measurements'? Would GST be advantageous over QPT in some scenarios?


1 Answer 1


"Can gate set tomography be applied to a quantum channel"

Yes, because gates are just unitary channels, if you wanted you could just let a qubit idle and undergo decay/dephasing processes instead of logic gates, and still perform whatever tomography you want. In general, any ideal unitary is actually a channel because gates aren't perfect (see equation 1, page 5 of the paper you linked).

"or multiple quantum channels"

Gate set tomography by definition produces a self-consistent estimation of a set of gates, and since gates are physically just channels, then, yes, GST can estimate multiple channels. And thus, the same advantages (i.e. the "only" advantages) still apply.

"Would GST be advantageous over QPT in some scenarios?"

The main advantage of GST over QPT is that GST is self-consistent, in the following sense. Imagine that you want to characterize the bit-flip gate, $X$. So you use QPT, and prepare a complete set of input states, and measurement in a complete output basis, to reconstruct the process map of $X$. Mathematically, this is \begin{align} p_{ij} = Tr(M_i X(\rho_j)) \end{align} where $X(\cdot)$ denotes the real channel $X$ that occurs when you try to run the ideal unitary gate $X_{ideal}$. We estimate the probabilities $p_{ij}$ from our measurements, and must assume prior knowledge of $\rho_j$ and $M_i$ so that the only unknown is $X$. One of the states in your complete input set is $|1\rangle$, which you can prepare from a fixed "fiducial" state $|0\rangle$ by applying $X$. But wait, you assumed perfect knowledge of all input states, and thereby $X$, but you were intending to characterize $X$ from the beginning, which means that you don't actually know it... right?

In reality, people usually assume known input states and measurement operators (and even assume that they are perfect) to constrain $X$, even though it is logically inconsistent. For example, if your state preparation is bad, and you can only prepare $\rho_0 = p|0\rangle\langle 0| + (1-p)\rho_{bad}$, then assuming your fiducial state is $|0\rangle$ will cause your estimate of $X$ to be bad, not because your gate itself is bad, but because other error in your system will be misattributed to $X$. GST estimates the fiducial initial state and final measurement (i.e. $\rho_0$ and $M_0$) along with a whole set of gates. The pro is the self-consistency just described. The con is that everything is sort of "set" dependent, in that if you tell me your $X$ gate, you should also tell me the rest of the GST output, since they are all characterized together, and thus are only defined relative to each other. Moreover, two different researchers comparing their gates is actually not well-defined since they have no way of knowing whether they are in the same gauge or not, as Eq. 26 page 10 explains.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.