Recently, I have seen many papers involving the concept of gauge freedom, but I have not yet got a good understanding of it. For example, in the paper: https://quantum-journal.org/papers/q-2021-10-05-557/pdf/, Why Gate set models may have gauge freedoms? How does gauge freedom come into being? And in what context we will use "up to gauge freedom"?

It seems that due to the gauge freedom, it will cause unavoidable gauge ambiguity, or a kind of degeneracy, then we need to fix the gauge to break the degeneracy.

Another example, it is said that a well-known issue in benchmarking is that not everything about quantum noise is learnable due to the existence of gauge freedom.


1 Answer 1


In this context, gauge freedom refers to the fact that many quantum states $\rho$, channels $\phi$, and POVMs $(E_i)_i$ may generate the same outcome statistics, described by the distribution $$ p(i) = (E_i|\phi|\rho)\,. $$ Here, I'm using an `operator bra-ket' notation, in the sense that $(A|\phi|B):=\mathrm{tr}(A^\dagger \phi(B))$.

More precisely, the above defined distribution $p$ has a gauge invariance in the following sense: Consider an invertible superoperator $T$ and the transformation $\rho \mapsto T^{-1}(\rho)$, $E_i \mapsto T^{\dagger}(E_i)$ and $\phi \mapsto T^{-1}\circ\phi\circ T$. Clearly, $p(i)$ is left invariant under such a transformation. But we require a bit more, namely that the transformed objects are again physical. For this, it is certainly sufficient that $T$ is a positive trace-preserving map and we know that such maps are of the form $T(\rho) = U\rho U^\dagger$ or $T(\rho)=U\rho^T U^\dagger$. Note that in the limited setting above, the set of gauge transformations is however larger, as we do not require that $T(\rho)$ is a quantum state for all $\rho$ but for a specific $\rho$, etc.

Note that this concept of gauge invariance is analogous to the one for matrix product states.

In the mentioned protocols, we usually execute a sequence of operations such that the relevant distribution is $$ p(i) = (E_i|\phi(g_m)\dots\phi(g_1)|\rho)\,, $$ where $\phi(g_k)$ is the quantum channel performed by the (noisy) gate $g_k\in\mathcal{G}$ and $\mathcal{G}\subset U(d)$ is the used gate set. In such a setting, the gauge transformation has to act in parallel on all channels, $\phi(g_k)\mapsto T^{-1}\circ \phi(g_k) \circ T$, which is a more restrictive setting as above. This is discussed to some extent in Nielsen et al..

This means that, in practice, a protocol like GST cannot distinguish between gauge-equivalent solutions based on the data alone. Hence, the solution is only valid "up to gauge freedom". To resolve this, the gauge often has to be fixed `by hand'. Some ways of doing this are e.g. discussed in the Nielsen et al. paper.

We also find a similar gauge freedom in randomized benchmarking, as it is also based on the execution of gate sequences. Here, however, we are not interested in recovering the channels $\phi(g_k)$, but instead it is claimed that we obtain an average gate fidelity of our gate set. Being derived from gauge-invariant data, this quantity is gauge-invariant, however, it is straightforward to see that the average gate fidelity is not. So what do we recover here? One can show that the result of RB is the average gate fidelity of the gate set in a special gauge - the depolarising gauge - which however may result in unphysical $\phi(g_k)$. For me details, see Helsen et al., Sec. IX for a concise discussion.

Finally, about this:

it is said that a well-known issue in benchmarking is that not everything about quantum noise is learnable due to the existence of gauge freedom.

Gauge freedom may be an issue in RB if you insist on interpreting decay rates as average gate fidelities. But that's it. Moreover, gauge freedom does not prevent you from learning everything about noise. It is an issue in self-consistent protocols. In principle, if you have perfect control over state preparation and measurement, you can do quantum process tomography and know everything (you basically fix the gauge there!).


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