# Why does making a quantum circuit more noise resilient make it easier to simulate classically?

I was reading Quantum Computing in the NISQ era and beyond (John Preskill, 2018) but I didn't get this point on pages 9-10:

There is a substantial opportunity for experimentalists and theorists, working together over the next few years, to find better ways of making quantum circuits noise resilient, and so extend the computational reach of NISQ technology. We should be wary, though, of a potential tradeoff — making a quantum circuit more noise resilient may also make it easier to simulate classically.

How does making a quantum circuit more noise resilient may also make it easier to simulate classically? What noise mitigation techniques could imply this?

Simulating a noiseless quantum circuit classically is as simple as calculating:

$$\tag{1} |\psi(t)\rangle = e^{-\frac{\textrm{i}}{\hbar }Ht} |\psi(0)\rangle,$$

where $$|\psi(t)\rangle$$ and $$|\psi(0)\rangle$$ are vectors each with $$2^n$$ elements (for an $$n$$-qubit circtuit) and $$H$$ is a Hermitian matrix with $$2^n \times 2^n$$ elements.

When you add noise, the qubits of the circuit can (and usually do) get entangled with their environment, meaning that a wavefunction is no longer good enoough to describe the statistics of the system of qubits, and a density matrix is needed instead. Even for very weak noise where a Markovian master equation is good enough, that Markovian master equation will be significantly more complicated than Eq. 1, and will involve the density matrix $$\rho(t)$$ which has $$2^n \times 2^n$$ elements rather than only the $$2^n$$ for a wavefunction. If you want to see what the dynamics of $$\rho(t)$$ looks like for a Markovian master equation, you can see my answer here.

As the coupling gets stronger, a weak-coupling Master equation or Redfield equation is not enough, and you may need to do something more accurate such as numerically calculating the double Feynman integral I described in my answer to: Simulating a quantum circuit with decoherence and noise.

The answer by Ron Cohen explains that as noise increases, more qubits may be needed for quantum error correction (QEC), but you're correct that QEC is usually not implemented in NISQ devices, and therefore the point is that a noise-resilient quantum circuit will be easier simulated by the unitary evolution of Eq. 1 or by a weak-coupling Markovian master equation rather than a numerical Feynman integral or other non-Markovian approach.

If your quantum circuit mitigates noise, it can also mitigate the approximations in a simulation. It can allow you to use a noisier more-approximate cheaper-to-run simulation.

Of course, the noise you get from using an approximate simulation may be totally different from the noise you get from using hardware. The error mitigation may fix one, but not the other. But you have to check to be sure. Otherwise your fancy error mitigation is a double edged sword: instead of only getting you better results on hardware, it may also act as way to lower the cost of simulating the circuit. Although I have to say I find it funny to consider it bad to improve the cost of simulating things...

As noise goes bigger, so bigger codes are needed to fix it.

In a quantum computer, every 1 logical qubit, is encoded into n physical qubits, in order to find the errors. you can make smaller n if noise is smaller, and this way the total circuit is smaller.

Smaller circuits are easier to simulate because you need 2^N floating point registers (N total qubits) to simulate 2^N amplitudes of N qubits.

• Thank you for the answer. However, it seems to me that Preskill is not thinking about quantum error correction at this point since it's talking about NISQ era. Dec 27, 2021 at 14:37