I would like to understand why two qubit gates are generally more noisy.
If I am interested in single qubit gates, I can have an effective modelling as a master equation:
$$\dot{\rho}=\frac{1}{i \hbar} [H,\rho]+\gamma \mathcal{D}[\sigma_-] \rho +\gamma^* \mathcal{D}[\sigma_z] \rho$$
The rate of noise is roughly: $\gamma+\gamma^*$
With $H=\frac{\hbar \Omega}{2} \sigma_y $ for a $y$ rotation in the interaction picture.
For a two qubit gate, I would expect a simple model like:
$$\dot{\rho}=\frac{1}{i \hbar} [H,\rho]+\gamma \left(\mathcal{D}[\sigma^1_-]+\mathcal{D}[\sigma^2_-]\right) \rho +\gamma^* \left( \mathcal{D}[\sigma^1_z] + \mathcal{D}[\sigma^2_z]\right) \rho$$
With $H=\frac{\hbar g}{2} \left( \sigma_1^+ \sigma_2^-+\sigma_1^- \sigma_2^+ \right)$
My question is:
Why are two qubit gates generally noisier ?
Indeed here with this simple model I would expect the rate noise to simply be $2(\gamma+\gamma^*)$. Thus twice as the one for the single qubit gate. But at the same time we have two qubits so "per qubit" the rate of noise is the same.
Are two qubits gate noisier because they rotate slower ?
I would like a justified answer (with references) in the context of superconducting qubits.
