# Why are two qubits gates more noisy?

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.

• I would suggest that perhaps you are missing a noise term on the coupling between the qubits. That interaction isn't always on and therefore should also be a source for decoherence or loss in fidelity of your operation. Though, this thought comes from the realm of spin qubits. Mar 9, 2020 at 1:17
• @user245427 probably but I would like to see if it is indeed the case with some reference. Because I still expect the noise to be kinda local also so.. maybe the main reason is that the gates are slow to rotate for 2 qubits... I dont know Mar 9, 2020 at 8:59