Relationship between coupling Strength in Cross-Resonance Gates and Coefficients in Hamiltonian Tomography

Question

In the effective Hamiltonian for the cross-resonance (CR) gate, the interaction term is written as:

$$ \tilde{H}_{\rm eff}^{\rm CR} = - \frac{\Delta_{12}}{2}\sigma_1^z + \frac{\Omega(t)}{2} \left(\sigma_2^x - \frac{J}{2\Delta_{12}} \sigma_1^z \sigma_2^x \right), $$

where J represents the coupling strength between the control and target qubits.

During Hamiltonian tomography, the Hamiltonian is expressed as:

$$ \hat{H} = \frac{\hat{Z} \otimes \hat{A}}{2} + \frac{\hat{I} \otimes \hat{B}}{2} = a_{x} \hat{Z}\hat{X} + a_{y} \hat{Z}\hat{Y} + a_{z} \hat{Z}\hat{Z} + b_{x} \hat{I}\hat{X} + b_{y} \hat{I}\hat{Y} + b_{z} \hat{I}\hat{Z} $$

Here, $𝑎_{x}$ is the coefficient corresponding to the 𝑍X interaction term.

Is the coupling strength J in the effective Hamiltonian same as to the coefficient $𝑎_{x}$ obtained from Hamiltonian tomography?

If not, what is the relationship. (For context i am trying to obtain the qubit-qubit coupling on IBM quantum computers)

Source: https://github.com/Qiskit/textbook/blob/main/notebooks/quantum-hardware-pulses/hamiltonian-tomography.ipynb

Answer 1

In theory, yes. If the system is ideal, there's only one interacting term($\sigma^z_1\sigma^x_2$) in CR gate. And $a_x$ will be the only non-zero interacting term in Hamiltonian tomography.

However, in real cases(both numerical simulation and experiment), the answer is NO. To know the reason, we need to know how a CR gate is realized in a real quantum computer.

In a real quantum computer, to perform a CR gate, the driving term is not $\frac{\Omega(t)}{2}\frac{J}{2\Delta_12}\sigma^z_1\sigma^x_2$ but $\frac{\Omega(t)}{2}\sin(\omega_d t)\sigma^x_1$. This $\sigma^x_1$ term is rotating with frequency $\omega_d = \omega_{11}-\omega_{10}$(Note this does not equal to $\omega_{01}$), and it's resonanting with states $|11\rangle$ and $|10\rangle$. If $Q1$ is in state $|1\rangle$, $Q2$ will be excited. And if not, the system stays unchanged(see this paper for details). The effect of this single-qubit driving term is the same as $\sigma^z_1\sigma^x_2$, so it is known as $ZX$ interacting term. But it's not this term we are really driving! I think that's the reason why this gate is called 'cross-resonant' instead of '$ZX$' (like) gate.

In numerical simulation, we do not simulate the ideal Hamiltonian $\tilde{H}^{CR}_{eff}$ directly. The full Hamiltonian is used. For example, in superconducting transmon qubits, the Hamiltonian used in simulation will be: $$ H = H_0 + H_{drive} $$ $$ H_0 = \sum_{i=1,2}\left(\omega_i \hat{a}_i^{\dagger}\hat{a}_i + \frac{\alpha_i}{2}\hat{a}_i^{\dagger}\hat{a}_i^{\dagger}\hat{a}_i\hat{a}_i \right) + J(\hat{a}_1+\hat{a}_1^{\dagger})(\hat{a}_2+\hat{a}_2^{\dagger})$$ $$ H_{drive} = \frac{\Omega(t)}{2}\sin(\omega_d t+\phi)(\hat{a}_1 + \hat{a}_1^{\dagger}) $$ $\hat{a},\hat{a}^{\dagger}$ are annihillation and creation operators. And this is the simplest situation in superconducting transmon qubits where two qubits are directly coupled and only two qubits are considered. In this simplest case, if you change to rotating frame and try to derive the $\sigma_1^z\sigma_2^x$ term, you will find that so many (high frequency, high order) terms are ignored. These ignored terms can be detected by Hamiltonian tomography and these are the 'quantum errors'.

In experiment, the 'quantum errors' will be much more complicated. Because there are usually more than two qubits and they are not always directly coupled.Besides, there are classical issues that will further affect the Hamiltonian tomography result. For example, the crosstalk. This means the drive is not only on $Q1$ but also on $Q2$(the amplitude is much smaller but not zero).

The Hamiltonian tomography result will include all these quantum and classical 'errors'. So $a_x$ is usually not equal to $J$ in real situations.