Quantum Gate Fidelity Calculator
Estimate the fidelity of single-qubit, CNOT, or multi-qubit gates by combining decoherence decay and control-field errors. Use this when benchmarking qubit performance or comparing quantum hardware designs.
About this calculator
Gate fidelity F quantifies how closely a real quantum gate matches its ideal unitary operation, ranging from 0 (complete failure) to 1 (perfect). Decoherence reduces fidelity exponentially with gate duration: the decoherence factor is exp(−t_gate / T_coherence). Control imperfections add a multiplicative penalty: (1 − ε/100), where ε is the percentage control-field error. For a single-qubit gate, F = exp(−t/T) · (1 − ε/100). A CNOT gate involves two qubits and an additional hardware overhead factor of 0.95, giving F_CNOT = [exp(−t/T) · (1 − ε/100)]² · 0.95. Three-qubit gates extend this to the cube with a 0.90 overhead factor. These models reflect the dominant error channels in superconducting and trapped-ion hardware. Keeping t_gate ≪ T_coherence is the key engineering requirement for fault-tolerant quantum computing.
How to use
Suppose you are running a CNOT gate with t_gate = 100 ns = 1×10⁻⁷ s, T_coherence = 50 µs = 5×10⁻⁵ s, and a control error of ε = 1%. First compute the base factor: exp(−1×10⁻⁷ / 5×10⁻⁵) = exp(−0.002) ≈ 0.9980. Apply the control penalty: 0.9980 · (1 − 0.01) = 0.9980 · 0.99 = 0.9880. For a CNOT, square it and multiply by 0.95: F = 0.9880² · 0.95 = 0.9761 · 0.95 ≈ 0.927. This means roughly a 7.3% infidelity, which is above the ~1% threshold typically required for fault tolerance.
Frequently asked questions
What coherence time is needed to achieve high quantum gate fidelity?
A commonly used rule of thumb is that the gate time should be at least 1000 times shorter than the coherence time to keep the decoherence contribution to infidelity below 0.1%. For example, if a gate takes 100 ns, the qubit should have a coherence time of at least 100 µs. In practice, state-of-the-art superconducting qubits achieve T₂ times of 100–500 µs with single-qubit gate times of 20–50 ns, giving comfortable margins. Multi-qubit gates are slower and therefore more sensitive to coherence time limits.
How does control-field error contribute to gate infidelity in quantum processors?
Control-field errors arise from imprecise microwave or laser pulses used to rotate qubits. A 1% amplitude error in the driving field typically translates directly into a ~1% error in the rotation angle, reducing fidelity by roughly the same fraction per gate. These errors can be systematic (always over- or under-rotating) or stochastic (random pulse-to-pulse fluctuations). Systematic errors can be partially corrected with composite pulse sequences such as DRAG or CORPSE, while stochastic errors set a floor on achievable fidelity that requires hardware improvements.
Why does a CNOT gate have lower fidelity than a single-qubit gate under the same conditions?
A CNOT gate involves an entangling interaction between two qubits, which generally takes longer to execute than a single-qubit rotation because it relies on a weaker two-qubit coupling. The longer gate time increases the decoherence penalty exponentially. Additionally, both qubits must remain coherent simultaneously, so errors from each qubit compound multiplicatively. The model here captures this with a squared base-fidelity factor and an extra 0.95 hardware overhead, reflecting realistic CNOT implementations in superconducting architectures where two-qubit gate fidelities are typically 1–5% lower than single-qubit gate fidelities.