I am a bit confused with the definition of Rabi amplitude and how it relates to experimental values. If I understand correctly, we can show a rabi driving (waveform) with the following formula:

$\Omega \sin(\omega t + \phi) \tag{1}$

in which $\Omega$, $\omega$ and $\phi$ are mw driving strength/amplitude (in unit of Hz), qubit frequency and phase of mw driving. However, in experiment, mw drive amplitude is often limited between 0 and 1. When we want to create a waveform on the hardware, for instance, we use the basic sine function:

$A \sin(\omega t + \phi) \tag{2}$

with the proper pulse length, this function can perform a $\pi$-rotation; i.e. $P_{0\rightarrow1}=\sin^2(\omega_1 t/2)$ where $t=\pi/\omega_1$

My question is how to get to formula (1) from formula (2). In another words, in formula (2), the amplitude is unitless (values between 0 and 1); but in formula (1), the amplitude is in Hz.

For instance: for a two-level system with $\omega$=2 GHz (transition between two qubit states), we need a rabi amplitude of 100 MHz to perform a $\pi$-rotation. But in practice, the amplitude of mw is only limited between 0-1; so, we need to tune the pulse duration instead to achieve a $\pi$-rotation.

Based on the above example and my understanding, rabi amplitude is basically is a fitting parameter rather than a experimental parameter. In practice, what it matters is the pulse duration (and of course mw amplitude (A)).


1 Answer 1


When you're working with actual hardware, there's a few different units you can use to describe your "A". Based on being limited between 0 and 1, I'm guessing you're either talking about the output voltage of an arbitrary waveform generator (AWG), which is typically between $0V$ or $1V$, or a digital/TTL output which is $0V/5V$ for $A=0/1$. In most practical realizations, that microwave (MW) output is usually thought of as an amount of RF power in dBm. The power out of a MW source is typically low, so it is often amplified and then directed to the qubit by a series of wires, waveguides, and other devices. Those induce some sort of MW field at your qubit, and your qubit's Hamiltonian picks up a term of the form $\Omega \sin(\omega t + \phi)$ based on some coupling parameter (e.g. the gyromagnetic ratio for electron spins).

So in the end, $\Omega$ is determined by the coupling of your qubit to a microwave field, which is itself determined by the devices you're using to create that field and how much power you're feeding into those devices.

  • 1
    $\begingroup$ Thanks @chris-e for the answer; that clears my confusion. I missed the part where the Hamiltonian will be written like $H =- \mu B = - (\hbar / 2) \gamma B_1 \sigma$ which this is equivalent to $H = - (\hbar / 2) \omega_1 \sigma$. (here) $\endgroup$
    – Rex
    Commented Sep 25, 2023 at 5:00

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