# Unit vanishes in the Quantum Cramer-Rao Bound?

The Quantum Cramer-Rao Bound states that the precision we can achieve is bounded below by: $$(\Delta \theta)^2\ge\frac{1}{mF_Q[\varrho,H]},$$ where $$m$$ is the number of independent repetitions, and $$F_{\mathrm{Q}}[\varrho, H]$$ is the quantum Fisher information.

The question is, the left side of the bound is the variance of the parameter $$\theta$$, so it might have a unit while the right side is Quantum Fisher Information which does not possess a unit.

• Maybe this answer will help math.stackexchange.com/q/1892921 May 9, 2021 at 17:05
• But Heisenberg Limit states that $(\Delta \theta)^2\ge\frac{1}{N^2}$, where N is the number of resources (maybe N qubits). There is obviously no unit in N. May 10, 2021 at 1:36

You are correct: the units must indeed match. If we take a standard evolution with unitary $$U=\exp(-i H \theta)$$, then the units of $$H$$ and $$\theta$$ must match such that $$H\theta$$ is unitless. For a pure state $$\rho_\theta=U|\psi\rangle\langle\psi|U^{\dagger}$$ with unitary evolution, the quantum Fisher information takes the form $$F_Q[\psi,H]=\langle\psi|H^2|\psi\rangle-\langle\psi|H|\psi\rangle^2.$$ The quantum Fisher information has the same units as $$H^2$$. We already know that $$H\theta$$ is unitless: this implies that $$H^2\theta^2$$ is also unitless, so the units in the quantum Cram'er-Rao bound work out just fine.
The crucial point is realizing the conventions for the units when we write things like $$\exp(-i H\theta)$$. Does $$\theta$$ have dimensions of time? Have we set $$\hbar$$ to unity? Are we measuring energy in terms $$\hbar\omega$$ for some energy level? These are all crucial considerations. For example, the $$N^2$$ in the Heisenberg limit is really saying something about energy squared, and that is related with $$\hbar$$ to time squared, so things will work out, but one must be careful about all of these details.