Can we measure the ground state energy?

Question

If we consider a quantum harmonic oscillator, the ground state energy $\hbar\omega/2$ is typically stated to be not measurable, as energies are always measured as relative values (energy differences). However, if we consider a measurement of $\Delta^2 X$ and $\Delta^2 P$ on the ground state, then we can apparently obtain an absolute value for the ground state energy. Where is the flaw in the reasoning above?

COMMENT 1: it is true that one might take independent measurements of $m$ (mass) and $\omega$ (frequency), and calculate from that $\hbar\omega/2$. However, is this a valid approach? Intuitively, this seems a rather indirect method, but isn't also a measurement of $\Delta^2 X$ and $\Delta^2 P$ an indirect measurement of the ground state energy based on other observables?

COMMENT 2: it is correct to say that $X$, $P$ and energy are incompatible observables. However, this does not prevent the possibility of measuring $\Delta^2 X$ and $\Delta^2 P$. In fact, such measurements are routinely performed experimentally. The idea consist of preparing an arbitrary number of times the oscillator in the desired state, and then measure either $X$ or $P$ in each experimental realisation. From the probability distribution of the measurement results variances are then estimated.

Answer 1

A few facts

Using the notation from this other post, the Hamiltonian of a harmonic oscillator can be expressed as $$H = \frac{1}{2} \alpha u^2 + \frac{1}{2} \beta v^2 = \hbar \omega \left(\frac{1}{2} + a^\dagger a \right)$$ with $[u,v] = i \gamma$. The ground state energy is $E_0 = \hbar \omega/2$ with $\hbar \omega = \gamma \sqrt{\alpha \beta}$, and the variance in the degrees of freedom are $$\langle u^2 \rangle_0 = \gamma \sqrt{\beta / \alpha}/2 \quad \langle v^2 \rangle = \gamma \sqrt{\alpha / \beta}/2 \, .$$

Interpretation

Notice that the ground state energy can be written in terms of the variances of the degrees of freedom: $$E_0 = \frac{\gamma}{2}\sqrt{\alpha \beta} = \alpha \langle u^2 \rangle_0 = \beta \langle v^2 \rangle_0 \, .$$ For a mechanical oscillator where $u$ is the position $x$, $v$ is the momentum $p$, $\alpha = 1 / k$ and $\beta = 1 / m$, we can re-express $E_0$ as $$E_0 = \frac{1}{2} k \langle x^2 \rangle_0 + \frac{1}{2} \frac{\langle p^2 \rangle_0}{m} \tag{$\star$}$$ which makes sense as the sum of the ground state "energy" in position and the ground state "energy" in momentum [1].

Warning

There's not really any energy in the "ground state energy" in the sense that we cannot extract work from a harmonic oscillator in its ground state. In a sense, one reason we call it "ground state energy" is because of Equation $(\star)$ which shows that the Hamiltonian term $\hbar \omega / 2$ is equal to the sum of the "energies" associated with the variances of position and momentum in the ground state.

[1] Remember that classically the potential energy is $(1/2)kx^2$ and the kinetic energy is $p^2 / 2m$.

Answer 2

Neither $X$, nor $P$ commute with the oscillator energy. You cannot measure them while having the oscillator in the ground state.

Answer 3

You can in principle measure the ground state energy by measuring the mass of the system in the ground state.