# Can we measure the ground state energy?

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.

• Are you satisfied with any of the answers? If not, what else are you looking for? Commented May 4, 2022 at 0:08

## 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 $$p^2 / 2m$$.

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

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