# If energy is only defined up to a constant, can we really claim that ground state energy has an absolute value?

Sorry if this is really naive, but we learned in Newtonian physics that the total energy of a system is only defined up to an additive constant, since you can always add a constant to the potential energy function without changing the equation of motion (since force is negative the gradient of the potential energy).

Then in Quantum Mechanics we showed how the ground state of a system with potential energy $V(x) = \frac{1}{2} m \omega^{2} x^{2}$ has an energy $E_{0}=\frac{1}{2} \hbar \omega$.

But if we add a constant to $V(x)$ won't that just shift the ground state energy by the same constant? So in what sense can we actually say that the ground state energy has an absolute value (as opposed to just a relative value)? Is there some way to measure it?

I ask this in part because I have heard that Dark Energy might be the ground state energy of quantum fields, but if this energy is only defined up to a constant, how can we say what it's value is?

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The clearest example of vacuum energy being meaningful is the Casimir effect, but its interpretation is somewhat tricky. – Jerry Schirmer Apr 10 '11 at 8:19

In non-relativistic and non-gravitational physics (both conditions have to be satisfied simultaneously for the following proposition to hold), energy is only defined up to an arbitrary additive shift. In this restricted context, the choice of the additive shift is an unphysical, unobservable convention.

Special relativity

However, in special relativity, energy is the time component of a 4-vector and it matters a great deal whether it is zero or nonzero. In particular, the energy of the empty Minkowski space has to be exactly zero because if it were nonzero, the state wouldn't be Lorentz-invariant: Lorentz transformations would transform the nonzero energy (time component of a vector) to a nonzero momentum (spatial components).

General relativity

In general relativity, the additive shifts to energy also matter because energy is a source of spacetime curvature. A uniform shift of energy density in the Universe is known as the cosmological constant, and it will curve the vacuum. So it's important to know what it is - and it is not just a convention. Also, in general relativity, the argument from the previous paragraph may be circumvented: dark energy, regardless of its value, preserves the Lorentz (or de Sitter or anti de Sitter, which are equally large) symmetry because the stress energy tensor is proportional to the metric tensor (because $p=-\rho$). However, as long as there is gravity, the additive shift matters.

In practice, we don't measure the zero-point energy by its gravitational effects, and the value of the cosmological constant remains largely mysterious. So I surely have a different, more observationally relevant answer.

Casimir energy, comparison of situations

The additive shifts to the energy are also important when one can compare the energy in two different situations. In particular, the Casimir effect may be measured. The Casimir force arises because in between two metallic plates, the electromagnetic field has to be organized to standing waves - because of the different boundary conditions. By summing the $\hbar\omega/2$ zero-point energies of these standing waves (each wavelength produces a harmonic oscillator), and by subtracting a similar "continuous" calculation in the absence of the metallic plates, one may discover that the total zero-point energy depends on the distance of the metallic plates if they're present, and experiments have verified that the corresponding force $dE/dr$ exists and numerically agrees with the prediction.

There are many other contexts in which the zero-point energy may be de facto measured. For example, there exist metastable states that behave like the harmonic oscillator for several low-lying states. The energy of these metastable states may be compared with the energy of the free particle at infinity, and the result is $V_{\rm local\,minimum}-V_{\infty}+\hbar\omega/2$. This is somewhat analogous to calculating the energies of the bound state in a Hydrogen atom - which may be measured (think about the ionization energy).

So yes, whenever one adds either special relativity or gravity or comparisons of configurations where the structure and frequencies of the harmonic oscillators differ, the additive shift becomes physical and measurable.

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Thanks for this! I still have some nagging uncertainties, probably because I got so used to Newtonian mechanics, but hopefully those will clear up when I read your answer (and Marek's below) a few more times. – user3035 Apr 10 '11 at 20:58

It's quite correct that you can additively shift energy, even in quantum mechanics, and one can always make the ground state carry zero energy. Nevertheless, you can still measure some other energy even in the ground state: the kinetic energy. Because $T = {p^2 \over 2m}$ the expectation of kinetic energy in a given energy state is essentially its uncertainty in momentum (because the average value of momentum is zero). So even in the ground state of the oscillator there is some intrinsic movement present (of course only in this sense, the state is still stationary w.r.t. evolution), notwithstanding that it has zero energy.

From another point of view, consider your potential $V(x) = {1\over 2} m\omega^2 x^2 - E_0$: it will intersect the $x$-axis. But the ground state energy lies at $E=0$. So it's not found at the bottom of your potential (as one would expect for a ground state in classical physics). This relative position of $E_0$ and $V(x)$ is independent of any shifts in energy.

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Right, @Marek, $E_0-V_{\rm min}$ is independent of conventions. However, one may always imagine that $V(x)$ was different by $\hbar\omega/2$ than we thought and we will produce the same energy levels. Of course, then we must ask whether $E_0$ and $V_{\rm min}$ may be measured independently. It depends what tools we have to measure them. You have to assume that we can - $V_{\rm min}$ may be measured by localizing the electron, except that then it has a huge kinetic energy. – Luboš Motl Apr 10 '11 at 8:39
Note that if you calculate the energy as $T(p)+V(x)$ out of measured values of $x$ and $p$, the uncertainty principle makes the error of the energy exceed $\hbar\omega/2$ or so, anyway. In this sense, $V_{\rm min}$ cannot be measured separately from $E_0$. – Luboš Motl Apr 10 '11 at 8:40
@Luboš: well, measuring these values is certainly a problem. Nevertheless, QM tells us that $E > V_{\rm min}$ for any bound state localized around $V_{\rm min}$, right? It's no problem that it's not verifiable. Theory can (and must) surely produce lots of results we can never measure. – Marek Apr 10 '11 at 9:02
Dear @Marek, right, physics contains many important non-measurable concepts. But one must distinguish whether a quantity is unmeasurable just "directly" - but it has physical consquences - from the case when it's unmeasurable in principle. In the latter case, it's literally unphysical. In non-relativistic non-gravitating quantum mechanics with a fixed potential etc., the additive energy shift is unmeasurable even in principle because it may be incorporated into a redefinition of $V$. This is not the case in SR; GR; or when we may change $V$ or $H$ and compare the energies. – Luboš Motl Apr 10 '11 at 10:15
The question whether "it's verifiable" was really the original question of the OP. If it were not verifiable even in principle - and in non-relativistic non-gravitating QM with a fixed potential, it's not - then the OP would be right that we can't really claim that there is a physical zero-point energy because it depends on the way how we write it. – Luboš Motl Apr 10 '11 at 10:17

## protected by Qmechanic♦Oct 14 '13 at 21:52

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