# Are quantum fluctuations in space real or virtual?

According to the uncertainty principle, the uncertainty in energy $\Delta E$ in a region of otherwise empty space $\Delta x$ is approximately

$$\Delta E \sim \frac{hc}{\Delta x}.$$

Thus there should be enormous amounts of energy in empty space. This is the basis of the cosmological constant problem.

Perhaps the solution is simply that as such fluctuations obey the uncertainty principle they are virtual rather than real?

Thus, according to an inertial observer, the fluctuations have no real measurable effects such as exponential spatial expansion. According to him local spacetime remains flat and stable. The observed non-deceleration of the Universe on a global scale could be due to the fact that we need to apply the uncertainty principle to curved spacetime rather than to flat spacetime as described above.

• Virtual particles are just a tool to do perturbation theory; you can formulate everything without them and the results are the same. In any case, they never enter in the derivation of the vacuum energy at all; you don't even need perturbation theory for vacuum energy, it's there even for a free field theory. Jan 14, 2018 at 20:50
• Jan 14, 2018 at 20:59
• Assuming your spacetime has the stationary Killing vectors to define "energy" to begin with! (Hint: FLRW does not!) Jan 15, 2018 at 1:06
• Vacuum fluctuations are always virtual only. See my synopsis of related concepts at at physicsforums.com/insights/physics-virtual-particles Jan 17, 2018 at 16:07

1) The cosmological constant problem arises from the sum of the ground state energies of all atoms in the universe - "vacuum energy".

Quantum and classical mechanics do not care about shifts in absolute energy, since any constant added to potentials results in the same classical force ($\mathbf{F} = -\nabla U$) and in a phase factor for the wavefunction ($H+H_0 \rightarrow \psi\cdot e^{iH_0 t/\hbar}$, where $\psi$ solved $H$). Quantum mechanics in particular does not care about the real value of the ground state energy, since all transitions only depend on the difference between energy levels. For this reason often mathematically tricks are used to either set the energy of the ground state to 0, or to change its value just to make the commutation relations look prettier.
But in GR, spacetime curvature is linearly proportional to the stress-energy tensor ($T_{\mu \nu}$), which incorporates the actual, real value of the ground state.
And if you sum up this value for all atoms in the universe, you get an energy density so ginormous that it should have cosmological implications - hence its appearance in Einstein's field equations, under the cosmological constant $\Lambda$ as, mathematically, they amount to the same term.
This can be measured from expansion-of-the-universe observations and it massively disagrees - hence the "problem".

So no uncertainty principle required.

2) The "real" uncertainty principle with statistical interpretation in quantum mechanics is the $\Delta x \Delta p$ one. The $\Delta E \Delta t$ cannot be used directly in quantum mechanics because $t$ (time) is not an operator. You can derive the formula for Ehrenfest theorem but it has a slightly different significance, and it's not strictly speaking uncertainty anymore.

You're using a mixed version so you cannot talk about uncertainty, in the strict quantum sense of the word.

3) As mentioned in the comments, virtual particles and virtual fluctuations are all a lie. They are not real and cannot, do not cause physical phenomena. They appear because of approximations made when trying to quantify results from theories that cannot be solved exactly (e.g. by perturbative methods). But you still get the cosmological constant problem even if you restrict yourself to theories that can be solved exactly (i.e. single electron atoms in quantum mechanics*), where virtual particles would not even enter the picture.

* = modulo Lamb shift.