Can one obtain free energy from the vacuum?

The energy is borrowed from the Heisenberg Uncertainty Principle to create virtual particles and has to be paid back in a very short time.

$\Delta{t} \geq \frac{\hbar}{2\Delta{E}}$

This is why virtual particles live for very short times (i.e pop in and out of existence). We cannot manipulate this energy.

Whether you can extract energy from this or not (and I strongly suspect not) the Casimir effect is a consequence of vacuum fluctuations.

Essentially when two metallic plates are very close to each other, the wavelengths of virtual particles that can be created between the plates is restricted and hence there are fewer particles between the plates and outside, where no restriction occurs. This creates a pressure on the plates and pushes them together. Not a source of free energy, but an interesting (and experimentally verified) result of vacuum fluctuations.

Another well-known effect due to virtual particles is Hawking radiation. This says that when virtual particles created across the event horizon of a black hole, one can escape and the other fall in, essentially turning a virtual particle into a real particle, since it's antiparticle is inside the black hole. This is not free energy however, as the energy required comes from the mass of the black hole, causing it to (very, very) slowly evaporate over time.

So in short, no, we can't get free energy from vacuum fluctuations, but that's not to say it it doesn't have some very interesting effects.


Just like in Chemistry and Thermodynamics, we never get anything for free.

On a mechanistic level, it's important to recognize that zero-point (vacuum) energy represents the lowest energy state waveform. I remember thinking that because the EM fields are everywhere and quantized, that there was some sort of magic taking place. Realistically, zero-point energy is more like a spring between objects (very good analogy Slereah) than something unknown. It is a mathematical consequence of very well understood relations.

The second matter of virtual particles requires a little more thought. When we expand (any) Schroedinger equation into its particle form (using the number operator (two opposing ladder operators) which changes the differential equation to a sum of a series of integrals), we get an infinite number of these terms (the virtual particles). They are a distraction at this level.

Finally, Nuclear_Wizard explained the Casimir effect brilliantly, and I won't add to that except to say that this effect represents the primary experiment validating the theory. To my knowledge and at this time, the other experiments have had a great deal of background noise to wade through (they're looking for the LOWEST energy level) and have not given nearly the experimental validation that the two plates experiment has.

One critical point that is fairly unrelated to your question (but somehow the accepted "right" answer), there is no time operator. The derivation of that relationship depends on somewhat specific circumstances relating to the decay of wavestates. If we look at a graph of the energies of incoming particles, there will be some sort of distribution - the particles don't all have the exact same energy level. If we repeat the experiment and instead look at the time of arrival (which gets us to the lifetime of the wavestate by subtraction) of particles (different setup, identically prepared source), I will also get a distribution. The Heisenberg equation (in that form) relates the uncertainty (width) of those distributions where the location of the peak represents the quantity of those distributions. There is no borrowing energy but only for a short time - the Heisenberg inequality represents a fundamental property of the Fourier Transform and NOT a vacuum energy lender.