Is the Vacuum Energy Density Sign Correct for Bosons?

[gerald V]

My head is spinning when it comes to the sign of the vacuum energy density and the cosmological constant. The cosmological term can be put at the left or the right side of Einsteins equation, energy density is not pressure and energy density is not action density.

There is a historic theory with a cosmological term arising naturally, that is the Born-Infeld theory describing the electromagnetic field. If this term was kept, the Born-Infeld Lagrangian would read (one minor term supressed)

$L = - b^2 \sqrt{1 + \frac{1}{2b^2} F_{\mu \nu} F^{\mu \nu}}$

So the vacuum action density is $-b^2$, which is negative assumed that $b$ is real. I am aware that first there is no obvious relation to the results from quantum physics. Second this term has to be counterbalanced (as Born-Infeld did by simple subtraction), since for the approximate expansion of the square root to work, $b^2$ has to be large.

Question:
If this cosmological term was regarded as if it was the result from current quantum physics, would it have the correct sign for bosons?



2

 

[Demystifier]

Negative constant term in the Lagrangian corresponds to positive energy. That's because Lagrangian is kinetic energy minus potential energy, and constant energy is a kind of potential energy.

 

[gerald V]

Thank you very much. I was aware of what you are saying. But, is this the correct sign for bosons? Does current quantum physics think the vacuum energy density of bosons is positive (and huge, namely of Planckian order of magnitude; forget observation), and is there a simple argument why?

 

[Demystifier]

But, is this the correct sign for bosons? Does current quantum physics think the vacuum energy density of bosons is positive (and huge, namely of Planckian order of magnitude; forget observation), and is there a simple argument why?

Yes, vacuum energy of free bosons is positive. A simple argument is that free bosons behave like a bunch of harmonic oscillators, and it is well known that ground state energy of a quantum harmonic oscillator is positive, $\hbar\omega/2$.