What is the exact meaning of the word vacuum? Is it just a state of very low pressure or is it nothingness (as in there is nothing)? Also, when we say space is vacuum – it must be referring to pressure as space has light travelling (which means photons) besides the big masses of comets, planets, stars.
Pressure – What is Meant by Vacuum?
pressurespacevacuum
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If you have a harmonic oscillator in x, the ground state wave function is a Gaussian;
$$ H = {p^2\over 2} + {\omega^2 x^2\over 2} $$
$$ \psi_0(x) = e^{ - {\omega x^2\over 2}} $$
If you have two independent oscillators x,y;
$$ H = {p_x^2\over 2} + {p_y^2\over 2} + {\omega_1^2 x^2\over 2} + {\omega_2^2 y^2\over 2} $$
the ground state is a product:
$$ \psi_0(x,y) = e^{-{\omega_1 x^2\over 2}} e^{-{\omega_2 y^2\over 2}} $$
So there is no entanglement in the ground state between x and y. But if you look at it in a rotated basis (and $\omega_1 \ne \omega_2$), there is entanglement.
For a scalar quantum field in a spatial lattice in finite volume (time is still continuous), you have (if you Fourier transform in space) a bunch of decoupled harmonic oscillators (the sum on k is over non-redundant k's for a real scalar field, this is half the full space $k_x>0$):
$$ H = \sum_k {1\over 2} \dot{\phi_k}^2 + {k^2+m^2\over 2} \phi^2 $$
Which is a bunch of decoupled oscillators, so the ground state is;
$$ \psi_0(\phi_k) = \prod_k e^{-{\sqrt{k^2+m^2} |\phi_k|^2\over 2}} $$
That's not entangled in terms of $\phi_k$, but in terms of the $\phi_x$ (on the lattice), it is entangled. The vacuum wave-function Gaussian can be expressed here as:
$$ \psi_0(\phi) = e^{-\int_{x,y} \phi(x) J(x-y) \phi(y)} $$
Where $J(x-y) = {1\over 2} \sqrt{\nabla^2 + m^2} $ is not the propagator, it is this weird nonlocal square-root operator.
The vacuum for bosonic field theories is a statistical distribution, it is a probability distribution, which is the probability of finding a field configuration $\phi$ in a Monte Carlo simulation at any one imaginary time slice in a simulation (when you make the t-coordinate long). This is one interpretation of the fact that it is real and positive. The correlations in this probability distribution are the vacuum correlations, and for free fields they are simple to compute.
The axiomatic field theory material is not worth reading in my opinion. It is obfuscatory and betrays ignorance of the foundational ideas of the field, including Monte Carlo and path-integral.
General vacuum wave function for bosonic fields
In any path integral for bosonic fields with a real action (PT invariant theory), and this includes pure Yang-Mills theory and theories with fermions integrated out, the vacuum wave-function is the exact same thing as the probability distribution of the field values in the Euclidean time formulation of the theory. This is true outside of perturbation theory, and it makes it completely ridiculous that the rigorous mathematical theory doesn't exist. The reason is that the limits of probability distributions on fields as the lattice becomes fine are annoying to define in measure theory, since they become measures on distributions.
To see this, note that at t=0, neither the imaginary time nor the real time theory has any time evolution factors, so they are equivalent. So in an unbounded imaginary box in time, the expected values in the Euclidean theory at one time slice are equal to the equal time vacuum expectation values in the Lorentzian theories.
This gives you a Monte Carlo definition of the vacuum wave function of any PT invariant bosonic field theory, free or not. This is the major insight on ground states due to Feynman, described explicitly in the path integral and in the work on the ground state of liquid He4 in the 1950s (this is also a bosonic system, so the ground state is a probability distribution). It is used to describe the 2+1 Yang-Mills vacuum in the 1981 by Feynman (his last published paper), and this work is extended to compute the string tension by Karbali and Nair about a decade ago.
There is no non-zero vacuum energy density (at least in Wightman QFT), it is a myth. All attempts to extract physics from this myth have failed:
- Measured Casimir force very likely has nothing to do with vacuum fluctuations, the explanation that involves vacuum fluctuations is one that works with an effective QFT model that isn't really the fundamental QFT under consideration.
- Cosmological constant is predicted by some 120 orders of magnitude off.
Take free QFT for example. It is only well-defined mathematically on the Fock space when energy and momentum operators are normal-ordered:
$$ E = \sum_i \omega_i a_i^{\dagger} a_i. $$
If you add the metaplectic correction $1/2 \sum_i$ of ground state energies of the oscillator, you will get an infinite value which obviously doesn't make sense.
Actually, you can add a finite constant term to $E$ that corresponds to some ordering ambiguity, but it is always possible to redefine the states such that its value is zero (and hence the vacuum is Poincare-invariant) without changing any of the real predictions of the theory. The same wouldn't be possible if the ordering term was infinite (and perhaps cut off at high energy, like the vacuum energy fairy tales make you believe), because such an operator wouldn't act on the Fock space.
Best Answer
Strictly speaking vacuum is the state of lowest energy. That means no matter or radiation (photons or any other particles).
Note that space is not a perfect vacuum. Also note that, technically, a gas of planets and comets etc. has a pressure (there is usually little reason to care about it though). There is also radiation pressure due to the photons.
People often use the term vacuum loosely to refer to anything less than atmospheric pressure. This is the sense people use when they say space is a vacuum.
EDIT (Re the comments):
Yes, there is a minimum energy. Imagine that you start with vacuum. There is nothing there by definition. Now create some particle. This necessarily takes some energy (at least $mc^2$ where $m$ is the mass of the particle), so the state with a particle in it has more energy. Now the value of the vacuum energy is a subtle thing. Without gravity only energy differences matter, so you can always set the vacuum energy to zero. But with gravity it is tricky, because all energy gravitates. Indeed, physicists now believe that empty space has an energy density, now known as dark energy.
Now people will tell you a big song and dance about quantum fluctuations and zero point energy, but this is only one side of the coin, and only comes in when you try to actually calculate the vacuum energy from a more basic theory (quantum field theory). The basic picture is really simple though: vacuum energy is just a number - some physical constant that we could go out and measure. Now if you check very carefully all the laws we know then you'll find that gravity is the only place the vacuum energy comes in, so for most purposes you can forget about vacuum energy. (People also mention the Casimir effect around this point, but that is another thing entirely.)
On the other question: whether true vacuum is achievable theoretically. Well, it depends what you mean "theoretically." If you mean "in the mind of a theoretical physicist" then sure, it's possible. ;) But if you mean there is some way to build a box and make a perfect vacuum inside of it then no, you can't, because the box will always have some finite temperature and hence blackbody radiation will fill the cavity. You can make it arbitrarily close to true vacuum by cooling the box, but you could never actually reach it.