# What goes on in a vacuum

1. May 29, 2005

### BoTemp

Generalized uncertainty principle: ((dA)*(dB)) >= 1/2*|<[A,B]>|
dA= standard deviation of the operator A, <Q>= <psi|Q|psi>

Now, I know people say that there can't be such thing as a true vacuum,
particles must be created and destroyed all the time, or else we'd know the
exact position & momentum, which would violate the heisenberg uncertainty
principle, a special case of the general uncertainty principle above
([pos,mom]= hbar*i). Normally, <psi|psi> is required to be 1, but in a true
vacuum, devoid of anything, psi=0. This would (trivially) satisfy the Schrodinger
equation, and though it wouldn't be normalizable, I don't see why it would have
to be, since by assumption there's nothing to be found there. In the above
equation, we'd get (dA)*(dB) >= 0. Presumably, we could measure both position
and momentum to be 0 every time.

So why can't a true vacuum exist?

2. May 29, 2005

### dextercioby

Nope.We can't.Vacuum state is normalized,therefore,even for the vacuum state,the uncertainty position-momentum holds.

Daniel.

3. May 29, 2005

### BoTemp

Can you explain that a bit more? I don't really see how a vacuum state is
normalized, if it's 0. psi = 0 solves the Schrodinger equation, allows for perfect
certainty in all measurements (0), but isn't normalizable. Basically, I'm asking
why it being non-normalizable is a problem.

4. May 29, 2005

### dextercioby

Nope.Vacuum state is defined as the Fock space vector which verifies

$$\langle 0|0\rangle =1$$

$$\hat{a}|0\rangle =0$$

for both bosonic & fermionic fields.

Daniel.

5. May 30, 2005

### BoTemp

Is $$\hat{a}$$ a symbol for any operator, or is it a specific operator?

6. May 30, 2005

### dextercioby

The annihilation operator for a spin 0 bosonic field.The notation is generic,though.

Daniel.

7. May 30, 2005

### BoTemp

Why must that necessarily be true? |0> can't be found as a function of
position, so there's no way to check it that way. My understanding is that
$$\langle \psi|\psi\rangle =1$$ is based on the fact that were we
to search the entire universe, we would necessarily find the particle. But in
the case of a vacuum, we wouldn't find anything. If

$$\langle \psi|\psi\rangle =0$$

then of course $$\psi = 0$$ mathematically, but I don't really see
anything wrong with that.

8. May 30, 2005

### dextercioby

Nah.It all comes from mathematics.Hilbert spaces,Fock spaces.Physics comes with probability conservation and necessity of unitary evolution.

The underlying structure is really mathematics.

Daniel.

9. May 31, 2005

### vanesch

Staff Emeritus
You are already assuming we work with a specific quantum system, namely a single point particle. The vacuum state of any specific quantum system is the state (or the states, in the case of a degenerate vacuum) of lowest energy (eigenstate of the hamiltonian).

Now, if you have a free particle, then the vacuum state doesn't correspond to psi(x) = 0, but to psi(x) = constant (you'll need box normalization here), which is the state of momentum 0.
Note that the state "there is no particle" which you probably identify with the "vacuum" state, is simply not a part of this quantum system, because in its very construction one has assumed the existence of a particle. So the "no particle" state is not a possible state of the system under consideration, in the same way it is impossible to give you the x,y and z coordinates of the vacuum in Newtonian mechanics.
If you work with a harmonic oscillator, then the vacuum state is the state which corresponds to the lowest energy, namely 1/2 hbar omega (a gaussian in position representation). In this case too, the vacuum state doesn't correspond to psi(x) = 0.

But usually, vacuum states are used when the quantum system is not a (set of) point particles, but rather a field. Then, there is no simple position representation psi(x) anymore. You just write out the theory, with the hamiltonian, and you look at the hilbert space vector which is the eigenstate with lowest eigenvalue for the hamiltonian. That's by definition the vacuum state.

cheers,
Patrick.

10. May 31, 2005

### BoTemp

Okay, I think I see what you're saying. We can't describe a space without any particles because the machinery of quantum mechanics is based on particle(s) existing. But does that prove that a space without particles
cannot exist? All it proves is that QM can't talk about it. Unless the theory
has been shown to be "complete", something which I don't think any physical theory could ever be proven to be, there still could be systems, like this one,
that QM won't work for.

Also, for a free particle, none of the energy states aren't normalizable:
$$\psi = A e^\frac{\pm i2mEx}{\hbar^2} +B e^\frac{\pm i2mEy}{\hbar^2} + C e^\frac{\pm i2mEz}{\hbar^2}$$ (I'm not sure about the constants
since this is from memory/extrapolation, but the gen. form should be right)

The psi above isn't normalizable; the only way $$\langle \psi|\psi\rangle$$ is finite is if A=B=C=0, and then $$\langle \psi|\psi\rangle =0$$. So one could use that to argue that QM can't describe particles in
a uniform potential of 0 (although such a thing doesn't exist in the universe
so it's not a big hole). So how can we be sure QM can be applied in any
way to a space with no potential and no particles (from now on I'll avoid
calling that a vacuum, since that definition isn't problematic in QM at all.)?

11. Jun 1, 2005

### dextercioby

Nope,those functions are normalizable in a rigged Hilbert space (a.k.a."Hilbertian triade").QM and QFT work with rigged Hilbert spaces only.So normalization is not a problem.Incidentally,the fact that a state vector is not normalizable in the Hilbert space of states,but in a larger space (the dual of the nuclear subspace),doesn't mean that free particles are not physical,it means they are not described by a plane wave,but by wave packets which are normalizable in $\tilde{M}$.

Daniel.

12. Jun 1, 2005

### vanesch

Staff Emeritus
No that's not what I said. QFT can describe a state without particles (indeed, the vacuum state). But then you don't have a wavefunction psi(x) anymore, which presupposes a single-particle position basis ; you have other descriptions (Fock space for instance) in which psi(x) doesn't really make much sense.

What I said was: IF YOU ASSUME PSI(x) to exist, you work with a VERY SPECIFIC QUANTUM SYSTEM, namely a one-particle quantum system. In THAT system, the notion of "no particle" is not present. But only because you set it up explicitly that way !

I tried to make an analogy with classical configuration space: imagine a 2-particle classical configuration space, which is R^6 (with tuples (x1,y1,z1,x2,y2,z2) ). So ALL configurations of the classical system correspond to a point in this R^6 configuration space. Now try to find me a point in R^6 that corresponds to: "there is only one particle" ; or "there are no particles" or "there are 3 particles". You can't. Each possible tuple of 6 numbers corresponds to two particles being somewhere in space.
Does that mean that classical physics cannot describe the situation "there is only one particle" ; or "there are no particles" or "there are 3 particles" ?
No, absolutely not. We just looked at A SPECIFIC EXAMPLE OF A CLASSICAL theory, which was built up around "there are 2 particles". What you don't put in, you can't get out, so you shouldn't be surprised that there are no possibilities, in this specific example, of describing 0, 1 or 3 particles.

In the same way, psi(x) only makes sense in the case of a single-particle quantum theory. So if you insist on having a psi(x), you are in this specific case, and then you shouldn't be surprised that you cannot describe 0 particles, or 2 particles, for that matter.

cheers,
Patrick.

13. Jun 4, 2005

### BoTemp

Ahhhhh, now I gotcha. Sorry for misunderstanding, that makes a certain amount of
sense. Would anybody happen to know a good introductory QFT text? I've only taken one semester of QM and I'm looking for more info. Thanks.

14. Jun 5, 2005

### Anomalous

Is it possible that space is infact matter but a different kinda something like waves but not waves. A matter not made of any particles but a continious lump. There reason I am asking this question is because it is said that space bends near blackholes, which means if we an instantly make the BH disapear then space would return to its normal dimensions in that region, so thats make me ask this question , can any body put some light on it ?