# Interacting vacuum

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1. Aug 15, 2015

### Gedankenspiel

Hi all,

I have a question about the ground state of an interacting quantum field theory.

The state space in non-interacting QFT is a space where each field mode (with specified momentum p) has some occupation number n. These modes are interpreted as n particles with momentum p. The vacuum is the state where all occupation numbers are zero.

In my understanding, if we add interactions, the ground state of the Hamiltonian is no longer the non-interacting vacuum. This means that the interacting vacuum has n-particle states mixed in (n>0). Shouldn't we then be able to measure these particles, which would show up even without any particle coming in? Or can we perform something like a Bogoliubov transformation, such that the real, measurable particles are actually the quasi-particles of the theory? But does the energy-momentum relation E2 = m2 + p2 then still hold in this case?

2. Aug 15, 2015

### Avodyne

True.
No. In the interacting theory, it makes no sense to use the free-field theory definition of "particle".
Yes, it's guaranteed by Lorentz invariance.

3. Aug 15, 2015

### Gedankenspiel

OK, let me be more careful with the word "particle".

Is it correct, that the ground state of the interacting theory has states with occupation number n>0 mixed in?
If the state space of the free field theory is the same as in the interacting theory, this must necessarily be the case.
If not, what is the state space then and why does it change by adding an interaction? The state space of a single non-relativistic oscillator does not change by adding e.g. a quartic term to the potential.

What is the definition of "particle" in the interacting theory if it is different from the free field theory?

4. Aug 15, 2015

### Avodyne

Depends. If you have a lattice regulator in place, yes. If you've removed the lattice and gotten a Lorentz invariant theory (usually not possible in 3+1D), then no. According to Haag's theorem, the ground state of the interacting theory is orthogonal to every finite-n state of the free theory (including n=0). But Haag's theorem is only relevant to theories that don't actually exist.

A one-particle state is a state with three-momentum $\vec p$ and energy $\sqrt{\vec p^2+m^2}$, where $m$ is the mass of the particle.

5. Aug 20, 2015

### Gedankenspiel

So on a lattice there are states with nonvanishing occupation number mixed into the vacuum. But they are not to be interpreted as particles.

I wonder how the energy-momentum relation of the noninteracting theory can still hold for although both the Hamiltonian and momentun operators change by adding the interaction.