# Vacuum in Qft

1. Feb 26, 2007

### alphaone

I recently studied supersymmetry breaking and read there that for Supersymmetry breaking we have the energy of the vacuum state >0. However what I do not really see is why such a vacuum would not break Poincare invariance as well as the energy is part of the momentum 4-vector and so transforms non-trivially under Poincare transformations.

2. Feb 27, 2007

### strangerep

What's to stop the vacuum state being both a Lorentz scalar, and an
energy eigenvector with non-zero eigenvalue?

3. Feb 28, 2007

### alphaone

Well I was wondering if the vacuum is an energy eigenvector with non zero eigenvalue then as always the energy of this state changes according to the fundamental representation, as the the 4-momentum operator changes according to it. But then it seems to me that the state is not a Lorentz scalar anymore as the eigenvalue is frame-dependent(we multiply the energy by a gamma factor when boosting, which would have no effect if the energy was 0). I am not even sure whether I am not even able to boost to another frame in which another state would have a smaller energy eigenvalue(meaning that it was not the vacuum in this frame)! However I might be wrong so please correct me if this argument id flaud.

4. Jul 10, 2007

### meopemuk

I think your reasoning is correct. If there is a reference frame in which energy E is non-zero and momentum is (supposedly) zero, then by boosting this state to various moving reference frames you'll obtain states with arbitrary momenta and energies E' greater than E. In relativistic QM this set of states is normally attributed to a particle with mass m = E/c^2. The unique vacuum state is normally associated with p=0, E=0.

5. Jul 10, 2007

### smallphi

The 'vacuum' state is usually manually set to energy=0 by normal ordering. If you don't do normal ordering of the operators, you get infinity.

6. Jul 15, 2007

### kharranger

I believe you can think of the vacuum in this case as carrying a projective representation of the Lorentz group. The state is a not part of a 4-vector, but it is not invariant under a lorentz transformation; it changes by a phase. In building perturbation theory about this state you would in effect remove this phase by defining all energies relative to the ground state.

7. Jul 15, 2007

### meopemuk

All representations of the Poincare group (= Lorentz group plus translations) that are relevant to physics are projective representations. There is a theorem (due to Bargmann, if I remember correctly) that projective representations of the Poincare group are equivalent to ordinary unitary representations. Another theorem (Wigner, 1939) provides a classification of the simplest (irreducible) unitary representations of the Poincare group. There are representations of three types (if we ignore spin), that can be classified according to allowed values of momentum p, energy E and mass m connected to each other by the usual relativistic formula

$$m = \frac{1}{c^2} \sqrt{E^2 - \mathbf{p}^2 c^2}$$

These three types are normally associated with massive particles, massless particles, and vacuum:

massive particles: $\mathbf{p} \in R^3; \ E > 0; \ m > 0$
massless particles: $\mathbf{p} \in R^3; \ E > 0; \ m = 0$
vacuum: $\mathbf{p} = 0; \ E = 0; \ m = 0$

So, if the condition says explicitly that $E \neq 0$, I don't see how it can be a vacuum in the usual sense.

8. Jul 16, 2007

### Haelfix

Be careful, Supersymmetry is a space-time symmetry (it bypasses the Coleman Mandula theorem), so the relevant group is the Super Poincare group.

In fact with global supersymmetry you have identically zero canonical vacuum energy, given by the antibracket of the supercharges. However, SuSy must be broken at some scale. So instead we redo the calculation and parametrize our ignorance on the breaking mechanism (hence the degrees of freedom go up roughly by 100 in the mSSm). In some of the more popular ways of SuSy breaking (say in the context of SUGRA), you are left with the vacuum as a difference of two terms: The Kahler potential and a scalar field.

This is the cosmological constant problem. These two terms must delicately cancel to unbelievably precise accuracy, through some finetuning of 64 orders of magnitude worth (depending on how you count and as given by current experimental bounds) away from their natural values.

Last edited: Jul 16, 2007