Vacuum expectation value and lorenz (trans) invariance

1. Apr 14, 2013

center o bass

Hi! I've seen it stated that because of Lorenz and translational invariance

$$\langle 0| \phi(x) |0 \rangle$$

has to be a constant and I wondered how to formally verify this?

2. Apr 14, 2013

The_Duck

For any spacetime displacement $a$, there is some unitary operator $T(a)$ that implements translations by $a$. That is, $T(a) | \psi \rangle$ is the same state as $| \psi \rangle$ but shifted by $a$, and also $\phi(x-a) = T(a)^\dagger \phi(x) T(a)$ for any local operator $\phi$. If we assume that the vacuum state $| 0 \rangle$ is translationally invariant, then we have

$T(a) | 0 \rangle = | 0 \rangle$

and as a consequence

$\langle 0 | \phi(x) | 0 \rangle = \langle 0 | T(a)^\dagger \phi(x+a) T(a) | 0 \rangle = \langle 0 | \phi(x+a) | 0 \rangle$

for any spacetime displacement $a$. So the expectation value is independent of x.

3. Apr 14, 2013

center o bass

Ah, thanks! Btw: can you give me any reasons (intuitions) on why one postulate the vacuum state to be translational and poincare invariant?

The books I've read have not been clear that this even is a postulate.

4. Apr 14, 2013

The_Duck

If the Hamiltonian has a certain symmetry, the default expectation is usually that the ground state will be invariant under that symmetry. For example, we would be surprised to find that the ground state of the hydrogen atom was not rotationally invariant. If the ground state is not invariant then the symmetry is "spontaneously broken." In this case there must actually be many possible ground states, which can be transformed into each other by the symmetry. I imagine it must be possible to construct relativistic field theories with spontaneously broken Lorentz invariance but I haven't seen one myself.

One heuristic reason to expect that translational invariance will not be spontaneously broken is that the spatial derivative terms in the Lagrangian which look like $| \vec{\nabla} \phi |^2$ tell you that spatial variation in the value of $\phi$ costs energy, so it is not surprising that the lowest energy state is translationally invariant.

5. May 1, 2013

DarMM

It could be the case that the field theory is in a thermal state, in which case the state with the lowest energy in the Hilbert space, the thermal vacuum, would not be PoincarĂ© invariant. However if you look at all possible vacua, then in Minkowski space, the translation invariant ones have the lowest energy, in fact the other vacua cost "infinite energy" and hence are pretty much eliminated.

In curved spacetimes, we lack a symmetry to select out one of the vacua over the others like this, so several vacuum states are possible. However even there they must look locally like the PoincarĂ©-invariant vacuum (this is the microlocal condition), or otherwise the spacetime develops pathologies.

So although you can't see why a priori, a few calculations show you that they are the only choice which is well defined.

6. May 3, 2013

vanhees71

One should, however, note that of course you can write also QFT for a thermalized system in a manifestly covariant way. You just make explicit that usually one measures energies in the restframe of the heatbath. Taking $u=(1,\vec{v})/\sqrt{1-\vec{v}^2}$ as the four-vector velocity of the heatbath in any inertial reference frame, you can write the Stat. Op. in the Lorentz-covariant way
$$\hat{R}=\frac{1}{Z} \exp(-\beta u \cdot \hat{P}),$$
where $\hat{P}$ is the operator of total four-momentum of the system, which is a vector operator wrt. to Poincare transformations, and $\beta$ is the inverse temperature, which is a scalar by definition. There where early attempts to define temperature as some non-covariant quantity, but that's as misleading as defining a "relativistic mass" instead of using the invariant mass. Concerning temperature, it's defined as being measured by a thermodmeter that's at rest with respect to the heat-bath frame.