Poincare invariance of the vacuum

Question

When quantizing the free scalar field, we define positive frequency modes according to

$$\frac{\partial}{\partial t} \phi_{\omega}=-i \omega \phi_{\omega}. $$

In the mode expansion we then separate positive and negative frequency solutions,

$$\phi = \int dw \ a_w \phi_{\omega} + a^{\dagger}_w \phi^*_{\omega},$$

and consequently define the vacuum state $|0\rangle$ as being annihilated by the operators corresponding to the coefficients of the positive frequency modes,

$$ a_w |0\rangle.$$

This means that if we quantize the field using a different time $t'$, the definitions do not necessarily agree. I wonder how I can show that for those cases, where $t$ and $t'$ are related by a Poincare transformation, the definitions of the vacuum state match?

Answer 1

A way to check if the two vacuums are matched is to calculate the Bogoliubov coefficients associated to your Poincarè transformation and check if each of the two vacuum is still devoid of modes when seen from the other transformed observer.

You should indeed find that both the vacuum states are empty when viewed from each of the observers.

Being $a_\omega$ and $b_\Omega$ the two different annihilation operators in your different times, you can always put:

$$ a_\omega = \alpha^* b_\Omega + \beta b^\dagger_\Omega$$

$$a_\omega^\dagger= \beta^* b_\Omega + \alpha b^\dagger_\Omega $$

Then defined the Number Operators:

$$N^{(a)} _\omega = a_\omega ^\dagger a_\omega $$

$$N^{(b)} _\Omega = b_\Omega ^\dagger b_\Omega $$

And the Number Density operators:

$$n^{(a)} _\omega = N^{(a)} _\omega /V$$

$$n^{(b)} _\Omega = N^{(b)} _\Omega /V$$

Your two vacua then are equivalent only if you find out that each vacuum is empty for the number density operator for the other modes:

$$ _b\langle 0 | n^{(a)} _\omega |0\rangle_b = 0$$

And

$$ _a\langle 0 | n^{(b)} _\Omega |0\rangle_a = 0$$

You need to evaluate such VEVs in order to find whether the two vacua are the same or not.