Lorentz Transformation of ladder operators Hello I am a beginner of QFT. I have some questions about the behavior of the momentum ladder operators under Lorentz Transformation.
For simplicity, let's consider the momentum ladder operators of a real scalar singlet $\phi$. We know that the ladder operators $a_{p}$ transform according to 
$$U(\Lambda)a^\dagger_{p}U^{-1}(\Lambda)=a^\dagger_{\Lambda p} , $$
where $U(\Lambda)$ is the unitary operator representing a given Lorentz Transformation $\Lambda$ (e.g Peskin p.59).
I tried to derive this identity in the following manner. Let's consider a typical situation in special relativity. Alice has a set of inertia coordinate $x$, Bob has another set of inertia coordinate $x^\prime = \Lambda x$. If a classical particle is observed to have momentum $p$ in the frame of Alice, it will be observed to have momentum $\Lambda p$ by Bob. In quantum mechanics, Alice observes the particle to be in a state $|p\rangle$ while Bob observes the particle to be in a state $|\Lambda p\rangle$, and the two single particle states are related by a unitary operator $U(\Lambda)$ representing a given Lorentz Transformation $\Lambda$. In summary, 
$$|\Lambda p\rangle=U(\Lambda)|p\rangle . $$
This then implies 
$$ a^\dagger_{\Lambda p}|0\rangle=U(\Lambda)a^\dagger _{p}|0\rangle , $$ 
where $|0\rangle$ on both sides is the ground state of the free field theory. We are then led to the conclusion that 
$$ a^\dagger_{\Lambda p}=U(\Lambda)a^\dagger_{p} , $$ 
in contrast to the standard result. Is there something wrong in my reasoning?
I observe that if we assume the vacuum ket $|0\rangle$ to be Lorentz invariant, i.e 
$$ U(\Lambda)|0\rangle=|0\rangle , $$
then we may write
$$ a^\dagger_{\Lambda p}|0\rangle =
  U(\Lambda)a^\dagger_{p}|0\rangle =
  U(\Lambda)a^\dagger_{p}U^{-1}(\Lambda) U(\Lambda)|0\rangle =
  U(\Lambda)a^\dagger_{p}U^{-1}(\Lambda)|0\rangle . $$
Then the standard result 
$$ a^\dagger_{\Lambda p} = U(\Lambda)a_{p}U^{-1}(\Lambda) $$ 
is recovered. Is the assumption of the Lorentz invariance of the vacuum state of free field theory correct?
 A: Since you're trying to prove the transformation properties of these operators by acting on the vacuum, I believe there is a way to reach the conclusion of both the invariance of the vacuum and the relation
$$a^\dagger_{\Lambda p}=U(\Lambda)a^\dagger_pU^{-1}(\Lambda) ,$$
without assuming a priori that either of them holds. You can do this as following.
Consider a 2-particle state $|\vec{p}, \vec{q}\rangle$. Then, using the fact that
$$|\Lambda p\rangle=U(\Lambda)|p\rangle , $$
we get (by the same reasoning explained in the question)
$$U(\Lambda)a^\dagger_\vec{p}a^\dagger_{\vec{q}}|0\rangle=a^\dagger_{\Lambda\vec{p}}a^\dagger_{\Lambda\vec{q}}|0\rangle.$$
Introducing $U^{-1}(\Lambda)U(\Lambda)$ between the two operators on the left-hand side, we get
$$[U(\Lambda)a^\dagger_\vec{p}U^{-1}(\Lambda)]U(\Lambda)a^\dagger_{\vec{q}}|0\rangle=a^\dagger_{\Lambda\vec{p}}a^\dagger_{\Lambda\vec{q}}|0\rangle.$$
Now, for all the creation operators to transform in a similar way, we need to introduce $U^{-1}(\Lambda)U(\Lambda)$ again as follows
$$[U(\Lambda)a^\dagger_\vec{p}U^{-1}(\Lambda)][U(\Lambda)a^\dagger_{\vec{q}}U^{-1}(\Lambda)]U(\Lambda)|0\rangle=a^\dagger_{\Lambda\vec{p}}a^\dagger_{\Lambda\vec{q}}|0\rangle.$$
Thus, the only way to get a consistent result (one that applies the same for eevry creation operator), we need
$$U(\Lambda)|0\rangle=|0\rangle,$$
i.e. the vacuum is Lorentz invariant. The transformation of the operators -modulo multipliers that come from a Lorentz covariant normalization choice for 1-particle states- follows immediately as
$$U(\Lambda)a^\dagger_\vec{p}U^{-1}(\Lambda)=a^\dagger_{\Lambda\vec{p}}$$
To get the correct factor of $\sqrt{\dfrac{E_{\Lambda\vec{p}}}{E_{\vec{p}}}}$ that Peskin puts in front of $a^\dagger_{\Lambda\vec{p}}$, you can either follow a similar process as above or just assume that the vacuum is Lorentz invariant (which makes intuitive sense) and proceed with your proof, but now use a Lorentz-covariant normalization
$$|\vec{p}\rangle=\sqrt{2E_\vec{p}}a^\dagger_\vec{p}|0\rangle$$
A: You have more or less answered your own question. Yes indeed, the vacuum must be Lorentz invariant, otherwise the Lorentz invariance would be spontaneously broken.
