Lorentz transformations for scalar fields in QFT --- Peskin and Schroder I'm struggling to understand the following logic from Peskin and Schroeder, page 23:
The book defines
$$ |\vec{p} \rangle = \sqrt{2E_p} a^\dagger_p |0 \rangle $$
so that the inner product $\langle \vec{p}| \vec{q}\rangle$ is a Lorentz invariant object. It then says that this above equation, which is a 'normalization condition', implies that
$$ U(\Lambda) |\vec{p}\rangle = | \Lambda \vec{p} \rangle$$
or, if we prefer to think of this transformation as acting on the operator $a^\dagger_p$, we can write
$$ U(\Lambda) a^\dagger_p U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_p}} a^\dagger_{\Lambda \vec{p}} $$
I don't understand how these two equations are arrived at. The book seems to suggest that it's obvious, but clearly I'm missing something. Thanks.
 A: The first statement
$| p \rangle$ is an eigenstate of $\hat{P}^{\mu}$ operators with eigenvalue $p^{\mu}$. It's natural then that 
$$
U(\Lambda )| p \rangle
$$
is also eigenstate of $\hat{P}^{\mu}$:
$$
\hat{P}^{\mu}U(\Lambda )| p \rangle = U(\Lambda )\left[ U(\Lambda^{-1})\hat{P}^{\mu}U(\Lambda )\right]| p \rangle = U(\Lambda )\Lambda^{\mu}_{\ \nu}\hat{P}^{\nu}| p \rangle = \Lambda^{\mu}_{\ \nu}p^{\nu}U(\Lambda )| p \rangle .
$$
In general (for non-scalar case when $| p \rangle = | p ,\sigma \rangle$) from this follows that 
$$
U(\Lambda )| p, \sigma  \rangle = \sum_{\sigma {'}}C_{\sigma {'} \sigma}| \Lambda p, \sigma {'}\rangle 
$$
and for all states $| p, \sigma \rangle $ states $U(\Lambda )| p, \sigma \rangle$ also belongs to the Hilbert space of eigenstates, so if in the beginning $| p, \sigma \rangle $ belongs to some definite orbit of the Lorentz group, then $U(\Lambda )| p, \sigma \rangle$ will also belong to this orbit: 
$$
| p, \sigma \rangle = |\mathbf p , \sigma \rangle_{p_{0} = f(\mathbf p , m)} \Rightarrow U(\Lambda )| p, \sigma \rangle = \sum_{\sigma {'}}C_{\sigma {'} \sigma}| \mathbf {\Lambda p}, \sigma {'}\rangle_{p_{0} = f(\mathbf p , m)}.
$$
The invariance condition of $\langle p | q \rangle $ (since $\langle p| q\rangle = C(p) \delta (\mathbf p - \mathbf q) $ lorentz-invariance condition means that $C(p) \sim E_{p})$ means that $C_{\sigma \sigma {'}}$ satisfies $C_{\sigma \sigma{''}}C^{\dagger}_{\sigma {''}\sigma{'}} = \delta_{\sigma \sigma {'}}$. In simplest scalar case this directly leads to
$$
\tag 1 U(\Lambda )| \mathbf p \rangle = e^{i\alpha}| \mathbf {U(\Lambda ) p}\rangle ,
$$
where $\alpha$ is some parameter of transformation given by $\Lambda$. It can be set to zero.
The second statement 
It automatically follows from the first statement (given by eq. $(1)$) if we use definition
$$
| \mathbf p \rangle = \sqrt{2E_{p}}a^{\dagger}(\mathbf p) | \rangle 
$$
and eq. $(1)$, we'll get
$$
U(\Lambda )| \mathbf p \rangle = U(\Lambda )\left[\sqrt{2E_{p}} a^{\dagger}(\mathbf p)|\rangle\right] = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ]U(\Lambda ) | \rangle = 
$$
$$
= \left| U(\Lambda ) | \rangle = | \rangle \right| = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ] | \rangle = \sqrt{2E_{\Lambda p}}\hat{a}^{\dagger}(\mathbf {\Lambda p })| \rangle \Rightarrow
$$
$$
U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_{p}}}\hat{a}^{\dagger}(\mathbf {\Lambda p }).
$$
