Wave Packets in QFT In Peskin's book, it says that:
$$|\phi\rangle =\int{\frac{\mathrm d^3k}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{k}}}}}\phi\left(\vec{k}\right)\left|\vec{k}\right\rangle \tag{4.65}$$
Also, Eq(2.39) says that:
$$(1)_\mathrm{1-particle}=\int{\frac{\mathrm d^3p}{(2\pi)^3}\left|\vec{p}\right\rangle\frac{1}{2E_{\vec{p}}}\left\langle\vec{p}\right|} \tag{2.39}$$
So, I tried:
$$|\phi\rangle=\int{\frac{\mathrm d^3p}{(2\pi)^3}\left|\vec{k}\right\rangle\frac{1}{2E_{\vec{k}}}\left\langle\vec{k}\bigg|\phi\right\rangle}=\int{\frac{\mathrm d^3p}{(2\pi)^3}\frac{1}{2E_{\vec{k}}}}\phi\left(\vec{k}\right)\left|\vec{k}\right\rangle,$$
which is different from Peskin's Eq(4.65) by a square root in the denominator. 
Why is my equation wrong? 
 A: There are different conventions according to the choice of the measure in the momentum space. (I disregard further conventions related to the powers of $2\pi$). Essentially everything depends on where you prefer to have  a simple form for the action of Lorentz group (compare (1)-(1') and (3)-(3') below). 
If one assumes that $$\left[a_{\vec{k}},a^*_{\vec{k}'}\right] =  \delta^3\left({\vec{k}-\vec{k}'}\right)I\:,$$
namely
$$\left\langle \vec{k}\bigg|  \vec{k'} \right\rangle = \delta^3\left({\vec{k}-\vec{k}'}\right)\:,$$
then the unitary action of the Lorentz transformation is
$$U_\Lambda \left|\vec{k}\right\rangle = \sqrt{\frac{E_{\vec{\Lambda k}}}{E_{\vec{k}}}}\left|\vec{\Lambda k}\right\rangle \:.\tag{1}$$
Using the fact that
$$\frac{\mathrm d\vec{k}}{E_{\vec{k}}} = \frac{\mathrm d \vec{\Lambda k}}{E_{\vec{\Lambda k}}}\tag{2}$$
one easily sees that, assuming 
$$|\phi\rangle=\int{\frac{\mathrm d^3k}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_{\vec{k}}}}}\phi(\vec{k})\left|\vec{k}\right\rangle\:,$$
it arises
$$\left(U_\Lambda \phi\right)\left(\vec {k}\right) = \phi\left(\vec{\Lambda^{-1}k}\right) \tag{3}$$
as expected.
Moreover
$$\phi\left(\vec{k}\right) = \sqrt{2E_{\vec{k}}} (2\pi)^{3/2} \langle \vec{k}| \phi \rangle\:.$$
Now the projector onto the one particle space is the standard
$$P_\mathrm{1-particle}=\int \mathrm d\vec{p}\left|\vec{p}\right\rangle\left\langle\vec{p}\right|$$
In particular,  using (1) and (2), one easily finds
$$U_\Lambda P_\mathrm{1-particle} U^*_\Lambda = P_\mathrm{1-particle}$$
which must be evidently true since the one particle space is relativistically invariant and would be false in the presence of the factor $1/E_p$.
If one instead assumes that $$\left[a_{\vec{k}},a^*_{\vec{k}'}\right] =  2E_{\vec{k}}\delta^3\left({\vec{k}-\vec{k}'}\right)I\:,$$
namely
$$\left\langle \vec{k}\bigg|  \vec{k'} \right\rangle = 2E_{\vec{k}} \delta^3\left({\vec{k}-\vec{k}'}\right)\:,$$
then the unitary action of the Lorentz transformation is
$$U_\Lambda \left|\vec{k}\right\rangle = \left|\vec{\Lambda k}\right\rangle \:.\tag{1'}$$
Using the fact that
$$\frac{\mathrm d\vec{k}}{E_{\vec{k}}} = \frac{\mathrm d \vec{\Lambda k}}{E_{\vec{\Lambda k}}}\tag{2}$$
one easily sees that, assuming 
$$|\phi\rangle=\int{\frac{\mathrm d^3k}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_{\vec{k}}}}}\phi(\vec{k})\left|\vec{k}\right\rangle$$
it arises
$$\left(U_\Lambda \phi\right)\left(\vec {k}\right) = \sqrt{\frac{E_{\vec{\Lambda^{1} k}}}{E_{\vec{k}}}}\phi\left(\vec{\Lambda^{-1}k}\right) \tag{3'}\:.$$
Moreover
$$\phi\left(\vec{k}\right) = \left(\sqrt{2E_{\vec{k}}}\right)^{-1} (2\pi)^{3/2} \langle \vec{k}| \phi \rangle\:.$$
Now the projector onto the one particle space is 
$$P_\mathrm{1-particle}=\int \mathrm d\vec{p}\left|\vec{p}\right\rangle\frac{1}{2E_{\vec{k}}}\left\langle\vec{p}\right|$$
In particular,  using (1') and (2), one easily finds
$$U_\Lambda P_\mathrm{1-particle} U^*_\Lambda = P_\mathrm{1-particle}\:.$$
ADDENDUM
I think that the second choice (that of the textbook you mention) is quite ineffective and seems  an unclear  mixing of the former and of a better choice I go to describe. 
Let us assume again that $$\left[a_{\vec{k}},a^*_{\vec{k}'}\right] =  2E_{\vec{k}}\delta^3\left({\vec{k}-\vec{k}'}\right)I\:,$$
namely
$$\left\langle \vec{k}\bigg|  \vec{k'} \right\rangle = 2E_{\vec{k}} \delta^3\left({\vec{k}-\vec{k}'}\right)\:,$$
so that the unitary action of the Lorentz transformation is still
$$U_\Lambda \left|\vec{k}\right\rangle = \left|\vec{\Lambda k}\right\rangle \:.\tag{1'}$$
But this time we  assume 
$$|\phi\rangle=\int{\mathrm d^3k\frac{1}{2E_{\vec{k}}}}\phi(\vec{k})\left|\vec{k}\right\rangle\:.$$
With these choices it arises
$$\left(U_\Lambda \phi\right)\left(\vec {k}\right) = \phi\left(\vec{\Lambda^{-1}k}\right) \tag{3}$$
and also 
$$\phi\left(\vec{k}\right) =   \left\langle \vec{k}\bigg| \phi \right\rangle\:,$$
and eventually
$$P_\mathrm{1-particle}=\int \mathrm d\vec{p}\left|\vec{p}\right\rangle\frac{1}{2E_{\vec{k}}}\left\langle\vec{p}\right|$$
A: Depend on the normalization of $|\vec{k}\rangle$. The identity operator revels that the normalization of a given state is acomplished or not. The $1/2E_p$ in the equation (2.39) tells you that the state are not normalized indeed and
$$
\sqrt{\langle\vec{k}|\vec{k}\rangle}=(2\pi)^{3/2}\sqrt{2E_k}.
$$
Here the definition of $\phi (k)$ is in terms of the normalized $k$-state.
