Fourier and inverse fourier transform in QFT According to my lecture notes, the inverse Fourier transform of an operator $\phi(p)$ is given by
$$\phi(x)=\int \frac {d^4p}{(2\pi)^4}\phi(p)e^{-ip\cdot x}.$$
As @WenChern pointed out below, Peskin has a somewhat different formula on p.20:
$$\phi(\mathbf{x},t) = \int \frac {d^3 \mathbf{p}}{(2\pi)^3} \phi(\mathbf{p},t) e^{i\mathbf{p} \cdot \mathbf{x}}.$$
I am trying to see how these two formulas are equivalent and what would be the corresponding expansions of $\phi(p)$ oand $\phi(\mathbf p, t)$ as a Fourier transform of $\phi(x)$. Also, I'd like to know why one does not take these integrals over the mass shell in the same way one does in the following other definition of $\phi(x)$:
$$\phi(x)=\int \frac {d^3\mathbf p} {(2\pi)^32E_{\mathbf{p}}}[a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}]\biggr\vert_{p_0 = E_{\mathbf p}}.$$
 A: Hint:  
1. $\phi(x,t)$ at different times are not independent.  
2. $\int{d^4p\delta(p^2-m^2)}=\int{d^4p\frac{\delta(p^0-E_p)}{2p^0}}$. The left side of this equation is Lorentz invariant.
This time your question is much clearer.  
If $\phi(x)$ is an arbitrary function of $x$, there's nothing confusing. If $\phi(x)$ is constrained by the Klein-Gordon equation, we have  
$0=(\square+m^2)\phi(x)=\int{\frac{dp^4}{(2\pi)^4}(m^2-p^2)\phi(p)e^{-ip\cdot x}}$.  
Since $e^{-ip\cdot x}$s are linearly independent,  $\phi(p)$ must vanish everywhere except on the mass shell $p^2=m^2$. Then the most general form of $\phi(p)$ should be  
$\phi(p)=\frac{2\pi}{\sqrt{2E_{\mathbf p}}}[\delta(p^0-E_{\mathbf p})a_{\mathbf p}+\delta(p^0+E_{\mathbf p})b_{\mathbf{-p}}^{\dagger}]$ .  
Thus  
$\phi(x)=\int{\frac{dp^4}{(2\pi)^4}\phi(p)e^{-ip\cdot x}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-ip\cdot x}+b_{\mathbf{p}}^{\dagger}e^{ip\cdot x}]}$.  
Obviously this is just the last equation in your question.  
Then the inverse Fourier transforms are  
$\phi(p)=\int{d^4x\phi(x)e^{ip\cdot x}}$,  
and  
$\phi(\mathbf p,t)\equiv \frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]=\int{d^3\mathbf x\phi(x)e^{-i\mathbf{p\cdot x}}}$.
Due to the limitation fo the length of characters, I add the comments below.  
The first identity in the last line is the definition of $\phi(\mathbf p,t)$. The second identity in it is the inverse 3-dimensional Fourier transform of $\phi(x)=\int{\frac{d\mathbf p^3}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf p}}}[a_{\mathbf p}e^{-iE_{\mathbf p}t}+b_{\mathbf{-p}}^{\dagger}e^{iE_{\mathbf p}t}]e^{i\mathbf{p\cdot x}}}$. Direct comparison of $\phi(\mathbf p,t)$ and the general form of $\phi(p)$ shows that $\phi(p)$ contains aditional delta functions, while  $\phi(\mathbf p,t)$ is free of delta functions. Beides, since $\phi(p)$ is the 4-dimensional Fourier transform of $\phi(x)$, it is not a function of $t$. I don't think that $\phi(p)$ can be understood as "a particle whith 4-momentum $p$". It onlly make sense mathematically. The square root is just a matter of convention which can be absorbed by $a_{\mathbf p}$ and $b_{\mathbf p}$ (see, Peskin p21).
