= Solution
The <Fock vacuum> condition and the <creation operator> <commutator> give
$$
\langle\mathbf p|\mathbf q\rangle
=\sqrt{2E_{\mathbf p}\,2E_{\mathbf q}}\,
\langle0|a_{\mathbf p}a^\dagger_{\mathbf q}|0\rangle
=\boxed{2E_{\mathbf p}(2\pi)^3\delta^3(\mathbf p-\mathbf q)}.
$$
Equality of the square-root product and $2E_{\mathbf p}$ holds on the support of the <Dirac delta function>.
The positive-energy <mass shell> has the <Lorentz invariant> measure
$$
d^4p\,\theta(p^0)\delta(p^2-m^2)=\frac{d^3p}{2E_{\mathbf p}}.
$$
Here both $p^2$ and the four-dimensional <volume element> are invariant, and positive <energy> is preserved by a proper orthochronous <Lorentz transformation>. Consequently the three-dimensional <Jacobian determinant> is $J=E_{\boldsymbol\Lambda p}/E_{\mathbf p}$. Transforming the <Dirac delta function> yields
$$
\delta^3(\boldsymbol\Lambda p-\boldsymbol\Lambda q)
=\frac{E_{\mathbf p}}{E_{\boldsymbol\Lambda p}}\delta^3(\mathbf p-\mathbf q),
$$
so its factor exactly cancels that of $2E_{\boldsymbol\Lambda p}$. Thus the displayed <inner product> is invariant under simultaneous transformation of both <momenta>. This is also why the one-particle completeness measure is $d^3p/((2\pi)^3 2E_{\mathbf p})$.<Quantum Field Theory by David Tong>[Section 2.4.1, relativistic normalization and the positive-energy mass-shell measure]{ref}
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