Integral to compute decay rate of meson to two mesons

I follow Zee, 2 ed., Appendix to Section II.6. I have implemented a big box normalization to impose momentum quantization on my meson field. Using $ tilde{a} $ as the box-normalized annihilation operator, I have meson fields in the form

$$ hatvarphi(x) = frac{1}{sqrt{V}}sum_k frac{1}{sqrt{2omega_k}} tilde{a}(k) e^{-ikx}~~.$$

Here, the sum is over all combinations of the three quantum numbers describing the allowed momenta in a box of volume $L^3$. To implement decay in the form $varphitoeta+xi$ , I use the toy Lagrangian $mathcal{L}=g,eta^dagger(x)xi^dagger(x)varphi(x)$. I label the incident momentum of the $phi$ meson $k$ and the outgoing momenta are $p$ and $q$. The amplitude of the transition is

$$ langle vec p,vec q|e^{-ihat HT}|vec krangle= langle vec p,vec q|e^{-iint d^4x,mathcal{L}}|vec krangle~~. $$

I expand to first order as

begin{align}
langle vec p,vec q|e^{-ihat HT}|vec krangle&=underbrace{langle vec p,vec q|vec krangle}_{0} -ilangle vec p,vec q| int !d^4x,g,hateta^dagger(x)hatxi^dagger(x)hatvarphi(x)|vec krangle
&=-iglangle vec p,vec q|int!d^4x,frac{1}{V^{frac{3}{2}}}sum_ksum_psum_q frac{1}{sqrt{8omega_komega_pomega_q}} tilde{a}^dagger_eta(p) e^{ipx}tilde{a}^dagger_xi(q) e^{iqx}tilde{a}_varphi(k) e^{-ikx}|vec krangle
&=-igfrac{1}{V^{frac{3}{2}}}sum_ksum_psum_q frac{1}{sqrt{8omega_komega_pomega_q}} underbrace{int!d^4x,e^{ix(p+q-k)}}_{(2pi)^4delta^{(4)}(p+q-k)}langle vec p,vec q|tilde{a}^dagger_eta(p), tilde{a}^dagger_xi(q) , tilde{a}_varphi(k) |vec krangle~~.
end{align}

I see how the $delta$-function will get rid of the sums over $p,q,k$ but I do not see how we get

$$ langle vec p,vec q|tilde{a}^dagger_eta(p) , tilde{a}^dagger_xi(q) ,tilde{a}_varphi(k) |vec krangle=1~~. $$

Zee finishes the above by writing

begin{align}
langle vec p,vec q|e^{-ihat HT}|vec krangle&=-igfrac{1}{V^{frac{3}{2}}}frac{1}{sqrt{8omega_komega_pomega_q}} (2pi)^4delta^{(4)}(p+q-k)~~,
end{align}

but I do not see why the bra-ket part disappears when the expression is simplified with the $delta$.