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

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

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 $\varphi\to\eta+\xi$ , 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

$$ \langle \vec p,\vec q|e^{-i\hat HT}|\vec k\rangle= \langle \vec p,\vec q|e^{-i\int d^4x\,\mathcal{L}}|\vec k\rangle~~. $$

I expand to first order as

\begin{align} \langle \vec p,\vec q|e^{-i\hat HT}|\vec k\rangle&=\underbrace{\langle \vec p,\vec q|\vec k\rangle}_{0} -i\langle \vec p,\vec q| \int \!d^4x\,g\,\hat\eta^\dagger(x)\hat\xi^\dagger(x)\hat\varphi(x)|\vec k\rangle \\ &=-ig\langle \vec p,\vec q|\int\!d^4x\,\frac{1}{V^{\frac{3}{2}}}\sum_k\sum_p\sum_q \frac{1}{\sqrt{8\omega_k\omega_p\omega_q}} \tilde{a}^\dagger_\eta(p) e^{ipx}\tilde{a}^\dagger_\xi(q) e^{iqx}\tilde{a}_\varphi(k) e^{-ikx}|\vec k\rangle\\ &=-ig\frac{1}{V^{\frac{3}{2}}}\sum_k\sum_p\sum_q \frac{1}{\sqrt{8\omega_k\omega_p\omega_q}} \underbrace{\int\!d^4x\,e^{ix(p+q-k)}}_{(2\pi)^4\delta^{(4)}(p+q-k)}\langle \vec p,\vec q|\tilde{a}^\dagger_\eta(p)\, \tilde{a}^\dagger_\xi(q) \, \tilde{a}_\varphi(k) |\vec k\rangle~~. \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

$$ \langle \vec p,\vec q|\tilde{a}^\dagger_\eta(p) \, \tilde{a}^\dagger_\xi(q) \,\tilde{a}_\varphi(k) |\vec k\rangle=1~~. $$

Zee finishes the above by writing

\begin{align} \langle \vec p,\vec q|e^{-i\hat HT}|\vec k\rangle&=-ig\frac{1}{V^{\frac{3}{2}}}\frac{1}{\sqrt{8\omega_k\omega_p\omega_q}} (2\pi)^4\delta^{(4)}(p+q-k)~~, \end{align}

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


1 Answer 1


This is simply because $\tilde{a}_\phi(k)|\vec{k}\rangle$ annhilates the momentum state, producing $\tilde{a}_\phi(k)|\vec{k}\rangle \sim |0\rangle$, while $\langle \vec{p},\vec{q}|\tilde{a}_\eta^\dagger(p)\tilde{a}_\xi^\dagger(p) = [\tilde{a}_\eta(p)\tilde{a}_\xi(p)|\vec{p},\vec{q}\rangle]^\dagger \sim (|0\rangle)^\dagger = \langle 0|$, since the operators annihilate their respective particles, producing the vacuum up to some normalistion constant. Thus all we're left with is $\langle0|0\rangle$, and the normalisation constant is usually chosen ad hoc so as to make this $1$.

  • $\begingroup$ For the bra part, I assume it can be written as \begin{align} \sqrt{2}\langle \vec p,\vec q|a_\eta^\dagger(p) a_\xi^\dagger(q)&=\langle\vec p|\langle\vec q| a_\eta^\dagger a_\xi^\dagger-\langle\vec q|\langle\vec p| a_\eta^\dagger a_\xi^\dagger\\ &=\langle\vec p|\langle\vec q| a_\eta^\dagger a_\xi^\dagger-\langle\vec q|\langle0| a_\xi^\dagger\\ &=\langle\vec p|\langle\vec q| a_\eta^\dagger a_\xi^\dagger-\langle\vec q|\langle0| 0\\ \end{align} but how should the first RHS term be treated? $\endgroup$ Commented Dec 21, 2020 at 16:28
  • $\begingroup$ Since the annihilation operators of two different particles commute, meaning $a^\dagger_\eta a^\dagger_\xi=a^\dagger_\xi a^\dagger_\eta$, that makes me thing the first term should go to zero as well. What am I doing wrong? I assume I am not constructing the two particle state correctly. How should write it to act with the operators directly on single particle states? $\endgroup$ Commented Dec 21, 2020 at 16:39
  • 1
    $\begingroup$ Hint: in a tensor product state, $\hat{O}_1\hat{O}_2|x_1, x_2\rangle$ is really short for $(\hat{O}_1 \otimes \mathbb{I})(\mathbb{I} \otimes \hat{O}_2)(|x_1\rangle\otimes|x_2\rangle)$ $\endgroup$ Commented Dec 22, 2020 at 10:06

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