0
$\begingroup$

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 $\delta$.

$\endgroup$
2
$\begingroup$

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 $1$.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Dec 21 '20 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? $\endgroup$ Dec 21 '20 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 $(\hat{O}_1 \otimes \mathbb{I})(\mathbb{I} \otimes \hat{O}_2)(|x_1\rangle\otimes|x_2\rangle)$ $\endgroup$ Dec 22 '20 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.