# 物理代写|量子场论代写Quantum field theory代考|The LSZ reduction formula

#### Doug I. Jones

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## 物理代写|量子场论代写Quantum field theory代考|The LSZ reduction formula

We derived a formula for the differential cross section for $2 \rightarrow n$ scattering of asymptotic states, Eq. (5.22):
$$d \sigma=\frac{1}{\left(2 E_1\right)\left(2 E_2\right)\left|\vec{v}1-\vec{v}_2\right|}|\mathcal{M}|^2 d \Pi{\mathrm{LIPS}},$$
where $d \Pi_{\mathrm{LIPS}}$ is the Lorentz-invariant phase space, and $\mathcal{M}$, which is shorthand for $\langle f|\mathcal{M}| i\rangle$, is the $S$-matrix element with an overall momentum-conserving $\delta$-function factored out:
$$\langle f|S-\mathbb{1}| i\rangle=i(2 \pi)^4 \delta^4(\Sigma p) \mathcal{M}$$
The state $|i\rangle$ is the initial state at $t=-\infty$, and $\langle f|$ is the final state at $t=+\infty$. More precisely, using the operators $a_p^{\dagger}(t)$, which create particles with momentum $p$ at time $t$, these states are
$$|i\rangle=\sqrt{2 \omega_1} \sqrt{2 \omega_2} a_{p_1}^{\dagger}(-\infty) a_{p_2}^{\dagger}(-\infty)|\Omega\rangle,$$
where $|\Omega\rangle$ is the ground state, with no particles, and
$$|f\rangle=\sqrt{2 \omega_3} \cdots \sqrt{2 \omega_n} a_{p_3}^{\dagger}(\infty) \cdots a_{p_n}^{\dagger}(\infty)|\Omega\rangle$$
We are generally interested in the case where some scattering actually happens, so let us assume $|f\rangle \neq|i\rangle$, in which case the $\mathbb{1}$ does not contribute. Then the $S$-matrix is
$$\langle f|S| i\rangle=2^{n / 2} \sqrt{\omega_1 \omega_2 \omega_3 \cdots \omega_n}\left\langle\Omega\left|a_{p_3}(\infty) \cdots a_{p_n}(\infty) a_{p_1}^{\dagger}(-\infty) a_{p_2}^{\dagger}(-\infty)\right| \Omega\right\rangle .$$
This expression is not terribly useful as is. We would like to relate it to something we can compute with our Lorentz-invariant quantum fields $\phi(x)$.
Recall that we defined the fields as a sum over creation and annihilation operators:
$$\phi(x)=\phi(\vec{x}, t)=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left[a_p(t) e^{-i p x}+a_p^{\dagger}(t) e^{i p x}\right],$$
where $\omega_p=\sqrt{\vec{p}^2+m^2}$. We also start to use the notation $\phi(x)=\phi(\vec{x}, t)$ as well, for simplicity. These are Heisenberg picture operators which create states at some particular time. However, the creation and annihilation operators at time $t$ are in general different from those at some other time $t^{\prime}$. An interacting Hamiltonian will rotate the basis of creation and annihilation operators, which encodes all the interesting dynamics. For example, if $H$ is time independent, $a_p(t)=e^{i H\left(t-t_0\right)} a_p\left(t_0\right) e^{-i H\left(t-t_0\right)}$, just as $\phi(x)=e^{i H\left(t-t_0\right)} \phi\left(\vec{x}, t_0\right) e^{-i H\left(t-t_0\right)}$, where $t_0$ is some arbitrary reference time where we have matched the interacting fields onto the free fields. We will not need to use anything at all in this section about $a_p(t)$ and $\phi(\vec{x}, t)$ except that these operators have some ability to annihilate fields at asymptotic times: $\langle\Omega|\phi(\vec{x}, t= \pm \infty)| p\rangle=C e^{i \vec{p} \vec{p}}$ for some constant $C$, as was shown for free fields in Eq. (2.76).

## 物理代写|量子场论代写Quantum field theory代考|Discussion

The LSZ reduction says that to calculate an $S$-matrix element, multiply the time-ordered product of fields by some $\square+m^2$ factors and Fourier transform. If the fields $\phi(x)$ were free fields, they would satisfy $\left(\square+m^2\right) \phi(x, t)=0$ and so the $\left(\square_i+m^2\right)$ terms would give zero. However, as we will see, when calculating amplitudes, there will be factors of propagators $\frac{1}{\square+m^2}$ for the one-particle states. These blow up as $\left(\square+m^2\right) \rightarrow 0$. The LSZ formula guarantees that the zeros and infinities in these terms cancel, leaving a non-zero result. Moreover, the $\square+m^2$ terms will kill anything that does not have a divergence, that will be anything but the exact initial and final state we want. ${ }^3$ That is the whole point of the LSZ formula: it isolates the asymptotic states by adding a carefully constructed zero to cancel everything that does not correspond to the state we want.

It is easy to think that LSZ is totally trivial, but it is not. The projections are the only thing that tells us what the initial states are (the things created from the vacuum at $t=-\infty$ ) and what the final states are (the things that annihilate into the vacuum at $t=+\infty$ ). Initial and final states are distinguished by the $\pm i$ in the phase factors. The time ordering is totally physical: all the creation of the initial states happens before the annihilation of the final states. In fact, because this is true not just for free fields, all the crazy stuff that happens at intermediate times in an interacting theory must be time-ordered too. But the great thing is that we do not need to know which are the initial states and which are the final states anymore when we do the hard part of the computation. We just have to calculate time-ordered products, and the LSZ formula sorts out what is being scattered for us.

# 量子场论代考

## 物理代写|量子场论代写Quantum field theory代考|The LSZ reduction formula

$$d \sigma=\frac{1}{\left(2 E_1\right)\left(2 E_2\right)\left|\vec{v}1-\vec{v}2\right|}|\mathcal{M}|^2 d \Pi{\mathrm{LIPS}},$$ 其中$d \Pi{\mathrm{LIPS}}$是洛伦兹不变相空间，$\mathcal{M}$是$\langle f|\mathcal{M}| i\rangle$的简写，是分解出整体动量守恒$\delta$函数的$S$ -矩阵元素:
$$\langle f|S-\mathbb{1}| i\rangle=i(2 \pi)^4 \delta^4(\Sigma p) \mathcal{M}$$

$$|i\rangle=\sqrt{2 \omega_1} \sqrt{2 \omega_2} a_{p_1}^{\dagger}(-\infty) a_{p_2}^{\dagger}(-\infty)|\Omega\rangle,$$
$|\Omega\rangle$是基态，没有粒子，那么
$$|f\rangle=\sqrt{2 \omega_3} \cdots \sqrt{2 \omega_n} a_{p_3}^{\dagger}(\infty) \cdots a_{p_n}^{\dagger}(\infty)|\Omega\rangle$$

$$\langle f|S| i\rangle=2^{n / 2} \sqrt{\omega_1 \omega_2 \omega_3 \cdots \omega_n}\left\langle\Omega\left|a_{p_3}(\infty) \cdots a_{p_n}(\infty) a_{p_1}^{\dagger}(-\infty) a_{p_2}^{\dagger}(-\infty)\right| \Omega\right\rangle .$$

$$\phi(x)=\phi(\vec{x}, t)=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left[a_p(t) e^{-i p x}+a_p^{\dagger}(t) e^{i p x}\right],$$

## 物理代写|量子场论代写Quantum field theory代考|Discussion

LSZ简化说，要计算一个$S$ -矩阵元素，将场的时间顺序乘积乘以一些$\square+m^2$因子和傅里叶变换。如果字段$\phi(x)$是自由字段，它们将满足$\left(\square+m^2\right) \phi(x, t)=0$，因此$\left(\square_i+m^2\right)$项将给出零。然而，正如我们将看到的，当计算振幅时，对于单粒子态，将会有传播因子$\frac{1}{\square+m^2}$的因素。这些被炸成$\left(\square+m^2\right) \rightarrow 0$。LSZ公式保证这些项中的零和无穷大相互抵消，留下一个非零的结果。此外，$\square+m^2$项会杀死任何没有散度的东西，也就是除了我们想要的初始和最终状态之外的任何东西。${ }^3$这就是LSZ公式的全部意义:它通过添加一个精心构造的零来消除所有不符合我们想要的状态的东西，从而隔离渐近状态。

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