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

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

The variables $s, t$ and $u$ are called Mandelstam variables. They are a great shorthand, used almost exclusively in $2 \rightarrow 2$ scattering and in $1 \rightarrow 3$ decays, although there are generalizations for more momenta. For $2 \rightarrow 2$ scattering, with initial momenta $p_1$ and $p_2$ and final momenta $p_3$ and $p_4$, they are defined by
\begin{aligned} & s \equiv\left(p_1+p_2\right)^2=\left(p_3+p_4\right)^2, \ & t \equiv\left(p_1-p_3\right)^2=\left(p_2-p_4\right)^2, \ & u \equiv\left(p_1-p_4\right)^2=\left(p_2-p_3\right)^2 . \end{aligned}
These satisfy
$$s+t+u=\sum m_j^2$$
where $m_j$ are the invariant masses of the particles.
As we saw in the previous example, $s, t$ and $u$ correspond to particular diagrams where momentum in the propagator has invariant $p_\mu^2=s, t$ or $u$. This correspondence is summarized in Box 7.2. The $s$-channel is an annihilation process. In the $s$-channel, the intermediate state has $p_\mu^2=s>0$. The $t$ – and $u$-channels are scattering diagrams and have $t<0$ and $u<0 . s, t$ and $u$ are great because they are Lorentz invariant. So we compute $\mathcal{M}(s, t, u)$ in the center-of-mass frame, and then we can easily find out what it is in any other frame, for example the frame of the lab in which we are doing the experiment. We will use $s, t$ and $u$ a lot.

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

Suppose we have an interaction with derivatives in it, such as
$$\mathcal{L}{\text {int }}=\lambda \phi_1\left(\partial\mu \phi_2\right)\left(\partial_\mu \phi_3\right)$$
where three different scalar fields are included for clarity. In momentum space, these $\partial_\mu$ ‘s give factors of momenta. But now remember that
$$\phi(x)=\int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}}\left(a_p e^{-i p x}+a_p^{\dagger} e^{i p x}\right) .$$
So, if the particle is being created (emerging from a vertex) it gets a factor of $i p_\mu$, and if it is being destroyed (entering a vertex) it gets a factor of $-i p_\mu$. So, we get a minus for incoming momentum and a plus for outgoing momentum. In this case, it is quite important to keep track of whether momentum is flowing into or out of the vertex.
For example, take the diagram
Label the initial momenta $p_1^\mu$ and $p_2^\mu$ and the final momenta $p_1^{\prime \mu}$ and $p_2^{\prime \mu}$. The exchanged momentum is $k^\mu=p_1^\mu+p_2^\mu=p_1^{\prime \mu}+p_2^{\prime \mu}$. Then this diagram gives
$$i \mathcal{M}=(i \lambda)^2\left(-i p_2^\mu\right)\left(i k^\mu\right) \frac{i}{k^2}\left(i p_2^{\prime \nu}\right)\left(-i k^\nu\right)=-i \lambda^2 \frac{\left[p_2 \cdot p_1+\left(p_2\right)^2\right]\left[p_2^{\prime} \cdot p_1^{\prime}+\left(p_2^{\prime}\right)^2\right]}{\left(p_1+p_2\right)^2} .$$
As a cross check, we should get the same answer if we use a different Lagrangian related to the one we used by integration by parts:
$$\mathcal{L}{\text {int }}=-\lambda \phi_3\left[\left(\partial\mu \phi_1\right)\left(\partial_\mu \phi_2\right)+\phi_1 \square \phi_2\right] .$$

# 量子场论代考

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

\begin{aligned} & s \equiv\left(p_1+p_2\right)^2=\left(p_3+p_4\right)^2, \ & t \equiv\left(p_1-p_3\right)^2=\left(p_2-p_4\right)^2, \ & u \equiv\left(p_1-p_4\right)^2=\left(p_2-p_3\right)^2 . \end{aligned}

$$s+t+u=\sum m_j^2$$

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

$$\mathcal{L}{\text {int }}=\lambda \phi_1\left(\partial\mu \phi_2\right)\left(\partial_\mu \phi_3\right)$$

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

$$i \mathcal{M}=(i \lambda)^2\left(-i p_2^\mu\right)\left(i k^\mu\right) \frac{i}{k^2}\left(i p_2^{\prime \nu}\right)\left(-i k^\nu\right)=-i \lambda^2 \frac{\left[p_2 \cdot p_1+\left(p_2\right)^2\right]\left[p_2^{\prime} \cdot p_1^{\prime}+\left(p_2^{\prime}\right)^2\right]}{\left(p_1+p_2\right)^2} .$$

$$\mathcal{L}{\text {int }}=-\lambda \phi_3\left[\left(\partial\mu \phi_1\right)\left(\partial_\mu \phi_2\right)+\phi_1 \square \phi_2\right] .$$

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