Propagator, Interaction, and Scattering

Following our previous introduction of the quantized fields, the next question is about their dynamics, which should be different from the one of point particle and wavefunction in quantum mechanics, and how to describe the physics process and interaction? Though we have the motion equation of fields, we prefer to deal with it using Lagrangian later.

We will only analyze the scalar field $\varphi$ as example, with $\mathcal L_0= \frac12\partial_\mu\varphi\partial^\mu\varphi -\frac12m^2\varphi^2,~S[\varphi]=\int d^4x\,\mathcal L(\varphi,\partial_\mu\varphi)$, and motion equation $(\square + m^2)\varphi = 0$, Fourier expansion $\varphi(x) = \int \widetilde{d^3 p}[a(\mathbf{p}) e^{ipx} + a^{\dagger}(\mathbf{p}) e^{-ipx})]$. Similarly, many calculation details will be skipped, and some convention may not be unified for now (so you may see different $\pm$ in $\mathcal L$ and other formulas).

Excitement and Dynamics

After quantization, the $a(\mathbf{p}), a^{\dagger}(\mathbf{p}),\varphi(x)$ become an operator, for example, $|\mathbf p_1,\mathbf p_2,\ldots\rangle = a^\dagger_{\mathbf p_1} a^\dagger_{\mathbf p_2}\cdots|0\rangle$ is a n-particle excited state with momentum $\{\mathbf{p_i}\}$. We should notice that this is still the same field but in different state, which means all same particle are generated from the identical field. And the field itself can link the vacuum and excited state: $\varphi(x)|0\rangle = \int \frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}} e^{ipx} |\mathbf p\rangle$, which is an amplitude distribution of all possible single-particle mode. Another example is, if the field operator act on a 1-particle state $\phi(x)|p\rangle = e^{-ipx}|0\rangle + \int\widetilde{d^3k}\, e^{ikx}|k,p\rangle$, it will gives a combination of vacuum and 2-particle state.

• Like what we have learned in quantum mechanics, the interaction of a quantum state is the evolution of amplitude distribution on different momentum mode.

Consider the propagation of the field state, a natural question is how the excited state of fields in different spacetime correlate with each other? Let’s first consider the amplitude overlap between $|\varphi_y\rangle=\varphi(y)|0\rangle, ~|\varphi_x\rangle=\varphi(x)|0\rangle$, where $\varphi^{\dagger}=\varphi$ for real scalar field. Thus $\langle \varphi_x | \varphi_y \rangle = \langle 0| \varphi(x)\varphi(y) | 0 \rangle$, and we can also write it using the spacetime transition generator and Heisenberg picture $\varphi(x) = e^{iPx} \varphi(0) e^{-iPx} \rightarrow \langle 0 |\varphi(0) e^{-iP(x-y)} \varphi(0) | 0\rangle$. Then we expand it with mode $\langle0|\phi(x)\phi(y)|0\rangle = \sum_n \langle0|\phi(x)|n\rangle \langle n|\phi(y)|0\rangle = \int \frac{d^3p}{(2\pi)^3 2E_{\mathbf p}} e^{-ip\cdot(x-y)}$, and read it as each mode is generated at $y$, then evolves and overlap with the excited state of the same field at $x$. How does the field in different spacetime position connect gives its dynamics, or called translation amplitude, like in quantum mechanics $K(x_f,t_f;x_i,t_i) = \langle x_f| e^{-iH(t_f-t_i)} |x_i\rangle$.

Then we introduce time ordering as we don’t need to ask $y$ generated before $x$, and it gives $D_F(x-y) \equiv \langle0| T\{\phi(x)\phi(y)\} |0\rangle= \int\frac{d^3p}{(2\pi)^3} \frac{e^{i\mathbf p\cdot\mathbf r}}{2E_{\mathbf p}} [ \theta(t)e^{-iE_{\mathbf p}t} + \theta(-t)e^{+iE_{\mathbf p}t} ]$. Using the contour integral of the complex function, we can verify that $\frac1{2E} \left[ \theta(t)e^{-iEt} + \theta(-t)e^{+iEt} \right] = \int_{-\infty}^{\infty} \frac{dp^0}{2\pi} \frac{i\,e^{-ip^0t}} {(p^0)^2-E^2+i\epsilon}$, and derive $D_F(x-y) = \int\frac{d^4p}{(2\pi)^4} \frac{i}{p^2-m^2+i\epsilon} e^{-ip\cdot(x-y)}$. Notice that $(\Box_x+m^2)D_F(x-y) = -i\delta^{(4)}(x-y)$ is the Green’s function of Klein-Gordon operator, which means the excitement at $y$ can serve as the source to have an influence at $x$. This shows us that the propagator can serve as the inverse of dynamic operator $D_F \sim (\Box+m^2)^{-1}$. In momentum space, we can derive $D_F(p) = \frac{i}{p^2-m^2+i\epsilon}$, the contribution of each mode, with the Fourier transformation $D_F(x-y) = \int\frac{d^4p}{(2\pi)^4} e^{-ip\cdot(x-y)} D_F(p)$.

• Of course, the free particle should be on-shell $p^2=m^2$. In the propagator, we might integrate over the $p$, which is just a variable and only have physical mass on the pole $p^2=m^2$, the internal $p$ in the propagator is not an observable physical state.

Lagrangian and Interaction

To change the state of the field, there needs external source or interaction. Source can be introduced in the Lagrangian $\mathcal L_0= -\frac12\partial_\mu\varphi\partial^\mu\varphi -\frac12m^2\varphi^2 + J(x)\varphi(x)$, which derive the motion equation $(\square + m^2)\varphi (x) = J(x)$. With propagator, we can write the influenced field as $\varphi(x) = \varphi_{0}(x) + \int d^4 y G_R(x-y) J(y)$. For another field, there must be a cross term in the Lagrangian to couple in the motion equation: $\mathcal L_{int} = -\frac{g}{2}\varphi \chi^2$, which derives $(\Box+m_\phi^2)\phi = -\frac g2\chi^2,~ (\Box+m_\chi^2)\chi = -g\phi\chi$, and the structure of one field become the source of another field in motion equation. The coupling between these field operators then allows the transition between different Fock states like $\varphi \rightarrow \chi\chi$.

We will use the interaction picture and canonical derivation with Hamiltonian to proceed the scattering process $\varphi \varphi \rightarrow \varphi \varphi$ here. Let’s consider $\mathcal L_{int} = -\frac{\lambda}{4!} \varphi^4$ first, note that the coefficient is defined for later convenience. In interaction picture, we have $H_{I} = \int d^3 x \frac{\lambda}{4!} \varphi^4$, $i\frac{d}{dt}|\Psi_I(t)\rangle = H_I(t)|\Psi_I(t)\rangle$, and $U_I(t,t_0) = T\exp \left[ -i\int_{t_0}^{t}dt'\,H_I(t') \right]$, which can be expanded as Dyson series. For the scattering process, we set the asymptotic boundary with $|i \rangle = \lim_{t \rightarrow -\infty}|p_1,p_2 \rangle$ and $|f \rangle = \lim_{t \rightarrow + \infty}|p_3,p_4 \rangle$.

In this way, we introduce the S-matrix to describe the scattering by $S = U_{I}(+\infty, -\infty) = T\exp\left[i\int d^4x\,\mathcal L_{\mathrm{int}}(x)\right]$, which can be written as $S_{fi} = \langle f| U_{I}(+ \infty,- \infty) | i \rangle = {}_{\rm out}\langle f,+\infty|i, -\infty\rangle_{\rm in}$, and expanded as $S = 1 -i\frac{\lambda}{4!} \int d^4x\,\phi^4(x) +O(\lambda^2)$ in perturbation theory. We can write $\langle p_3p_4| S^{(1)} |p_1p_2\rangle = -i\frac{\lambda}{4!} \int d^4x\, \langle p_3p_4| \phi^4(x) |p_1p_2\rangle$, and there must be 2 $a(\mathbf{p})$ for in and 2 $a^{\dagger}(\mathbf{p})$ for out, which yields $4!$ equivalent permutations and gives $\langle f|S^{(1)}|i\rangle_c = -i\lambda (2\pi)^4 \delta^{(4)}(P_i-P_f)$. In our normalization and conservation, it means that the scattering amplitude is $i \mathcal{M} = - i \lambda$, the left dynamic term in the transition amplitude besides $(2\pi)^4 \delta^{(4)}(P_i-P_f)$. Consider the general $\mathcal{L}_{int} = - \frac{g}{n!} \varphi^n$, the first-order expansion of S-matrix is $S^{(1)} =-\frac{ig}{n!}\int d^4x ~ \phi^n(x)$. This corresponds to the interaction point of $n$ field operators, or called vertex, which similarly yields the vertex factor $-ig$.

The $\varphi \varphi \rightarrow \varphi \varphi$ process can also correspond to another interaction term $\mathcal L_{int} = -\frac{g}{3!}\varphi^3$ in simply tree level with an internal propagator line. This requires $\langle p_3p_4|S^{(1)}_3 S^{(1)}_3|p_1p_2\rangle$ for 2 interaction points and 4 external scalar lines in Feynman diagram, which corresponds to the second-order term in the Dyson expansion $S^{(2)} = \frac{(-ig)^2}{2!(3!)^2} \int d^4x\,d^4y\, T\{\phi^3(x)\phi^3(y)\} = \frac{1}{2!}T\{S^{(1)}S^{(1)}\}$. In the scattering matrix, we can write $\langle p_3p_4| T\{\phi^3(x)\phi^3(y)\} |p_1p_2\rangle = \langle0| a_{p_4}a_{p_3}\, T\{\phi^3(x)\phi^3(y)\}\, a_{p_1}^\dagger a_{p_2}^\dagger |0\rangle$. For a giving incoming external attachment, $\langle0|\phi(x)|p\rangle=e^{-ipx}$ gives the plane-wave factor, while other components are organized into other contraction and Feynman diagrams. Then the left terms is $\langle0| T\{\phi(x)\phi(y)\} |0\rangle = D_F(x-y)$, which is the propagator we introduced before! For example, the s-channel, where internal $q = p_1 + p_2,~ q^2 = s$, corresponds to $i\mathcal M_s = (-ig)^2 \frac{i}{s-m^2+i\epsilon}$ in $S^{(2)}$, as well as $t=(p_1-p_3)^2,~ u=(p_1-p_4)^2$. Then we can write the amplitude $i\mathcal M = (-ig)^2 \left[ \frac{i}{s-m^2+i\epsilon} + \frac{i}{t-m^2+i\epsilon} + \frac{i}{u-m^2+i\epsilon} \right]$.

• We skip the Wick theorem and contraction here.

Cross section and Experiment

To derive the cross section, we need to deal with the $\delta(P_i - P_f)$ in $\langle f | S-1 | i \rangle$. Let’s work in the finite space $V$ and time $T$, we get $|\langle f|S-1|i\rangle|^2 = (2\pi)^4VT \delta^{(4)}(P_f-P_i) |\mathcal M|^2$. Consider the infinitesimal momentum region in final state $d^3 p_{f1}...d^3 p_{fn}$, the corresponding probability is $dP = \frac{ |\langle f|S-1|i\rangle|^2 }{ \langle i|i\rangle\langle f|f\rangle } \prod_f \frac{Vd^3p_f}{(2\pi)^3} = \frac{T}{V} \frac{1}{4E_1E_2} |\mathcal M|^2 d\Phi_n$ under our convention $\langle p'|p\rangle = 2E_{\mathbf p}(2\pi)^3 \delta^{(3)}(\mathbf p-\mathbf p')$. The $d\Phi_n = (2\pi)^4 \delta^{(4)} \left( \sum_i p_i - \sum_f p_f \right) \prod_f \frac{d^3p_f}{(2\pi)^3\,2E_f}$ is called Lorentz-invariant phase space (LIPS), and you can identify the $|\mathcal{M}|^2$ as the weight of a certain final-state configuration. Under time unit we get the transition rate $dR = \frac{dP}{T}$, and under space unit, we define the particle beam density as $\frac{1}{V}$ (only 1 particle in V), and the relative velocity $v_{rel}$ of 2 beams is used to define the incident flux $\Phi_{in} = \frac{v_{rel}}{V}$, which yields the cross section $d\sigma = \frac{dR}{\Phi_{in}} = \frac{1}{4E_1E_2v_{\rm rel}} |\mathcal M|^2 d\Phi_n$.

Generally, we can write $E_1E_2v_{\rm rel} = \sqrt{ (p_1\cdot p_2)^2-m_1^2m_2^2 }$, and $\overline{|\mathcal{M}|^2}$ to average over initial states and sum over final states, as well as different channels’ amplitudes $\mathcal M_{\rm tot} = \sum_{\text{all diagrams}} \mathcal M_i$, and express $d\sigma = \frac{1}{\text{flux}} \overline{|\mathcal M|^2} d\Phi_n$, where flux is the beam normalization. For scattering process like ($p_1+p_2\rightarrow p_3+p_4$), we define $\mathbf p_1+\mathbf p_2=0,~ \mathbf p_3+\mathbf p_4=0,~ E_{CM} = \sqrt{s}$, and write $d\Phi_2 = (2\pi)^4 \delta^{(4)}(p_1+p_2-p_3-p_4) \frac{d^3p_3}{(2\pi)^3 2E_3} \frac{d^3p_4}{(2\pi)^3 2E_4} = \frac1{16\pi^2} \frac{|\mathbf p_f|}{\sqrt s} d\Omega_{p_f}$, where $p_i=p_1,p_f = p_3$, then we can give $\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2s} \frac{|\mathbf p_f|}{|\mathbf p_i|} \overline{|\mathcal M|^2}$. For $\varphi \varphi \rightarrow \varphi \varphi$, we need to add $\frac{1}{2!}$ to avoid repeated calculation of $p_3 \leftrightarrow p_4$, and with $\mathbf{p}_i = \mathbf{p}_f$ we can write $\frac{d\sigma}{d\Omega} = \frac{1}{2!} \frac{\overline{|\mathcal M|^2}} {64\pi^2s}$. Specially, you should write $i\mathcal M_{\rm tot} = -i\lambda + (-ig)^2 \left[ \frac{i}{s-m^2+i\epsilon} + \frac{i}{t-m^2+i\epsilon} + \frac{i}{u-m^2+i\epsilon} \right]$ based on our previously defined $\mathcal L_{\rm int} = -\frac{g}{3!}\phi^3 -\frac{\lambda}{4!}\phi^4$.

For decay process $(p \rightarrow p_1 + ... +p_n)$, we only have 1 particle in the initial state and have $\langle i | i \rangle=2E_i V$. And we can write $dP = \frac{T}{2E_i} |\mathcal M|^2 d\Phi_n$, where $E_i = m_i$ in rest frame, and yield decay rate $d\Gamma = \frac{dP}{T} = \frac{1}{2E_A}|\mathcal M|^2d\Phi_n$. For $\varphi \rightarrow \chi\chi$ in $\mathcal L_{\rm int} = -\frac g2\Phi\chi^2$, we have $|\mathcal M|^2=g^2$ and $|\mathbf p_f| = \frac M2\sqrt{1-\frac{4m^2}{M^2}}$. As there is not $(\theta,\phi)$ dependent, we can just calculate the decay rate over $\Omega_{p_f}$, and multiply with the permutation factor $\frac{1}{2!}$ to get $\Gamma_{\Phi\to\chi\chi} = \frac{g^2}{32\pi M} \sqrt{1-\frac{4m^2}{M^2}}$. This decay rate corresponds to $\frac{dN}{N(t)}=-\Gamma dt$ in experiment, and multi channels should also sum up $\Gamma_{\rm total} = \sum_f\Gamma_f$, where $\mathrm{BR}(i\to f) = \frac{\Gamma_f}{\Gamma_{\rm total}}$ (this is different from sum over different Feynman diagrams for amplitude). For a specific process in experiment, we use the production cross section, branch ratio, acceptance and efficiency to calculate the expected signal events: $N_{\rm signal} \approx \mathcal L_{\rm int}\, \sigma(pp\to H+X)\, \mathrm{BR}(H\to f)\, A\varepsilon$.

We’d like to introduce Feynman Rules for calculation, which assign corresponding mathematical factors to the amplitude components in Feynman diagrams of certain Lagrangian conveniently. In our previous derivation, we have $-i\lambda \sim -\frac{\lambda}{4!}\phi^4$ for 4-vertex, $-ig \sim -\frac{g}{3!}\phi^3$ for 3-vertex, $\langle0|T\phi(x)\phi(y)|0\rangle \sim \frac{i}{p^2-m^2+i\epsilon}$ for internal propagator line; and 1 for external line of on-shell incoming or outgoing particle, each vertex requires momentum conservation, sum over the amplitudes for the same process, with a symmetry factor for identical process. In later loop correction, we will introduce the integral over internal momentum in loops. And different Feynman Rules will also be derived for different Lagrangian.

Another way to deal with physical process is using path integral, where we use the Action and generating functional. We will introduce it in other sections as $\mathcal L \rightarrow S[\phi] \rightarrow Z[J] \rightarrow \text{correlation functions} \rightarrow \text{LSZ} \rightarrow \mathcal M$.