B.3 Interactions

B.3.1 The interaction picture and Dyson’s formula

For treating interactions that are small perturbations to the free theory, it is most useful to employ the interaction picture of QM, a hybrid of the Schrödinger and Heisenberg pictures. Recall that in the Schrödinger picture, operators are fixed while states evolve with time, and vice versa in the Heisenberg picture. In the interaction picture, we split the Hamiltonian into the free ($H_0$) and interaction terms ($H_{\mathrm{int}}$), defining operators to evolve with the former and states with the latter.

The upshot of this in QFT is that the S-matrix element can be written according to Dyson’s formula:

where $T$ is the same time-ordering operator from Section 3.1.4 and $H_I$ is the time-evolved interaction Hamiltonian in the interaction picture:

Assuming a small $H_{\mathrm{int}}$, Dyson’s formula can be Taylor expanded as:

The first term in the expansion is the free field term, which we ignore.⁵ The $n$th term after that is of order $g^n$, where $g$ is the coupling constant of the interaction term. Thus, this offers a prescription for calculating the S-matrix element up to any fixed order in the interaction strength.

Note that $|i⟩$ and $|f⟩$ are particle momentum eigenstates of the free theory. We can justify this intuitively by thinking of them as the states long before and after the interaction, when the interaction term is negligible. Formally, there is in fact a complicated formula relating the free and interacting eigenstates; however, the proportionality factors cancel rather beautifully in the S-matrix element, allowing us to focus on only “connected” and “amputated” Feynman diagrams between the free eigenstates, defined in the next section. This is illustrated (literally) for the vacuum states in Peskin and Schroeder [81] Chapter 4, and justified more generally by the LSZ reduction formula.

B.3.2 First-order examples and the matrix element $M$

Let us look at the $n=1$ and $n=2$ S-matrix element terms from Eq. 3 for our scalar Yukawa theory (Eq. 3.2.3):

For given initial and final $N$-particle momentum states, these can be calculated manually by plugging in the field expansions (Eq. B.2.2 and B.2.15). For example, the first-order term ${⟨f|S|i⟩}^{(1)}$ is non-zero only for processes like:

• Meson decay $\varphi \to {\psi }^{†}\psi$: $|i⟩=\sqrt{2E_p}{a}_p^{†}|0⟩$, $|f⟩=\sqrt{4E_{q_1}{E}_{q_2}}{b}_{q_1}^{†}{c}_{q_2}^{†}|0⟩$; and
• Nucleon-antinucleon annihilation ${\psi }^{†}\psi \to \varphi$: $|i⟩=\sqrt{4E_{q_1}{E}_{q_2}}{b}_{q_1}^{†}{c}_{q_2}^{†}|0⟩$, $|f⟩=\sqrt{2E_p}{a}_p^{†}|0⟩$.

The amplitude for these can be calculated to be:

with the simple matrix element $M=-g$. Generally, however, calculating $M$ each time using the field expansions can be quite cumbersome. This is especially true at higher orders, which require Wick’s theorem [431] to treat time-ordered fields. We can avoid this by using Feynman diagrams, and their associated rules, which allow us to simply read off a matrix element from a drawing of the process.

B.3.3 Feynman diagrams

The conventions for Feynman diagrams in this dissertation are as follows:

1.

Time and momentum always flow from left to right. Thus, the left-most particles represent the initial, and the right-most the final states. Momentum arrows are shown in Figure 3.1 explicitly but need not be.

2.

Mesons are plotted as dotted and nucleons as solid lines.

3.

Nucleon lines have arrows representing particle-flow. For external (i.e., initial or final state) nucleons they point in the direction of momentum for particles and opposite for antiparticles. Again, in general, particles need not be explicitly labeled as in in Figure 3.1 since the linestyles and particle-flow arrows suffice.

As discussed above, only connected and amputated diagrams contribute to the S-matrix element, and we will focus on these. Connected means that every part of the diagrams is connected to at least one external line, and amputated means that there are no loops on external lines. Examples of disconnected and un-amputated diagrams are shown in Figure B.1. Interestingly, disconnected and un-amputated diagrams contribute to the vacuum and one-particle states, respectively, differing in the interacting versus free theory.

Example: nucleon scattering

Nucleon-nucleon scattering is the process: $\mathit{\psi \psi }\to \mathit{\psi \psi }$. The lowest order at which this can occur is of $O(g^2)$, as it requires at least two interaction vertices. The possible second-order diagrams are shown in Figure B.2. We interpret them as nucleons interacting via the exchange of a meson. As the nucleons are identical, we require two diagrams, for the two permutations of the two final states.

Using the first two Feynman rules, we find

for both diagrams. What remains is to enforce momentum conservation at each vertex. For the left-most diagram, we see $k=q_{f1}-q_{i1}=q_{f2}-q_{i2}$, while for the right-most $k=q_{f2}-q_{i1}=q_{f1}-q_{i2}$. Thus, the total matrix element is

where we have left out the $\mathit{i𝜀}$ term as there is no integral to perform.

Generally, we have to be careful with the relative signs of the matrix elements of different diagrams, corresponding to either constructive or destructive interference. (In fact, Peskin and Schroeder list “Figure out the overall sign of the diagram” as a Feynman rule.) In this case, we can reason physically that since nucleons are bosons, the amplitude will be symmetric under interchange of the two final states, and hence the two diagrams should be summed.

Mandelstam variables

To build some intuition for Mandelstam variables, let us sit in the center of mass (COM) frame, and define our coordinate frame such that incoming particles collide along the $z$-axis and scatter in the $y$-$z$ plane:

Then,

Thus, $s$ is the total energy in the COM frame squared — hence, we usually refer to the COM energy as $\sqrt{s}$ — while $t$ and $u$ are a measure of how much momentum is exchanged between the scattered particles. For example, if $𝜃=0$, both particles continue in the same direction and $t=0$, while if $𝜃=\pi$, they completely reverse direction and the momentum transfer along the collision axis is maximized at $\sqrt{\left|t\right|}=2p$.