Fermion and Yukawa Tree Amplitudes

Tree amplitudes with fermions are built along oriented fermion lines. External particles contribute $u$ or $\bar u$, antiparticles contribute $v$ or $\bar v$, and matrices are multiplied in their order along the line. Relative minus signs come from reordering fermionic operators and from exchanging identical external fermions; spin sums are applied only after the amplitude has been assembled. A scalar Yukawa decay provides a normalization and threshold check in one line of algebra.

Required background. Connected Tree Diagrams and Amputated Amplitudes supplies graph construction. Plane Waves, Spin Sums, and Bilinears supplies the external spinors and completeness relations. Fermion Signs and Closed Loops supplies the operator-order sign rule.

Read an open fermion line in order

For a Dirac fermion and a real scalar, take

$$\mathcal L_{\mathrm{int}}=-y\phi\bar\psi\psi,$$

with real $y$ and vertex $-iy$. Write $p\!\!\!/\equiv\gamma^\mu p_\mu$; the propagator is

$$\frac{i(p\!\!\!/+m)}{p^2-m^2+i0}.$$

With the chapter’s relativistic normalization, the external wave functions are

• $u_s(p)$ for an incoming fermion and $\bar u_s(p)$ for an outgoing fermion;
• $\bar v_s(p)$ for an incoming antifermion and $v_s(p)$ for an outgoing antifermion.

These assignments include the LSZ external-line factors; no external fermion propagator remains. Put the row spinor on the left and the column spinor on the right, then traverse the line from the row-spinor end toward the column-spinor end. Factors encountered first remain leftmost, so the noncommuting gamma-matrix order is fixed rather than guessed from the visual direction of the line.

Choose a definite external-operator order before doing contractions. Along an open line, begin at one external spinor and multiply every vertex and propagator in sequence until the other external spinor is reached. For $\psi(p)+\phi(k)\to\psi(p')+\phi(k')$, the two tree orderings give

$$\begin{aligned} i\mathcal M_s &=\bar u(p')(-iy) \frac{i(p\!\!\!/+k\!\!\!/+m)}{(p+k)^2-m^2+i0} (-iy)u(p),\\ i\mathcal M_u &=\bar u(p')(-iy) \frac{i(p\!\!\!/-k'\!\!\!/+m)}{(p-k')^2-m^2+i0} (-iy)u(p). \end{aligned}$$

There is no arbitrary extra minus sign between these two graphs: their open fermion ordering is the same. A relative sign does arise when two identical external fermions are exchanged, because restoring the chosen external-state order requires an odd permutation. Srednicki gives the external-line and ordering rules in Srednicki 2007, § 45, printed pp. 282–291. Matching external-spinor and perturbative scattering conventions are given in Schwartz 2014, §§ 13.1–13.3, printed pp. 225–233 and Weinberg 1995, §§ 5.5 and 6.3, printed pp. 219–228 and 280–285.

The arrow on a Dirac line tracks conserved fermion number when the interaction has it. It is not a literal momentum arrow. Crossing an outgoing fermion to an incoming antifermion changes the external wave function and its analytic assignment; it is not implemented by replacing $p\to-p$ inside an unchanged $u$-spinor.

The two-ordering result has an immediate soft-momentum check. In the first numerator, the incoming Dirac equation gives

$$(p\!\!\!/+k\!\!\!/+m)u(p) =(2m+k\!\!\!/)u(p),$$

whereas the second numerator is most simply reduced from the left or after using momentum conservation. Applying an on-shell equation to an internal momentum would be wrong: $(p+k)^2$ and $(p-k')^2$ are generically off shell and their denominators must remain.

Scalar decay into a fermion pair

Let a scalar of mass $M$ decay into a Dirac pair of mass $m$, with $M>2m$. With $P=p_1+p_2$,

$$i\mathcal M=-iy\,\bar u(p_1)v(p_2), \qquad \mathcal M=-y\,\bar u(p_1)v(p_2).$$

Summing over final spins and using

$$\sum_s u_s(p)\bar u_s(p)=p\!\!\!/+m, \qquad \sum_s v_s(p)\bar v_s(p)=p\!\!\!/-m,$$

gives

$$\begin{aligned} \sum_{s_1,s_2}|\mathcal M|^2 &=y^2\operatorname{tr}\!\left[ (p_1\!\!\!/+m)(p_2\!\!\!/-m)\right]\\ &=4y^2(p_1\cdot p_2-m^2)\\ &=2y^2(M^2-4m^2). \end{aligned}$$

The last line uses $2p_1\cdot p_2=M^2-2m^2$. It is nonnegative exactly in the allowed decay region and vanishes at threshold. With the standard two-body phase space, define

$$\beta=\sqrt{1-\frac{4m^2}{M^2}}, \qquad |\mathbf p|=\frac{M\beta}{2}.$$

For two distinguishable final particles and no initial-spin average,

$$\Gamma =\frac{|\mathbf p|}{8\pi M^2} \sum_{s_1,s_2}|\mathcal M|^2,$$

and therefore

$$\Gamma(\phi\to\psi\bar\psi) =\frac{y^2M}{8\pi} \left(1-\frac{4m^2}{M^2}\right)^{3/2}.$$

The power $3/2$ combines the phase-space velocity with the additional helicity/bilinear suppression in the squared amplitude. This is an incisive check: changing the $v$-spin sum to $p\!\!\!/+m$ produces the wrong threshold behavior. The displayed result is for one colorless Dirac species; $N_c$ identical color copies multiply the spin-summed width by $N_c$. Spin-sum and trace methods are developed in Srednicki 2007, §§ 46–48, printed pp. 292–302.

What changes for identical fermions

If two identical fermions appear in the initial or final state, construct a labeled amplitude first and then antisymmetrize under their exchange. For example, two graphs related by swapping identical outgoing fermions enter as

$$\mathcal M_{\mathrm{antisym}} =\mathcal M(p_3,p_4)-\mathcal M(p_4,p_3).$$

This sign is fixed by Fock-state ordering, not by diagram aesthetics. Closed fermion loops carry their own minus sign, but they do not occur in a tree graph. Chiral gauge vertices, loop traces, and flavor phenomenology require their respective specialist treatments; this page fixes only the Yukawa and Dirac tree grammar.

Mandelstam Channels and Tree-Level Crossing develops the state and operator changes required when a fermion is crossed. Cross Sections and Decay Rates develops the general phase-space and averaging formula used in the decay check above.

Check your understanding

Redo the decay trace for $m=0$, then replace the antifermion completeness relation by the incorrect plus sign and identify where the threshold zero is lost. State whether any initial spin average is present.

Solution

For $m=0$, $2p_1\cdot p_2=M^2$, hence $\sum|\mathcal M|^2=4y^2p_1\cdot p_2=2y^2M^2$. If one incorrectly uses $\sum v\bar v=p\!\!\!/+m$, the trace becomes $4y^2(p_1\cdot p_2+m^2)=2y^2M^2$, even for nonzero $m$; it no longer vanishes at $M=2m$. The scalar parent has one spin state, so there is no initial-spin average.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014, §§ 13.1–13.3, printed pp. 225–233. doi:10.1017/9781139540940.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007, §§ 45–48, printed pp. 282–302. doi:10.1017/CBO9780511813917.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, §§ 5.5 and 6.3, printed pp. 219–228 and 280–285. doi:10.1017/CBO9781139644167.