Editing Feynman diagram (section)

=== Spin {{sfrac|1|2}}; "photons" and "ghosts" ===

==== Spin {{sfrac|1|2}}: Grassmann integrals ====
The field path integral can be extended to the Fermi case, but only if the notion of integration is expanded. A [[Berezin integral|Grassmann integral]] of a free Fermi field is a high-dimensional [[determinant]] or [[Pfaffian]], which defines the new type of Gaussian integration appropriate for Fermi fields.

The two fundamental formulas of Grassmann integration are:

:<math> \int e^{M_{ij}{\bar\psi}^i \psi^j}\, D\bar\psi\, D\psi= \mathrm{Det}(M)\,, </math>

where {{mvar|M}} is an arbitrary matrix and {{math|''ψ'', {{overline|''ψ''}}}} are independent Grassmann variables for each index {{mvar|i}}, and

:<math> \int e^{\frac12 A_{ij} \psi^i \psi^j}\, D\psi = \mathrm{Pfaff}(A)\,,</math>

where {{mvar|A}} is an antisymmetric matrix, {{mvar|ψ}} is a collection of Grassmann variables, and the {{sfrac|1|2}} is to prevent double-counting (since {{math|''ψ<sup>i</sup>ψ<sup>j</sup>'' {{=}} −''ψ<sup>j</sup>ψ<sup>i</sup>''}}).

In matrix notation, where {{mvar|{{overline|ψ}}}} and {{mvar|{{overline|η}}}} are Grassmann-valued row vectors, {{mvar|η}} and {{mvar|ψ}} are Grassmann-valued column vectors, and {{mvar|M}} is a real-valued matrix:

:<math> Z = \int e^{\bar\psi M \psi + \bar\eta \psi + \bar\psi \eta}\, D\bar\psi\, D\psi = \int e^{\left(\bar\psi+\bar\eta M^{-1}\right)M \left(\psi+ M^{-1}\eta\right) - \bar\eta M^{-1}\eta}\, D\bar\psi\, D\psi = \mathrm{Det}(M) e^{-\bar\eta M^{-1}\eta}\,,</math>

where the last equality is a consequence of the translation invariance of the Grassmann integral. The Grassmann variables {{mvar|η}} are external sources for {{mvar|ψ}}, and differentiating with respect to {{mvar|η}} pulls down factors of {{mvar|{{overline|ψ}}}}.

:<math> \left\langle\bar\psi \psi\right\rangle = \frac{1}{Z} \frac{\partial}{\partial \eta} \frac{\partial}{\partial \bar\eta} Z |_{\eta=\bar\eta=0} = M^{-1}</math>

again, in a schematic matrix notation. The meaning of the formula above is that the derivative with respect to the appropriate component of {{mvar|η}} and {{mvar|{{overline|η}}}} gives the matrix element of {{math|''M''<sup>−1</sup>}}. This is exactly analogous to the bosonic path integration formula for a Gaussian integral of a complex bosonic field:

:<math> \int e^{\phi^* M \phi + h^* \phi + \phi^* h } \,D\phi^*\, D\phi = \frac{e^{h^* M^{-1} h} }{ \mathrm{Det}(M)}</math>
:<math> \left\langle\phi^* \phi\right\rangle = \frac{1}{Z} \frac{\partial}{\partial h} \frac{\partial}{\partial h^*}Z |_{h=h^*=0} = M^{-1} \,.</math>

So that the propagator is the inverse of the matrix in the quadratic part of the action in both the Bose and Fermi case.

For real Grassmann fields, for [[Majorana fermion]]s, the path integral is a Pfaffian times a source quadratic form, and the formulas give the square root of the determinant, just as they do for real Bosonic fields. The propagator is still the inverse of the quadratic part.

The free Dirac Lagrangian:

:<math> \int \bar\psi\left(\gamma^\mu \partial_{\mu} - m \right) \psi </math>

formally gives the equations of motion and the anticommutation relations of the Dirac field, just as the Klein Gordon Lagrangian in an ordinary path integral gives the equations of motion and commutation relations of the scalar field. By using the spatial Fourier transform of the Dirac field as a new basis for the Grassmann algebra, the quadratic part of the Dirac action becomes simple to invert:

:<math> S= \int_k \bar\psi\left( i\gamma^\mu k_\mu - m \right) \psi\,. </math>

The propagator is the inverse of the matrix {{mvar|M}} linking {{math|''ψ''(''k'')}} and {{math|''{{overline|ψ}}''(''k'')}}, since different values of {{mvar|k}} do not mix together.

:<math> \left\langle\bar\psi(k') \psi (k) \right\rangle = \delta (k+k')\frac{1} {\gamma\cdot k - m}  = \delta(k+k')\frac{\gamma\cdot k+m }{ k^2 - m^2} </math>

The analog of Wick's theorem matches {{mvar|ψ}} and {{mvar|{{overline|ψ}}}} in pairs:

:<math> \left\langle\bar\psi(k_1) \bar\psi(k_2) \cdots \bar\psi(k_n) \psi(k'_1) \cdots \psi(k_n)\right\rangle = \sum_{\mathrm{pairings}} (-1)^S \prod_{\mathrm{pairs}\; i,j} \delta\left(k_i -k_j\right) \frac{1}{\gamma\cdot k_i - m}</math>

where S is the sign of the permutation that reorders the sequence of {{mvar|{{overline|ψ}}}} and {{mvar|ψ}} to put the ones that are paired up to make the delta-functions next to each other, with the {{mvar|{{overline|ψ}}}} coming right before the {{mvar|ψ}}. Since a {{math|''ψ'', ''{{overline|ψ}}''}} pair is a commuting element of the Grassmann algebra, it does not matter what order the pairs are in. If more than one {{math|''ψ'', ''{{overline|ψ}}''}} pair have the same {{mvar|k}}, the integral is zero, and it is easy to check that the sum over pairings gives zero in this case (there are always an even number of them). This is the Grassmann analog of the higher Gaussian moments that completed the Bosonic Wick's theorem earlier.

The rules for spin-{{sfrac|1|2}} Dirac particles are as follows: The propagator is the inverse of the Dirac operator, the lines have arrows just as for a complex scalar field, and the diagram acquires an overall factor of −1 for each closed Fermi loop. If there are an odd number of Fermi loops, the diagram changes sign. Historically, the −1 rule was very difficult for Feynman to discover. He discovered it after a long process of trial and error, since he lacked a proper theory of Grassmann integration.

The rule follows from the observation that the number of Fermi lines at a vertex is always even. Each term in the Lagrangian must always be Bosonic. A Fermi loop is counted by following Fermionic lines until one comes back to the starting point, then removing those lines from the diagram. Repeating this process eventually erases all the Fermionic lines: this is the Euler algorithm to 2-color a graph, which works whenever each vertex has even degree. The number of steps in the Euler algorithm is only equal to the number of independent Fermionic homology cycles in the common special case that all terms in the Lagrangian are exactly quadratic in the Fermi fields, so that each vertex has exactly two Fermionic lines. When there are four-Fermi interactions (like in the Fermi effective theory of the [[weak nuclear interaction]]s) there are more {{mvar|k}}-integrals than Fermi loops. In this case, the counting rule should apply the Euler algorithm by pairing up the Fermi lines at each vertex into pairs that together form a bosonic factor of the term in the Lagrangian, and when entering a vertex by one line, the algorithm should always leave with the partner line.

To clarify and prove the rule, consider a Feynman diagram formed from vertices, terms in the Lagrangian, with Fermion fields. The full term is Bosonic, it is a commuting element of the Grassmann algebra, so the order in which the vertices appear is not important. The Fermi lines are linked into loops, and when traversing the loop, one can reorder the vertex terms one after the other as one goes around without any sign cost. The exception is when you return to the starting point, and the final half-line must be joined with the unlinked first half-line. This requires one permutation to move the last {{mvar|{{overline|ψ}}}} to go in front of the first {{mvar|ψ}}, and this gives the sign.

This rule is the only visible effect of the exclusion principle in internal lines. When there are external lines, the amplitudes are antisymmetric when two Fermi insertions for identical particles are interchanged. This is automatic in the source formalism, because the sources for Fermi fields are themselves Grassmann valued.

==== Spin 1: photons ====
The naive propagator for photons is infinite, since the Lagrangian for the A-field is:

:<math> S = \int \tfrac14 F^{\mu\nu} F_{\mu\nu} = \int -\tfrac12\left(\partial^\mu A_\nu \partial_\mu A^\nu - \partial^\mu A_\mu \partial_\nu A^\nu \right)\,.</math>

The quadratic form defining the propagator is non-invertible. The reason is the [[gauge invariance]] of the field; adding a gradient to {{mvar|A}} does not change the physics.

To fix this problem, one needs to fix a gauge. The most convenient way is to demand that the divergence of {{mvar|A}} is some function {{mvar|f}}, whose value is random from point to point. It does no harm to integrate over the values of {{mvar|f}}, since it only determines the choice of gauge. This procedure inserts the following factor into the path integral for {{mvar|A}}:

:<math> \int \delta\left(\partial_\mu A^\mu - f\right) e^{-\frac{f^2}{2} }\, Df\,. </math>

The first factor, the delta function, fixes the gauge. The second factor sums over different values of {{mvar|f}} that are inequivalent gauge fixings. This is simply

:<math> e^{- \frac{\left(\partial_\mu A_\mu\right)^2}{2}}\,.</math>

The additional contribution from gauge-fixing cancels the second half of the free Lagrangian, giving the Feynman Lagrangian:

:<math> S= \int \partial^\mu A^\nu \partial_\mu A_\nu </math>

which is just like four independent free scalar fields, one for each component of {{mvar|A}}. The Feynman propagator is:

:<math> \left\langle A_\mu(k) A_\nu(k') \right\rangle = \delta\left(k+k'\right) \frac{g_{\mu\nu}}{ k^2 }.</math>

The one difference is that the sign of one propagator is wrong in the Lorentz case: the timelike component has an opposite sign propagator. This means that these particle states have negative norm—they are not physical states. In the case of photons, it is easy to show by diagram methods that these states are not physical—their contribution cancels with longitudinal photons to only leave two physical photon polarization contributions for any value of {{mvar|k}}.

If the averaging over {{mvar|f}} is done with a coefficient different from {{sfrac|1|2}}, the two terms do not cancel completely. This gives a covariant Lagrangian with a coefficient <math>\lambda</math>, which does not affect anything:

:<math> S= \int \tfrac12\left(\partial^\mu A^\nu \partial_\mu A_\nu - \lambda \left(\partial_\mu A^\mu\right)^2\right)</math>

and the covariant propagator for QED is:

:<math>\left \langle A_\mu(k) A_\nu(k') \right\rangle =\delta\left(k+k'\right)\frac{g_{\mu\nu} - \lambda\frac{k_\mu k_\nu }{ k^2} }{ k^2}.</math>

==== Spin 1: non-Abelian ghosts ====
To find the Feynman rules for non-Abelian gauge fields, the procedure that performs the gauge fixing must be carefully corrected to account for a change of variables in the path-integral.

The gauge fixing factor has an extra determinant from popping the delta function:

:<math> \delta\left(\partial_\mu A_\mu - f\right) e^{-\frac{f^2}{2}} \det M </math>

To find the form of the determinant, consider first a simple two-dimensional integral of a function {{mvar|f}} that depends only on {{mvar|r}}, not on the angle {{mvar|θ}}. Inserting an integral over {{mvar|θ}}:

:<math> \int f(r)\, dx\, dy = \int f(r) \int d\theta\, \delta(y) \left|\frac{dy}{d\theta}\right|\, dx\, dy </math>

The derivative-factor ensures that popping the delta function in {{mvar|θ}} removes the integral. Exchanging the order of integration,

:<math> \int f(r)\, dx\, dy = \int d\theta\, \int f(r) \delta(y) \left|\frac{dy}{d\theta}\right|\, dx\, dy </math>

but now the delta-function can be popped in {{mvar|y}},

:<math> \int f(r)\, dx\, dy = \int d\theta_0\, \int f(x) \left|\frac{dy}{d\theta}\right|\, dx\,. </math>

The integral over {{mvar|θ}} just gives an overall factor of 2{{pi}}, while the rate of change of {{mvar|y}} with a change in {{mvar|θ}} is just {{mvar|x}}, so this exercise reproduces the standard formula for polar integration of a radial function:

:<math> \int f(r)\, dx\, dy = 2\pi \int f(x) x\, dx </math>

In the path-integral for a nonabelian gauge field, the analogous manipulation is:

:<math> \int DA \int \delta\big(F(A)\big) \det\left(\frac{\partial F}{\partial G}\right)\, DG e^{iS} = \int DG \int \delta\big(F(A)\big)\det\left(\frac{\partial F}{ \partial G}\right) e^{iS} \,</math>

The factor in front is the volume of the gauge group, and it contributes a constant, which can be discarded. The remaining integral is over the gauge fixed action.

:<math> \int \det\left(\frac{\partial F}{ \partial G}\right)e^{iS_{GF}}\, DA \,</math>

To get a covariant gauge, the gauge fixing condition is the same as in the Abelian case:

:<math> \partial_\mu A^\mu = f \,,</math>

Whose variation under an infinitesimal gauge transformation is given by:

:<math> \partial_\mu\, D_\mu \alpha \,,</math>

where {{mvar|α}} is the adjoint valued element of the Lie algebra at every point that performs the infinitesimal gauge transformation. This adds the Faddeev Popov determinant to the action:

:<math> \det\left(\partial_\mu\, D_\mu\right) \,</math>

which can be rewritten as a Grassmann integral by introducing ghost fields:

:<math> \int e^{\bar\eta \partial_\mu\, D^\mu \eta}\, D\bar\eta\, D\eta \,</math>

The determinant is independent of {{mvar|f}}, so the path-integral over {{mvar|f}} can give the Feynman propagator (or a covariant propagator) by choosing the measure for {{mvar|f}} as in the abelian case. The full gauge fixed action is then the Yang Mills action in Feynman gauge with an additional ghost action:

:<math> S= \int \operatorname{Tr} \partial_\mu A_\nu \partial^\mu A^\nu + f^i_{jk} \partial^\nu A_i^\mu A^j_\mu A^k_\nu + f^i_{jr} f^r_{kl} A_i A_j A^k A^l + \operatorname{Tr} \partial_\mu \bar\eta \partial^\mu \eta + \bar\eta A_j \eta \,</math>

The diagrams are derived from this action. The propagator for the spin-1 fields has the usual Feynman form. There are vertices of degree 3 with momentum factors whose couplings are the structure constants, and vertices of degree 4 whose couplings are products of structure constants. There are additional ghost loops, which cancel out timelike and longitudinal states in {{mvar|A}} loops.

In the Abelian case, the determinant for covariant gauges does not depend on {{mvar|A}}, so the ghosts do not contribute to the connected diagrams.