Quantum Field Theory (QFT)

Field Operators¶

The free (non-interacting) fields in the interaction picture are expressed using the creation and anihilation operators below, also the corresponding non-interacting Hamiltonian is shown.

The general idea behind the machinery is that the field operator $\hat\psi(\mathbf{x}) = \sum_k \psi_k(\mathbf{x}) c_k$ is constructed as a sum (or an integral, depending on if the index $k$ is discrete or continuous) of single-particle wave functions (i.e. solutions of the noninteracting equation of motion) multiplied by the creation/anihilation operators ( $c_k$ or $c_k^\dagger$ ) that create/destroy the particle in the given single-particle state. Note that the noninteracting equation of motion usually means that we set all potentials (interactions) as zero, but in principle it can be any equation that we can solve exactly.

The coefficients $\psi_k(\mathbf{x})$ don’t depend on time (so neither the field operators in the Schrödinger picture), but we work in the interaction picture, where the creation/anihilation operators depend on time, and the time dependence is put into the exponentials below (but the integration is still done over the spatial components of $p$ only).

Scalar bosons:

\phi_I(x) = \int {\mathrm{d}^3 p\over (2\pi)^3} {1\over\sqrt{2 E_{\mathbf{p}}}} \left(a_{\mathbf{p}} e^{-ip\cdot x} + a_{\mathbf{p}}^\dag e^{ip\cdot x}\right)

(1)

\pi_I(x) = {\partial_t}\phi_I(x) = \int {\mathrm{d}^3 p\over (2\pi)^3}(-i) {\sqrt{E_{\mathbf{p}}\over2}} \left(a_{\mathbf{p}} e^{-ip\cdot x} - a_{\mathbf{p}}^\dag e^{ip\cdot x}\right)

(2)

where:

\left[a_{\mathbf{p}}, a_{\mathbf{q}}^{\dag}\right] = (2\pi)^3\delta^{(3)}({\mathbf{p}} - {\mathbf{q}})

(3)

(all other commutators are equal to zero). The equal-time commutation relations for $\phi$ and $pi$ are then:

\left[\phi({\mathbf{x}}), \pi({\mathbf{y}})\right] = i\delta^{(3)}({\mathbf{x}} - {\mathbf{y}})

(4)

(all other commutators are equal to zero).

The Hamiltonian is

H = \int{\mathrm{d}^3p\over(2\pi)^3} E_{\mathbf{p}} a_{\mathbf{p}}^\dag a_{\mathbf{p}}

(5)

Fermions:

\psi_I(x) = \int {\mathrm{d}^3 p\over (2\pi)^3} {1\over\sqrt{2 E_{\mathbf{p}}}} \sum_{s=1}^2 \left(b_{\mathbf{p}}^s u^s({\mathbf{p}}) e^{-ip\cdot x} + d_{\mathbf{p}}^{s\dag} v^s({\mathbf{p}}) e^{ip\cdot x}\right)

(6)

\bar\psi_I(x) = \psi_I^\dag(x)\gamma^0 = \int {\mathrm{d}^3 p\over (2\pi)^3} {1\over\sqrt{2 E_{\mathbf{p}}}} \sum_{s=1}^2 \left(d_{\mathbf{p}}^s \bar v^s({\mathbf{p}}) e^{-ip\cdot x} + b_{\mathbf{p}}^{s\dag} \bar u^s({\mathbf{p}}) e^{ip\cdot x}\right)

(7)

where

u^s({\mathbf{p}}) = \begin{pmatrix}\sqrt{{\mathbf{p}}\cdot\sigma}\xi^s \\ \sqrt{{\mathbf{p}}\cdot\bar\sigma}\xi^s\end{pmatrix}

(8)

v^s({\mathbf{p}}) = \begin{pmatrix}\sqrt{{\mathbf{p}}\cdot\sigma}\eta^s \\ -\sqrt{{\mathbf{p}}\cdot\bar\sigma}\eta^s\end{pmatrix}

(9)

\sum_{s=1}^2 u^s({\mathbf{p}})\bar u^s({\mathbf{p}}) = \slashed{p} + m

(10)

\sum_{s=1}^2 v^s({\mathbf{p}})\bar v^s({\mathbf{p}}) = \slashed{p} - m

(11)

\left\{b_{\mathbf{p}}^r, b_{\mathbf{q}}^{s\dag}\right\} = \left\{d_{\mathbf{p}}^r, d_{\mathbf{q}}^{s\dag}\right\} = (2\pi)^3\delta^{(3)}({\mathbf{p}} - {\mathbf{q}}) \delta^{rs}

(12)

(all other anticommutators are equal to zero). The equal-time anticommutation relations for $\psi$ and $\psi^\dagger$ are then:

\left\{\psi_a({\mathbf{x}}), \psi_b^\dag({\mathbf{y}})\right\} = \delta^{(3)}({\mathbf{x}} - {\mathbf{y}}) \delta_{ab}

(13)

\left\{\psi_a({\mathbf{x}}), \psi_b({\mathbf{y}})\right\} = \left\{\psi_a^\dag({\mathbf{x}}), \psi_b^\dag({\mathbf{y}})\right\} = 0

(14)

The Hamiltonian is

H = \int{\mathrm{d}^3p\over(2\pi)^3}\sum_{s=1}^2 E_{\mathbf{p}} \left(b_{\mathbf{p}}^{s\dag}b_{\mathbf{p}}^{s} +d_{\mathbf{p}}^{s\dag}d_{\mathbf{p}}^{s}\right)

(15)

and the total charge:

Q = \int{\mathrm{d}^3p\over(2\pi)^3}\sum_{s=1}^2 \left(b_{\mathbf{p}}^{s\dag}b_{\mathbf{p}}^{s} -d_{\mathbf{p}}^{s\dag}d_{\mathbf{p}}^{s}\right)

(16)

So the $b$ -type particles and $d$ -type particles are identical except the charge. In QED, we identify the $b$ -type particles as electrons and the $d$ -type particles as positrons.

Vector bosons:

A_\mu(x) = \int {\mathrm{d}^3 p\over (2\pi)^3} {1\over\sqrt{2 E_{\mathbf{p}}}} \sum_{r=0}^3 \left(a_{\mathbf{p}}^r\epsilon_\mu^r({\mathbf{p}}) e^{-ip\cdot x} + a_{\mathbf{p}}^{r\dag}\epsilon_\mu^{r*}({\mathbf{p}}) e^{ip\cdot x}\right)

(17)

where

\left[a_{\mathbf{p}}^r, a_{\mathbf{q}}^{s\dag}\right] = (2\pi)^3\delta^{(3)}({\mathbf{p}} - {\mathbf{q}}) \delta^{rs}

(18)

The equal-time commutation relations for $A_\mu$ are then:

\left[A_\mu({\mathbf{x}}), A_\nu^\dag({\mathbf{y}})\right] = \delta^{(3)}({\mathbf{x}} - {\mathbf{y}}) \delta_{\mu\nu}

(19)

Calculating Scattering Amplitudes using Green Functions¶

We are interested in calculating the following scattering amplitudes:

\langle f|i \rangle

(20)

where the initial $|i\rangle$ and final $|f\rangle$ states are created by creation operators of the fields from the previous section. For example

|i\rangle = b_1^\dag b_2^\dag|\Omega\rangle

(21)

|f\rangle = b_{1'}^\dag b_{2'}^\dag|\Omega\rangle

(22)

Depending on the particular creation and anihilation operators, it can be shown that they can be replaced by:

a^\dag_{\mathbf{k}}\,{}_\text{in} \to i\int\mathrm{d}^4 x e^{ikx}\left(\partial^2+m^2\right)\phi(x) ={k^2-m^2\over i}\tilde\phi(-k) ={1\over\tilde D(k)}\tilde\phi(-k)

(23)

a_{\mathbf{k}}\,{}_\text{out} \to i\int\mathrm{d}^4 x e^{-ikx}\left(\partial^2+m^2\right)\phi(x) ={k^2-m^2\over i}\tilde\phi(k) ={1\over\tilde D(k)}\tilde\phi(k)

(24)

b^{s\dag}_{\mathbf{k}}\,{}_\text{in} \to i\int\mathrm{d}^4 x\bar\psi(x) \left(i\overleftarrow{\slashed{\partial}}+m\right)u^s({\mathbf{k}})e^{ikx} =\tilde{\bar\psi}(-k){-\slashed{k} - m\over i}u^s({\mathbf{k}}) =\tilde{\bar\psi}(-k){1\over\tilde S(-k)}u^s({\mathbf{k}})

(25)

b^s_{\mathbf{k}}\,{}_\text{out} \to i\int\mathrm{d}^4 x e^{-ikx}\bar u^s({\mathbf{k}})\left(-i\slashed{\partial}+m\right)\psi(x) =\bar u^s({\mathbf{k}}){\slashed{k} - m\over i}\psi(k) =\bar u^s({\mathbf{k}}){1\over\tilde S(k)}\tilde\psi(k)

(26)

d^{s\dag}_{\mathbf{k}}\,{}_\text{in} \to -i\int\mathrm{d}^4 x e^{ikx}\bar v^s({\mathbf{k}})\left(-i\slashed{\partial}+m\right)\psi(x) =-\bar v^s({\mathbf{k}}){\slashed{k} - m\over i}\psi(-k) =-\bar v^s({\mathbf{k}}){1\over\tilde S(k)}\tilde\psi(-k)

(27)

d^s_{\mathbf{k}}\,{}_\text{out} \to -i\int\mathrm{d}^4 x\bar\psi(x) \left(i\overleftarrow{\slashed{\partial}}+m\right)v^s({\mathbf{k}})e^{-ikx} =-\tilde{\bar\psi}(k){-\slashed{k} - m\over i}v^s({\mathbf{k}}) =-\tilde{\bar\psi}(k){1\over\tilde S(-k)}v^s({\mathbf{k}})

(28)

a^{r\dag}_{\mathbf{k}}\,{}_\text{in} \to i\epsilon_\mu^{r*}({\mathbf{k}}) \int\mathrm{d}^4 x e^{ikx} \partial^2 A^\mu(x) = \epsilon_\mu^{r*}({\mathbf{k}}){k^2\over i} \tilde A^\mu(-k)

(29)

a^r_{\mathbf{k}}\,{}_\text{out} \to i\epsilon_\mu^r({\mathbf{k}}) \int\mathrm{d}^4 x e^{-ikx} \partial^2 A^\mu(x) = \epsilon_\mu^r({\mathbf{k}}){k^2\over i} \tilde A^\mu(k)

(30)

where the "in" is the operator for $t\to -\infty$ and "out" for $t\to\infty$ . The fields $\phi(x)$ , $\psi(x)$ , $\bar\psi(x)$ and $A^\mu(x)$ have to be time ordered. On the left hand side is a position space representation, the two expressions on the right hand side are the momentum representation (the last expression is written using the propagators), e.g. a Fourier transform, which is essentially just the following substitutions:

\partial^2\to -k^2

(31)

i\slashed{\partial}\to\slashed{k}

(32)

e^{\pm ikx}\phi(x) \to \tilde\phi(\mp k)

(33)

{k^2 - m^2\over i} \to {1\over\tilde D(k)}

(34)

{\pm\slashed{k} - m\over i} \to {1\over\tilde S(\pm k)}

(35)

both representations are of course equivalent (but the momentum one is easier to use, since the formulas are shorter).

For our example we get in the position space:

\langle f|i \rangle = \langle \Omega| b_{{\mathbf{p}}_{2'}} b_{{\mathbf{p}}_{1'}} b_{{\mathbf{p}}_{1}}^\dag b_{{\mathbf{p}}_{2}}^\dag |\Omega \rangle = \langle \Omega|T\, b_{{\mathbf{p}}_{2'}} b_{{\mathbf{p}}_{1'}} b_{{\mathbf{p}}_{1}}^\dag b_{{\mathbf{p}}_{2}}^\dag |\Omega \rangle=

(36)

= i^4 \int \mathrm{d}^4 x_1\mathrm{d}^4 x_2\mathrm{d}^4 x_{1'}\mathrm{d}^4 x_{2'}

(37)

e^{-ip_{1'}x_{1'}}\left[\bar u^{s_{1'}}({\mathbf{k}}_{1'}) \left(-i\slashed{\partial}_{1'} +m\right)\right]_{\alpha_{1'}}

(38)

e^{-ip_{2'}x_{2'}}\left[\bar u^{s_{2'}}({\mathbf{k}}_{2'}) \left(-i\slashed{\partial}_{2'} +m\right)\right]_{\alpha_{2'}}

(39)

\langle \Omega|T\, \psi_{\alpha_{2'}}(x_{2'}) \psi_{\alpha_{1'}}(x_{1'}) \bar\psi_{\alpha_1}(x_1) \bar\psi_{\alpha_2}(x_2) |\Omega \rangle

(40)

\left[\left(i\overleftarrow{\slashed{\partial}_1}+m\right) u^{s_1}({\mathbf{p}}_1)\right]_{\alpha_1} \,e^{ip_1x_1}

(41)

\left[\left(i\overleftarrow{\slashed{\partial}_2}+m\right) u^{s_2}({\mathbf{p}}_2)\right]_{\alpha_2} \,e^{ip_2x_2}

(42)

where the $\alpha_1$ , $\alpha_2$ , $\alpha_{1'}$ and $\alpha_{2'}$ spinor indices were introduced to show how the matrices should be multiplied. The vacuum amplitude is called a 4 point interacting Green function in position space:

G^{(4)}_{\alpha_{1'} \alpha_{2'} \alpha_1 \alpha_2} (x_{1'}, x_{2'}, x_1, x_2) = \langle \Omega|T\, \psi_{\alpha_{2'}}(x_{2'}) \psi_{\alpha_{1'}}(x_{1'}) \bar\psi_{\alpha_1}(x_1) \bar\psi_{\alpha_2}(x_2) |\Omega \rangle

(43)

we can also take a Fourier transform to get the Green function in momentum space:

\tilde G^{(n)}(p_1,\dots, p_n) = \int\prod_{i=1}^n \mathrm{d}^4 x_i e^{-i p_i x_i}\, G^{(n)}(x_1, \dots, x_n)

(44)

then the scattering amplitude becomes (resuming the previous calculation):

\langle f|i \rangle = \cdots = i^4

(45)

\left[\bar u^{s_{1'}}({\mathbf{k}}_{1'}) \left(-\slashed{p}_{1'} +m\right)\right]_{\alpha_{1'}}

(46)

\left[\bar u^{s_{2'}}({\mathbf{k}}_{2'}) \left(-\slashed{p}_{2'} +m\right)\right]_{\alpha_{2'}}

(47)

\tilde G^{(4)}_{\alpha_{1'} \alpha_{2'} \alpha_1 \alpha_2} (p_{1'}, p_{2'}, -p_1, -p_2)

(48)

\left[\left({\slashed{p}_1}+m\right) u^{s_1}({\mathbf{p}}_1)\right]_{\alpha_1}

(49)

\left[\left({\slashed{p}_2}+m\right) u^{s_2}({\mathbf{p}}_2)\right]_{\alpha_2}

(50)

We can get the same result much faster if we use the momentum space from the beginning:

\langle f|i \rangle = \langle \Omega| b_{{\mathbf{p}}_{2'}} b_{{\mathbf{p}}_{1'}} b_{{\mathbf{p}}_{1}}^\dag b_{{\mathbf{p}}_{2}}^\dag |\Omega \rangle = \langle \Omega|T\, b_{{\mathbf{p}}_{2'}} b_{{\mathbf{p}}_{1'}} b_{{\mathbf{p}}_{1}}^\dag b_{{\mathbf{p}}_{2}}^\dag |\Omega \rangle=

(51)

= \langle \Omega|T\, \bar u^{s_{2'}}({\mathbf{p}}_{2'}){1\over\tilde S({\mathbf{p}}_{2'})} \tilde\psi({\mathbf{p}}_{2'}) \bar u^{s_{1'}}({\mathbf{p}}_{1'}){1\over\tilde S({\mathbf{p}}_{1'})} \tilde\psi({\mathbf{p}}_{1'}) \tilde{\bar\psi}(-{\mathbf{p}}_1){1\over\tilde S(-{\mathbf{p}}_1)} u^{s_1}({\mathbf{p}}_1) \tilde{\bar\psi}(-{\mathbf{p}}_2){1\over\tilde S(-{\mathbf{p}}_2)} u^{s_2}({\mathbf{p}}_2) |\Omega \rangle=

(52)

= \left[\bar u^{s_{2'}}({\mathbf{p}}_{2'}){1\over\tilde S({\mathbf{p}}_{2'})} \right]_{\alpha_{2'}} \left[\bar u^{s_{1'}}({\mathbf{p}}_{1'}){1\over\tilde S({\mathbf{p}}_{1'})} \right]_{\alpha_{1'}}

(53)

\langle \Omega|T\, \tilde\psi_{\alpha_{2'}}({\mathbf{p}}_{2'}) \tilde\psi_{\alpha_{1'}}({\mathbf{p}}_{1'}) \tilde{\bar\psi}_{\alpha_1}(-{\mathbf{p}}_1) \tilde{\bar\psi}_{\alpha_2}(-{\mathbf{p}}_2) |\Omega \rangle

(54)

\left[{1\over\tilde S(-{\mathbf{p}}_1)} u^{s_1}({\mathbf{p}}_1) \right]_{\alpha_1} \left[{1\over\tilde S(-{\mathbf{p}}_2)} u^{s_2}({\mathbf{p}}_2) \right]_{\alpha_1}

(55)

This is called Lehmann-Symanzik-Zimmermann (LSZ) reduction formula. One obtains similar expressions for other fields as well (if there were different creation operators between the initial and final states). All that remains is to calculate the interacting Green functions (for which we need to know the interaction Lagrangian). But first couple more examples:

Example 1¶

$\nu_e$ - $e$ elastic scattering:

\nu_e(k) + e(p) \to \nu_e(k') + e(p')

(56)

So the initial and final states are:

|i\rangle = b_{\mathbf{k}}^\dag b_{\mathbf{p}}^\dag|\Omega\rangle

(57)

|f\rangle = b_{\mathbf{k}'}^\dag b_{\mathbf{p}'}^\dag|\Omega\rangle

(58)

and we get:

\langle f|i \rangle = \langle \Omega| b_{{\mathbf{p}'}} b_{{\mathbf{k}'}} b_{{\mathbf{k}}}^\dag b_{{\mathbf{p}}}^\dag |\Omega \rangle =\langle \Omega|T\, b_{\mathbf{p}'} b_{\mathbf{k}'} b_{\mathbf{k}}^\dag b_{\mathbf{p}}^\dag |\Omega \rangle=

(59)

= \left[\bar u({\mathbf{p}'}){1\over\tilde S({\mathbf{p}'})} \right] \left[\bar u({\mathbf{k}'}){1\over\tilde S({\mathbf{k}'})} \right]

(60)

\langle \Omega|T\, \tilde\psi({\mathbf{p}'}) \tilde\psi({\mathbf{k}'}) \tilde{\bar\psi}(-{\mathbf{k}}) \tilde{\bar\psi}(-{\mathbf{p}}) |\Omega \rangle

(61)

\left[{1\over\tilde S(-{\mathbf{k}})} u({\mathbf{k}}) \right] \left[{1\over\tilde S(-{\mathbf{p}})} u({\mathbf{p}}) \right]

(62)

We only multiply the matrices with the same momentum, i.e. $\left[\bar u^s(\mathbf{p'})\frac{1}{\tilde S(\mathbf{p'})}\right]$ with $\tilde\psi(\mathbf{p'})$ , $\left[\bar u^s(\mathbf{k'})\frac{1}{\tilde S(\mathbf{k'})}\right]$ with $\tilde\psi(\mathbf{k'})$ and so on. Also we don’t write the spin anymore, e.g. $u(\mathbf{k})$ should in fact be $u^{s_k}(\mathbf{k})$ and so on.

Example 2¶

Muon decay:

\mu(P) \to e(p) + \bar\nu_e(k') + \nu_\mu(k)

(63)

So the initial and final states are:

|i\rangle = b_{\mathbf{P}}^\dag|\Omega\rangle

(64)

|f\rangle = b_{\mathbf{p}}^\dag d_{\mathbf{k}'}^\dag b_{\mathbf{k}}^\dag|\Omega\rangle

(65)

and we get:

\langle f|i \rangle = \langle \Omega| b_{{\mathbf{k}}} d_{{\mathbf{k}'}} b_{{\mathbf{p}}} b_{{\mathbf{P}}}^\dag |\Omega \rangle

(66)

= \left[\bar u({\mathbf{k}}){1\over\tilde S({\mathbf{k}})} \right] \left[\bar u({\mathbf{p}}){1\over\tilde S({\mathbf{p}})} \right]

(67)

\langle \Omega|T\, \tilde\psi({\mathbf{k}}) \tilde\psi({\mathbf{p}}) \tilde{\bar\psi}({\mathbf{k}'}) \tilde{\bar\psi}(-{\mathbf{P}}) |\Omega \rangle

(68)

\left[-{1\over\tilde S(-{\mathbf{k}'})} v({\mathbf{k}'}) \right] \left[{1\over\tilde S(-{\mathbf{P}})} u({\mathbf{P}}) \right]

(69)

Example 3¶

$e^+ + e^-$ scattering:

e^-(p_1)+e^+(p_2) \to e^-(k_1)+e^+(k_2)

(70)

Initial and final states:

|i\rangle = b_{{\mathbf{p}}_1}^{t\dag} d_{{\mathbf{p}}_2}^{u\dag}|\Omega\rangle

(71)

|f\rangle = b_{{\mathbf{k}}_1}^{r\dag} d_{{\mathbf{k}}_2}^{s\dag}|\Omega\rangle

(72)

And we get:

\langle f|i \rangle= \langle \Omega| b^r_{{\mathbf{k}}_1} d^s_{{\mathbf{k}}_2} b^{t\dag}_{{\mathbf{p}}_1} d^{u\dag}_{{\mathbf{p}}_2} |\Omega \rangle =

(73)

=\langle \Omega|T b^r_{{\mathbf{k}}_1} d^s_{{\mathbf{k}}_2} b^{t\dag}_{{\mathbf{p}}_1} d^{u\dag}_{{\mathbf{p}}_2} |\Omega \rangle =

(74)

=\langle \Omega|T \left[\bar u^r({\mathbf{k}}_1){1\over\tilde S(k_1)}\tilde \psi(k_1)\right] \left[-\tilde{\bar\psi}(k_2){1\over\tilde S(-k_2)}v^s({\mathbf{k}}_2)\right] \left[\tilde{\bar\psi}(-p_1){1\over\tilde S(-p_1)}u^t({\mathbf{p}}_1)\right] \left[-\bar v^u({\mathbf{p}}_2){1\over\tilde S(p_2)}\tilde \psi(-p_2)\right] |\Omega \rangle =

(75)

= \left[\bar u^r({\mathbf{k}}_1){1\over\tilde S(k_1)}\right] \left[\bar v^u({\mathbf{p}}_2){1\over\tilde S(p_2)}\right]

(76)

\langle \Omega|T \tilde \psi(k_1) \tilde{\bar\psi}(k_2) \tilde{\bar\psi}(-p_1) \tilde \psi(-p_2) |\Omega \rangle

(77)

\left[{1\over\tilde S(-k_2)}v^s({\mathbf{k}}_2)\right] \left[{1\over\tilde S(-p_1)}u^t({\mathbf{p}}_1)\right]

(78)

Example 4¶

$H(p)\to Z(k)+Z(l)$ decay. Initial and final states:

|i\rangle = a^\dag_{\mathbf{p}}|\Omega\rangle

(79)

|f\rangle = a^{r\dag}_{\mathbf{k}}a^{s\dag}_{\mathbf{l}}|\Omega\rangle

(80)

and we get:

\langle f|i \rangle= \langle \Omega|a^r_{\mathbf{k}} a^s_{\mathbf{l}} a^\dag_{\mathbf{p}}|\Omega \rangle = \langle \Omega|T a^r_{\mathbf{k}} a^s_{\mathbf{l}} a^\dag_{\mathbf{p}}|\Omega \rangle =

(81)

= \langle \Omega|T \epsilon^{r*}_\mu({\mathbf{k}}){k^2\over i} \tilde A^\mu(k) \epsilon^{s*}_\nu({\mathbf{l}}){l^2\over i} \tilde A^\nu(l) {1\over\tilde D(p)}\tilde\phi(-p) |\Omega \rangle =

(82)

= {\epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s*}_\nu({\mathbf{l}}) \over {i\over k^2}{i\over l^2}\tilde D(p)} \langle \Omega|T \tilde A^\mu(k)\tilde A^\nu(l)\tilde \phi(-p)|\Omega \rangle

(83)

Example 5¶

$e^+ e^- \to W^+ W^-$ scattering:

e^-(k) + e^+(-l) \to W^-(p) + W^+(r)

(84)

So the initial and final states are:

|i\rangle = b_{\mathbf{k}}^\dag d_{-{\mathbf{l}}}^\dag|\Omega\rangle

(85)

|f\rangle = a_{\mathbf{p}}^{\lambda \dag} a_{\mathbf{r}}^{\mu \dag}|\Omega\rangle

(86)

and we get:

\langle f|i \rangle = \langle \Omega| a_{\mathbf{r}}^\mu a_{\mathbf{p}}^\lambda b_{\mathbf{k}}^\dag d_{-{\mathbf{l}}}^\dag |\Omega \rangle

(87)

= \epsilon^\mu_\alpha({\mathbf{r}}) {r^2\over i} \epsilon^\lambda_\beta({\mathbf{p}}) {p^2\over i} \left[-\bar v(-{\mathbf{l}}){1\over\tilde S(-{\mathbf{l}})} \right]

(88)

\langle \Omega|T\, \tilde A^\alpha({\mathbf{r}}) \tilde A^\beta({\mathbf{p}}) \tilde{\bar\psi}(-{\mathbf{k}}) \tilde\psi({\mathbf{l}}) |\Omega \rangle

(89)

\left[{1\over\tilde S(-{\mathbf{k}})} u({\mathbf{k}}) \right]

(90)

Example 6¶

$W^+ W^- \to W^+ W^-$ scattering:

W^-(k) + W^+(p) \to W^-(r) + W^+(l)

(91)

So the initial and final states are:

|i\rangle = a_{\mathbf{k}}^{\mu \dag} a_{\mathbf{p}}^{\nu \dag}|\Omega\rangle

(92)

|f\rangle = a_{\mathbf{r}}^{\alpha \dag} a_{\mathbf{l}}^{\beta \dag}|\Omega\rangle

(93)

and we get:

\langle f|i \rangle = \langle \Omega| a_{\mathbf{l}}^\beta a_{\mathbf{r}}^\alpha a_{\mathbf{k}}^{\mu \dag} a_{\mathbf{p}}^{\nu \dag} |\Omega \rangle

(94)

= \epsilon^{\mu*}_\rho({\mathbf{k}}) {k^2\over i} \epsilon^{\nu*}_\sigma({\mathbf{p}}) {p^2\over i} \langle \Omega|T\, \tilde A^\kappa({\mathbf{r}}) \tilde A^\lambda({\mathbf{l}}) \tilde A^\sigma(-{\mathbf{p}}) \tilde A^\rho(-{\mathbf{k}}) |\Omega \rangle \epsilon^\alpha_\kappa({\mathbf{r}}) {r^2\over i} \epsilon^\beta_\lambda({\mathbf{l}}) {l^2\over i}

(95)

Evaluation of the Interacting Green Functions¶

The interacting Green functions can be evaluated using the formula:

G^{(n)}(x_1, \dots, x_n) = \langle \Omega|T\, \phi_H(x_1) \dots \phi_H(x_n) |\Omega \rangle =

(96)

={\langle 0|T\, \phi_I(x_1) \dots \phi_I(x_n) S|0 \rangle\over \langle 0|S|0 \rangle}

(97)

where

S=U_I(\infty, -\infty) = T\exp\left(-{i\over\hbar}\int_{-\infty}^{\infty}H_1(t)\mathrm{d} t \right) = T\exp\left(-{i\over\hbar}\int\mathrm{d}^4 x \mathcal{H}_1(x) \right)

(98)

$\phi_H$ is a field in the Heisenberg picture ( $\phi(\mathbf{x}, t)=e^{iHt}\phi(\mathbf{x}, 0)e^{-iHt}$ ) and $\phi_I$ is a field in the interaction picture ( $\phi(\mathbf{x}, t)=e^{iH_0t}\phi(\mathbf{x}, 0)e^{-iH_0t}$ ), where the Hamiltonian is $H=H_0 + H_1$ and the vacua (ground states) are $H_0|0\rangle = 0$ and $H|\Omega\rangle = 0$ .

This can be proven by evaluating the right hand side:

{\langle 0|T\, \phi_I(x_1) \dots \phi_I(x_n) S|0 \rangle\over \langle 0|S|0 \rangle} = {\langle 0|T\, \phi_I(x_1) \dots \phi_I(x_n) U_I(\infty, -\infty)|0 \rangle\over \langle 0|U_I(\infty, 0)U_I(0, -\infty)|0 \rangle}

(99)

= {\langle 0|U_I(\infty, t_1) \phi_I(x_1) U_I(t_1, t_2) \dots U_I(t_{n-1}, t_n)\phi_I(x_n) U_I(t_n, -\infty)|0 \rangle\over \langle 0|U_I(\infty, 0)U_I(0, -\infty)|0 \rangle}

(100)

= {\langle 0|U_I(\infty, 0) \phi_H(x_1) \dots \phi_H(x_n) U_I(0, -\infty)|0 \rangle\over \langle 0|U_I(\infty, 0)U_I(0, -\infty)|0 \rangle}=

(101)

= {\langle 0|\Omega \rangle\langle \Omega|T \phi_H(x_1) \dots \phi_H(x_n) |\Omega \rangle\langle \Omega|0 \rangle\over \langle 0|\Omega \rangle\langle \Omega|\Omega \rangle\langle \Omega|0 \rangle}=

(102)

= {\langle \Omega|T \phi_H(x_1) \dots \phi_H(x_n) |\Omega \rangle\over \langle \Omega|\Omega \rangle}=

(103)

= \langle \Omega|T \phi_H(x_1) \dots \phi_H(x_n) |\Omega \rangle

(104)

where we used the following relations:

U_I(t_{k-1}, t_k) \phi_I(x_k) U_I(t_k, t_{k+1})= U_I(t_{k-1}, 0) U_I^\dag(t_k, 0) \phi_I(x_k) U_I(t_k, 0) U_I(0, t_{k+1})= U_I(t_{k-1}, 0) \phi_H(x_k) U_I(0, t_{k+1})

(105)

U_I(0, -\infty)|0\rangle= U_I(0, -\infty)\left[|\Omega\rangle\langle\Omega|+\sum_{n\neq0}|n\rangle\langle n| \right]|0\rangle = |\Omega\rangle\langle \Omega|0 \rangle+\lim_{t\to-\infty} \sum_{n\neq0}e^{i E_n t}|n\rangle\langle n|0 \rangle = |\Omega\rangle\langle \Omega|0 \rangle

(106)

\langle \Omega|\Omega \rangle=1

(107)

single: S-matrix pair: evolution; operator

Evolution Operator, S-Matrix Elements¶

The evolution operator $U$ is defined by the equations:

|\phi(t_2)\rangle=U(t_2, t_1)|\phi(t_1)\rangle

(108)

i\hbar{\partial U(t, t_1)\over\partial t} = H(t)U(t, t_1)

(109)

U(t_1, t_1) = 1

(110)

We are interested in calculating the S matrix elements:

\langle f|U(\infty, -\infty)|i \rangle=\langle f|S|i \rangle=S_{fi}

(111)

so we first calculate $U(\infty, -\infty)$ . Integrating the equation for the evolution operator:

U(t_2, t_1)=U(t_1, t_1)-{i\over\hbar}\int_{t_1}^{t_2} H(t)U(t, t_1)\mathrm{d} t =1-{i\over\hbar}\int_{t_1}^{t_2} H(t)U(t, t_1)\mathrm{d} t

(112)

Now:

S=U(\infty, -\infty) =1-{i\over\hbar} \int_{-\infty}^{\infty} H(t')U(t', -\infty)\mathrm{d} t'=

(113)

=1+\left(-{i\over\hbar}\right)\int_{-\infty}^{\infty} H(t')U(t', -\infty)\mathrm{d} t' +\left(-{i\over\hbar}\right)^2\int_{-\infty}^{\infty} \int_{-\infty}^{t'} H(t')H(t'')U(t'', -\infty)\mathrm{d} t'\mathrm{d} t''=

(114)

=\cdots= \sum_{n=0}^\infty \left(-{i\over\hbar}\right)^n {1\over n!} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\cdots T\{H(t_1)H(t_2)\cdots\}\mathrm{d} t_1\mathrm{d} t_2\cdots=

(115)

= T\exp\left(-{i\over\hbar}\int_{-\infty}^{\infty}H(t)\mathrm{d} t \right) = T\exp\left(-{i\over\hbar}\int_{-\infty}^{\infty}\mathrm{d}^4 x \mathcal{H}(x) \right)

(116)

If $L$ doesn’t contain derivatives of the fields, then $H = -L$ so:

U(\infty, -\infty) = T\exp\left({i\over\hbar}\int_{-\infty}^{\infty}\mathrm{d}^4 x \mathcal{L}(x) \right)

(117)

Let’s write $S=1+iT$ and $|i\rangle=|k_1\cdots k_m\rangle$ , $|f\rangle=|p_1\cdots p_n\rangle$ . As a first step now, let’s investigate a scalar field, e.g. $L=-{\lambda\over4}\phi^4$ (e.g. a Higgs self interaction term above), we’ll look at other fields later:

\langle f|S|i \rangle= \langle f|iT|i \rangle= \langle p_1\cdots p_n|iT|k_1\cdots k_m \rangle= {1\over\tilde D(k_1)\cdots\tilde D(k_m)} {1\over\tilde D(p_1)\cdots\tilde D(p_n)}

(118)

\int\mathrm{d}^4 x_1\cdots \mathrm{d}^4 x_m e^{-i(k_1 x_1+\cdots + k_m x_m)} \int\mathrm{d}^4 y_1\cdots \mathrm{d}^4 y_n e^{+i(p_1 y_1+\cdots + p_n y_n)} G(x_1, \cdots, x_m, y_1, \cdots, y_m)

(119)

where

G(x_1, \cdots, x_n)= \langle \Omega|T\{\phi(x_1)\cdots\phi(x_n)\}|\Omega \rangle =

(120)

{\langle 0|T\{\phi_I(x_1)\cdots\phi_I(x_n)\exp\left({i\over\hbar}\int_{-\infty}^{\infty}\mathrm{d}^4 x \mathcal{L}(x) \right)\}|0 \rangle \over \langle 0|T\exp\left({i\over\hbar}\int_{-\infty}^{\infty}\mathrm{d}^4 x \mathcal{L}(x) \right)|0 \rangle }

(121)

This is called the LSZ formula. Now we use the Wick contraction, get some terms like $D_{23}D_{34}$ integrate things out, this will give the delta function and $\tilde D(p)$ 's and that’s it.

Let’s see how it goes for $L=-{\lambda\over4}\phi^4$ for the process $k_1+k_2\to p_1+p_2$ :

\langle p_1 p_2|S|k_1 k_2 \rangle = {\int\mathrm{d}^4 x_1\mathrm{d}^4 x_2 e^{-i(k_1x_1+k_2x_2)} \int\mathrm{d}^4 y_1\mathrm{d}^4 y_2 e^{i(p_1y_1+p_2y_2)} \over \tilde D(k_1)\tilde D(k_2) \tilde D(p_1)\tilde D(p_2)}

(122)

{\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(y_1)\phi_I(y_2)\exp\left( -{i\lambda\over4\hbar} \int\mathrm{d}^4 x \phi_I^4(x) \right)\}|0 \rangle \over \langle 0|T\exp\left(-{i\lambda\over4\hbar}\int\mathrm{d}^4 x \phi_I^4(x) \right)|0 \rangle }=

(123)

= {\int\mathrm{d}^4 x_1\mathrm{d}^4 x_2 e^{-i(k_1x_1+k_2x_2)} \int\mathrm{d}^4 y_1\mathrm{d}^4 y_2 e^{i(p_1y_1+p_2y_2)} \over \tilde D(k_1)\tilde D(k_2) \tilde D(p_1)\tilde D(p_2)}

(124)

\left[ { \langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(y_1)\phi_I(y_2)\}|0 \rangle \over \langle 0|T\exp\left(-{i\lambda\over4\hbar}\int\mathrm{d}^4 x \phi_I^4(x) \right)|0 \rangle }\right. +

(125)

+{ \left(-{i\lambda\over4\hbar}\right)\int\mathrm{d}^4 x \langle 0|T\{\phi_I(x_1)\phi_I(x_2) \phi_I(y_1)\phi_I(y_2) \phi_I^4(x)\}|0 \rangle \over \langle 0|T\exp\left(-{i\lambda\over4\hbar}\int\mathrm{d}^4 x \phi_I^4(x) \right)|0 \rangle } +

(126)

\left. +{ \left(-{i\lambda\over4\hbar}\right)^2\int\mathrm{d}^4 x\,\mathrm{d}^4 y \langle 0|T\{\phi_I(x_1)\phi_I(x_2) \phi_I(y_1)\phi_I(y_2) \phi_I^4(x)\phi_I^4(y)\}|0 \rangle \over \langle 0|T\exp\left(-{i\lambda\over4\hbar}\int\mathrm{d}^4 x \phi_I^4(x) \right)|0 \rangle } +\cdots\right]=

(127)

= { 1 \over \tilde D(k_1)\tilde D(k_2) \tilde D(p_1)\tilde D(p_2)}

(128)

\left[ (2\pi)^4 \delta^{(4)}(p_1+p_2)(2\pi)^4 \delta^{(4)}(k_1+k_2)\tilde D(p_1) \tilde D(k_1)+\right.

(129)

(-i\lambda)6(2\pi)^4\delta^{(4)}(p_1+p_2-k_1-k_2)\tilde D(k_1)\tilde D(k_2) \tilde D(p_1)\tilde D(p_2)+

(130)

\left. (-i\lambda)(\text{disconnected terms with not enough $\tilde D(\cdots)$s})+(-i\lambda)^2(\cdots)+\cdots\right]=

(131)

= (2\pi)^4\delta^{(4)}(p_1+p_2-k_1-k_2)\left[6(-i\lambda)+ 3(-i\lambda)^2\int{\mathrm{d}^4 k\over (2\pi)^4}\tilde D(k)\tilde D(p1+p2-k) +(-i\lambda)^3(\cdots)+\cdots\right]

(132)

The denominator cancels with the disconnected terms. We used the Wick contractions (see below for a thorough explanation+derivation):

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(y_1)\phi_I(y_2)\}|0 \rangle= D(x_1-x_2)D(y_1-y_2)+D(x_2-y_1)D(x_1-y_2)+D(x_2-y_2)D(x_1-y_1)

(133)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2) \phi_I(y_1)\phi_I(y_2) \phi_I^4(x)\}|0 \rangle= D(x_1-x)D(x_2-x)D(y_1-x)D(y_2-x)+\text{disconnected}

(134)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2) \phi_I(y_1)\phi_I(y_2) \phi_I^4(x)\phi_I^4(y)\}|0 \rangle= D(x_1-x)D(x_2-x)D(y_1-y)D(y_2-y)D(x-y)D(x-y)

(135)

+\text{disconnected}

(136)

Where the "disconnected" terms are $D(x_1-y_1)D(x_2-y_2)D(x-x)D(x-x)$ and similar. When they are integrated over, they do not generate enough $\tilde D(p_1)$ propagators to cancel the propagators from the LSZ formula, which will cause the terms to vanish.

For the $L=\phi^3(x)$ theory, one also needs the following contractions:

\langle 0|T\{\phi_I(x_1)\phi_I(x_2) \phi_I(y_1)\phi_I(y_2) \phi_I^3(x)\}|0 \rangle = 0

(137)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2) \phi_I(y_1)\phi_I(y_2) \phi_I^3(x)\phi_I^3(y)\}|0 \rangle = D(x_1-x)D(x_2-x)D(x-y)D(y_1-y)D(y_2-y)

(138)

Thus it is clear that the only difference from the above is the factor $D(x-y)$ which after integrating changes to $\tilde D(p_1+p_2)$ and this ends up in the final result.

One always gets the delta function in the result, so we define the matrix element $M_{fi}$ by:

S_{fi} = (2\pi)^4\delta^{(4)}(p_1+p_2+\cdots - k_1 - k_2 - \cdots) i \mathcal{M}_{fi}

(139)

fermions, vector bosons

Propagators for Scalar Bosons, Fermions and Vector Bosons¶

The only nonzero contractions that can occur are the propagators below. All other contractions are zero.

Propagator for a scalar boson is:

\langle 0|T\{\phi_I(x)\phi_I(y)\}|0 \rangle\equiv D(x-y)= \int {\mathrm{d}^4 p\over (2\pi)^4}\tilde D(p) e^{-ip(x-y)}

(140)

with

\tilde D(p) = {i\over p^2-m^2+i\epsilon}

(141)

For fermions (Feynman propagator):

\langle 0|T\{\psi_I(x)\bar\psi_I(y)\}|0 \rangle\equiv S(x-y)= \int {\mathrm{d}^4 p\over (2\pi)^4}\tilde S(p) e^{-ip(x-y)}

(142)

with

\tilde S(p) = {i\over \slashed{p} - m +i\epsilon}= {i(\slashed{p}+m)\over p^2-m^2+i\epsilon}

(143)

For vector bosons:

\langle 0|T\{A_\mu(x)A_\nu(y)\}|0 \rangle\equiv D_{\mu\nu}(x-y)= \int {\mathrm{d}^4 p\over (2\pi)^4}\tilde D_{\mu\nu}(p) e^{-ip(x-y)}

(144)

with

\tilde D_{\mu\nu}(p) = i{-g_{\mu\nu}+{p_\mu p_\nu\over m^2}\over p^2-m^2+i\epsilon}

(145)

For massless bosons:

\tilde D_{\mu\nu}(p) = i{-g_{\mu\nu}\over p^2+i\epsilon}

(146)

Wick Theorem

Wick Theorem¶

As seen above, we need to be able to calculate

\langle 0|T\{\phi_I(x_1)\cdots\phi_I(x_n)\}|0 \rangle

(147)

The Wick theorem says, that this is equal to all possible contractions of fields (all fields need to be contracted), where a contraction is defined as:

\langle 0|T\{\phi_I(x)\phi_I(y)\}|0 \rangle\equiv D(x-y)= \int {\mathrm{d}^4 p\over (2\pi)^4}\tilde D(p) e^{-ip(x-y)}

(148)

with

\tilde D(p) = {i\over p^2-m^2+i\epsilon}

(149)

A few lowest possibilities:

\langle 0|T\{\phi_I(x_1)\}|0 \rangle= 0

(150)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\}|0 \rangle= D_{12}

(151)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\}|0 \rangle= 0

(152)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\}|0 \rangle= \text{disconnected}

(153)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\phi_I(x)\}|0 \rangle= 0

(154)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\phi_I^2(x)\}|0 \rangle= \text{disconnected}

(155)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\phi_I^3(x)\}|0 \rangle= 0

(156)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\phi_I^4(x)\}|0 \rangle= 4!\,D(x_1-x)D(x_2-x)D(x_3-x)D(x_4-x)+\text{disconnected}

(157)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\phi_I^3(x)\phi_I^3(y)\}|0 \rangle=

(158)

=D(x_1-x)D(x_2-x)D(x-y)D(x_3-y)D(x_4-y) + \text{disconnected}

(159)

\langle 0|T\{\phi_I(x_1)\phi_I(x_2)\phi_I(x_3)\phi_I(x_4)\phi_I^4(x)\phi_I^4(y)\}|0 \rangle=

(160)

=D(x_1-x)D(x_2-x)D(x-y)D(x-y)D(x_3-y)D(x_4-y) + \text{disconnected}

(161)

For the last two equations, not all possibilities of the connected graphs are listed (and also the combinatorial factor is omitted).

Feynman rules

Evaluating the Interacting Green Functions¶

The Green functions below can either be evaluated using the Wick theorem, or using Feynman diagrams and the corresponding Feynman rules.

Example 1¶

$\mathcal{L}_{ZZH} = \lambda Z_\mu Z^\mu H$ , in the first order:

\langle \Omega|T \tilde A^\mu(k)\tilde A^\nu(l)\tilde \phi(-p)|\Omega \rangle =i\lambda (2\pi)^4 \delta(k+l-p) \tilde D^\mu{}_\alpha(k) \tilde D^{\nu\alpha}(l)\tilde D(p)

(162)

Example 2¶

$\mathcal{L}_{ee\gamma}=-\lambda \bar e\gamma^\mu e A_\mu$ , in the second order:

\langle \Omega|T\tilde\psi(k_1)\tilde{\bar\psi}(k_2)\tilde{\bar\psi}(-p_1) \tilde\psi(-p_2) |\Omega \rangle =

(163)

=(-i\lambda)^2(2\pi)^4\delta(k_1+k_2-p_1-p_2)\left[ \tilde S(k_1)\gamma^\mu\tilde S(-k_2) D_{\mu\nu}(k_1+k_2) \tilde S(p_2)\gamma^\nu\tilde S(-p_1) +\right.

(164)

\left. + \tilde S(k_1)\gamma^\mu\tilde S(-p_1) D_{\mu\nu}(k_1-p_1) \tilde S(p_2)\gamma^\nu\tilde S(-k_2)\right]

(165)

Example 3¶

$\mathcal{L}_{\mathrm{int}}={g\over 2\sqrt{2}} \bar\nu_e \gamma^\mu (1-\gamma_5) e W_\mu^+ + \text{h.c.}$ , in the second order:

\langle \Omega|T\tilde A^\alpha(r) \tilde A^\beta(p) \tilde{\bar\psi}(-k)\tilde{\psi}(-l) |\Omega \rangle =

(166)

=\int \mathrm{d}^4x\,\mathrm{d}^4 y \langle0|T\tilde A^\alpha(r) \tilde A^\beta(p) \tilde{\bar\psi}(-k)\tilde{\psi}(-l)

(167)

i{g\over 2 \sqrt 2}\bar\nu_e(x) \gamma^\mu (1-\gamma_5) e(x) W_\mu^+(x)

(168)

i{g\over 2 \sqrt 2}\bar\nu_e(y) \gamma^\nu (1-\gamma_5) e(y) W_\nu^+(y)

(169)

|0\rangle =

(170)

=\left(i{g\over 2 \sqrt 2}\right)^2 \int \mathrm{d}^4x\,\mathrm{d}^4 y \,\mathrm{d}\hat r\,\mathrm{d}\hat p\,\mathrm{d}\hat k\,\mathrm{d}\hat l\, e^{i(\hat rr + \hat pp - \hat kk - \hat ll)}

(171)

\langle0|T A^\alpha(\hat r) A^\beta(\hat p) {\bar\psi}(\hat k){\psi}(\hat l)

(172)

\bar\nu_e(x) \gamma^\mu (1-\gamma_5) e(x) W_\mu^+(x)

(173)

\bar\nu_e(y) \gamma^\nu (1-\gamma_5) e(y) W_\nu^+(y)

(174)

|0\rangle =

(175)

=\left(i{g\over 2 \sqrt 2}\right)^2 \int \mathrm{d}^4x\,\mathrm{d}^4 y \,\mathrm{d}\hat r\,\mathrm{d}\hat p\,\mathrm{d}\hat k\,\mathrm{d}\hat l\, e^{i(\hat rr + \hat pp - \hat kk - \hat ll)}

(176)

D^\alpha{}_\mu(\hat r - x) D^\mu{}_\nu(\hat p - y) S(\hat l - x) \gamma^\mu (1-\gamma_5)S(x-y)\gamma^\nu(1-\gamma_5) S(\hat k - y) =

(177)

=\left(i{g\over 2 \sqrt 2}\right)^2 \int \mathrm{d}^4x\,\mathrm{d}^4 y \,\mathrm{d}\hat r\,\mathrm{d}\hat p\,\mathrm{d}\hat k\,\mathrm{d}\hat l\, e^{i((\hat r+x)r + (\hat p+y)p - (\hat k+y)k - (\hat l+x)l)}

(178)

D^\alpha{}_\mu(\hat r) D^\mu{}_\nu(\hat p) S(\hat l) \gamma^\mu (1-\gamma_5)S(x-y)\gamma^\nu(1-\gamma_5) S(\hat k) =

(179)

=\left(i{g\over 2 \sqrt 2}\right)^2 \int \mathrm{d}^4x\,\mathrm{d}^4 y \, e^{i(xr + yp - yk - xl)}

(180)

\tilde D^\alpha{}_\mu(r) \tilde D^\mu{}_\nu(p) \tilde S(l) \gamma^\mu (1-\gamma_5)S(x-y)\gamma^\nu(1-\gamma_5) \tilde S(k) =

(181)

=\left(i{g\over 2 \sqrt 2}\right)^2 \int \mathrm{d}^4x\,\mathrm{d}^4 y \, e^{i((x+y)r + yp - yk - (x+y)l)}

(182)

\tilde D^\alpha{}_\mu(r) \tilde D^\mu{}_\nu(p) \tilde S(l) \gamma^\mu (1-\gamma_5)S(x)\gamma^\nu(1-\gamma_5) \tilde S(k) =

(183)

=\left(i{g\over 2 \sqrt 2}\right)^2 \int \mathrm{d}^4 y \, e^{i(yr + yp - yk - yl)}

(184)

\tilde D^\alpha{}_\mu(r) \tilde D^\mu{}_\nu(p) \tilde S(l) \gamma^\mu (1-\gamma_5)\tilde S(r-l)\gamma^\nu(1-\gamma_5) \tilde S(k) =

(185)

= \left(i{g\over 2 \sqrt 2}\right)^2 (2\pi)^4\delta(r+p-k-l) \tilde D^\alpha{}_\mu(r) \tilde D^\beta{}_\nu(p)\tilde S(l) \gamma^\mu(1-\gamma_5) \tilde S(r-l) \gamma^\nu(1-\gamma_5)\tilde S(k)

(186)

ZZH interaction¶

Let’s calculate the $\mathcal{L}_{ZZH}=\lambda Z_\mu Z^\mu H$ interaction in the SM, where $\lambda={g^2\over4\cos\theta_W}$ . Consider $H(p)\to Z(k)+Z(l)$ :

\langle f|S|i \rangle= \langle f|iT|i \rangle= \langle k l|iT|p \rangle= \langle \Omega|a^r_{\mathbf{k}} a^s_{\mathbf{l}} a^\dag_{\mathbf{p}}|\Omega \rangle = \langle \Omega|T a^r_{\mathbf{k}} a^s_{\mathbf{l}} a^\dag_{\mathbf{p}}|\Omega \rangle =

(187)

= \langle \Omega|T \epsilon^{r*}_\mu({\mathbf{k}}){k^2\over i} \tilde A^\mu(k) \epsilon^{s*}_\nu({\mathbf{l}}){l^2\over i} \tilde A^\nu(l) {1\over\tilde D(p)}\tilde\phi(-p) |\Omega \rangle =

(188)

= {\epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s*}_\nu({\mathbf{l}}) \over {i\over k^2}{i\over l^2}\tilde D(p)} \langle \Omega|T \tilde A^\mu(k)\tilde A^\nu(l)\tilde \phi(-p)|\Omega \rangle

(189)

= {\epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s*}_\nu({\mathbf{l}}) \over {i\over k^2}{i\over l^2}\tilde D(p)} i\lambda (2\pi)^4 \delta(k+l-p) \tilde D^\mu{}_\alpha(k) \tilde D^{\nu\alpha}(l)\tilde D(p)

(190)

= {\epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s*}_\nu({\mathbf{l}}) \over {i\over k^2}{i\over l^2}\tilde D(p)} i\lambda (2\pi)^4 \delta(k+l-p) {-ig^\mu{}_\alpha\over k^2} {-ig^{\nu\alpha}\over l^2} \tilde D(p)

(191)

= \epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s*}_\nu({\mathbf{l}}) i\lambda (2\pi)^4 \delta(k+l-p) g^\mu{}_\alpha g^{\nu\alpha}

(192)

= \epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s*}_\nu({\mathbf{l}}) i\lambda (2\pi)^4 \delta(k+l-p) g^{\mu\nu}

(193)

= i\lambda (2\pi)^4 \delta(k+l-p) \epsilon^{r*}_\mu({\mathbf{k}})\epsilon^{s\mu*}({\mathbf{l}})

(194)

where we used the fact, that the first order contribution of the $\lambda Z_\mu Z^\mu H$ interaction to the interacting Green function is:

\langle \Omega|T \tilde A^\mu(k)\tilde A^\nu(l)\tilde \phi(-p)|\Omega \rangle =i\lambda (2\pi)^4 \delta(k+l-p) \tilde D^\mu{}_\alpha(k) \tilde D^{\nu\alpha}(l)\tilde D(p)

(195)

eeH interaction¶

This is only approximate, it will be fixed soon.

Let’s calculate the $\mathcal{L}_{eeH}=-\lambda \bar e e H$ interaction in the SM, where $\lambda={h_e\over\sqrt{2}}$ . Consider $H(p)\to e^-(k)+e^+(l)$ :

\langle f|S|i \rangle= \langle f|iT|i \rangle= \langle k l|iT|p \rangle= {\bar u(k) v(l)\over\tilde S(k)\tilde S(l)} {1\over\tilde D(p)}

(196)

\int\mathrm{d}^4 x_1 e^{-i p x_1} \int\mathrm{d}^4 y_1 \mathrm{d}^4 y_2 e^{+i(k y_1+l y_2)} \langle 0|T\{\bar e(y_1) e(y_2) H(x_1)\}|0 \rangle =

(197)

= {\bar u(k) v(l)\over\tilde S(k)\tilde S(l)} {1\over\tilde D(p)}

(198)

\int\mathrm{d}^4 x_1 e^{-i p x_1} \int\mathrm{d}^4 y_1 \mathrm{d}^4 y_2 e^{+i(k y_1+l y_2)}\int\mathrm{d}^4 x (-i\lambda) S(y_1-x)S(y_2-x)D(x_1-x) =

(199)

=(-i\lambda)(2\pi)^4\delta^{(4)}(p-k-l)\bar u(k) v(l)

(200)

where we used the fact, that the only nonzero element of the Green function is

\int\mathrm{d}^4 x \langle 0|T\{\bar e(y_1) e(y_2) H(x_1)\bar e(x)e(x) H(x)\}|0 \rangle

(201)

ee gamma interaction¶

This is only approximate, it will be fixed soon.

Let’s calculate the $\mathcal{L}_{ee\gamma}=-\lambda \bar e\gamma^\mu e A_\mu$ interaction in the SM, where $\lambda=g\sin\theta_W$ . Consider $\gamma(p)\to e^-(k)+e^+(l)$ :

\langle f|S|i \rangle= \langle f|iT|i \rangle= \langle k l|iT|p \rangle= {\bar u(k) v(l)\over\tilde S(k)\tilde S(l)} {\varepsilon_\mu(p)\over\tilde D_{\alpha\beta}(p)}

(202)

\int\mathrm{d}^4 x_1 e^{-i p x_1} \int\mathrm{d}^4 y_1 \mathrm{d}^4 y_2 e^{+i(k y_1+l y_2)} \langle 0|T\{\bar e(y_1) e(y_2) A^\mu(x_1)\}|0 \rangle =

(203)

= {\bar u(k) v(l)\over\tilde S(k)\tilde S(l)} {\varepsilon_\mu(p)\over\tilde D_{\alpha\beta}(p)}

(204)

\int\mathrm{d}^4 x_1 e^{-i p x_1} \int\mathrm{d}^4 y_1 \mathrm{d}^4 y_2 e^{+i(k y_1+l y_2)}\int\mathrm{d}^4 x (-i\lambda) S(y_2-x) \gamma^\mu S(y_1-x) D^\alpha_\mu(x_1-x) =

(205)

=(2\pi)^4\delta^{(4)}(p-k-l)\bar u(k)(-i\lambda)\gamma^\mu v(l)\varepsilon_\mu(p)

(206)

where we used the fact, that the only nonzero element of the Green function is

\int\mathrm{d}^4 x \langle 0|T\{\bar e(y_1) e(y_2) A^\alpha(x_1)\bar e(x)\gamma^\mu e(x) A_\mu(x)\}|0 \rangle =

(207)

=\pm S(y_2-x) \gamma^\mu S(y_1-x) D^\alpha_\mu(x_1-x)

(208)

eeee interaction¶

Let’s calculate the $\mathcal{L}_{ee\gamma}=-\lambda \bar e\gamma^\mu e A_\mu$ interaction in the SM, where $\lambda=g\sin\theta_W$ . Consider $e^-(p_1)+e^+(p_2)\to\gamma(q)\to e^-(k_1)+e^+(k_2)$ :