Quantum fields: a new approach

A need for confirmation

In this talk we apply the theoretical work that has been done so far. Using these new calculation techniques, we attempt to recover Compton scattering.

This choice is symbolic of movement from semi-classical particle physics, to a full acceptance of the quantum theory [5].

Invariant amplitude

The processes which contribute to this interaction can be written as a scattering amplitude, expressed as the sum of Feynman diagrams.

By applying the Feynman rules to these diagrams and grouping terms, we obtain \[ i \mathcal{M}=i {\color{#b5e48c} e^{2}} {\color{#ffe66d} \epsilon_{\mu \lambda^\prime}^{\ast} \left(k^{\prime}\right)} {\color{#ffe66d} \epsilon_{\nu \lambda}\left(k\right)} {\color{#fe6d73} \bar{u}^{s^{\prime}}\left(p^{\prime}\right) } \left( {\color{#5adbff} \dfrac{% numerator {\color{#b5e48c} \gamma^{\mu}}(\not{p}+k+m) {\color{#b5e48c} \gamma^{\nu}}} {(p+k)^{2}-m^{2}}}% denominator + {\color{#5adbff} \dfrac{% numerator {\color{#b5e48c} \gamma^{\nu}} \left(\not{p}-k^{\prime}+m\right) {\color{#b5e48c} \gamma^{\mu}}} {\left(p-k^{\prime}\right)^{2}-m^{2}}% denominator } \right) {\color{#fe6d73} u^{s}(p)} \]

Invariant amplitude cont’d

This unwieldy expression can be reduced a little by expanding the binomials in the denominator, and observing for the numerator \[ \begin{aligned} (\not{p}+m) \gamma^{\nu} u^{s}(p) &= \left(2 p^{\nu} - \gamma^{\nu}\not{p} + \gamma^{\nu} m\right) u^{s}(p) \\ &=2 p^{\nu} u^{s}(p)-\gamma^{\nu}\underbrace{(\not{p}-m) u^{s}(p)}_{ \text{Dirac equation} \implies 0 } \\ &=2_{} p^{\nu} u^{s}(p) \end{aligned} \]

which yields the invariant amplitude in the simpler form \[ i \mathcal{M}= -i e^{2} \epsilon_{\mu \lambda^\prime}^{\ast} \left(k^{\prime}\right) \epsilon_{\nu\lambda}(k) \bar{u}^{s^\prime}\left(p^{\prime}\right) \left( \frac{\gamma^{\mu} \not{k}\gamma^{\nu}+2 \gamma^{\mu} p^{\nu}} {2 p \cdot k} + \frac{-\gamma^{\nu} \not{k}^{\prime} \gamma^{\mu}+2 \gamma^{\nu} p^{\mu}} {-2 p \cdot k^{\prime}}\right) u^{s}(p)\text{.} \]

Lab frame

Inertial frame in which the electron is at rest. This will also be the assumed rest frame for our particle detectors, hence lab frame.

Differential cross section

Cross section, \(\sigma\), used to characterise interaction strength. Analogous to effective cross sectional area of particle through beam, while capturing relativistic and quantum dynamics information.

\[ \begin{aligned} \mathrm{d} \sigma=& \frac{1}{2 E_{\mathcal{A}} 2 E_{\mathcal{B}}\left|v_{\mathcal{A}}-v_{\mathcal{B}}\right|}\left(\prod_{f} \frac{\mathrm{d}^{3} p_{f}}{(2 \pi)^{3}} \frac{1}{2 E_{f}}\right) \\ &\left|\mathcal{M}\left(p_{\mathcal{A}}, p_{\mathcal{B}} \rightarrow\left\{p_{f}\right\}\right)\right|^{2}(2 \pi)^{4} \delta^{(4)}\left(p_{\mathcal{A}}+p_{\mathcal{B}}-\sum p_{f}\right) \end{aligned} \]

Constructed from phase space volume, invariant amplitude \(\mathcal{M}\), and momentum conserving Dirac delta terms.

Phase space integral

The phase space volume for a two particle collision may be defined by the integral,

\[ \int \mathrm{d} \Pi_{2}=\int \frac{\mathrm{d}^{3} k^{\prime}} {(2 \pi)^{3} 2 E_{k^{\prime}}} \frac{\mathrm{d}^{3} p^{\prime}}{(2 \pi)^{3} 2 E_{p^{\prime}}}(2 \pi)^{4} \delta^{(4)}\left(p+k-k^{\prime}-p^{\prime}\right) \text{.} \]

By making careful substitutions, and using fundamental theorem of calculus, can extract derivative of cross section wrt specific quantities, eg. \(\mathrm{d}\sigma / \mathrm{d}\cos\theta\).

The phase space of an interaction carries its kinematic information.

Spin averages and polarisation sums

In the unpolarised case, we don’t know the spins of particles. The electron orientations are random, so we average over initial particles, and sum over final particles.

\[ \langle |\mathcal{M}|^2 \rangle = \dfrac{1}{4} \sum_\text{spins} | \mathcal{M} |^2 \]

We apply the replacement \(\sum^2_{\lambda=1} \epsilon^\ast_{\mu\lambda} \epsilon_{\nu\lambda} \to - g_{\mu\nu}\), and the property of Dirac spinors, \(\sum_{s=1}^{2} u^{s}(p) \bar{u}^{s}(p) = (\not{p}+m)\), to our spin averaged invariant amplitude, to obtain

\[ \begin{aligned} \frac{1}{4} \sum_{\text {spins}}|\mathcal{M}|^{2}= \dfrac{e^{4}}{4} \bigg[ &\left(\not{p}^{\prime} + m\right) \left( \dfrac{% numerator \gamma^{\mu} \not{k} \gamma^{\nu} + 2 \gamma^{\mu} p^{\nu}} {2 p \cdot k}% denominator + \dfrac{% numerator \gamma^{\nu} \not{k}^{\prime} \gamma^{\mu} -2 \gamma^{\nu} p^{\mu}} {2 p \cdot k^{\prime}}% denominator \right) \\ & \left(\not{p} + m\right) \left( \dfrac{% numerator \gamma_{\nu} \not{k} \gamma_{\mu} + 2 \gamma_{\mu} p_{\nu}} {2 p \cdot k}% denominator + \dfrac{% numerator \gamma_{\mu} \not{k}^{\prime} \gamma_{\nu} - 2 \gamma_{\nu} p_{\mu}} {2 p \cdot k^{\prime}} % denominator \right) \bigg] \end{aligned} \]

Working out the traces

Contracting covariant and contravariant indices yields the spin averaged invariant amplitude as a sum of four traces.

\[ \frac{1}{4} \sum_{\text {spins}}|\mathcal{M}|^{2} = \dfrac{e^4}{4} \left[ \dfrac {\operatorname{Tr} A} {\left(2 p \cdot k\right)^2} + \dfrac {\operatorname{Tr} B} {\left(2 p \cdot k\right) \left(2 p \cdot k^{\prime} \right)} + \dfrac {\operatorname{Tr} C} {\left(2 p \cdot k^{\prime} \right)\left(2 p \cdot k\right)} + \dfrac {\operatorname{Tr} D} {\left(2 p \cdot k^{\prime} \right)^2} \right] \]

Clifford algebra provides simplifications, eg. half the traces contain odd number of \(\gamma^\mu\) matrices, so are identically zero. The traces are also related, such that \(\operatorname{Tr} B = \operatorname{Tr} C\), and similarly \(\operatorname{Tr} D=\operatorname{Tr} A \; \left(k \leftrightarrow-k^{\prime}\right)\). Applying this gives

\[ \begin{aligned} \frac{1}{4} \sum_{\text {spins }}|\mathcal{M}|^{2}= 2 e^{4} \left[ \frac{m^2 - u}{s - m^2} + \frac{s - m^2}{m^2 - u}\right. +&\left. 2 m^{2}\left( \frac{2}{s - m^2} - \frac{2}{m^2 - u} \right) \right. \\ +& \left. m^{4}\left( \frac{2}{s - m^2} - \frac{2}{m^2 - u} \right)^{2} \right] \end{aligned} \]

Phase space

To evaluate the phase space integral introduced earlier \[ \int \mathrm{d} \Pi_{2}= \int \frac{\mathrm{d}^3 k^{\prime}}{(2 \pi)^{3} 2 E_{k^{\prime}}} \frac{\mathrm{d}^3 p^{\prime}}{(2 \pi)^{3} 2 E_{p^{\prime}}} (2 \pi)^{4} \delta^{(4)}\left(p + k - k^{\prime} - p^{\prime}\right) \]

The momentum conserving Dirac delta reduces the calculation to a two dimensional integral, rather than a six dimensional one. We split into spatial and temporal momentum components \[ \delta^{(4)}\left(p + k - k^{\prime} - p^{\prime}\right) = \delta\left( E_\mathbf{p} + \omega - E_\mathbf{p}^\prime - \omega^\prime \right) \delta^{(3)}\left( | \mathbf{p} | + | \mathbf{k} | - | \mathbf{p}^\prime | - | \mathbf{k}^\prime | \right) \]

This may then be evaluated for a specific integration measure, for instance we can substitute the solid angle \[ \mathrm{d}^3 p = | \mathbf{p} |^2 \mathrm{d}p \; \mathrm{d}\Omega \]

In centre-of-momentum frame

In the centre-of-momentum frame, 4-momenta of incoming and outgoing particles are \[ \begin{array}{cccc} p = (E, -\omega\mathbf{\hat{z}}) & k = (\omega, \omega \mathbf{\hat{z}}) & p^\prime = (E, \mathbf{p}^\prime) & k^\prime = (\omega, \omega \sin{\theta}, 0, \omega \cos{\theta}) & \end{array} \] and the phase space integral takes the form \[ \int \mathrm{d} \Pi_{2} = \frac{1}{16 \pi^{2}} \int \mathrm{d} \Omega \frac{\omega}{E_{\text{CoM}}} \]

In lab frame

Whereas in the lab frame, the 4-momenta are given by \[ \begin{array}{cccc} p = (m, 0) & k = (\omega, \omega\mathbf{\hat{z}}) & p^\prime = (E^\prime, \mathbf{p}^\prime) & k^\prime = (\omega^{\prime}, \omega^{\prime} \sin{\theta}, 0, \omega^{\prime} \cos{\theta}) & \end{array} \]

and the phase space integral is obtained as

\[ \int \mathrm{d} \Pi_{2} = \dfrac{1}{16 \pi^2} \int \mathrm{d}\Omega \frac{\omega^{\prime \; 2}}{m \omega} \]

Function of squared momentum transfer

We may now bring our equations for the phase space integral and the invariant amplitude together. Substituting the integration measure, \(\mathrm{d} t = -2\omega^2\mathrm{d}\Omega / 2\pi\) and writing in Mandelstam variables in the centre-of-momentum frame, we obtain \[ \begin{aligned} \dfrac{\mathrm{d}\sigma\left(s, t\right)}{\mathrm{d}t} = \dfrac{2\pi\alpha^2}{\left(s - m^2\right)^2} \left[ \frac{s + t - m^2}{s - m^2} + \frac{s - m^2}{s + t - m^2}\right. +&\left. 2 m^{2}\left( \frac{2}{s - m^2} - \frac{2}{s + t - m^2} \right) \right. \\ +& \left. m^{4}\left( \frac{2}{s - m^2} - \frac{2}{s + t - m^2} \right)^{2} \right] \end{aligned} \]

Expressing in terms of scattering angle, \(t = \frac{ -2\left(\frac{s-m^{2}}{2 m}\right)^{2}(1-\cos \theta) } { 1+\left(\frac{s-m^{2}}{2 m^{2}}\right)(1-\cos \theta) }\), and applying the range of \(\cos\theta\) to this equation yields the domain for \(\mathrm{d}\sigma / \mathrm{d}t\) \[-\dfrac{\left(s - m^2\right)^2}{s} \leq t \leq 0\]

Function of angle

We perform the same procedure in the lab frame, using integration measure \(\mathrm{d}\Omega = -2\pi\mathrm{d}(\cos\theta)\), to obtain the angular dependence.

\[ \begin{aligned} \dfrac{\mathrm{d}\sigma}{\mathrm{d}\cos\theta} = \dfrac{1}{4m\omega} \dfrac{1}{8\pi} \dfrac{\omega^{\prime \; 2}}{\omega m} 2 e^{4} \left[ \frac{m^2 - u}{s - m^2} + \frac{s - m^2}{m^2 - u}\right. +&\left. 2 m^{2}\left( \frac{2}{s - m^2} - \frac{2}{m^2 - u} \right) \right. \\ +& \left. m^{4}\left( \frac{2}{s - m^2} - \frac{2}{m^2 - u} \right)^{2} \right] \end{aligned} \]

Writing in terms of photon momenta obtains the famous Klein-Nishina formula [6]

\[ \dfrac{\mathrm{d}\sigma}{\mathrm{d}\cos\theta} = \frac{\pi \alpha^{2}}{m^{2}} \left(\frac{\omega^{\prime}}{\omega}\right)^{2} \left[\frac{\omega^{\prime}}{\omega}+\frac{\omega}{\omega^{\prime}}-\sin ^{2} \theta\right] \]

QED prediction for \(\rm{d}\sigma/\rm{d}\cos\theta\)

QED prediction for \(\rm{d}\sigma/\rm{d}t\)

QED prediction for \(\rm{d}\sigma/\rm{d}t\)

QED prediction for \(\rm{d}\sigma/\rm{d}t\)