Quantum field theory 1, lecture 16

Transition amplitude

Quite generally, one can define for the non-trivial part of an S-matrix \[\langle \beta;\text{out}|\alpha;\text{in}\rangle = (2\pi)^4 \delta^{(4)} (p^{\text{out}}-p^{\text{in}}) \, i \, {\cal T}_{\beta\alpha}.\] Together with the trivial part from “no scattering”, one can write \[S_{\beta\alpha} = \delta_{\beta\alpha} + (2\pi)^4 \delta^{(4)} (p^{\text{out}}-p^{\text{in}}) \, i \, {\cal T}_{\beta\alpha}.\] By comparison of expressions we find for the \(2\to2\) scattering of non-relativistic bosons at lowest order in \(\lambda\) simply \[{\cal T} = -2\lambda,\] independant of momenta. More generally, the transition amplitude \({\cal T}\) is expected to depend on the momenta of incoming and outgoing particles.

Diagrammatic representation

To keep the overview over a calculation it is sometimes useful to introduce a graphical representation. For the perturbation series discussed above we may represent incoming particles by and similarly outgoing particles by These functional derivatives are acting on the partition function \(Z[J]\). The partition function in \(\eqref{eq:partitionFunctionPerturbativeSeries}\) can be written in a perturbative series with the interaction term represented by One can the let the functional derivatives act on the sources and at the end evaluate everything at \(J=0\). While the diagrammatic representation is useful, it is only an auxiliary tool to organize the algebra. With a bit of experience one can work well with it.

From the S-matrix to a cross-section

Transition propability

Let us start from an S-matrix element in the form \[\langle \beta;\text{out}|\alpha;\text{in}\rangle = (2\pi)^4 \delta^{(4)} (p^{\text{out}}-p^{\text{in}}) \, i \, {\cal T}\] with transition amplitude \({\cal T}\) which may depend on the momenta itself. (For \(2 \to 2\) scattering of non-relativistic bosons, and at lowest order in \(\lambda\), we found simply \({\cal T} = - 2\lambda\).) Let us now discuss how one can relate S-matrix elements to actual scattering cross-sections that can be measured in an experiment. We start by writing the transition probability from a state \(\alpha\) to a state \(\beta\) as \[P= \frac{|\langle \beta;\text{out}|\alpha;\text{in}\rangle|^2}{\langle \beta;\text{out}|\beta;\text{out}\rangle\langle \alpha;\text{in}|\alpha;\text{in}\rangle}.\]

Transition rate

The numerator contains a factor \[\left[(2\pi)^4 \delta^{(4)} (p^{\text{out}}-p^{\text{in}})\right]^2 = (2\pi)^4 \delta^{(4)} (p^{\text{out}}-p^{\text{in}}) (2\pi)^4 \delta^{(4)}(0).\] This looks ill defined but becomes meaningful in a finite volume \(V\) and for finite time interval \(\Delta T\). In fact \[(2\pi)^4 \delta^4 (0) = \int d^4 x \, e^{i0 x} = V\Delta T.\] For the transition rate \(\dot{P} = \frac{P}{\Delta T}\) we can therefore write \[\dot{P} = \frac{V(2\pi)^4 \delta^{(4)} (p^{\text{out}}-p^{\text{in}})|{\cal T}|^2}{\langle \beta; \text{out}|\beta; \text{out}\rangle \langle \alpha; \text{in}| \alpha; \text{in}\rangle}.\]

Normalization of incoming and outgoing states

Moreover, for incoming and outgoing two-particle states, their normalization is obtained from \[\begin{split} \langle \mathbf{p}_1,\mathbf{p}_2;\text{in}| \mathbf{p}_1,\mathbf{p}_2; \text{in}\rangle &= \lim_{\mathbf{q}_j \to \mathbf{p}_j}\langle \mathbf{p}_1,\mathbf{p}_2;\text{in}| \mathbf{q}_1,\mathbf{q}_2; \text{in}\rangle\\ &= \lim_{\mathbf{q}_j \to \mathbf{p}_j}\left[(2\pi)^6 \left(\delta^{(3)}(\mathbf{p}_1 - \mathbf{q}_1) \delta^{(3)}(\mathbf{p}_2 - \mathbf{q}_2)+\delta^{(3)}(\mathbf{p}_1 - \mathbf{q}_2)\delta^{(3)}(\mathbf{p}_2 - \mathbf{q}_1)\right)\right]\\ &= \left[(2\pi)^3 \delta^{(3)}(0)\right]^2 \\ &= V^2. \end{split}\]

Counting of momentum states

In a finite volume \(V= L^3\), and with periodic boundary conditions, the final momenta are of the form \[\mathbf{p} = \frac{2\pi}{L}(m,n,l),\] with some integer numbers \(m,n,l\). We can count final states according to \[\sum_{m,n,l} = \sum_{m,n,l} \Delta m \Delta n \Delta l = L^3 \sum_{m,n,l}\frac{\Delta p_1 \Delta p_2 \Delta p_3}{(2\pi)^3}.\] In the continuum limit this becomes \[V \int \frac{d^3p}{(2\pi)^3}.\] The differential transition rate has one factor \(V d^3p/(2\pi)^3\) for each final state particle.

Differential transition rate

For \(2\to 2\) scattering, \[d\dot{P} = (2\pi)^4 \delta^{(4)}(p^{\text{out}}-p^{\text{in}}) |{\cal T}|^2 \frac{1}{V} \frac{d^3 q_1}{(2\pi)^3}\frac{d^3 q_2}{(2\pi)^3}.\] This can be integrated to give the transition rate into a certain region of momentum states.

Flux of incoming particles

We can go from the transition probability to a cross-section by dividing through the flux of incoming particles \[{\cal F} = \frac{1}{V} v = \frac{2 |\mathbf{p}_1|}{m V}.\] Here we have a density of one particle per volume \(V\) and the relative velocity of the two particles is \(v= 2|\mathbf{p}_1| / m\), in the center-of-mass frame where \(|\mathbf{p}_1| = |\mathbf{p}_2|,\) for identical particles with equal mass \(m\).

Differential cross-section

This cancels the last factor \(V\) and we find for the differential cross-section \[d\sigma = \frac{|{\cal T}|^2 m}{2|\mathbf{p}_1|}(2\pi)^4 \delta^{(4)}(p^\text{out}-p^\text{in}) \frac{d^3 q_1}{(2\pi)^3}\frac{d^3 q_2}{(2\pi)^3}.\]

Phase space integrals

In the center-of-mass frame one has also \(\mathbf{p}^{\text{in}} = \mathbf{p}_1 + \mathbf{p}_2 = 0\) and accordingly \[\delta^{(4)}(p^\text{out}-p^\text{in}) = \delta (E^\text{out}-E^\text{in}) \, \delta^{(3)}(\mathbf{q}_1 + \mathbf{q}_2).\] The three-dimensional part can be used to perform the integral over \(\mathbf{q}_2\). In doing these integrals over final state momenta, a bit of care is needed because the two final state particles are indistinguishable. An outgoing state \(|\mathbf{q}_1, \mathbf{q}_2; \text{out} \rangle\) equals the state \(|\mathbf{q}_2, \mathbf{q}_1;\text{out}\rangle\). Therefore, in order to count only really different final states, one must divide by a factor \(2\) if one simply integrates \(d^3 q_1\) and \(d^3 q_2\) independently. Keeping this in mind, we find for the differential cross-section after doing the integral over \(\mathbf{q}_2\), \[d\sigma = \frac{|{\cal T}|^2 m}{2 |\mathbf{p}_1|(2\pi)^2} \delta (E^{\text{out}}-E^{\text{in}}) d^3q_1.\]

Magnitude and solid angle

We can now use \[d^3 \mathbf{q}_1 = |\mathbf{q}_1|^2 d|\mathbf{q}_1|\; d\Omega_{q_1}\] where \(d\Omega_{q_1}\) is the differential solid angle element. Moreover \[E^{\text{out}} = \frac{\mathbf{q}_1^2}{2m} + \frac{\mathbf{q}_2^2}{2m} + 2V_0 = \frac{\mathbf{q}_1^2}{m} + 2V_0 ,\] and \[\frac{dE^\text{out}}{d|\mathbf{q}_1|} = 2\frac{|\mathbf{q}_1|}{m}.\] With this, and using the familiar relation \(\delta(f(x))=\delta(x-x_0) / |f^\prime(x_0)|\), one can perform the integral over the magnitude \(|\mathbf{q}_1|\) using the Dirac function \(\delta (E^{\text{out}}-E^{\text{in}})\). This yields \(|\mathbf{q}_1|=|\mathbf{p}_1|\) and \[d\sigma = \frac{|{\cal T}|^2 m^2}{ 16\pi^2}\;d\Omega_{q_1}.\]

Total cross-section

For the simple case where \({\cal T}\) is independent of the solid angle \(\omega_{q_1}\), we can calculate the total cross-section. Here we must now take into account that only half of the solid angle \(4\pi\) corresponds to physically independent configurations. The total cross-sections is therefore \[\sigma = \frac{|{\cal T}|^2 m^2}{8 \pi}.\] In a final step we use \({\cal T} = -2\lambda\) to lowest order in \(\lambda\) (equivalent to the Born approximation in quantum mechanics) and find here the cross-section \[\sigma = \frac{\lambda^2 m^2}{2\pi}.\]

Dimensions

Let us check the dimensions. The action \[S = \int dt\;d^3x \left\{\varphi^*\left(i\partial_t + \tfrac{\boldsymbol{\nabla}^2}{2m}-V_0\right) \varphi - \tfrac{\lambda}{2}(\varphi^*\varphi)^2\right\}\] must be dimensionless. The field \(\varphi\) must have dimension \[= \text{length}^{-\tfrac{3}{2}}.\] The interaction strength \(\lambda\) must accordingly have dimension \[= \frac{\text{length}^3}{\text{time}}.\] Because \[\left[\tfrac{\boldsymbol{\nabla}^2}{2m}\right] = \frac{1}{\text{time}},\] one has \([m] = \frac{\text{time}}{\text{length}^2}\) and therefore \([\lambda m] = \text{length}.\) It follows that indeed \[= \text{length}^2\] as appropriate for a cross-section.

Non-relativistic fermions

So far we have discussed bosonic fields and bosonic particles as their excitations. Let us now turn to fermions. Fermions as quantum particles differ in two central aspects from bosons. First, they satisfy fermionic statistics. Wave functions for several particles are anti-symmetric under the exchange of particles and occupation numbers of modes can only be 0 or 1. Second, fermionic particles have half integer spin, i. e. \(1/2\), \(3/2\), and so on, in contrast to bosonic particles which have integer spin \(0\), \(1\), \(2\) and so on. Both these aspects lead to interesting new developments. Half-integer spin in the context of relativistic theories leads to a new and deeper understanding of space-time symmetries and fermionic statistics leads to a new kind of functional integral based on anti-commuting numbers. The latter appears already for functional integral representations of non-relativistic quantum fields. We will start with this second-aspect and then turn to aspects of space-time symmetry for relativistic theories later on.

Pauli spinors

Pauli spinor fields

In non-relativistic quantum mechanics, particles with spin \(1/2\) are described by a variant of Schrödinger’s equation with two-component fields. The fields are so-called Pauli spinors with components describing spin-up and spin-down parts with respect to some axis. One can write this as \[\Psi(t,\mathbf{x}) = \begin{pmatrix} \psi_\uparrow (t,\mathbf{x}) \\ \psi_\downarrow (t,\mathbf{x}) \end{pmatrix}\] We also use the notation \(\psi_a (t,\mathbf{x})\) where \(a= 1,2\) and \[\psi_1(t,\mathbf{x})= \psi_\uparrow (t,\mathbf{x}),\quad\quad\quad \psi_2(t,\mathbf{x})= \psi_\downarrow (t,\mathbf{x}).\]

Pauli equation

The Pauli equation is a generalisation of Schrödinger’s equation (neglecting spin-orbit coupling), \[\left[\left(-i\partial_t - \tfrac{\boldsymbol{\nabla}^2}{2m}+V_0\right)\mathbb{1}+\mu_B \, \boldsymbol{\sigma} \cdot \mathbf{B}\right]\Psi(t, \mathbf{x}) = 0 ,\] or equivalently \[\left[\left(-i\partial_t - \tfrac{\vec \nabla^2}{2m}+V_0\right)\delta_{ab}+\mu_B \, \boldsymbol{\sigma}_{ab} \cdot \mathbf{B}\right]\psi_b(t, \mathbf{x}) = 0.\] Here we use the Pauli matrices \[\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\] and \(\mathbf{B} = (B_1, B_2, B_3)\) is the magnetic field, while \(\mu_B\) is the magneton that quantifies the magnetic moment.

Attempt for an action

Based on this, one would expect that the quadratic part of an action for a non-relativistic field describing spin-\(1/2\) particles is of the form \[S_2 \overset{?}{=} \int dt d^3 x \left\{-\Psi^\dagger \left[\left(-i\partial_t - \frac{\boldsymbol{\nabla}^2}{2m}+V_0\right)\mathbb{1} + \mu_B \, \boldsymbol{\sigma} \cdot \mathbf{B}\right]\Psi\right\}\] However, we also need to take care of fermionic (anti-symmetric) exchange symmetry, such that for fermionic states \[|\mathbf{p}_1, \mathbf{p}_2; \text{in} \rangle = -|\mathbf{p}_2, \mathbf{p}_1; \text{in} \rangle.\] To this aspect we turn next.