## Notes so far for phy2403, Quantum Field Theory I

September 23, 2018 phy2403 No comments

Here’s an aggregate collection of notes for QFT I that I’ll update as the term progresses.  It contains very rough notes for the first 4 lectures (6 hours of material).

These notes also contain my (ungraded) problem set I solution from a few years ago.  I was originally going to take the class and had started preparing for it with some independent reading of the Professor’s class notes (Prof. Luke), and by doing problems he’d posted for a previous year.  I ended up changing my plans and took something else instead, probably to satisfy the graduation requirements for the M.Eng program.  It’s been so long since I’ve done those problems that I don’t even remember all the material that the problems covered (some of which we haven’t gotten to yet in class).

## PHY2403H Quantum Field Theory. Lecture 4: Scalar action, least action principle, Euler-Lagrange equations for a field, canonical quantization. Taught by Prof. Erich Poppitz

### DISCLAIMER: Very rough notes from class. May have some additional side notes, but otherwise probably barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Principles (cont.)

• Lorentz (Poincar\’e : Lorentz and spacetime translations)
• locality
• dimensional analysis
• gauge invariance

These are the requirements for an action. We postulated an action that had the form
\label{eqn:qftLecture4:20}
\int d^d x \partial_\mu \phi \partial^\mu \phi,

called the “Kinetic term”, which mimics $$\int dt \dot{q}^2$$ that we’d see in quantum or classical mechanics. In principle there exists an infinite number of local Poincar\’e invariant terms that we can write. Examples:

• $$\partial_\mu \phi \partial^\mu \phi$$
• $$\partial_\mu \phi \partial_\nu \partial^\nu \partial^\mu \phi$$
• $$\lr{\partial_\mu \phi \partial^\mu \phi}^2$$
• $$f(\phi) \partial_\mu \phi \partial^\mu \phi$$
• $$f(\phi, \partial_\mu \phi \partial^\mu \phi)$$
• $$V(\phi)$$

It turns out that nature (i.e. three spatial dimensions and one time dimension) is described by a finite number of terms. We will now utilize dimensional analysis to determine some of the allowed forms of the action for scalar field theories in $$d = 2, 3, 4, 5$$ dimensions. Even though the real world is only $$d = 4$$, some of the $$d < 4$$ theories are relevant in condensed matter studies, and $$d = 5$$ is just for fun (but also applies to string theories.)

With $$[x] \sim \inv{M}$$ in natural units, we must define $$[\phi]$$ such that the kinetic term is dimensionless in d spacetime dimensions

\label{eqn:qftLecture4:40}
\begin{aligned}
[d^d x] &\sim \inv{M^d} \\
[\partial_\mu] &\sim M
\end{aligned}

so it must be that
\label{eqn:qftLecture4:60}
[\phi] = M^{(d-2)/2}

It will be easier to characterize the dimensionality of any given term by the power of the mass units, that is

\label{eqn:qftLecture4:80}
\begin{aligned}
[\text{mass}] &= 1 \\
[d^d x] &= -d \\
[\partial_\mu] &= 1 \\
[\phi] &= (d-2)/2 \\
[S] &= 0.
\end{aligned}

Since the action is
\label{eqn:qftLecture4:100}
S = \int d^d x \lr{ \LL(\phi, \partial_\mu \phi) },

and because action had dimensions of $$\Hbar$$, so in natural units, it must be dimensionless, the Lagrangian density dimensions must be $$[d]$$. We will abuse language in QFT and call the Lagrangian density the Lagrangian.

## $$d = 2$$

Because $$[\partial_\mu \phi \partial^\mu \phi ] = 2$$, the scalar field must be dimension zero, or in symbols
\label{eqn:qftLecture4:120}
[\phi] = 0.

This means that introducing any function $$f(\phi) = 1 + a \phi + b\phi^2 + c \phi^3 + \cdots$$ is also dimensionless, and
\label{eqn:qftLecture4:140}
[f(\phi) \partial_\mu \phi \partial^\mu \phi ] = 2,

for any $$f(\phi)$$. Another implication of this is that the a potential term in the Lagrangian $$[V(\phi)] = 0$$ needs a coupling constant of dimension 2. Letting $$\mu$$ have mass dimensions, our Lagrangian must have the form
\label{eqn:qftLecture4:160}
f(\phi) \partial_\mu \phi \partial^\mu \phi + \mu^2 V(\phi).

An infinite number of coupling constants of positive mass dimensions for $$V(\phi)$$ are also allowed. If we have higher order derivative terms, then we need to compensate for the negative mass dimensions. Example (still for $$d = 2$$).
\label{eqn:qftLecture4:180}
\LL =
f(\phi) \partial_\mu \phi \partial^\mu \phi + \mu^2 V(\phi) + \inv{{\mu’}^2}\partial_\mu \phi \partial_\nu \partial^\nu \partial^\mu \phi + \lr{ \partial_\mu \phi \partial^\mu \phi }^2 \inv{\tilde{\mu}^2}.

The last two terms, called \underline{couplings} (i.e. any non-kinetic term), are examples of terms with negative mass dimension. There is an infinite number of those in any theory in any dimension.

### Definitions

• Couplings that are dimensionless are called (classically) marginal.
• Couplings that have positive mass dimension are called (classically) relevant.
• Couplings that have negative mass dimension are called (classically) irrelevant.

In QFT we are generally interested in the couplings that are measurable at long distances for some given energy. Classically irrelevant theories are generally not interesting in $$d > 2$$, so we are very lucky that we don’t live in three dimensional space. This means that we can get away with a finite number of classically marginal and relevant couplings in 3 or 4 dimensions. This was mentioned in the Wilczek’s article referenced in the class forum [1]\footnote{There’s currently more in that article that I don’t understand than I do, so it is hard to find it terribly illuminating.}

Long distance physics in any dimension is described by the marginal and relevant couplings. The irrelevant couplings die off at low energy. In two dimensions, a priori, an infinite number of marginal and relevant couplings are possible. 2D is a bad place to live!

## $$d = 3$$

Now we have
\label{eqn:qftLecture4:200}
[\phi] = \inv{2}

so that
\label{eqn:qftLecture4:220}
[\partial_\mu \phi \partial^\mu \phi] = 3.

A 3D Lagrangian could have local terms such as
\label{eqn:qftLecture4:240}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \mu^{3/2} \phi^3 + \mu’ \phi^4
+ \lr{\mu”}{1/2} \phi^5
+ \lambda \phi^6.

where $$m, \mu, \mu”$$ all have mass dimensions, and $$\lambda$$ is dimensionless. i.e. $$m, \mu, \mu”$$ are relevant, and $$\lambda$$ marginal. We stop at the sixth power, since any power after that will be irrelevant.

## $$d = 4$$

Now we have
\label{eqn:qftLecture4:260}
[\phi] = 1

so that
\label{eqn:qftLecture4:280}
[\partial_\mu \phi \partial^\mu \phi] = 4.

In this number of dimensions $$\phi^k \partial_\mu \phi \partial^\mu$$ is an irrelevant coupling.

A 4D Lagrangian could have local terms such as
\label{eqn:qftLecture4:300}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \mu \phi^3 + \lambda \phi^4.

where $$m, \mu$$ have mass dimensions, and $$\lambda$$ is dimensionless. i.e. $$m, \mu$$ are relevant, and $$\lambda$$ is marginal.

## $$d = 5$$

Now we have
\label{eqn:qftLecture4:320}
[\phi] = \frac{3}{2},

so that
\label{eqn:qftLecture4:340}
[\partial_\mu \phi \partial^\mu \phi] = 5.

A 5D Lagrangian could have local terms such as
\label{eqn:qftLecture4:360}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \sqrt{\mu} \phi^3 + \inv{\mu’} \phi^4.

where $$m, \mu, \mu’$$ all have mass dimensions. In 5D there are no marginal couplings. Dimension 4 is the last dimension where marginal couplings exist. In condensed matter physics 4D is called the “upper critical dimension”.

From the point of view of particle physics, all the terms in the Lagrangian must be the ones that are relevant at long distances.

## Least action principle (classical field theory).

Now we want to study 4D scalar theories. We have some action
\label{eqn:qftLecture4:380}
S[\phi] = \int d^4 x \LL(\phi, \partial_\mu \phi).

Let’s keep an example such as the following in mind
\label{eqn:qftLecture4:400}
\LL = \underbrace{\inv{2} \partial_\mu \phi \partial^\mu \phi}_{\text{Kinetic term}} – \underbrace{m^2 \phi – \lambda \phi^4}_{\text{all relevant and marginal couplings}}.

The even powers can be justified by assuming there is some symmetry that kills the odd powered terms.

fig. 1. Cylindrical spacetime boundary.

We will be integrating over a space time region such as that depicted in fig. 1, where a cylindrical spatial cross section is depicted that we allow to tend towards infinity. We demand that the field is fixed on the infinite spatial boundaries. The easiest way to demand that the field dies off on the spatial boundaries, that is
\label{eqn:qftLecture4:420}
\lim_{\Abs{\Bx} \rightarrow \infty} \phi(\Bx) \rightarrow 0.

The functional $$\phi(\Bx, t)$$ that obeys the boundary condition as stated extremizes $$S[\phi]$$.

Extremizing the action means that we seek $$\phi(\Bx, t)$$
\label{eqn:qftLecture4:440}
\delta S[\phi] = 0 = S[\phi + \delta \phi] – S[\phi].

How do we compute the variation?
\label{eqn:qftLecture4:460}
\begin{aligned}
\delta S
&= \int d^d x \lr{ \LL(\phi + \delta \phi, \partial_\mu \phi + \partial_\mu \delta \phi) – \LL(\phi, \partial_\mu \phi) } \\
&= \int d^d x \lr{ \PD{\phi}{\LL} \delta \phi + \PD{(\partial_mu \phi)}{\LL} (\partial_\mu \delta \phi) } \\
&= \int d^d x \lr{ \PD{\phi}{\LL} \delta \phi
+ \partial_\mu \lr{ \PD{(\partial_mu \phi)}{\LL} \delta \phi}
– \lr{ \partial_\mu \PD{(\partial_mu \phi)}{\LL} } \delta \phi
} \\
&=
\int d^d x
\delta \phi
\lr{ \PD{\phi}{\LL}
– \partial_\mu \PD{(\partial_mu \phi)}{\LL} }
+ \int d^3 \sigma_\mu \lr{ \PD{(\partial_\mu \phi)}{\LL} \delta \phi }
\end{aligned}

If we are explicit about the boundary term, we write it as
\label{eqn:qftLecture4:480}
\int dt d^3 \Bx \partial_t \lr{ \PD{(\partial_t \phi)}{\LL} \delta \phi }
=
\int d^3 \Bx \evalrange{ \PD{(\partial_t \phi)}{\LL} \delta \phi }{t = -T}{t = T}
– \int dt d^2 \BS \cdot \lr{ \PD{(\spacegrad \phi)}{\LL} \delta \phi }.

but $$\delta \phi = 0$$ at $$t = \pm T$$ and also at the spatial boundaries of the integration region.

This leaves
\label{eqn:qftLecture4:500}
\delta S[\phi] = \int d^d x \delta \phi
\lr{ \PD{\phi}{\LL} – \partial_\mu \PD{(\partial_mu \phi)}{\LL} } = 0 \forall \delta \phi.

That is

\label{eqn:qftLecture4:540}
\boxed{
\PD{\phi}{\LL} – \partial_\mu \PD{(\partial_mu \phi)}{\LL} = 0.
}

This are the Euler-Lagrange equations for a single scalar field.

Returning to our sample scalar Lagrangian
\label{eqn:qftLecture4:560}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \inv{2} m^2 \phi^2 – \frac{\lambda}{4} \phi^4.

This example is related to the Ising model which has a $$\phi \rightarrow -\phi$$ symmetry. Applying the Euler-Lagrange equations, we have
\label{eqn:qftLecture4:580}
\PD{\phi}{\LL} = -m^2 \phi – \lambda \phi^3,

and
\label{eqn:qftLecture4:600}
\begin{aligned}
\PD{(\partial_\mu \phi)}{\LL}
&=
\PD{(\partial_\mu \phi)}{} \lr{
\inv{2} \partial_\nu \phi \partial^\nu \phi } \\
&=
\inv{2} \partial^\nu \phi
\PD{(\partial_\mu \phi)}{}
\partial_\nu \phi
+
\inv{2} \partial_\nu \phi
\PD{(\partial_\mu \phi)}{}
\partial_\alpha \phi g^{\nu\alpha} \\
&=
\inv{2} \partial^\mu \phi
+
\inv{2} \partial_\nu \phi g^{\nu\mu} \\
&=
\partial^\mu \phi
\end{aligned}

so we have
\label{eqn:qftLecture4:620}
\begin{aligned}
0
&=
\PD{\phi}{\LL} -\partial_\mu
\PD{(\partial_\mu \phi)}{\LL} \\
&=
-m^2 \phi – \lambda \phi^3 – \partial_\mu \partial^\mu \phi.
\end{aligned}

For $$\lambda = 0$$, the free field theory limit, this is just
\label{eqn:qftLecture4:640}
\partial_\mu \partial^\mu \phi + m^2 \phi = 0.

Written out from the observer frame, this is
\label{eqn:qftLecture4:660}
(\partial_t)^2 \phi – \spacegrad^2 \phi + m^2 \phi = 0.

With a non-zero mass term
\label{eqn:qftLecture4:680}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi = 0,

is called the Klein-Gordan equation.

If we also had $$m = 0$$ we’d have
\label{eqn:qftLecture4:700}
\lr{ \partial_t^2 – \spacegrad^2 } \phi = 0,

which is the wave equation (for a massless free field). This is also called the D’Alembert equation, which is familiar from electromagnetism where we have
\label{eqn:qftLecture4:720}
\begin{aligned}
\lr{ \partial_t^2 – \spacegrad^2 } \BE &= 0 \\
\lr{ \partial_t^2 – \spacegrad^2 } \BB &= 0,
\end{aligned}

in a source free region.

## Canonical quantization.

\label{eqn:qftLecture4:740}
\LL = \inv{2} \dot{q} – \frac{\omega^2}{2} q^2

This has solution $$\ddot{q} = – \omega^2 q$$.

Let
\label{eqn:qftLecture4:760}
p = \PD{\dot{q}}{\LL} = \dot{q}

\label{eqn:qftLecture4:780}
H(p,q) = \evalbar{p \dot{q} – \LL}{\dot{q}(p, q)}
= p p – \inv{2} p^2 + \frac{\omega^2}{2} q^2 = \frac{p^2}{2} + \frac{\omega^2}{2} q^2

In QM we quantize by mapping Poisson brackets to commutators.
\label{eqn:qftLecture4:800}
\antisymmetric{\hatp}{\hat{q}} = -i

One way to represent is to say that states are $$\Psi(\hat{q})$$, a wave function, $$\hat{q}$$ acts by $$q$$
\label{eqn:qftLecture4:820}
\hat{q} \Psi = q \Psi(q)

With
\label{eqn:qftLecture4:840}
\hatp = -i \PD{q}{},

so
\label{eqn:qftLecture4:860}
\antisymmetric{ -i \PD{q}{} } { q} = -i

Let’s introduce an explicit space time split. We’ll write
\label{eqn:qftLecture4:880}
L = \int d^3 x \lr{
\inv{2} (\partial_0 \phi(\Bx, t))^2 – \inv{2} \lr{ \spacegrad \phi(\Bx, t) }^2 – \frac{m^2}{2} \phi
},

so that the action is
\label{eqn:qftLecture4:900}
S = \int dt L.

The dynamical variables are $$\phi(\Bx)$$. We define
\label{eqn:qftLecture4:920}
\begin{aligned}
\pi(\Bx, t) = \frac{\delta L}{\delta (\partial_0 \phi(\Bx, t))}
&=
\partial_0 \phi(\Bx, t) \\
&=
\dot{\phi}(\Bx, t),
\end{aligned}

called the canonical momentum, or the momentum conjugate to $$\phi(\Bx, t)$$. Why $$\delta$$? Has to do with an implicit Dirac function to eliminate the integral?

\label{eqn:qftLecture4:940}
\begin{aligned}
H
&= \int d^3 x \evalbar{\lr{ \pi(\bar{\Bx}, t) \dot{\phi}(\bar{\Bx}, t) – L }}{\dot{\phi}(\bar{\Bx}, t) = \pi(x, t) } \\
&= \int d^3 x \lr{ (\pi(\Bx, t))^2 – \inv{2} (\pi(\Bx, t))^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m}{2} \phi^2 },
\end{aligned}

or
\label{eqn:qftLecture4:960}
H
= \int d^3 x \lr{ \inv{2} (\pi(\Bx, t))^2 + \inv{2} (\spacegrad \phi(\Bx, t))^2 + \frac{m}{2} (\phi(\Bx, t))^2 }

In analogy to the momentum, position commutator in QM
\label{eqn:qftLecture4:1000}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

we “quantize” the scalar field theory by promoting $$\pi, \phi$$ to operators and insisting that they also obey a commutator relationship
\label{eqn:qftLecture4:980}
\antisymmetric{\pi(\Bx, t)}{\phi(\By, t)} = -i \delta^3(\Bx – \By).

# References

[1] Frank Wilczek. Fundamental constants. arXiv preprint arXiv:0708.4361, 2007. URL https://arxiv.org/abs/0708.4361.

## PHY2403H Quantum Field Theory. Lecture 3: Lorentz transformations and a scalar action. Taught by Prof. Erich Poppitz

September 18, 2018 phy2403 No comments , ,

### DISCLAIMER: Very rough notes from class. Some additional side notes, but otherwise barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz.

## Determinant of Lorentz transformations

We require that Lorentz transformations leave the dot product invariant, that is $$x \cdot y = x’ \cdot y’$$, or
\label{eqn:qftLecture3:20}
x^\mu g_{\mu\nu} y^\nu = {x’}^\mu g_{\mu\nu} {y’}^\nu.

Explicitly, with coordinate transformations
\label{eqn:qftLecture3:40}
\begin{aligned}
{x’}^\mu &= {\Lambda^\mu}_\rho x^\rho \\
{y’}^\mu &= {\Lambda^\mu}_\rho y^\rho
\end{aligned}

such a requirement is equivalent to demanding that
\label{eqn:qftLecture3:500}
\begin{aligned}
x^\mu g_{\mu\nu} y^\nu
&=
{\Lambda^\mu}_\rho x^\rho
g_{\mu\nu}
{\Lambda^\nu}_\kappa y^\kappa \\
&=
x^\mu
{\Lambda^\alpha}_\mu
g_{\alpha\beta}
{\Lambda^\beta}_\nu
y^\nu,
\end{aligned}

or
\label{eqn:qftLecture3:60}
g_{\mu\nu}
=
{\Lambda^\alpha}_\mu
g_{\alpha\beta}
{\Lambda^\beta}_\nu

multiplying by the inverse we find
\label{eqn:qftLecture3:200}
\begin{aligned}
g_{\mu\nu}
{\lr{\Lambda^{-1}}^\nu}_\lambda
&=
{\Lambda^\alpha}_\mu
g_{\alpha\beta}
{\Lambda^\beta}_\nu
{\lr{\Lambda^{-1}}^\nu}_\lambda \\
&=
{\Lambda^\alpha}_\mu
g_{\alpha\lambda} \\
&=
g_{\lambda\alpha}
{\Lambda^\alpha}_\mu.
\end{aligned}

This is now amenable to expressing in matrix form
\label{eqn:qftLecture3:220}
\begin{aligned}
(G \Lambda^{-1})_{\mu\lambda}
&=
(G \Lambda)_{\lambda\mu} \\
&=
((G \Lambda)^\T)_{\mu\lambda} \\
&=
(\Lambda^\T G)_{\mu\lambda},
\end{aligned}

or
\label{eqn:qftLecture3:80}
G \Lambda^{-1}
=
(G \Lambda)^\T.

Taking determinants (using the normal identities for products of determinants, determinants of transposes and inverses), we find
\label{eqn:qftLecture3:100}
det(G)
det(\Lambda^{-1})
=
det(G) det(\Lambda),

or
\label{eqn:qftLecture3:120}
det(\Lambda)^2 = 1,

or
$$det(\Lambda)^2 = \pm 1$$. We will generally ignore the case of reflections in spacetime that have a negative determinant.

Smart-alec Peeter pointed out after class last time that we can do the same thing easier in matrix notation
\label{eqn:qftLecture3:140}
\begin{aligned}
x’ &= \Lambda x \\
y’ &= \Lambda y
\end{aligned}

where
\label{eqn:qftLecture3:160}
\begin{aligned}
x’ \cdot y’
&=
(x’)^\T G y’ \\
&=
x^\T \Lambda^\T G \Lambda y,
\end{aligned}

which we require to be $$x \cdot y = x^\T G y$$ for all four vectors $$x, y$$, that is
\label{eqn:qftLecture3:180}
\Lambda^\T G \Lambda = G.

We can find the result \ref{eqn:qftLecture3:120} immediately without having to first translate from index notation to matrices.

## Field theory

The electrostatic potential is an example of a scalar field $$\phi(\Bx)$$ unchanged by SO(3) rotations
\label{eqn:qftLecture3:240}
\Bx \rightarrow \Bx’ = O \Bx,

that is
\label{eqn:qftLecture3:260}
\phi'(\Bx’) = \phi(\Bx).

Here $$\phi'(\Bx’)$$ is the value of the (electrostatic) scalar potential in a primed frame.

However, the electrostatic field is not invariant under Lorentz transformation.
We postulate that there is some scalar field
\label{eqn:qftLecture3:280}
\phi'(x’) = \phi(x),

where $$x’ = \Lambda x$$ is an SO(1,3) transformation. There are actually no stable particles (fields that persist at long distances) described by Lorentz scalar fields, although there are some unstable scalar fields such as the Higgs, Pions, and Kaons. However, much of our homework and discussion will be focused on scalar fields, since they are the easiest to start with.

We need to first understand how derivatives $$\partial_\mu \phi(x)$$ transform. Using the chain rule
\label{eqn:qftLecture3:300}
\begin{aligned}
\PD{x^\mu}{\phi(x)}
&=
\PD{x^\mu}{\phi'(x’)} \\
&=
\PD{{x’}^\nu}{\phi'(x’)}
\PD{{x}^\mu}{{x’}^\nu} \\
&=
\PD{{x’}^\nu}{\phi'(x’)}
\partial_\mu \lr{
{\Lambda^\nu}_\rho x^\rho
} \\
&=
\PD{{x’}^\nu}{\phi'(x’)}
{\Lambda^\nu}_\mu \\
&=
\PD{{x’}^\nu}{\phi(x)}
{\Lambda^\nu}_\mu.
\end{aligned}

Multiplying by the inverse $${\lr{\Lambda^{-1}}^\mu}_\kappa$$ we get
\label{eqn:qftLecture3:320}
\PD{{x’}^\kappa}{}
=
{\lr{\Lambda^{-1}}^\mu}_\kappa \PD{x^\mu}{}

This should be familiar to you, and is an analogue of the transformation of the
\label{eqn:qftLecture3:340}
=

## Actions

We will start with a classical action, and quantize to determine a QFT. In mechanics we have the particle position $$q(t)$$, which is a classical field in 1+0 time and space dimensions. Our action is
\label{eqn:qftLecture3:360}
S
= \int dt \LL(t)
= \int dt \lr{
\inv{2} \dot{q}^2 – V(q)
}.

This action depends on the position of the particle that is local in time. You could imagine that we have a more complex action where the action depends on future or past times
\label{eqn:qftLecture3:380}
S
= \int dt’ q(t’) K( t’ – t ),

but we don’t seem to find such actions in classical mechanics.

### Principles determining the form of the action.

• relativity (action is invariant under Lorentz transformation)
• locality (action depends on fields and the derivatives at given $$(t, \Bx)$$.
• Gauge principle (the action should be invariant under gauge transformation). We won’t discuss this in detail right now since we will start with studying scalar fields.
Recall that for Maxwell’s equations a gauge transformation has the form
\label{eqn:qftLecture3:520}
\phi \rightarrow \phi + \dot{\chi}, \BA \rightarrow \BA – \spacegrad \chi.

Suppose we have a real scalar field $$\phi(x)$$ where $$x \in \mathbb{R}^{1,d-1}$$. We will be integrating over space and time $$\int dt d^{d-1} x$$ which we will write as $$\int d^d x$$. Our action is
\label{eqn:qftLecture3:400}
S = \int d^d x \lr{ \text{Some action density to be determined } }

The analogue of $$\dot{q}^2$$ is
\label{eqn:qftLecture3:420}
\begin{aligned}
\lr{ \PD{x^\mu}{\phi} }
\lr{ \PD{x^\nu}{\phi} }
g^{\mu\nu}
&=
(\partial_\mu \phi) (\partial_\nu \phi) g^{\mu\nu} \\
&= \partial^\mu \phi \partial_\mu \phi.
\end{aligned}

This has both time and spatial components, that is
\label{eqn:qftLecture3:440}
\partial^\mu \phi \partial_\mu \phi =

so the desired simplest scalar action is
\label{eqn:qftLecture3:460}
S = \int d^d x \lr{ \dotphi^2 – (\spacegrad \phi)^2 }.

The measure transforms using a Jacobian, which we have seen is the Lorentz transform matrix, and has unit determinant
\label{eqn:qftLecture3:480}
d^d x’ = d^d x \Abs{ det( \Lambda^{-1} ) } = d^d x.

## Question: Four vector form of the Maxwell gauge transformation.

Show that the transformation
\label{eqn:qftLecture3:580}
A^\mu \rightarrow A^\mu + \partial^\mu \chi

is the desired four-vector form of the gauge transformation \ref{eqn:qftLecture3:520}, that is
\label{eqn:qftLecture3:540}
\begin{aligned}
j^\nu
&= \partial_\mu {F’}^{\mu\nu} \\
&= \partial_\mu F^{\mu\nu}.
\end{aligned}

Also relate this four-vector gauge transformation to the spacetime split.

\label{eqn:qftLecture3:560}
\begin{aligned}
\partial_\mu {F’}^{\mu\nu}
&=
\partial_\mu \lr{ \partial^\mu {A’}^\nu – \partial_\nu {A’}^\mu } \\
&=
\partial_\mu \lr{
\partial^\mu \lr{ A^\nu + \partial^\nu \chi }
– \partial_\nu \lr{ A^\mu + \partial^\mu \chi }
} \\
&=
\partial_\mu {F}^{\mu\nu}
+
\partial_\mu \partial^\mu \partial^\nu \chi

\partial_\mu \partial^\nu \partial^\mu \chi \\
&=
\partial_\mu {F}^{\mu\nu},
\end{aligned}

by equality of mixed partials. Expanding \ref{eqn:qftLecture3:580} explicitly we find
\label{eqn:qftLecture3:600}
{A’}^\mu = A^\mu + \partial^\mu \chi,

which is
\label{eqn:qftLecture3:620}
\begin{aligned}
\phi’ = {A’}^0 &= A^0 + \partial^0 \chi = \phi + \dot{\chi} \\
\BA’ \cdot \Be_k = {A’}^k &= A^k + \partial^k \chi = \lr{ \BA – \spacegrad \chi } \cdot \Be_k.
\end{aligned}

The last of which can be written in vector notation as $$\BA’ = \BA – \spacegrad \chi$$.

## UofT QFT Fall 2018 Lecture 2. Units, scales, and Lorentz transformations. Taught by Prof. Erich Poppitz

September 17, 2018 phy2403 No comments , , ,

## Natural units.

\label{eqn:qftLecture2:20}
\begin{aligned}
[\Hbar] &= [\text{action}] = M \frac{L^2}{T^2} T = \frac{M L^2}{T} \\
&= [\text{velocity}] = \frac{L}{T} \\
& [\text{energy}] = M \frac{L^2}{T^2}.
\end{aligned}

Setting $$c = 1$$ means
\label{eqn:qftLecture2:240}
\frac{L}{T} = 1

and setting $$\Hbar = 1$$ means
\label{eqn:qftLecture2:260}
[\Hbar] = [\text{action}] = M L {\frac{L}{T}} = M L

therefore
\label{eqn:qftLecture2:280}
[L] = \inv{\text{mass}}

and
\label{eqn:qftLecture2:300}
[\text{energy}] = M {\frac{L^2}{T^2}} = \text{mass}\, \text{eV}

Summary

• $$\text{energy} \sim \text{eV}$$
• $$\text{distance} \sim \inv{M}$$
• $$\text{time} \sim \inv{M}$$

From:
\label{eqn:qftLecture2:320}
\alpha = \frac{e^2}{4 \pi {\Hbar c}}

which is dimensionless ($$1/137$$), so electric charge is dimensionless.

Some useful numbers in natural units

\label{eqn:qftLecture2:40}
\begin{aligned}
m_\txte &\sim 10^{-27} \text{g} \sim 0.5 \text{MeV} \\
m_\txtp &\sim 2000 m_\txte \sim 1 \text{GeV} \\
m_\pi &\sim 140 \text{MeV} \\
m_\mu &\sim 105 \text{MeV} \\
\Hbar c &\sim 200 \text{MeV} \,\text{fm} = 1
\end{aligned}

## Gravity

Interaction energy of two particles

\label{eqn:qftLecture2:60}
G_\txtN \frac{m_1 m_2}{r}

\label{eqn:qftLecture2:80}
[\text{energy}] \sim [G_\txtN] \frac{M^2}{L}

\label{eqn:qftLecture2:100}
[G_\txtN]
\sim
[\text{energy}] \frac{L}{M^2}

but energy x distance is dimensionless (action) in our units

\label{eqn:qftLecture2:120}
[G_\txtN]
\sim
{\text{dimensionless}}{M^2}

\label{eqn:qftLecture2:140}
\frac{G_\txtN}{\Hbar c} \sim \inv{M^2} \sim \frac{1}{10^{20} \text{GeV}}

Planck mass

\label{eqn:qftLecture2:160}
M_{\text{Planck}} \sim \sqrt{\frac{\Hbar c}{G_\txtN}}
\sim 10^{-4} g \sim \inv{\lr{10^{20} \text{GeV}}^2}

We can revisit the scale diagram from last lecture in terms of MeV mass/energy values, as sketched in fig. 1.

fig. 1. Scales, take II.

At the classical electron radius scale, we consider phenomena such as back reaction of radiation, the self energy of electrons. At the Compton wavelength we have to allow for production of multiple particle pairs. At Bohr radius scales we must start using QM instead of classical mechanics.

## Cross section.

Verbal discussion of cross section, not captured in these notes. Roughly, the cross section sounds like the number of events per unit time, related to the flux of some source through an area.

We’ll compute the cross section of a number of different systems in this course. The cross section is relevant in scattering such as the electron-electron scattering sketched in fig. 2.

fig. 2. Electron electron scattering.

We assume that QED is highly relativistic. In natural units, our scale factor is basically the square of the electric charge
\label{eqn:qftLecture2:180}
\alpha \sim e^2,

so the cross section has the form
\label{eqn:qftLecture2:200}
\sigma \sim \frac{\alpha^2}{E^2} \lr{ 1 + O(\alpha) + O(\alpha^2) + \cdots }

In gravity we could consider scattering of electrons, where $$G_\txtN$$ takes the place of $$\alpha$$. However, $$G_\txtN$$ has dimensions.

For electron-electron scattering due to gravitons

\label{eqn:qftLecture2:220}
\sigma \sim \frac{G_\txtN^2 E^2}{1 + G_\txtN E^2 + \cdots }

Now the cross section grows with energy. This will cause some problems (violating unitarity: probabilities greater than 1!) when $$O(G_\txtN E^2) = 1$$.

In any quantum field theories when the coupling constant is not-dimensionless we have the same sort of problems at some scale.

The point is that we can get far considering just dimensional analysis.

If the coupling constant has a dimension $$(1/\text{mass})^N\,, N > 0$$, then unitarity will be violated at high energy. One such theory is the Fermi theory of beta decay (electro-weak theory), which had a coupling constant with dimensions inverse-mass-squared. The relevant scale for beta decay was 4 Fermi, or $$G_\txtF \sim (1/{100 \text{GeV}})^2$$. This was the motivation for introducing the Higgs theory, which was motivated by restoring unitarity.

## Lorentz transformations.

The goal, perhaps not for today, is to study the simplest (relativistic) scalar field theory. First studied classically, and then consider such a quantum field theory.

How is relativity implemented when we write the Lagrangian and action?

Our first step must be to consider Lorentz transformations and the Lorentz group.

Spacetime (Minkowski space) is \R{3,1} (or \R{d-1,1}). Our coordinates are

\label{eqn:qftLecture2:340}
(c t, x^1, x^2, x^3) = (c t, \Br).

Here, we’ve scaled the time scale by $$c$$ so that we measure time and space in the same dimensions. We write this as

\label{eqn:qftLecture2:360}
x^\mu = (x^0, x^1, x^2, x^3),

where $$\mu = 0, 1, 2, 3$$, and call this a “4-vector”. These are called the space-time coordinates of an event, which tell us where and when an event occurs.

For two events whose spacetime coordinates differ by $$dx^0, dx^1, dx^2, dx^3$$ we introduce the notion of a space time \underline{interval}

\label{eqn:qftLecture2:380}
\begin{aligned}
ds^2
&= c^2 dt^2
– (dx^1)^2
– (dx^2)^2
– (dx^3)^2 \\
&=
\sum_{\mu, \nu = 0}^3 g_{\mu\nu} dx^\mu dx^\nu
\end{aligned}

Here $$g_{\mu\nu}$$ is the Minkowski space metric, an object with two indexes that run from 0-3. i.e. this is a diagonal matrix

\label{eqn:qftLecture2:400}
g_{\mu\nu} \sim
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}

i.e.
\label{eqn:qftLecture2:420}
\begin{aligned}
g_{00} &= 1 \\
g_{11} &= -1 \\
g_{22} &= -1 \\
g_{33} &= -1 \\
\end{aligned}

We will use the Einstein summation convention, where any repeated upper and lower indexes are considered summed over. That is \ref{eqn:qftLecture2:380} is written with an implied sum
\label{eqn:qftLecture2:440}
ds^2 = g_{\mu\nu} dx^\mu dx^\nu.

Explicit expansion:
\label{eqn:qftLecture2:460}
\begin{aligned}
ds^2
&= g_{\mu\nu} dx^\mu dx^\nu \\
&=
g_{00} dx^0 dx^0
+g_{11} dx^1 dx^1
+g_{22} dx^2 dx^2
+g_{33} dx^3 dx^3
&=
(1) dx^0 dx^0
+ (-1) dx^1 dx^1
+ (-1) dx^2 dx^2
+ (-1) dx^3 dx^3.
\end{aligned}

Recall that rotations (with orthogonal matrix representations) are transformations that leave the dot product unchanged, that is

\label{eqn:qftLecture2:480}
\begin{aligned}
(R \Bx) \cdot (R \By)
&= \Bx^\T R^\T R \By \\
&= \Bx^\T \By \\
&= \Bx \cdot \By,
\end{aligned}

where $$R$$ is a rotation orthogonal 3×3 matrix. The set of such transformations that leave the dot product unchanged have orthonormal matrix representations $$R^\T R = 1$$. We call the set of such transformations that have unit determinant the SO(3) group.

We call a Lorentz transformation, if it is a linear transformation acting on 4 vectors that leaves the spacetime interval (i.e. the inner product of 4 vectors) invariant. That is, a transformation that leaves
\label{eqn:qftLecture2:500}
x^\mu y^\nu g_{\mu\nu} = x^0 y^0 – x^1 y^1 – x^2 y^2 – x^3 y^3

unchanged.

Suppose that transformation has a 4×4 matrix form

\label{eqn:qftLecture2:520}
{x’}^\mu = {\Lambda^\mu}_\nu x^\nu

For an example of a possible $$\Lambda$$, consider the transformation sketched in fig. 3.

fig. 3. Boost transformation.

We know that boost has the form
\label{eqn:qftLecture2:540}
\begin{aligned}
x &= \frac{x’ + v t’}{\sqrt{1 – v^2/c^2}} \\
y &= y’ \\
z &= z’ \\
t &= \frac{t’ + (v/c^2) x’}{\sqrt{1 – v^2/c^2}} \\
\end{aligned}

(this is a boost along the x-axis, not y as I’d drawn),
or
\label{eqn:qftLecture2:560}
\begin{bmatrix}
c t \\
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
\inv{\sqrt{1 – v^2/c^2}} & \frac{v/c}{\sqrt{1 – v^2/c^2}} & 0 & 0 \\
\frac{v/c}{\sqrt{1 – v^2/c^2}} & \frac{1}{\sqrt{1 – v^2/c^2}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
c t’ \\
x’ \\
y’ \\
z’
\end{bmatrix}

Other examples include rotations ($${\lambda^0}_0 = 1$$ zeros in $${\lambda^0}_k, {\lambda^k}_0$$, and a rotation matrix in the remainder.)

Back to Lorentz transformations ($$\text{SO}(1,3)^+$$), let
\label{eqn:qftLecture2:600}
\begin{aligned}
{x’}^\mu &= {\Lambda^\mu}_\nu x^\nu \\
{y’}^\kappa &= {\Lambda^\kappa}_\rho y^\rho
\end{aligned}

The dot product
\label{eqn:qftLecture2:620}
g_{\mu \kappa}
{x’}^\mu
{y’}^\kappa
=
g_{\mu \kappa}
{\Lambda^\mu}_\nu
{\Lambda^\kappa}_\rho
x^\nu
y^\rho
=
g_{\nu\rho}
x^\nu
y^\rho,

where the last step introduces the invariance requirement of the transformation. That is

\label{eqn:qftLecture2:640}
\boxed{
g_{\nu\rho}
=
g_{\mu \kappa}
{\Lambda^\mu}_\nu
{\Lambda^\kappa}_\rho.
}

### Upper and lower indexes

We’ve defined

\label{eqn:qftLecture2:660}
x^\mu = (t, x^1, x^2, x^3)

We could also define a four vector with lower indexes
\label{eqn:qftLecture2:680}
x_\nu = g_{\nu\mu} x^\mu = (t, -x^1, -x^2, -x^3).

That is
\label{eqn:qftLecture2:700}
\begin{aligned}
x_0 &= x^0 \\
x_1 &= -x^1 \\
x_2 &= -x^2 \\
x_3 &= -x^3.
\end{aligned}

which allows us to write the dot product as simply $$x^\mu y_\mu$$.

We can also define a metric tensor with upper indexes

\label{eqn:qftLecture2:401}
g^{\mu\nu} \sim
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}

This is the inverse matrix of $$g_{\mu\nu}$$, and it satisfies
\label{eqn:qftLecture2:720}
g^{\mu \nu} g_{\nu\rho} = {\delta^\mu}_\rho

Exercise: Check:
\label{eqn:qftLecture2:740}
\begin{aligned}
g_{\mu\nu} x^\mu y^\nu
&= x_\nu y^\nu \\
&= x^\nu y_\nu \\
&= g^{\mu\nu} x_\mu y_\nu \\
&= {\delta^\mu}_\nu x_\mu y^\nu.
\end{aligned}

Class ended around this point, but it appeared that we were heading this direction:

Returning to the Lorentz invariant and multiplying both sides of
\ref{eqn:qftLecture2:640} with an inverse Lorentz transformation $$\Lambda^{-1}$$, we find
\label{eqn:qftLecture2:760}
\begin{aligned}
g_{\nu\rho}
{\lr{\Lambda^{-1}}^\rho}_\alpha
&=
g_{\mu \kappa}
{\Lambda^\mu}_\nu
{\Lambda^\kappa}_\rho
{\lr{\Lambda^{-1}}^\rho}_\alpha \\
&=
g_{\mu \kappa}
{\Lambda^\mu}_\nu
{\delta^\kappa}_\alpha \\
&=
g_{\mu \alpha}
{\Lambda^\mu}_\nu,
\end{aligned}

or
\label{eqn:qftLecture2:780}
\lr{\Lambda^{-1}}_{\nu \alpha} = \Lambda_{\alpha \nu}.

This is clearly analogous to $$R^\T = R^{-1}$$, although the index notation obscures things considerably. Prof. Poppitz said that next week this would all lead to showing that the determinant of any Lorentz transformation was $$\pm 1$$.

For what it’s worth, it seems to me that this index notation makes life a lot harder than it needs to be, at least for a matrix related question (i.e. determinant of the transformation). In matrix/column-(4)-vector notation, let $$x’ = \Lambda x, y’ = \Lambda y$$ be two four vector transformations, then
\label{eqn:qftLecture2:800}
x’ \cdot y’ = {x’}^T G y’ = (\Lambda x)^T G \Lambda y = x^T ( \Lambda^T G \Lambda) y = x^T G y.

so
\label{eqn:qftLecture2:820}
\boxed{
\Lambda^T G \Lambda = G.
}

Taking determinants of both sides gives $$-(det(\Lambda))^2 = -1$$, and thus $$det(\Lambda) = \pm 1$$.

## UofT QFT Fall 2018 phy2403 ; Lecture 1, What is a field? Taught by Prof. Erich Poppitz

September 14, 2018 math and physics play No comments

DISCLAIMER: Very rough notes from class. Some additional side notes, but otherwise barely edited.

## What is a field?

A field is a map from space(time) to some set of numbers. These set of numbers may be organized some how, possibly scalars, or vectors, …

One example is the familiar spacetime vector, where $$\Bx \in \mathbb{R}^{d}$$

\label{eqn:qftLecture1:20}
(\Bx, t) \rightarrow \mathbb{R}^{\lr{d,1}}

Examples of fields:

1. $$0 + 1$$ dimensional “QFT”, where the spatial dimension is zero dimensional and we have one time dimension. Fields in this case are just functions of time $$x(t)$$. That is, particle mechanics is a 0 + 1 dimensional classical field theory. We know that classical mechanics is described by the action
\label{eqn:qftLecture1:40}
S = \frac{m}{2} \int dt \xdot^2.

This is non-relativistic. We can make this relativistic by saying this is the first order term in the Taylor expansion
\label{eqn:qftLecture1:60}
S = – m c^2 \int dt \sqrt{ 1 – \xdot^2/c^2 }.

Classical field theory (of $$x(t)$$). The “QFT” of $$x(t)$$. i.e. QM.
All of you know quantum mechanics. If you don’t just leave. Not this way (pointing to the window), but this way (pointing to the door).
The solution of a quantum mechanical state is
\label{eqn:qftLecture1:80}
\bra{x} e^{-i H t/\,\hbar } \ket{x’},

which can be found by evaluating the “Feynman path integral”
\label{eqn:qftLecture1:100}
\sum_{\text{all paths x}} e^{i S[x]/\,\hbar}

This will be particularly useful for QFT, despite the fact that such a sum is really hard to evaluate (try it for the Hydrogen atom for example).
2. $$3 + 0$$ dimensional field theory, where we have 3 spatial dimensions and 0 time dimensions. Classical equilibrium static systems. The field may have a structure like
\label{eqn:qftLecture1:120}
\Bx \rightarrow \BM(\Bx),

for example, magnetization.
We can write the solution to such a system using the partition function
\label{eqn:qftLecture1:140}
Z \sim \sum_{\text{all} \BM(x)} e^{-E[\BM]/\kB T}.

For such a system the energy function may be like
\label{eqn:qftLecture1:160}
E[\BM] = \int d^3 \Bx \lr{ a \BM^2(\Bx) + b \BM^4(\Bx) + c \sum_{i = 1}^3 \lr{ \PD{x_i}{} \BM }
\cdot \lr{ \PD{x_i}{} \BM }
}.

There is an analogy between the partition function and the Feynman path integral, as both are summing over all possible energy states in both cases.
This will be probably be the last time that we mention the partition function and condensed matter physics in this term for this class.
3. $$3 + 1$$ dimensional field theories, with 3 spatial dimensions and 1 time dimension.
Example, electromagnetism with $$\BE(\Bx, t), \BB(\Bx, t)$$ or better use $$\BA(\Bx, t), \phi(\Bx, t)$$. The action is
\label{eqn:qftLecture1:180}
S = -\inv{16 \pi c} \int d^3 \Bx dt \lr{ \BE^2 – \BB^2 }.

This is our first example of a relativistic field theory in $$3 + 1$$ dimensions. It will take us a while to get there.

These are examples of classical field theories, such as fluid dynamics and general relativity. We want to consider electromagnetism because this is the place that we everything starts to fall apart (i.e. blackbody radiation, relating to the equilibrium states of radiating matter). Part of the resolution of this was the quantization of the energy states, where we studied the normal modes of electromagnetic radiation in a box. These modes can be considered an infinite number of radiating oscillators (the ultraviolet catastrophe). This was resolved by Planck by requiring those energy states to be quantized (an excellent discussion of this can be found in [1]. In that sense you have already seen quantum field theory.

For electromagnetism the classical description is not always good. Examples:

2. electron energy $$e^2/r_\txte$$ of a point charge diverges as $$r_\txte \rightarrow 0$$.
We can define the classical radius of the electron by
\label{eqn:qftLecture1:200}
\frac{e^2}{r^{\textrm{cl}}_{\txte}} \sim m_\txte c^2,

or
\label{eqn:qftLecture1:220}
r^{\textrm{cl}}_{\txte} \sim \frac{m_\txte c^2}{e^2} \sim 10^{-15} \text{m}

Don’t treat this very seriously, but it becomes useful at frequencies $$\omega \sim c/r_\txte$$, where $$r_\txte/c$$ is approximately the time for light to cross a distance $$r_\txte$$.
At frequencies like this, we should not believe the solutions that are obtained by classical electrodynamics.
In particular, self-accelerating solutions appear at these frequencies in classical EM. This is approximately $$\omega_\conj \sim 10^{23} Hz$$, or
\label{eqn:qftLecture1:240}
\begin{aligned}
\,\hbar \omega_\conj
&\sim \lr{ 10^{-21} \,\text{MeV s}} \lr{ 10^{23} \,\text{1/s} }\\
&\sim 100 \text{MeV}.
\end{aligned}

At such frequencies particle creation becomes possible.

## Scales

A (dimensionless) value that is very useful in determining scale is
\label{eqn:qftLecture1:260}
\alpha = \frac{e^2}{4 \pi \,\hbar c} \sim \inv{137},

called the fine scale constant, which relates three important scales relevant to quantum mechanics, as sketched in fig. 1.

fig. 1. Interesting scales in quantum mechanics.

• The Bohr radius (large end of the scale).
• The Compton wavelength of the electron.
• The classical radius of the electron.

A quick motivation for the Bohr radius was mentioned in passing in class while discussing scale, following the high school method of deriving the Balmer series ([2]).

That method assumes a circular electron trajectory ($$i = \Be_1 \Be_2$$)
\label{eqn:qftLecture1:280}
\begin{aligned}
\Br &= r \Be_1 e^{i \omega t} \\
\Bv &= \omega r \Be_2 e^{i \omega t} \\
\Ba &= -\omega^2 r \Be_1 e^{i \omega t} \\
\end{aligned}

The Coulomb force (in cgs units) on the electron is
\label{eqn:qftLecture1:300}
\BF = m\Ba = -m \omega^2 r \Be_1 e^{i \omega t} = \frac{-e (e)}{r^2} \Be_1 e^{i \omega t},

or
\label{eqn:qftLecture1:320}
m \lr{ \frac{v}{r}}^2 r = \frac{e^2}{r^2},

giving
\label{eqn:qftLecture1:340}
m v^2 = \frac{e^2}{r}.

The energy of the system, including both Kinetic and potential (from an infinite reference point) is
\label{eqn:qftLecture1:360}
\begin{aligned}
E
&= \inv{2} m v^2 – \frac{e^2}{r} \\
&= – \inv{2} m v^2 \sim \,\hbar \omega = \,\hbar \frac{v}{r},
\end{aligned}

or
\label{eqn:qftLecture1:380}
m v r \sim \,\hbar.

Eliminating $$v$$ using \ref{eqn:qftLecture1:340}, assuming a ground state radius $$r = a_0$$ gives

\label{eqn:qftLecture1:400}
a_0 \sim \frac{\hbar^2}{m e^2}.

The Bohr radius is of the order $$10^{-10} \text{m}$$.

### Compton wavelength.

When particle momentum starts approaching the speed of light, by the uncertainty relation ($$\Delta x \Delta p \sim \,\hbar$$) the variation in position must be of the order
\label{eqn:qftLecture1:420}
\lambda_\txtc \sim \frac{\hbar}{m_\txte c},

called the Compton wavelength.
Similarly, when the length scales are reduced to the Compton wavelength, the momentum increases to relativistic levels.
Because of the relativistic velocities at the Compton wavelength, particle creation and annihilation occurs and any theory has to account for multiple particle states.

### Relations.

Scaling the Bohr radius once by the fine structure constant, we obtain the Compton wavelength (after dropping factors of $$4\pi$$)
\label{eqn:qftLecture1:440}
\begin{aligned}
a_0 \alpha
&= \frac{\hbar^2}{m e^2}
\frac{e^2}{4 \pi \,\hbar c} \\
&= \frac{\hbar}{4 \pi m c} \\
&\sim
\frac{\hbar}{m c} \\
&= \lambda_\txtc.
\end{aligned}

Scaling once more, we obtain (after dropping another $$4\pi$$) the classical electron radius
\label{eqn:qftLecture1:n}
\begin{aligned}
\lambda_\txtc \alpha
&=
\frac{e^2}{4 \pi m c^2} \\
&\sim
\frac{e^2}{m c^2}.
\end{aligned}

# References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

[2] A.P. French and E.F. Taylor. An Introduction to Quantum Physics. CRC Press, 1998.

## Applied vanity press

Amazon’s createspace turns out to be a very cost effective way to get a personal color copy of large pdf (>250 pages) to markup for review. The only hassle was having to use their app to create cover art (although that took less time than commuting downtown to one of the cheap copy shops near the university.)

As a side effect, after I edit it, I’d have something I could actually list for sale.  Worldwide, I’d guess at least three people would buy it, that is, if they weren’t happy with the pdf version already available.

## The book.

A draft of my book: Geometric Algebra for Electrical Engineers, is now available. I’ve supplied limited distribution copies of some of the early drafts and have had some good review comments of the chapter I (introduction to geometric algebra), and chapter II (multivector calculus) material, but none on the electromagnetism content. In defense of the reviewers, the initial version of the electromagnetism chapter, while it had a lot of raw content, was pretty exploratory and very rough. It’s been cleaned up significantly and is hopefully now more reader friendly.

## Why I wrote this book.

I have been working on a part time M.Eng degree for a number of years. I wasn’t happy with the UofT ECE electromagnetics offerings in recent years, which have been inconsistently offered or unsatisfactory.  For example: the microwave circuits course which sounded interesting, and had an interesting text book, was mind numbing, almost entirely about Smith charts.  I had to go elsewhere to obtain the M.Eng degree requirements. That elsewhere was a project course.

I proposed a project to an electromagnetism project with the following goals

1. Perform a literature review of applications of geometric algebra to the study of electromagnetism.
2. Identify the subset of the literature that had direct relevance to electrical engineering.
3. Create a complete, and as compact as possible, introduction to the prerequisites required for a graduate or advanced undergraduate electrical engineering student to be able to apply geometric algebra to problems in electromagnetism. With those prerequisites in place, work through the fundamentals of electromagnetism in a geometric algebra context.

In retrospect, doing this project was a mistake. I could have done this work outside of an academic context without paying so much (in both time and money). Somewhere along the way I lost track of the fact that I enrolled on the M.Eng to learn (it provided a way to take grad physics courses on a part time schedule), and got side tracked by degree requirements. Basically I fell victim to a “I may as well graduate” sentiment that would have been better to ignore. All that coupled with the fact that I did not actually get any feedback from my “supervisor”, who did not even read my work (at least so far after one year), made this project-course very frustrating. On the bright side, I really like what I produced, even if I had to do so in isolation.

## Why geometric algebra?

Geometric algebra generalizes vectors, providing algebraic representations of not just directed line segments, but also points, plane segments, volumes, and higher degree geometric objects (hypervolumes.). The geometric algebra representation of planes, volumes and hypervolumes requires a vector dot product, a vector multiplication operation, and a generalized addition operation. The dot product provides the length of a vector and a test for whether or not any two vectors are perpendicular. The vector multiplication operation is used to construct directed plane segments (bivectors), and directed volumes (trivectors), which are built from the respective products of two or three mutually perpendicular vectors. The addition operation allows for sums of scalars, vectors, or any products of vectors. Such a sum is called a multivector.

The power to add scalars, vectors, and products of vectors can be exploited to simplify much of electromagnetism. In particular, Maxwell’s equations for isotropic media can be merged into a single multivector equation
\label{eqn:quaternion2maxwellWithGA:20}
\lr{ \spacegrad + \inv{c} \PD{t}{}} \lr{ \BE + I c \BB } = \eta\lr{ c \rho – \BJ },

where $$\spacegrad$$ is the gradient, $$I = \Be_1 \Be_2 \Be_3$$ is the ordered product of the three R^3 basis vectors, $$c = 1/\sqrt{\mu\epsilon}$$ is the group velocity of the medium, $$\eta = \sqrt{\mu/\epsilon}$$, $$\BE, \BB$$ are the electric and magnetic fields, and $$\rho$$ and $$\BJ$$ are the charge and current densities. This can be written as a single equation
\label{eqn:ece2500report:40}
\lr{ \spacegrad + \inv{c} \PD{t}{}} F = J,

where $$F = \BE + I c \BB$$ is the combined (multivector) electromagnetic field, and $$J = \eta\lr{ c \rho – \BJ }$$ is the multivector current.

Encountering Maxwell’s equation in its geometric algebra form leaves the student with more questions than answers. Yes, it is a compact representation, but so are the tensor and differential forms (or even the quaternionic) representations of Maxwell’s equations. The student needs to know how to work with the representation if it is to be useful. It should also be clear how to use the existing conventional mathematical tools of applied electromagnetism, or how to generalize those appropriately. Individually, there are answers available to many of the questions that are generated attempting to apply the theory, but they are scattered and in many cases not easily accessible.

Much of the geometric algebra literature for electrodynamics is presented with a relativistic bias, or assumes high levels of mathematical or physics sophistication. The aim of this work was an attempt to make the study of electromagnetism using geometric algebra more accessible, especially to other dumb engineering undergraduates like myself. In particular, this project explored non-relativistic applications of geometric algebra to electromagnetism. The end product of this project was a fairly small self contained book, titled “Geometric Algebra for Electrical Engineers”. This book includes an introduction to Euclidean geometric algebra focused on R^2 and R^3 (64 pages), an introduction to geometric calculus and multivector Green’s functions (64 pages), applications to electromagnetism (82 pages), and some appendices. Many of the fundamental results of electromagnetism are derived directly from the multivector Maxwell’s equation, in a streamlined and compact fashion. This includes some new results, and many of the existing non-relativistic results from the geometric algebra literature. As a conceptual bridge, the book includes many examples of how to extract familiar conventional results from simpler multivector representations. Also included in the book are some sample calculations exploiting unique capabilities that geometric algebra provides. In particular, vectors in a plane may be manipulated much like complex numbers, which has a number of advantages over working with coordinates explicitly.

## Followup.

In many ways this work only scratches the surface. Many more worked examples, problems, figures and computer algebra listings should be added. In depth applications of derived geometric algebra relationships to problems customarily tackled with separate electric and magnetic field equations should also be incorporated. There are also theoretical holes, topics covered in any conventional introductory electromagnetism text, that are missing. Examples include the Fresnel relationships for transmission and reflection at an interface, in depth treatment of waveguides, dipole radiation and motion of charged particles, bound charges, and meta materials to name a few. Many of these topics can probably be handled in a coordinate free fashion using geometric algebra. Despite all the work that is required to help bridge the gap between formalism and application, making applied electromagnetism using geometric algebra truly accessible, it is my belief this book makes some good first steps down this path.

The choice that I made to completely avoid the geometric algebra space-time-algebra (STA) is somewhat unfortunate. It is exceedingly elegant, especially in a relativisitic context. Despite that, I think that this was still a good choice from a pedagogical point of view, as most of the prerequisites for an STA based study will have been taken care of as a side effect, making that study much more accessible.

## Potential solutions to the static Maxwell’s equation using geometric algebra

When neither the electromagnetic field strength $$F = \BE + I \eta \BH$$, nor current $$J = \eta (c \rho – \BJ) + I(c\rho_m – \BM)$$ is a function of time, then the geometric algebra form of Maxwell’s equations is the first order multivector (gradient) equation
\label{eqn:staticPotentials:20}

While direct solutions to this equations are possible with the multivector Green’s function for the gradient
\label{eqn:staticPotentials:40}
G(\Bx, \Bx’) = \inv{4\pi} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3 },

the aim in this post is to explore second order (potential) solutions in a geometric algebra context. Can we assume that it is possible to find a multivector potential $$A$$ for which
\label{eqn:staticPotentials:60}

is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\label{eqn:staticPotentials:80}

which can be easily solved using the scalar Green’s function for the Laplacian
\label{eqn:staticPotentials:240}
G(\Bx, \Bx’) = -\inv{\Norm{\Bx – \Bx’} },

a beastie that may be easier to convolve than the vector valued Green’s function for the gradient.

It is immediately clear that some restrictions must be imposed on the multivector potential $$A$$. In particular, since the field $$F$$ has only vector and bivector grades, this gradient must have no scalar, nor pseudoscalar grades. That is
\label{eqn:staticPotentials:100}

This constraint on the potential can be avoided if a grade selection operation is built directly into the assumed potential solution, requiring that the field is given by
\label{eqn:staticPotentials:120}

However, after imposing such a constraint, Maxwell’s equation has a much less friendly form
\label{eqn:staticPotentials:140}

Luckily, it is possible to introduce a transformation of potentials, called a gauge transformation, that eliminates the ugly grade selection term, and allows the potential equation to be expressed as a plain old Laplacian. We do so by assuming first that it is possible to find a solution of the Laplacian equation that has the desired grade restrictions. That is
\label{eqn:staticPotentials:160}
\begin{aligned}
\end{aligned}

for which $$F = \spacegrad A’$$ is a grade 1,2 solution to $$\spacegrad F = J$$. Suppose that $$A$$ is any formal solution, free of any grade restrictions, to $$\spacegrad^2 A = J$$, and $$F = \gpgrade{\spacegrad A}{1,2}$$. Can we find a function $$\tilde{A}$$ for which $$A = A’ + \tilde{A}$$?

Maxwell’s equation in terms of $$A$$ is
\label{eqn:staticPotentials:180}
\begin{aligned}
J
\end{aligned}

or
\label{eqn:staticPotentials:200}

This non-homogeneous Laplacian equation that can be solved as is for $$\tilde{A}$$ using the Green’s function for the Laplacian. Alternatively, we may also solve the equivalent first order system using the Green’s function for the gradient.
\label{eqn:staticPotentials:220}

Clearly $$\tilde{A}$$ is not unique, as we can add any function $$\psi$$ satisfying the homogeneous Laplacian equation $$\spacegrad^2 \psi = 0$$.

In summary, if $$A$$ is any multivector solution to $$\spacegrad^2 A = J$$, that is
\label{eqn:staticPotentials:260}
A(\Bx)
= \int dV’ G(\Bx, \Bx’) J(\Bx’)
= -\int dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} },

then $$F = \spacegrad A’$$ is a solution to Maxwell’s equation, where $$A’ = A – \tilde{A}$$, and $$\tilde{A}$$ is a solution to the non-homogeneous Laplacian equation or the non-homogeneous gradient equation above.

### Integral form of the gauge transformation.

Additional insight is possible by considering the gauge transformation in integral form. Suppose that
\label{eqn:staticPotentials:280}
A(\Bx) = -\int_V dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \tilde{A}(\Bx),

is a solution of $$\spacegrad^2 A = J$$, where $$\tilde{A}$$ is a multivector solution to the homogeneous Laplacian equation $$\spacegrad^2 \tilde{A} = 0$$. Let’s look at the constraints on $$\tilde{A}$$ that must be imposed for $$F = \spacegrad A$$ to be a valid (i.e. grade 1,2) solution of Maxwell’s equation.
\label{eqn:staticPotentials:300}
\begin{aligned}
F
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
&=
\int_{\partial V} dA’ \ncap’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
\end{aligned}

Where $$\ncap’ = (\Bx’ – \Bx)/\Norm{\Bx’ – \Bx}$$, and the fundamental theorem of geometric calculus has been used to transform the gradient volume integral into an integral over the bounding surface. Operating on Maxwell’s equation with the gradient gives $$\spacegrad^2 F = \spacegrad J$$, which has only grades 1,2 on the left hand side, meaning that $$J$$ is constrained in a way that requires $$\spacegrad J$$ to have only grades 1,2. This means that $$F$$ has grades 1,2 if
\label{eqn:staticPotentials:320}
= \int_{\partial V} dA’ \frac{ \gpgrade{\ncap’ J(\Bx’)}{0,3} }{\Norm{\Bx – \Bx’} }.

The product $$\ncap J$$ expands to
\label{eqn:staticPotentials:340}
\begin{aligned}
\ncap J
&=
&=
\ncap \cdot (-\eta \BJ) + \gpgradethree{\ncap (-I \BM)} \\
&=- \eta \ncap \cdot \BJ -I \ncap \cdot \BM,
\end{aligned}

so
\label{eqn:staticPotentials:360}
=
-\int_{\partial V} dA’ \frac{ \eta \ncap’ \cdot \BJ(\Bx’) + I \ncap’ \cdot \BM(\Bx’)}{\Norm{\Bx – \Bx’} }.

Observe that if there is no flux of current density $$\BJ$$ and (fictitious) magnetic current density $$\BM$$ through the surface, then $$F = \spacegrad A$$ is a solution to Maxwell’s equation without any gauge transformation. Alternatively $$F = \spacegrad A$$ is also a solution if $$\lim_{\Bx’ \rightarrow \infty} \BJ(\Bx’)/\Norm{\Bx – \Bx’} = \lim_{\Bx’ \rightarrow \infty} \BM(\Bx’)/\Norm{\Bx – \Bx’} = 0$$ and the bounding volume is taken to infinity.

# References

## Generalizing Ampere’s law using geometric algebra.

The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,

generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\label{eqn:flux:40}

where the multivector fields and currents are
\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}

Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.

My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,

where $$d^2 \Bx = d\Ba \wedge d\Bb$$ is a two parameter bivector valued surface, and $$\boldpartial$$ is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.

The gotcha is that we actually want a surface integral with $$\spacegrad F$$. We can split the gradient into the vector derivative a normal component
\label{eqn:flux:160}

so
\label{eqn:flux:100}
=
\int_A d^2 \Bx \boldpartial F
+
\int_A d^2 \Bx \ncap \lr{ \ncap \cdot \spacegrad } F,

so
\label{eqn:flux:120}
\begin{aligned}
\oint_{\partial A} d\Bx F
&=
\int_A d^2 \Bx \lr{ J – \ncap \lr{ \ncap \cdot \spacegrad } F } \\
&=
\int_A dA \lr{ I \ncap J – \lr{ \ncap \cdot \spacegrad } I F }
\end{aligned}

This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The $$d\Bx F$$ product has all possible grades, as does the $$d^2 \Bx J$$ product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\label{eqn:flux:140}
\begin{aligned}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}

Let’s expand these explicitly in terms of the component fields and densities to check against the conventional relationships, and see if things look right. The line integrand expands to
\label{eqn:flux:180}
\begin{aligned}
d\Bx F
&=
d\Bx \lr{ \BE + I \eta \BH }
=
d\Bx \cdot \BE + I \eta d\Bx \cdot \BH
+
d\Bx \wedge \BE + I \eta d\Bx \wedge \BH \\
&=
d\Bx \cdot \BE
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
+ I \eta (d\Bx \cdot \BH),
\end{aligned}

the current integrand expands to
\label{eqn:flux:200}
\begin{aligned}
I \ncap J
&=
I \ncap
\lr{
\frac{\rho}{\epsilon} – \eta \BJ + I \lr{ c \rho_\txtm – \BM }
} \\
&=
\ncap I \frac{\rho}{\epsilon} – \eta \ncap I \BJ – \ncap c \rho_\txtm + \ncap \BM \\
&=
\ncap \cdot \BM
+ \eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
– \eta I (\ncap \cdot \BJ).
\end{aligned}

We are left with
\label{eqn:flux:220}
\begin{aligned}
\oint_{\partial A}
\lr{
d\Bx \cdot \BE + I \eta (d\Bx \cdot \BH)
}
&=
\int_A dA
\lr{
\ncap \cdot \BM – \eta I (\ncap \cdot \BJ)
} \\
\oint_{\partial A}
\lr{
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
}
&=
\int_A dA
\lr{
\eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
-\PD{n}{} \lr{ I \BE – \eta \BH }
}.
\end{aligned}

This is a crazy mess of dots, crosses, fields and sources. We can split it into one equation for each grade, which will probably look a little more regular. That is
\label{eqn:flux:240}
\begin{aligned}
\oint_{\partial A} d\Bx \cdot \BE &= \int_A dA \ncap \cdot \BM \\
\oint_{\partial A} d\Bx \cross \BH
&=
\int_A dA
\lr{
– \ncap \cross \BJ
+ \frac{ \ncap \rho_\txtm }{\mu}
– \PD{n}{\BH}
} \\
\oint_{\partial A} d\Bx \cross \BE &=
\int_A dA
\lr{
\ncap \cross \BM
+ \frac{\ncap \rho}{\epsilon}
– \PD{n}{\BE}
} \\
\oint_{\partial A} d\Bx \cdot \BH &= -\int_A dA \ncap \cdot \BJ \\
\end{aligned}

The first and last equations could have been obtained much more easily from Maxwell’s equations in their conventional form more easily. The two cross product equations with the normal derivatives are not familiar to me, even without the fictitious magnetic sources. It is somewhat remarkable that so much can be packed into one multivector equation:
\label{eqn:flux:260}
\oint_{\partial A} d\Bx F
=
I \int_A dA \lr{ \ncap J – \PD{n}{F} }.

# References

[1] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

## Solving Maxwell’s equation in freespace: Multivector plane wave representation

The geometric algebra form of Maxwell’s equations in free space (or source free isotopic media with group velocity $$c$$) is the multivector equation
\label{eqn:planewavesMultivector:20}
\lr{ \spacegrad + \inv{c}\PD{t}{} } F(\Bx, t) = 0.

Here $$F = \BE + I c \BB$$ is a multivector with grades 1 and 2 (vector and bivector components). The velocity $$c$$ is called the group velocity since $$F$$, or its components $$\BE, \BH$$ satisfy the wave equation, which can be seen by pre-multiplying with $$\spacegrad – (1/c)\PDi{t}{}$$ to find
\label{eqn:planewavesMultivector:n}
\lr{ \spacegrad^2 – \inv{c^2}\PDSq{t}{} } F(\Bx, t) = 0.

Let’s look at the frequency domain solution of this equation with a presumed phasor representation
\label{eqn:planewavesMultivector:40}
F(\Bx, t) = \textrm{Re} \lr{ F(\Bk) e^{-j \Bk \cdot \Bx + j \omega t} },

where $$j$$ is a scalar imaginary, not necessarily with any geometric interpretation.

Maxwell’s equation reduces to just
\label{eqn:planewavesMultivector:60}
0
=
-j \lr{ \Bk – \frac{\omega}{c} } F(\Bk).

If $$F(\Bk)$$ has a left multivector factor
\label{eqn:planewavesMultivector:80}
F(\Bk) =
\lr{ \Bk + \frac{\omega}{c} } \tilde{F},

where $$\tilde{F}$$ is a multivector to be determined, then
\label{eqn:planewavesMultivector:100}
\begin{aligned}
\lr{ \Bk – \frac{\omega}{c} }
F(\Bk)
&=
\lr{ \Bk – \frac{\omega}{c} }
\lr{ \Bk + \frac{\omega}{c} } \tilde{F} \\
&=
\lr{ \Bk^2 – \lr{\frac{\omega}{c}}^2 } \tilde{F},
\end{aligned}

which is zero if $$\Norm{\Bk} = \ifrac{\omega}{c}$$.

Let $$\kcap = \ifrac{\Bk}{\Norm{\Bk}}$$, and $$\Norm{\Bk} \tilde{F} = F_0 + F_1 + F_2 + F_3$$, where $$F_0, F_1, F_2,$$ and $$F_3$$ are respectively have grades 0,1,2,3. Then
\label{eqn:planewavesMultivector:120}
\begin{aligned}
F(\Bk)
&= \lr{ 1 + \kcap } \lr{ F_0 + F_1 + F_2 + F_3 } \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap F_1 + \kcap F_2 + \kcap F_3 \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap \cdot F_1 + \kcap \cdot F_2 + \kcap \cdot F_3
+
\kcap \wedge F_1 + \kcap \wedge F_2 \\
&=
\lr{
F_0 + \kcap \cdot F_1
}
+
\lr{
F_1 + \kcap F_0 + \kcap \cdot F_2
}
+
\lr{
F_2 + \kcap \cdot F_3 + \kcap \wedge F_1
}
+
\lr{
F_3 + \kcap \wedge F_2
}.
\end{aligned}

Since the field $$F$$ has only vector and bivector grades, the grades zero and three components of the expansion above must be zero, or
\label{eqn:planewavesMultivector:140}
\begin{aligned}
F_0 &= – \kcap \cdot F_1 \\
F_3 &= – \kcap \wedge F_2,
\end{aligned}

so
\label{eqn:planewavesMultivector:160}
\begin{aligned}
F(\Bk)
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap \cdot F_1 +
F_2 – \kcap \wedge F_2
} \\
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap F_1 + \kcap \wedge F_1 +
F_2 – \kcap F_2 + \kcap \cdot F_2
}.
\end{aligned}

The multivector $$1 + \kcap$$ has the projective property of gobbling any leading factors of $$\kcap$$
\label{eqn:planewavesMultivector:180}
\begin{aligned}
(1 + \kcap)\kcap
&= \kcap + 1 \\
&= 1 + \kcap,
\end{aligned}

so for $$F_i \in F_1, F_2$$
\label{eqn:planewavesMultivector:200}
(1 + \kcap) ( F_i – \kcap F_i )
=
(1 + \kcap) ( F_i – F_i )
= 0,

leaving
\label{eqn:planewavesMultivector:220}
F(\Bk)
=
\lr{ 1 + \kcap } \lr{
\kcap \cdot F_2 +
\kcap \wedge F_1
}.

For $$\kcap \cdot F_2$$ to be non-zero $$F_2$$ must be a bivector that lies in a plane containing $$\kcap$$, and $$\kcap \cdot F_2$$ is a vector in that plane that is perpendicular to $$\kcap$$. On the other hand $$\kcap \wedge F_1$$ is non-zero only if $$F_1$$ has a non-zero component that does not lie in along the $$\kcap$$ direction, but $$\kcap \wedge F_1$$, like $$F_2$$ describes a plane that containing $$\kcap$$. This means that having both bivector and vector free variables $$F_2$$ and $$F_1$$ provide more degrees of freedom than required. For example, if $$\BE$$ is any vector, and $$F_2 = \kcap \wedge \BE$$, then
\label{eqn:planewavesMultivector:240}
\begin{aligned}
\lr{ 1 + \kcap }
\kcap \cdot F_2
&=
\lr{ 1 + \kcap }
\kcap \cdot \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\lr{
\BE

\kcap \lr{ \kcap \cdot \BE }
} \\
&=
\lr{ 1 + \kcap }
\kcap \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\kcap \wedge \BE,
\end{aligned}

which has the form $$\lr{ 1 + \kcap } \lr{ \kcap \wedge F_1 }$$, so the solution of the free space Maxwell’s equation can be written
\label{eqn:planewavesMultivector:260}
\boxed{
F(\Bx, t)
=
\textrm{Re} \lr{
\lr{ 1 + \kcap }
\BE\,
e^{-j \Bk \cdot \Bx + j \omega t}
}
,
}

where $$\BE$$ is any vector for which $$\BE \cdot \Bk = 0$$.