## Maxwell’s equations, from Lagrangian, for electric sources.

This is the 3rd part in a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a Lagrangian representation.

[Click here for a PDF version of this series of posts, up to and including this one.]  The first and second parts are also available here on this blog.

<h2>Continuing…</h2>

Given the Lagrangian density

\label{eqn:fsquared:500}
\LL = F \cdot F + a \lr{ A \cdot J },

we may derive Maxwell’s equations from it, fixing the constant $$a$$ by doing so. We can do this three different ways, with direct variation with respect to the field components $$A_\mu$$, using the Euler-Lagrange equations, or with direct variation with respect to $$A = \gamma^\mu A_\mu$$, as a single four-vector field variable.

Let’s try this first with direct variation using the coordinate expansion of $$A$$. The action is
\label{eqn:fsquared:520}
S = \int d^4 x \lr{ -\inv{2} F_{\mu\nu} F^{\mu\nu} + a J^\mu A_\mu }.

The variational principle requires the action variation to be zero for all $$\delta A_\mu$$, where $$\delta A_\mu = 0$$ on the boundaries of the space. That is
\label{eqn:fsquared:540}
\begin{aligned}
0 &= \delta S \\
&= \int d^4 x \lr{ -\inv{2} \lr{ \delta F_{\mu\nu} } F^{\mu\nu} -\inv{2} F_{\mu\nu} \delta F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \delta F_{\mu\nu} } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \delta \lr{ \partial_\mu A_\nu – \partial_\nu A_\mu } } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \partial_\mu \delta A_\nu – \partial_\nu \delta A_\mu } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – \lr{ \lr{ \partial_\mu \delta A_\nu } F^{\mu\nu} – \lr{ \partial_\mu \delta A_\nu } F^{\nu\mu} } + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – 2 \lr{ \partial_\mu \delta A_\nu } F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ – 2 \partial_\mu \lr{ \delta A_\nu F^{\mu\nu} } + 2 \delta A_\nu \partial_\mu F^{\mu\nu} + a J^\mu \delta A_\mu } \\
&= \int d^4 x \lr{ 2 \delta A_\nu \partial_\mu F^{\mu\nu} + a J^\nu \delta A_\nu } \\
&= \int d^4 x \lr{ 2 \partial_\mu F^{\mu\nu} + a J^\nu } \delta A_\nu.
\end{aligned}

We have all the usual types of index gymnastics above, and dropped the complete derivative term since $$\delta A_\nu$$ is zero on the boundary by definition. Since the end result must be zero for all variations, we must have
\label{eqn:fsquared:560}
0 = 2 \partial_\mu F^{\mu\nu} + a J^\nu.

We also determine our constant $$a = -2$$.

Now, let’s do the same calculation using the Euler-Lagrange equations. We derive those by varying a general Lagrangian density, just as above
\label{eqn:fsquared:580}
\begin{aligned}
0
&=
\delta S \\
&= \int d^4 x \delta \LL(A_\mu, \partial_\nu A_\mu) \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} \delta A_\mu + \PD{(\partial_\nu A_\mu)}{\LL} \delta \partial_\nu A_\mu } \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} \delta A_\mu + \PD{(\partial_\nu A_\mu)}{\LL} \partial_\nu \delta A_\mu } \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} \delta A_\mu
+ \partial_\nu \lr{ \PD{(\partial_\nu A_\mu)}{\LL} \delta A_\mu }
– \lr{ \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL} } \delta A_\mu
} \\
&= \int d^4 x \lr{ \PD{A_\mu}{\LL} – \lr{ \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL} } } \delta A_\mu.
\end{aligned}

Since this is zero for all variations $$\delta A_\mu$$, we find the field Euler-Lagrange equations are
\label{eqn:fsquared:600}
\PD{A_\mu}{\LL} = \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL} .

We should be able to re-derive Maxwell’s equations from the Lagrangian using these field Euler-Lagrange equations, with a bit less work, since we’ve pre-calculated some of the variation. Let’s try that. Since we now know the value of the constant $$a$$, our Lagrangian is
\label{eqn:fsquared:620}
\LL = -\inv{2} F_{\mu\nu} F^{\mu\nu} – 2 J^\mu A_\mu.

On the LHS we have
\label{eqn:fsquared:640}
\begin{aligned}
\PD{A_\mu}{\LL}
&=
\PD{A_\mu}{} \lr{ – 2 J^\nu A_\nu } \\
&=
– 2 J^\mu.
\end{aligned}

For the RHS, let’s first calculate
\label{eqn:fsquared:660}
\begin{aligned}
\PD{(\partial_\nu A_\mu)}{\LL}
&=
\PD{(\partial_\nu A_\mu)}{}
\lr{
-\inv{2} F_{\alpha\beta} F^{\alpha\beta}
} \\
&=

\lr{
\PD{(\partial_\nu A_\mu)}{}
F_{\alpha\beta}
}
F^{\alpha\beta}
\\
&=

\lr{
\PD{(\partial_\nu A_\mu)}{}
\lr{
\partial_\alpha A_\beta – \partial_\beta A_\alpha
}
}
F^{\alpha\beta}
\\
&=
– F^{\nu\mu}
+ F^{\mu\nu} \\
&=
– 2 F^{\nu\mu}
.
\end{aligned}

We are left with
\label{eqn:fsquared:680}
-2 \partial_\nu F^{\nu\mu} = -2 J^\mu.

This is the source portion of Maxwell’s equation (after canceling $$-2’s$$), as expected.

Now let’s perform a (mostly) coordinate free evaluation of the variation. We should be able to vary $$A$$ directly without first expanding it in coordinates.

We write the field as a curl
\label{eqn:fsquared:700}

For completeness sake, before continuing, since we’ve not already done so, we should verify that this is equivalent to the tensor expansion of $$F$$ that we have been using. We find that by expanding the gradient and the field in coordinates
\label{eqn:fsquared:720}
\begin{aligned}
F
&= \lr{ \gamma^\mu \partial_\mu } \wedge \lr{ \gamma^\nu A_\nu } \\
&= \lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu \\
&= \inv{2} \lr{
\lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu
+
\lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu
} \\
&= \inv{2} \lr{
\lr{ \gamma^\mu \wedge \gamma^\nu } \partial_\mu A_\nu
+
\lr{ \gamma^\nu \wedge \gamma^\mu } \partial_\nu A_\mu
} \\
&= \inv{2} \
\lr{ \gamma^\mu \wedge \gamma^\nu }
\lr{
\partial_\mu A_\nu – \partial_\nu A_\mu
} \\
&= \inv{2} \
\lr{ \gamma^\mu \wedge \gamma^\nu } F_{\mu\nu},
\end{aligned}

as claimed.

We want to expand the gradient portion of $$\grad \wedge A$$, but leave the field as is. That is
\label{eqn:fsquared:740}
\grad \wedge A = \gamma^\mu \wedge \partial_\mu A.

The scalar part of $$F^2$$ is therefore
\label{eqn:fsquared:760}
\begin{aligned}
F \cdot F
&=
\lr{ \gamma^\mu \wedge \partial_\mu A } \cdot \lr{ \gamma^\nu \wedge \partial_\nu A } \\
&=
\gamma^\mu \cdot \lr{ \partial_\mu A \cdot \lr{ \gamma^\nu \wedge \partial_\nu A } } \\
&=
\lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \partial_\nu A }

\lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\partial_\nu A) }.
\end{aligned}

Our Lagrangian is now fully specified in terms of $$A$$ and it’s derivatives.
\label{eqn:fsquared:780}
\LL =
\lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \partial_\nu A }

\lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\partial_\nu A) }
– 2 J \cdot A.

Observe the symmetry, with respect to index swap, in the first two terms. This means that the variation is just
\label{eqn:fsquared:800}
\begin{aligned}
\delta \LL
&=
2 \lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \delta \partial_\nu A }

2 \lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\delta \partial_\nu A) }
– 2 J \cdot \delta A
\\
&=
2 \lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \partial_\nu \delta A }

2 \lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot (\partial_\nu \delta A) }
– 2 J \cdot \delta A
\\
&=
2 \partial_\nu \lr{ \lr{ \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \delta A } }
– 2 \partial_\nu \lr{ \lr{ \gamma^\mu \cdot \gamma^\nu } \lr{ (\partial_\mu A) \cdot \delta A } } \\
-2 \lr{ \partial_\nu \gamma^\nu \cdot \partial_\mu A } \lr{ \gamma^\mu \cdot \delta A }
+ 2 \lr{ \gamma^\mu \cdot \partial_\nu \gamma^\nu } \lr{ (\partial_\mu A) \cdot \delta A }
– 2 J \cdot \delta A \\
&=
2 (\delta A) \cdot \lr{
– \lr{ \grad \cdot \partial_\mu A } \gamma^\mu
+ \lr{ \gamma^\mu \cdot \grad } \partial_\mu A
– J
} \\
&=
2 (\delta A) \cdot \lr{
\grad \cdot \lr{ \gamma^\mu \wedge \partial_\mu A } – J
} \\
&=
2 (\delta A) \cdot \lr{
}.
\end{aligned}

The complete derivative term above was dropped, leaving us with the source part of Maxwell’s equation
\label{eqn:fsquared:n}

It makes sense that we should not have to resort to coordinates, and sure enough, we are able to avoid doing so.

There’s more to do that we will tackle in subsequent posts. Questions include, how do we express the Euler-Lagrange equations without resorting to coordinates? We also want to tackle the Lagrangian with magnetic source contributions.

## Square of electrodynamic field.

The electrodynamic Lagrangian (without magnetic sources) has the form
\label{eqn:fsquared:20}
\LL = F \cdot F + \alpha A \cdot J,

where $$\alpha$$ is a constant that depends on the unit system.
My suspicion is that one or both of the bivector or quadvector grades of $$F^2$$ are required for Maxwell’s equation with magnetic sources.

Let’s expand out $$F^2$$ in coordinates, as preparation for computing the Euler-Lagrange equations. The scalar and pseudoscalar components both simplify easily into compact relationships, but the bivector term is messier. We start with the coordinate expansion of our field, which we may write in either upper or lower index form
\label{eqn:fsquared:40}
F = \inv{2} \gamma_\mu \wedge \gamma_\nu F^{\mu\nu}
= \inv{2} \gamma^\mu \wedge \gamma^\nu F_{\mu\nu}.

The square is
\label{eqn:fsquared:60}
F^2 = F \cdot F + \gpgradetwo{F^2} + F \wedge F.

Let’s compute the scalar term first. We need to make a change of dummy indexes, for one of the $$F$$’s. It will also be convenient to use upper indexes in one factor, and lowers in the other. We find
\label{eqn:fsquared:80}
\begin{aligned}
F \cdot F
&=
\inv{4}
\lr{ \gamma_\mu \wedge \gamma_\nu } \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta }
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\lr{
{\delta_\nu}^\alpha {\delta_\mu}^\beta
– {\delta_\mu}^\alpha {\delta_\nu}^\beta
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\lr{
F^{\mu\nu} F_{\nu\mu}

F^{\mu\nu} F_{\mu\nu}
} \\
&=
-\inv{2}
F^{\mu\nu} F_{\mu\nu}.
\end{aligned}

Now, let’s compute the pseudoscalar component of $$F^2$$. This time we uniformly use upper index components for the tensor, and find
\label{eqn:fsquared:100}
\begin{aligned}
F \wedge F
&=
\inv{4}
\lr{ \gamma_\mu \wedge \gamma_\nu } \wedge \lr{ \gamma_\alpha \wedge \gamma_\beta }
F^{\mu\nu}
F^{\alpha\beta} \\
&=
\frac{I}{4}
\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta},
\end{aligned}

where $$\epsilon_{\mu\nu\alpha\beta}$$ is the completely antisymmetric (Levi-Civita) tensor of rank four. This pseudoscalar components picks up all the products of components of $$F$$ where all indexes are different.

Now, let’s try computing the bivector term of the product. This will require fancier index gymnastics.
\label{eqn:fsquared:120}
\begin{aligned}
&=
\inv{4}
\lr{ \gamma_\mu \wedge \gamma_\nu } \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
\inv{4}
\gamma_\mu \gamma_\nu \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
F^{\mu\nu}
F_{\alpha\beta}

\inv{4}
\lr{ \gamma_\mu \cdot \gamma_\nu} \lr{ \gamma^\alpha \wedge \gamma^\beta } F^{\mu\nu} F_{\alpha\beta}.
\end{aligned}

The dot product term is killed, since $$\lr{ \gamma_\mu \cdot \gamma_\nu} F^{\mu\nu} = g_{\mu\nu} F^{\mu\nu}$$ is the contraction of a symmetric tensor with an antisymmetric tensor. We can now proceed to expand the grade two selection
\label{eqn:fsquared:140}
\begin{aligned}
\gamma_\mu \gamma_\nu \lr{ \gamma^\alpha \wedge \gamma^\beta }
}
&=
\gamma_\mu \wedge \lr{ \gamma_\nu \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta } }
+
\gamma_\mu \cdot \lr{ \gamma_\nu \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } } \\
&=
\gamma_\mu \wedge
\lr{
{\delta_\nu}^\alpha \gamma^\beta

{\delta_\nu}^\beta \gamma^\alpha
}
+
g_{\mu\nu} \lr{ \gamma^\alpha \wedge \gamma^\beta }

{\delta_\mu}^\alpha \lr{ \gamma_\nu \wedge \gamma^\beta }
+
{\delta_\mu}^\beta \lr{ \gamma_\nu \wedge \gamma^\alpha } \\
&=
{\delta_\nu}^\alpha \lr{ \gamma_\mu \wedge \gamma^\beta }

{\delta_\nu}^\beta \lr{ \gamma_\mu \wedge \gamma^\alpha }

{\delta_\mu}^\alpha \lr{ \gamma_\nu \wedge \gamma^\beta }
+
{\delta_\mu}^\beta \lr{ \gamma_\nu \wedge \gamma^\alpha }.
\end{aligned}

Observe that I’ve taken the liberty to drop the $$g_{\mu\nu}$$ term. Strictly speaking, this violated the equality, but won’t matter since we will contract this with $$F^{\mu\nu}$$. We are left with
\label{eqn:fsquared:160}
\begin{aligned}
&=
\lr{
{\delta_\nu}^\alpha \lr{ \gamma_\mu \wedge \gamma^\beta }

{\delta_\nu}^\beta \lr{ \gamma_\mu \wedge \gamma^\alpha }

{\delta_\mu}^\alpha \lr{ \gamma_\nu \wedge \gamma^\beta }
+
{\delta_\mu}^\beta \lr{ \gamma_\nu \wedge \gamma^\alpha }
}
F^{\mu\nu}
F_{\alpha\beta} \\
&=
F^{\mu\nu}
\lr{
\lr{ \gamma_\mu \wedge \gamma^\alpha }
F_{\nu\alpha}

\lr{ \gamma_\mu \wedge \gamma^\alpha }
F_{\alpha\nu}

\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\mu\alpha}
+
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\alpha\mu}
} \\
&=
2 F^{\mu\nu}
\lr{
\lr{ \gamma_\mu \wedge \gamma^\alpha }
F_{\nu\alpha}
+
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\alpha\mu}
} \\
&=
2 F^{\nu\mu}
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\mu\alpha}
+
2 F^{\mu\nu}
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F_{\alpha\mu},
\end{aligned}

which leaves us with
\label{eqn:fsquared:180}
=
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F^{\mu\nu}
F_{\alpha\mu}.

I suspect that there must be an easier way to find this result.

We now have the complete coordinate expansion of $$F^2$$, separated by grade
\label{eqn:fsquared:200}
F^2 =
-\inv{2}
F^{\mu\nu} F_{\mu\nu}
+
\lr{ \gamma_\nu \wedge \gamma^\alpha }
F^{\mu\nu}
F_{\alpha\mu}
+
\frac{I}{4}
\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta}.

Tomorrow’s task is to start evaluating the Euler-Lagrange equations for this multivector Lagrangian density, and see what we get.

## More satisfying editing of classical mechanics notes.

I’ve purged about 30 pages of material related to field Lagrangian densities and Maxwell’s equation, replacing it with about 8 pages of new less incoherent material.

As before, I’ve physically ripped out all the pages that have been replaced, which is satisfying, and makes it easier to see what is left to review.

The new version is now reduced to 333 pages, close to a 100 page reduction from the original mess.  I may print myself a new physical copy, as I’ve moved things around so much that I have to search the latex to figure out where to make changes.

## Maxwell’s equation Lagrangian (geometric algebra and tensor formalism)

Maxwell’s equation using geometric algebra Lagrangian.

## Motivation.

In my classical mechanics notes, I’ve got computations of Maxwell’s equation (singular in it’s geometric algebra form) from a Lagrangian in various ways (using a tensor, scalar and multivector Lagrangians), but all of these seem more convoluted than they should be.
Here we do this from scratch, starting with the action principle for field variables, covering:

• Derivation of the relativistic form of the Euler-Lagrange field equations from the covariant form of the action,
• Derivation of Maxwell’s equation (in it’s STA form) from the Maxwell Lagrangian,
• Relationship of the STA Maxwell Lagrangian to the tensor equivalent,
• Relationship of the STA form of Maxwell’s equation to it’s tensor equivalents,
• Relationship of the STA Maxwell’s equation to it’s conventional Gibbs form.
• Show that we may use a multivector valued Lagrangian with all of $$F^2$$, not just the scalar part.

It is assumed that the reader is thoroughly familiar with the STA formalism, and if that is not the case, there is no better reference than [1].

## Theorem 1.1: Relativistic Euler-Lagrange field equations.

Let $$\phi \rightarrow \phi + \delta \phi$$ be any variation of the field, such that the variation
$$\delta \phi = 0$$ vanishes at the boundaries of the action integral
\label{eqn:maxwells:2120}
S = \int d^4 x \LL(\phi, \partial_\nu \phi).

The extreme value of the action is found when the Euler-Lagrange equations
\label{eqn:maxwells:2140}
0 = \PD{\phi}{\LL} – \partial_\nu \PD{(\partial_\nu \phi)}{\LL},

are satisfied. For a Lagrangian with multiple field variables, there will be one such equation for each field.

### Start proof:

To ease the visual burden, designate the variation of the field by $$\delta \phi = \epsilon$$, and perform a first order expansion of the varied Lagrangian
\label{eqn:maxwells:20}
\begin{aligned}
\LL
&\rightarrow
\LL(\phi + \epsilon, \partial_\nu (\phi + \epsilon)) \\
&=
\LL(\phi, \partial_\nu \phi)
+
\PD{\phi}{\LL} \epsilon +
\PD{(\partial_\nu \phi)}{\LL} \partial_\nu \epsilon.
\end{aligned}

The variation of the Lagrangian is
\label{eqn:maxwells:40}
\begin{aligned}
\delta \LL
&=
\PD{\phi}{\LL} \epsilon +
\PD{(\partial_\nu \phi)}{\LL} \partial_\nu \epsilon \\
&=
\PD{\phi}{\LL} \epsilon +
\partial_\nu \lr{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }

\epsilon \partial_\nu \PD{(\partial_\nu \phi)}{\LL},
\end{aligned}

which we may plug into the action integral to find
\label{eqn:maxwells:60}
\delta S
=
\int d^4 x \epsilon \lr{
\PD{\phi}{\LL}

\partial_\nu \PD{(\partial_\nu \phi)}{\LL}
}
+
\int d^4 x
\partial_\nu \lr{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }.

The last integral can be evaluated along the $$dx^\nu$$ direction, leaving
\label{eqn:maxwells:80}
\int d^3 x
\evalbar{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }{\Delta x^\nu},

where $$d^3 x = dx^\alpha dx^\beta dx^\gamma$$ is the product of differentials that does not include $$dx^\nu$$. By construction, $$\epsilon$$ vanishes on the boundary of the action integral so \ref{eqn:maxwells:80} is zero. The action takes its extreme value when
\label{eqn:maxwells:100}
0 = \delta S
=
\int d^4 x \epsilon \lr{
\PD{\phi}{\LL}

\partial_\nu \PD{(\partial_\nu \phi)}{\LL}
}.

The proof is complete after noting that this must hold for all variations of the field $$\epsilon$$, which means that we must have
\label{eqn:maxwells:120}
0 =
\PD{\phi}{\LL}

\partial_\nu \PD{(\partial_\nu \phi)}{\LL}.

### End proof.

Armed with the Euler-Lagrange equations, we can apply them to the Maxwell’s equation Lagrangian, which we will claim has the following form.

## Theorem 1.2: Maxwell’s equation Lagrangian.

Application of the Euler-Lagrange equations to the Lagrangian
\label{eqn:maxwells:2160}
\LL = – \frac{\epsilon_0 c}{2} F \cdot F + J \cdot A,

where $$F = \grad \wedge A$$, yields the vector portion of Maxwell’s equation
\label{eqn:maxwells:2180}
\grad \cdot F = \inv{\epsilon_0 c} J,

which implies
\label{eqn:maxwells:2200}
\grad F = \inv{\epsilon_0 c} J.

This is Maxwell’s equation.

### Start proof:

We wish to apply all of the Euler-Lagrange equations simultaneously (i.e. once for each of the four $$A_\mu$$ components of the potential), and cast it into four-vector form
\label{eqn:maxwells:140}
0 = \gamma_\nu \lr{ \PD{A_\nu}{} – \partial_\mu \PD{(\partial_\mu A_\nu)}{} } \LL.

Since our Lagrangian splits nicely into kinetic and interaction terms, this gives us
\label{eqn:maxwells:160}
0 = \gamma_\nu \lr{ \PD{A_\nu}{(A \cdot J)} + \frac{\epsilon_0 c}{2} \partial_\mu \PD{(\partial_\mu A_\nu)}{ (F \cdot F)} }.

The interaction term above is just
\label{eqn:maxwells:180}
\gamma_\nu \PD{A_\nu}{(A \cdot J)}
=
\gamma_\nu \PD{A_\nu}{(A_\mu J^\mu)}
=
\gamma_\nu J^\nu
=
J,

but the kinetic term takes a bit more work. Let’s start with evaluating
\label{eqn:maxwells:200}
\begin{aligned}
\PD{(\partial_\mu A_\nu)}{ (F \cdot F)}
&=
\PD{(\partial_\mu A_\nu)}{ F } \cdot F
+
F \cdot \PD{(\partial_\mu A_\nu)}{ F } \\
&=
2 \PD{(\partial_\mu A_\nu)}{ F } \cdot F \\
&=
2 \PD{(\partial_\mu A_\nu)}{ (\partial_\alpha A_\beta) } \lr{ \gamma^\alpha \wedge \gamma^\beta } \cdot F \\
&=
2 \lr{ \gamma^\mu \wedge \gamma^\nu } \cdot F.
\end{aligned}

We hit this with the $$\mu$$-partial and expand as a scalar selection to find
\label{eqn:maxwells:220}
\begin{aligned}
\partial_\mu \PD{(\partial_\mu A_\nu)}{ (F \cdot F)}
&=
2 \lr{ \partial_\mu \gamma^\mu \wedge \gamma^\nu } \cdot F \\
&=
– 2 (\gamma^\nu \wedge \grad) \cdot F \\
&=
&=
&=
– 2 \gamma^\nu \cdot \lr{ \grad \cdot F }.
\end{aligned}

Putting all the pieces together yields
\label{eqn:maxwells:240}
0
= J – \epsilon_0 c \gamma_\nu \lr{ \gamma^\nu \cdot \lr{ \grad \cdot F } }
= J – \epsilon_0 c \lr{ \grad \cdot F },

but
\label{eqn:maxwells:260}
\begin{aligned}
&=
&=
&=
\end{aligned}

so the multivector field equations for this Lagrangian are
\label{eqn:maxwells:280}
\grad F = \inv{\epsilon_0 c} J,

as claimed.

## Problem: Correspondence with tensor formalism.

Cast the Lagrangian of \ref{eqn:maxwells:2160} into the conventional tensor form
\label{eqn:maxwells:300}
\LL = \frac{\epsilon_0 c}{4} F_{\mu\nu} F^{\mu\nu} + A^\mu J_\mu.

Also show that the four-vector component of Maxwell’s equation $$\grad \cdot F = J/(\epsilon_0 c)$$ is equivalent to the conventional tensor form of the Gauss-Ampere law
\label{eqn:maxwells:320}
\partial_\mu F^{\mu\nu} = \inv{\epsilon_0 c} J^\nu,

where $$F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu$$ as usual. Also show that the trivector component of Maxwell’s equation $$\grad \wedge F = 0$$ is equivalent to the tensor form of the Gauss-Faraday law
\label{eqn:maxwells:340}
\partial_\alpha \lr{ \epsilon^{\alpha \beta \mu \nu} F_{\mu\nu} } = 0.

To show the Lagrangian correspondence we must expand $$F \cdot F$$ in coordinates
\label{eqn:maxwells:360}
\begin{aligned}
F \cdot F
&=
( \grad \wedge A ) \cdot
( \grad \wedge A ) \\
&=
\lr{ (\gamma^\mu \partial_\mu) \wedge (\gamma^\nu A_\nu) }
\cdot
\lr{ (\gamma^\alpha \partial_\alpha) \wedge (\gamma^\beta A_\beta) } \\
&=
\lr{ \gamma^\mu \wedge \gamma^\nu } \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta }
(\partial_\mu A_\nu )
(\partial^\alpha A^\beta ) \\
&=
\lr{
{\delta^\mu}_\beta
{\delta^\nu}_\alpha

{\delta^\mu}_\alpha
{\delta^\nu}_\beta
}
(\partial_\mu A_\nu )
(\partial^\alpha A^\beta ) \\
&=
– \partial_\mu A_\nu \lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
} \\
&=
– \partial_\mu A_\nu F^{\mu\nu} \\
&=
– \inv{2} \lr{
\partial_\mu A_\nu F^{\mu\nu}
+
\partial_\nu A_\mu F^{\nu\mu}
} \\
&=
– \inv{2} \lr{
\partial_\mu A_\nu

\partial_\nu A_\mu
}
F^{\mu\nu} \\
&=

\inv{2}
F_{\mu\nu}
F^{\mu\nu}.
\end{aligned}

With a substitution of this and $$A \cdot J = A_\mu J^\mu$$ back into the Lagrangian, we recover the tensor form of the Lagrangian.

To recover the tensor form of Maxwell’s equation, we first split it into vector and trivector parts
\label{eqn:maxwells:1580}

Now the vector component may be expanded in coordinates by dotting both sides with $$\gamma^\nu$$ to find
\label{eqn:maxwells:1600}
\inv{\epsilon_0 c} \gamma^\nu \cdot J = J^\nu,

and
\label{eqn:maxwells:1620}
\begin{aligned}
\gamma^\nu \cdot
&=
\partial_\mu \gamma^\nu \cdot \lr{ \gamma^\mu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta } \partial^\alpha A^\beta } \\
&=
\lr{
{\delta^\mu}_\alpha
{\delta^\nu}_\beta

{\delta^\nu}_\alpha
{\delta^\mu}_\beta
}
\partial_\mu
\partial^\alpha A^\beta \\
&=
\partial_\mu
\lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
} \\
&=
\partial_\mu F^{\mu\nu}.
\end{aligned}

Equating \ref{eqn:maxwells:1600} and \ref{eqn:maxwells:1620} finishes the first part of the job. For the trivector component, we have
\label{eqn:maxwells:1640}
0
= (\gamma^\mu \partial_\mu) \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } \partial_\alpha A_\beta
= \inv{2} (\gamma^\mu \partial_\mu) \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } F_{\alpha \beta}.

Wedging with $$\gamma^\tau$$ and then multiplying by $$-2 I$$ we find
\label{eqn:maxwells:1660}
0 = – \lr{ \gamma^\mu \wedge \gamma^\alpha \wedge \gamma^\beta \wedge \gamma^\tau } I \partial_\mu F_{\alpha \beta},

but
\label{eqn:maxwells:1680}
\gamma^\mu \wedge \gamma^\alpha \wedge \gamma^\beta \wedge \gamma^\tau = -I \epsilon^{\mu \alpha \beta \tau},

which leaves us with
\label{eqn:maxwells:1700}
\epsilon^{\mu \alpha \beta \tau} \partial_\mu F_{\alpha \beta} = 0,

as expected.

## Problem: Correspondence of tensor and Gibbs forms of Maxwell’s equations.

Given the identifications

\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,

and
\label{eqn:lorentzForceCovariant:1520}
F^{rs} = -\epsilon^{rst} B^t,

and
\label{eqn:maxwells:1560}
J^\mu = \lr{ c \rho, \BJ },

the reader should satisfy themselves that the traditional Gibbs form of Maxwell’s equations can be recovered from \ref{eqn:maxwells:320}.

The reader is referred to Exercise 3.4 “Electrodynamics, variational principle.” from [2].

## Problem: Correspondence with grad and curl form of Maxwell’s equations.

With $$J = c \rho \gamma_0 + J^k \gamma_k$$ and $$F = \BE + I c \BB$$ show that Maxwell’s equation, as stated in \ref{eqn:maxwells:2200} expand to the conventional div and curl expressions for Maxwell’s equations.

To obtain Maxwell’s equations in their traditional vector forms, we pre-multiply both sides with $$\gamma_0$$
\label{eqn:maxwells:1720}
\gamma_0 \grad F = \inv{\epsilon_0 c} \gamma_0 J,

and then select each grade separately. First observe that the RHS above has scalar and bivector components, as
\label{eqn:maxwells:1740}
\gamma_0 J
=
c \rho + J^k \gamma_0 \gamma_k.

In terms of the spatial bivector basis $$\Be_k = \gamma_k \gamma_0$$, the RHS of \ref{eqn:maxwells:1720} is
\label{eqn:maxwells:1760}
\gamma_0 \frac{J}{\epsilon_0 c} = \frac{\rho}{\epsilon_0} – \mu_0 c \BJ.

For the LHS, first note that
\label{eqn:maxwells:1780}
\begin{aligned}
&=
\gamma_0
\lr{
\gamma_0 \partial^0 +
\gamma_k \partial^k
} \\
&=
\partial_0 – \gamma_0 \gamma_k \partial_k \\
&=
\end{aligned}

We can express all the the LHS of \ref{eqn:maxwells:1720} in the bivector spatial basis, so that Maxwell’s equation in multivector form is
\label{eqn:maxwells:1800}
\lr{ \inv{c} \PD{t}{} + \spacegrad } \lr{ \BE + I c \BB } = \frac{\rho}{\epsilon_0} – \mu_0 c \BJ.

Selecting the scalar, vector, bivector, and trivector grades of both sides (in the spatial basis) gives the following set of respective equations
\label{eqn:maxwells:1840}

\label{eqn:maxwells:1860}
\inv{c} \partial_t \BE + I c \spacegrad \wedge \BB = – \mu_0 c \BJ

\label{eqn:maxwells:1880}
\spacegrad \wedge \BE + I \partial_t \BB = 0

\label{eqn:maxwells:1900}
I c \spacegrad \cdot B = 0,

which we can rewrite after some duality transformations (and noting that $$\mu_0 \epsilon_0 c^2 = 1$$), we have
\label{eqn:maxwells:1940}

\label{eqn:maxwells:1960}
\spacegrad \cross \BB – \mu_0 \epsilon_0 \PD{t}{\BE} = \mu_0 \BJ

\label{eqn:maxwells:1980}
\spacegrad \cross \BE + \PD{t}{\BB} = 0

\label{eqn:maxwells:2000}

which are Maxwell’s equations in their traditional form.

## Problem: Alternative multivector Lagrangian.

Show that a scalar+pseudoscalar Lagrangian of the following form
\label{eqn:maxwells:2220}
\LL = – \frac{\epsilon_0 c}{2} F^2 + J \cdot A,

which omits the scalar selection of the Lagrangian in \ref{eqn:maxwells:2160}, also represents Maxwell’s equation. Discuss the scalar and pseudoscalar components of $$F^2$$, and show why the pseudoscalar inclusion is irrelevant.

The quantity $$F^2 = F \cdot F + F \wedge F$$ has both scalar and pseudoscalar
components. Note that unlike vectors, a bivector wedge in 4D with itself need not be zero (example: $$\gamma_0 \gamma_1 + \gamma_2 \gamma_3$$ wedged with itself).
We can see this multivector nature nicely by expansion in terms of the electric and magnetic fields
\label{eqn:maxwells:2020}
\begin{aligned}
F^2
&= \lr{ \BE + I c \BB }^2 \\
&= \BE^2 – c^2 \BB^2 + I c \lr{ \BE \BB + \BB \BE } \\
&= \BE^2 – c^2 \BB^2 + 2 I c \BE \cdot \BB.
\end{aligned}

Both the scalar and pseudoscalar parts of $$F^2$$ are Lorentz invariant, a requirement of our Lagrangian, but most Maxwell equation Lagrangians only include the scalar $$\BE^2 – c^2 \BB^2$$ component of the field square. If we allow the Lagrangian to be multivector valued, and evaluate the Euler-Lagrange equations, we quickly find the same results
\label{eqn:maxwells:2040}
\begin{aligned}
0
&= \gamma_\nu \lr{ \PD{A_\nu}{} – \partial_\mu \PD{(\partial_\mu A_\nu)}{} } \LL \\
&= \gamma_\nu \lr{ J^\nu + \frac{\epsilon_0 c}{2} \partial_\mu
\lr{
(\gamma^\mu \wedge \gamma^\nu) F
+
F (\gamma^\mu \wedge \gamma^\nu)
}
}.
\end{aligned}

Here some steps are skipped, building on our previous scalar Euler-Lagrange evaluation experience. We have a symmetric product of two bivectors, which we can express as a 0,4 grade selection, since
\label{eqn:maxwells:2060}
\gpgrade{ X F }{0,4} = \inv{2} \lr{ X F + F X },

for any two bivectors $$X, F$$. This leaves
\label{eqn:maxwells:2080}
\begin{aligned}
0
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ (\grad \wedge \gamma^\nu) F }{0,4} \\
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ -\gamma^\nu \grad F + (\gamma^\nu \cdot \grad) F }{0,4} \\
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ -\gamma^\nu \grad F }{0,4} \\
&= J – \epsilon_0 c \gamma_\nu
\lr{
\gamma^\nu \cdot \lr{ \grad \cdot F } + \gamma^\nu \wedge \grad \wedge F
}.
\end{aligned}

However, since $$\grad \wedge F = \grad \wedge \grad \wedge A = 0$$, we see that there is no contribution from the $$F \wedge F$$ pseudoscalar component of the Lagrangian, and we are left with
\label{eqn:maxwells:2100}
\begin{aligned}
0
&= J – \epsilon_0 c (\grad \cdot F) \\
&= J – \epsilon_0 c \grad F,
\end{aligned}

which is Maxwell’s equation, as before.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Peeter Joot. Quantum field theory. Kindle Direct Publishing, 2018.

## PHY2403H Quantum Field Theory. Lecture 5: Klein-Gordon equation, Hamilton’s equations, SHOs, momentum space representation, raising and lowering operators. Taught by Prof. Erich Poppitz

### 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 I, taught by Prof. Erich Poppitz fall 2018.

## Canonical quantization

Last time we introduced a Lagrangian density associated with the Klein-Gordon equation (with a quadratic potential coupling)
\label{eqn:qftLecture5:20}
L = \int d^3 x
\lr{
\inv{2} \lr{\partial_0 \phi}^2 – \inv{2} \lr{\spacegrad \phi}^2 – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4
}.

This Lagrangian density was related to the action by
\label{eqn:qftLecture5:40}
S = \int dt L = \int dt d^3 x \LL,

with momentum canonically conjugate to the field $$\phi$$ defined as
\label{eqn:qftLecture5:60}
\Pi(\Bx, t) = \frac{\delta \LL}{\delta \phidot(\Bx, t) } = \PD{\phidot(\Bx, t)}{\LL}

The Hamiltonian defined as
\label{eqn:qftLecture5:80}
H = \int d^3 x \lr{ \Pi(\Bx, t) \phidot(\Bx, t) – \LL },

led to
\label{eqn:qftLecture5:680}
H
= \int d^3 x
\lr{ \inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4 }.

Like the Lagrangian density, we may introduce a Hamiltonian density $$\mathcal{H}$$ as
\label{eqn:qftLecture5:100}
H = \int d^3 x \mathcal{H}(\Bx, t).

For our Klein-Gordon system, this is
\label{eqn:qftLecture5:120}
\mathcal{H}(\Bx, t) =
\inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4.

### Canonical Commutation Relations (CCR)

:

We quantize the system by promoting our fields to Heisenberg-Picture (HP) operators, and imposing commutation relations
\label{eqn:qftLecture5:140}
\antisymmetric{\hat{\Pi}(\Bx, t)}{\hat{\phi}(\By, t)} = -i \delta^3 (\Bx – \By)

This is in analogy to
\label{eqn:qftLecture5:160}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

To choose a representation, we may map the $$\Psi$$ of QM $$\rightarrow$$ to a wave functional $$\Psi[\phi]$$
\label{eqn:qftLecture5:180}
\hat{\phi}(\By, t) \Psi[\phi] = \phi(\By, t) \Psi[\phi]

This is similar to the QM wave functions
\label{eqn:qftLecture5:200}
\begin{aligned}
\hat{q}_i \Psi(\setlr{q}) &= q_i \Psi(q) \\
\hat{p}_i \Psi(\setlr{q}) &= -i \PD{q_i}{} \Psi(p)
\end{aligned}

Our momentum operator is quantized by expressing it in terms of a variational derivative
\label{eqn:qftLecture5:220}
\hat{\Pi}(\Bx, t) = -i \frac{\delta}{\delta \phi(\Bx, t)}.

(Fixme: I’m not really sure exactly what is meant by using the variation derivative $$\delta$$ notation here), and to
quantize the Hamiltonian we just add hats, assuming that our fields are all now HP operators
\label{eqn:qftLecture5:240}
\hat{\mathcal{H}}(\Bx, t)
=
\inv{2} \hat{\Pi}^2 + (\spacegrad \hat{\phi})^2 + \inv{2} m^2 \hat{\phi}^2 + \frac{\lambda}{4} \hat{\phi}^4.

### QM SHO review

Recall the QM SHO had a Hamiltonian
\label{eqn:qftLecture5:260}
\hat{H} = \inv{2} \hat{p}^2 + \inv{2} \omega^2 \hat{q}^2,

where
\label{eqn:qftLecture5:280}
\antisymmetric{\hat{p}}{\hat{q}} = -i,

and that
HP time evolution operators $$O$$ satisfied
\label{eqn:qftLecture5:700}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}.

In particular
\label{eqn:qftLecture5:300}
\begin{aligned}
\ddt{\hat{p}}
&= i \antisymmetric{\hat{H}}{\hatp} \\
&= i \frac{\omega^2}{2} \antisymmetric{\hat{q}^2}{\hatp} \\
&= i \frac{\omega^2}{2} (2 i \hat{q}) \\
&= -i \omega^2 \hat{q},
\end{aligned}

and
\label{eqn:qftLecture5:320}
\begin{aligned}
\ddt{\hat{q}}
&= i \antisymmetric{\hat{H}}{\hat{q}} \\
&= i \inv{2} \antisymmetric{\hatp^2}{\hat{q}} \\
&= \frac{i}{2}(-2 i \hatp ) \\
&= \hatp.
\end{aligned}

Applying the time evolution operator twice, we find
\label{eqn:qftLecture5:340}
\frac{d^2}{dt^2}{\hat{q}}
= \ddt{\hat{p}}
= – \omega^2 \hat{q}.

We see that the Heisenberg operators obey the classical equations of motion.

Now we want to try this with the quantized QFT fields we’ve promoted to operators
\label{eqn:qftLecture5:360}
\begin{aligned}
\ddt{\hat{\Pi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\Pi}(\Bx, t)} \\
&=
i \int d^3 y \inv{2} \antisymmetric{ \lr{\spacegrad \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
+
i \int d^3 y \frac{m^2}{2} \antisymmetric{ \phihat(\By)^2 }{\hat{\Pi}(\Bx) }
+
i \frac{\lambda}{4} \int d^3 \antisymmetric{ \phihat(\By)^4 }{\hat{\Pi}(\Bx) }
\end{aligned}

Starting with the non-gradient commutators, and utilizing the HP field analogues of the relations $$\antisymmetric{\hat{q}^n}{\hatp} = n i \hat{q}^{n-1}$$, we find
\label{eqn:qftLecture5:780}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 2 i \phihat(\By) \delta^3(\Bx – \By)
= 2 i \phihat(\Bx).

\label{eqn:qftLecture5:740}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^4 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 4 i \phihat(\By)^3 \delta^3(\Bx – \By)
= 4 i \phihat(\Bx)^3.

For the gradient commutators, we have more work. Prof Poppitz blitzed through that, just calling it integration by parts. I had trouble seeing what he was doing, so here’s a more explicit dumb expansion required to calculate the commutator
\label{eqn:qftLecture5:720}
\begin{aligned}
\int d^3 y (\spacegrad \phihat(\By))^2 \hat{\Pi}(\Bx)
&=
\int d^3 y
&=
\int d^3 y
\lr{ \spacegrad (\phihat(\By) \hat{\Pi}(\Bx)) } \\
&=
\int d^3 y
\lr{ \spacegrad (\hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By)) } \\
&=
\int d^3 y
\Biglr{
+ i
} \\
&=
\int d^3 y
\Biglr{
\spacegrad \lr{ \hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By) } \cdot \spacegrad \phihat(\By)
+ i
} \\
&=
\int d^3 y
\hat{\Pi}(\Bx)
\lr{
}
+ 2 i
\int d^3 y
&=
\int d^3 y
+
2 i
\int d^3 y

2 i
\int d^3 y
\delta^3(\Bx – \By) \spacegrad^2 \phihat(\By) \\
&=
\int d^3 y
+
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

\end{aligned}

Here we take advantage of the fact that the derivative operators $$\spacegrad = \spacegrad_\By$$ commute with $$\hat{\Pi}(\Bx)$$, and use the identity
$$\spacegrad \cdot (a \spacegrad b) = (\spacegrad a) \cdot (\spacegrad b) + a \spacegrad^2 b$$, so the commutator is
\label{eqn:qftLecture5:800}
\begin{aligned}
&=
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

&=

\end{aligned}

where the boundary integral is presumed to be zero (without enough justification.) All the pieces can now be put back together
\label{eqn:qftLecture5:820}
\ddt{} \hat{\Pi}(\Bx, t)
=

m^2 \phihat(\Bx, t)

\lambda \phihat^3(\Bx, t).

Now, for the $$\phihat$$ time evolution, which is much easier
\label{eqn:qftLecture5:380}
\begin{aligned}
\ddt{\hat{\phi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\phi}(\Bx, t)} \\
&= i \inv{2} \int d^3 y \antisymmetric{\hat{\Pi}^2(\By)}{\hat{\phi}(\Bx)} \\
&= i \inv{2} \int d^3 y (-2 i) \hat{\Pi}(\By, t) \delta^3(\Bx – \By) \\
&= \hat{\Pi}(\Bx, t)
\end{aligned}

\label{eqn:qftLecture5:400}
\frac{d^2}{dt^2}{\hat{\phi}}(\Bx, t)
=
-m^2 \phi – \lambda \phihat^3.

That is
\label{eqn:qftLecture5:420}
\ddot{\phihat} – \spacegrad^2 \phihat + m^2 \phihat + \lambda \phihat^3 = 0,

which is the classical Euler-Lagrange equation, also obeyed by the Heisenberg operator $$\phi(\Bx, t)$$. When $$\lambda = 0$$ this is the Klein-Gordon equation.

## Momentum space representation.

Dropping hats, we now consider the momentum space representation of our operators, as determined by Fourier transform pairs
\label{eqn:qftLecture5:440}
\begin{aligned}
\phi(\Bx, t) &= \int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx} \tilde{\phi}(\Bp, t) \\
\tilde{\phi}(\Bp, t) &= \int d^3 x e^{-i \Bp \cdot \Bx} \phi(\Bx, t)
\end{aligned}

We can discover a representation of the delta function by applying these both in turn
\label{eqn:qftLecture5:480}
\tilde{\phi}(\Bp, t)
= \int d^3 x e^{-i \Bp \cdot \Bx} \int \frac{d^3 q}{(2 \pi)^3} e^{i \Bq \cdot \Bx} \tilde{\phi}(\Bq, t)

so
\label{eqn:qftLecture5:500}
\boxed{
\int d^3 x e^{i \BA \cdot \Bx} = (2 \pi)^3 \delta^3(\BA)
}

Also observe that $$\phi^\conj(\Bx, t) = \phi(\Bx, t)$$ iff $$\tilde{\phi}(\Bp, t) = \tilde{\phi}^\conj(-\Bp, t)$$.

We want the EOM for $$\tilde{\phi}(\Bp, t)$$ where the operator obeys the KG equation
\label{eqn:qftLecture5:520}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi(\Bx, t) = 0

Inserting the transform relation \ref{eqn:qftLecture5:440} we get
\label{eqn:qftLecture5:540}
\int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx}
\lr{
\ddot{\tilde{\phi}}(\Bp, t) + \lr{ \Bp^2 + m^2 }
\tilde{\phi}(\Bp, t)
}
= 0,

or
\label{eqn:qftLecture5:580}
\boxed{
\ddot{\tilde{\phi}}(\Bp, t) = – \omega_\Bp^2 \,\tilde{\phi}(\Bp, t),
}

where
\label{eqn:qftLecture5:560}
\omega_\Bp = \sqrt{ \Bp^2 + m^2 }.

The Fourier components of the HP operators are SHOs!

As we have SHO’s and know how to deal with these in QM, we use the same strategy, introducing raising and lowering operators
\label{eqn:qftLecture5:600}
\tilde{\phi}(\Bp, t) = \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{-i \omega_\Bp t } a_\Bp + e^{i \omega_\Bp t} a^\dagger_{-\Bp}
}

Observe that
\label{eqn:qftLecture5:840}
\begin{aligned}
\tilde{\phi}^\dagger(-\Bp, t)
&= \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{i \omega_\Bp t } a^\dagger_{-\Bp} + e^{-i \omega_\Bp t} a_{\Bp} } \\
&=
\tilde{\phi}(\Bp, t),
\end{aligned}

or
\label{eqn:qftLecture5:620}
\tilde{\phi}^\dagger(\Bp, t) = \tilde{\phi}(-\Bp, t),

so $$\phi(\Bp, t)$$ has a real representation in terms of $$a_\Bp$$.

We will find (Wednesday) that
\label{eqn:qftLecture5:640}
\antisymmetric{a_\Bq}{a^+_\Bp} = \delta^3(\Bp – \Bq) (2 \pi)^3.

These are equivalent to
\label{eqn:qftLecture5:660}
\antisymmetric{\hat{\Pi}(\By, t)}{\tilde{\phi}(\Bx, t)} = -i \delta^3(\Bx – \By)