## 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.

## Energy-momentum tensor for a scalar field

It is claimed in [1] (3.2.1) that the momentum components of the energy-momentum tensor was found to be

\label{eqn:noetherCurrentScalarField:20}
\Be_n \int d^3 x T^{0 n} = \int d^3 k \Bk a_k^\dagger a_k.

I don’t see this result anywhere, so let’s calculate it.

First, from the Noether current for the scalar field Lagrangian in question, what is the energy-momentum tensor explicitly?

\label{eqn:noetherCurrentScalarField:40}
\begin{aligned}
T^{\mu \nu}
&= \Pi^\mu \partial^\nu \phi – g^{\mu \nu} \LL \\
&= \Pi^\mu \partial^\nu \phi – g^{\mu \nu} \inv{2} \lr{ \partial_\alpha \phi \partial^\alpha \phi – \mu^2 \phi^2 } \\
&= \Pi^\mu \Pi^\nu – g^{\mu \nu} \inv{2} \lr{ \Pi_\alpha \Pi^\alpha – \mu^2 \phi^2 } \\
&= \Pi^\mu \Pi^\nu – \inv{2} g^{\mu \nu} g_{\alpha\beta} \Pi^\beta \Pi^\alpha + \inv{2} g^{\mu \nu} \mu^2 \phi^2.
\end{aligned}

Consider some special cases for the indexes. For $$\mu = \nu = 0$$, the result is the Hamiltonian density

\label{eqn:noetherCurrentScalarField:200}
\begin{aligned}
T^{00}
&= \Pi^0 \Pi^0 – \inv{2} g^{0 0} \Pi_\alpha \Pi^\alpha + \inv{2} g^{0 0} \mu^2 \phi^2 \\
&= \Pi^0 \Pi^0 – \inv{2} \Pi_\alpha \Pi^\alpha + \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^0 \Pi^0 – \inv{2} \Pi_n \Pi^n + \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \inv{2} (\spacegrad \phi)^2 + \inv{2} \mu^2 \phi^2,
\end{aligned}

where $$\Pi^2 = (\partial_0 \phi)^2 \ne \partial^2 \phi$$. For any $$\mu \ne \nu$$ the off diagonal metric elements are zero, leaving just
\label{eqn:noetherCurrentScalarField:220}
T^{\mu\nu} = \Pi^\mu \Pi^\nu.

Finally, when $$n \ne 0$$, the remaining diagonal terms are
\label{eqn:noetherCurrentScalarField:240}
\begin{aligned}
T^{nn}
&= \Pi^n \Pi^n – \inv{2} g^{n n} \Pi_\alpha \Pi^\alpha + \inv{2} g^{n n} n^2 \phi^2 \\
&= \Pi^n \Pi^n + \inv{2} \Pi_\alpha \Pi^\alpha – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \Pi^n \Pi^n – \inv{2} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \inv{2} \Pi^n \Pi^n – \inv{2} \sum_{m\ne n,0} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \sum_{m = n,0} \Pi^m \Pi^m – \inv{2} \sum_{m\ne n,0} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2.
\end{aligned}

The canonical momenta are

\label{eqn:noetherCurrentScalarField:60}
\Pi^\mu
=
\partial^\mu
\int \frac{d^3 k}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ a_k e^{-i k \cdot x} + a_k^\dagger e^{i k \cdot x} },

but
\label{eqn:noetherCurrentScalarField:80}
\begin{aligned}
\partial^\mu e^{i k \cdot x}
&=
\partial^\mu \exp\lr{ i k^\alpha x_\alpha } \\
&=
i k^\mu \exp\lr{ i k \cdot x },
\end{aligned}

so
\label{eqn:noetherCurrentScalarField:100}
\begin{aligned}
\Pi^\mu
&=
i
\int \frac{d^3 k k^\mu}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ – a_k e^{-i k \cdot x} + a_k^\dagger e^{i k \cdot x} } \\
&=
i
\int \frac{d^3 k k^\mu}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ – a_k e^{-i \omega_k t + \Bk \cdot \Bx} + a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx} }.
\end{aligned}

This gives
\label{eqn:noetherCurrentScalarField:120}
\begin{aligned}
\int d^3 x \Pi^\mu \Pi^\nu
&=
-\inv{2} \int d^3 x \inv{(2\pi)^3}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{ – a_k e^{-i \omega_k t + \Bk \cdot \Bx} + a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx} }
\lr{ – a_j e^{-i \omega_j t + \Bj \cdot \Bx} + a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx} } \\
&=
-\inv{2} \int d^3 x \inv{(2\pi)^3}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{
a_k a_j e^{-i (\omega_j + \omega_k) t + (\Bj + \Bk) \cdot \Bx}
– a_k a_j^\dagger e^{i (\omega_j – \omega_k) t – i (\Bj -\Bk) \cdot \Bx}
– a_k^\dagger a_j e^{-i (\omega_j -\omega_k) t – (\Bk – \Bj) \cdot \Bx}
+ a_k^\dagger a_j^\dagger e^{i (\omega_j + \omega_k) t – i (\Bj + \Bk) \cdot \Bx}
} \\
&=
-\inv{2}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{
a_k a_j e^{-i (\omega_j + \omega_k) t } \delta^3(\Bj + \Bk)
– a_k a_j^\dagger e^{i (\omega_j – \omega_k) t } \delta^3(\Bj -\Bk)
– a_k^\dagger a_j e^{-i (\omega_j -\omega_k) t } \delta^3 (\Bk – \Bj)
+ a_k^\dagger a_j^\dagger e^{i (\omega_j + \omega_k) t } \delta^3 (\Bj + \Bk)
}.
\end{aligned}

There are two cases here to consider. The first is $$\nu = 0$$, for which

\label{eqn:noetherCurrentScalarField:140}
\int d^3 x \Pi^\mu \Pi^0
=
-\inv{2}
\int d^3 k k^\mu
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
}.

For $$\nu \ne 0$$

\label{eqn:noetherCurrentScalarField:160}
\begin{aligned}
\int d^3 x \Pi^\mu \Pi^\nu
&=
-\inv{2}
\int d^3 k \frac{k^\mu k^\nu}{\omega_k}
\lr{
– a_k a_{-k} e^{- 2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
– a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
} \\
&=
\inv{2}
\int d^3 k \frac{k^\mu k^\nu}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.
\end{aligned}

Here’s a summary of these products

\label{eqn:noetherCurrentScalarField:300}
\int d^3 x \Pi^0 \Pi^0
=
-\inv{2}
\int d^3 k \omega_k
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},

\label{eqn:noetherCurrentScalarField:280}
\int d^3 x \Pi^n \Pi^0
= \int d^3 x \Pi^0 \Pi^n
=
-\inv{2}
\int d^3 k k^n
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},

\label{eqn:noetherCurrentScalarField:340}
\int d^3 x \Pi^m \Pi^n
=
\inv{2}
\int d^3 k \frac{k^m k^n}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.

For the mass term it was previously found that

\label{eqn:noetherCurrentScalarField:180}
\inv{2} \int d^3 x \mu^2 \phi^2
=
\frac{\mu^2}{4}
\int
d^3 k
\inv{ \omega_k }
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}.

The Hamiltonian component has been previously calculated, and resolves to

\label{eqn:noetherCurrentScalarField:360}
\int d^3 x T^{00}
=
\inv{2}
\int d^3 k
\omega_k
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.

The other diagonal components, for $$r \ne s \ne t$$ are
\label{eqn:noetherCurrentScalarField:380}
\begin{aligned}
\int d^3 x T^{rr}
&=
\int d^3 x
\lr{
\inv{2} \sum_{m = r,0} \Pi^m \Pi^m – \inv{2} \sum_{m = s,t} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2
} \\
&=
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
-\inv{4}
\int d^3 k \omega_k
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
} \\
&=
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2 – \omega_k^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
+
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2 + \omega_k^2}{\omega_k}
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
} \\
&=
\inv{2}
\int d^3 k \frac{ (k^r)^2 – \omega_k^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
+
\inv{2}
\int d^3 k \frac{ (k^r)^2}{\omega_k}
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.
\end{aligned}

This doesn’t have the nice cancelation that killed the time dependent terms in the Hamiltonian. Such cancellation also doesn’t appear in the off diagonal energy-momentum tensor components, which are

\label{eqn:noetherCurrentScalarField:400}
\begin{aligned}
\int d^3 x T^{n 0}
&=
\int d^3 x T^{n 0} \\
&=
-\inv{2}
\int d^3 k k^n
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},
\end{aligned}

and for $$m \ne n \ne 0$$
\label{eqn:noetherCurrentScalarField:420}
\int d^3 x T^{m n}
=
\inv{2}
\int d^3 k \frac{k^m k^n}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.

The \ref{eqn:noetherCurrentScalarField:400} result has time dependence that the stated result does not (but is linear in $$\Bk$$ as desired)? Did I miss something?

# References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL https://piazza.com/utoronto.ca/fall2015/phy2403f/resources. [Online; accessed 02-Jan-2016].

## PHY1520H Graduate Quantum Mechanics. Lecture 5: time evolution of coherent states, and charged particles in a magnetic field. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering \textchapref{{1}} [1] content.

### Coherent states (cont.)

A coherent state for the SHO $$H = \lr{ N + \inv{2} } \Hbar \omega$$ was given by

\label{eqn:qmLecture5:20}
a \ket{z} = z \ket{z},

where we showed that

\label{eqn:qmLecture5:40}
\ket{z} = c_0 e^{ z a^\dagger } \ket{0}.

In the Heisenberg picture we found

\label{eqn:qmLecture5:60}
\begin{aligned}
a_{\textrm{H}}(t) &= e^{i H t/\Hbar} a e^{-i H t/\Hbar} = a e^{-i\omega t} \\
a_{\textrm{H}}^\dagger(t) &= e^{i H t/\Hbar} a^\dagger e^{-i H t/\Hbar} = a^\dagger e^{i\omega t}.
\end{aligned}

Recall that the position and momentum representation of the ladder operators was

\label{eqn:qmLecture5:80}
\begin{aligned}
a &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} + i \hat{p} \sqrt{\inv{m \Hbar \omega}} } \\
a^\dagger &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} – i \hat{p} \sqrt{\inv{m \Hbar \omega}} },
\end{aligned}

or equivalently
\label{eqn:qmLecture5:100}
\begin{aligned}
\hat{x} &= \lr{ a + a^\dagger } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
\hat{p} &= i \lr{ a^\dagger – a } \sqrt{\frac{m \Hbar \omega}{2}}.
\end{aligned}

Given this we can compute expectation value of position operator

\label{eqn:qmLecture5:120}
\begin{aligned}
\bra{z} \hat{x} \ket{z}
&=
\sqrt{\frac{\Hbar}{ 2 m \omega}}
\bra{z}
\lr{ a + a^\dagger }
\ket{z} \\
&=
\lr{ z + z^\conj } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
&=
2 \textrm{Re} z \sqrt{\frac{\Hbar}{ 2 m \omega}} .
\end{aligned}

Similarly

\label{eqn:qmLecture5:140}
\begin{aligned}
\bra{z} \hat{p} \ket{z}
&=
i \sqrt{\frac{m \Hbar \omega}{2}}
\bra{z}
\lr{ a^\dagger – a }
\ket{z} \\
&=
\sqrt{\frac{m \Hbar \omega}{2}}
2 \textrm{Im} z.
\end{aligned}

How about the expectation of the Heisenberg position operator? That is

\label{eqn:qmLecture5:160}
\begin{aligned}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z}
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \bra{z} \lr{ a + a^\dagger } \ket{z} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ z e^{-i \omega t} + z^\conj e^{i \omega t}} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \lr{z + z^\conj} \cos( \omega t ) -i \lr{ z – z^\conj } \sin( \omega t) } \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \expectation{x(0)} \sqrt{ \frac{2 m \omega}{\Hbar}} \cos( \omega t ) -i \expectation{p(0)} i \sqrt{\frac{2 m \omega}{\Hbar} } \sin( \omega t) } \\
&=
\expectation{x(0)} \cos( \omega t ) + \frac{\expectation{p(0)}}{m \omega} \sin( \omega t) .
\end{aligned}

We find that the average of the Heisenberg position operator evolves in time in exactly the same fashion as position in the classical Harmonic oscillator. This phase space like trajectory is sketched in fig. 1.

fig. 1. phase space like trajectory

In the text it is shown that we have the same structure for the Heisenberg operator itself, before taking expectations

\label{eqn:qmLecture5:220}
\hat{x}_{\textrm{H}}(t)
=
{x(0)} \cos( \omega t ) + \frac{{p(0)}}{m \omega} \sin( \omega t).

Where the coherent states become useful is that we will see that the second moments of position and momentum are not time dependent with respect to the coherent states. Such states remain localized.

### Uncertainty

First note that using the commutator relationship we have

\label{eqn:qmLecture5:180}
\begin{aligned}
\bra{z} a a^\dagger \ket{z}
&=
\bra{z} \lr{ \antisymmetric{a}{a^\dagger} + a^\dagger a } \ket{z} \\
&=
\bra{z} \lr{ 1 + a^\dagger a } \ket{z}.
\end{aligned}

For the second moment we have

\label{eqn:qmLecture5:200}
\begin{aligned}
\bra{z} \hat{x}^2 \ket{z}
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{a + a^\dagger } \lr{a + a^\dagger } \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + a a^\dagger + a^\dagger a
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + 2 a^\dagger a + 1
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z^2 + {(z^\conj)}^2 + 2 z^\conj z + 1} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z + z^\conj }^2
+
\frac{\Hbar}{ 2 m \omega}.
\end{aligned}

We find

\label{eqn:qmLecture5:240}
\sigma_x^2 = \frac{\Hbar}{ 2 m \omega},

and

\label{eqn:qmLecture5:260}
\sigma_p^2 = \frac{m \Hbar \omega}{2}

so

\label{eqn:qmLecture5:280}
\sigma_x^2 \sigma_p^2 = \frac{\Hbar^2}{4},

or

\label{eqn:qmLecture5:300}
\sigma_x \sigma_p = \frac{\Hbar}{2}.

This is the minimum uncertainty.

### Quantum Field theory

In Quantum Field theory the ideas of isolated oscillators is used to model particle creation. The lowest energy state (a no particle, vacuum state) is given the lowest energy level, with each additional quantum level modeling a new particle creation state as sketched in fig. 2.

fig. 2. QFT energy levels

We have to imagine many oscillators, each with a distinct vacuum energy $$\sim \Bk^2$$ . The Harmonic oscillator can be used to model the creation of particles with $$\Hbar \omega$$ energy differences from that “vacuum energy”.

### Charged particle in a magnetic field

In the classical case ( with SI units or $$c = 1$$ ) we have

\label{eqn:qmLecture5:320}
\BF = q \BE + q \Bv \cross \BB.

Alternately, we can look at the Hamiltonian view of the system, written in terms of potentials

\label{eqn:qmLecture5:340}

\label{eqn:qmLecture5:360}
\BE = – \spacegrad \phi – \PD{t}{\BA}.

Note that the curl form for the magnetic field implies one of the required Maxwell’s equations $$\spacegrad \cdot \BB = 0$$.

Ignoring time dependence of the potentials, the Hamiltonian can be expressed as

\label{eqn:qmLecture5:380}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.

In this Hamiltonian the vector $$\Bp$$ is called the canonical momentum, the momentum conjugate to position in phase space.

It is left as an exercise to show that the Lorentz force equation results from application of the Hamiltonian equations of motion, and that the velocity is given by $$\Bv = (\Bp – q \BA)/m$$.

For quantum mechanics, we use the same Hamiltonian, but promote our position, momentum and potentials to operators.

\label{eqn:qmLecture5:400}
\hat{H} = \inv{2 m} \lr{ \hat{\Bp} – q \hat{\BA}(\Br, t) }^2 + q \hat{\phi}(\Br, t).

### Gauge invariance

Can we say anything about this before looking at the question of a particle in a magnetic field?

Recall that the we can make a gauge transformation of the form

\label{eqn:qmLecture5:420a}
\label{eqn:qmLecture5:420}
\BA \rightarrow \BA + \spacegrad \chi

\label{eqn:qmLecture5:440}
\phi \rightarrow \phi – \PD{t}{\chi}

Does this notion of gauge invariance also carry over to the Quantum Hamiltonian. After gauge transformation we have

\label{eqn:qmLecture5:460}
\hat{H}’
= \inv{2 m} \lr{ \hat{\Bp} – q \BA – q \spacegrad \chi }^2 + q \lr{ \phi – \PD{t}{\chi} }

Now we are in a mess, since this function $$\chi$$ can make the Hamiltonian horribly complicated. We don’t see how gauge invariance can easily be applied to the quantum problem. Next time we will introduce a transformation that resolves some of this mess.

## Question: Lorentz force from classical electrodynamic Hamiltonian

Given the classical Hamiltonian

\label{eqn:qmLecture5:381}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.

apply the Hamiltonian equations of motion

\label{eqn:qmLecture5:480}
\begin{aligned}
\ddt{\Bp} &= – \PD{\Bq}{H} \\
\ddt{\Bq} &= \PD{\Bp}{H},
\end{aligned}

to show that this is the Hamiltonian that describes the Lorentz force equation, and to find the velocity in terms of the canonical momentum and vector potential.

The particle velocity follows easily

\label{eqn:qmLecture5:500}
\begin{aligned}
\Bv
&= \ddt{\Br} \\
&= \PD{\Bp}{H} \\
&= \inv{m} \lr{ \Bp – a \BA }.
\end{aligned}

For the Lorentz force we can proceed in the coordinate representation

\label{eqn:qmLecture5:520}
\begin{aligned}
\ddt{p_k}
&= – \PD{x_k}{H} \\
&= – \frac{2}{2m} \lr{ p_m – q A_m } \PD{x_k}{}\lr{ p_m – q A_m } – q \PD{x_k}{\phi} \\
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
\end{aligned}

We also have

\label{eqn:qmLecture5:540}
\begin{aligned}
\ddt{p_k}
&=
\ddt{} \lr{m x_k + q A_k } \\
&=
m \frac{d^2 x_k}{dt^2} + q \PD{x_m}{A_k} \frac{d x_m}{dt} + q \PD{t}{A_k}.
\end{aligned}

Putting these together we’ve got

\label{eqn:qmLecture5:560}
\begin{aligned}
m \frac{d^2 x_k}{dt^2}
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
– q \PD{x_m}{A_k} \frac{d x_m}{dt} – q \PD{t}{A_k} \\
&=
q v_m \lr{ \PD{x_k}{A_m} – \PD{x_m}{A_k} } + q E_k \\
&=
q v_m \epsilon_{k m s} B_s + q E_k,
\end{aligned}

or

\label{eqn:qmLecture5:580}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&=
q \Be_k v_m \epsilon_{k m s} B_s + q E_k \\
&= q \Bv \cross \BB + q \BE.
\end{aligned}

## Question: Show gauge invariance of the magnetic and electric fields

After the gauge transformation of \ref{eqn:qmLecture5:420} show that the electric and magnetic fields are unaltered.

For the magnetic field the transformed field is

\label{eqn:qmLecture5:600}
\begin{aligned}
\BB’
&= \BB.
\end{aligned}

\label{eqn:qmLecture5:620}
\begin{aligned}
\BE’
&=
– \PD{t}{\BA’} – \spacegrad \phi’ \\
&=
– \PD{t}{}\lr{\BA + \spacegrad \chi} – \spacegrad \lr{ \phi – \PD{t}{\chi}} \\
&=
– \PD{t}{\BA} – \spacegrad \phi \\
&=
\BE.
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Question:

Given a gauge transformation of the free particle Hamiltonian to

\label{eqn:gaugeTx:20}
H = \inv{2 m} \BPi \cdot \BPi + e \phi,

where

\label{eqn:gaugeTx:40}
\BPi = \Bp – \frac{e}{c} \BA,

calculate $$m d\Bx/dt$$, $$\antisymmetric{\Pi_i}{\Pi_j}$$, and $$m d^2\Bx/dt^2$$, where $$\Bx$$ is the Heisenberg picture position operator, and the fields are functions only of position $$\phi = \phi(\Bx), \BA = \BA(\Bx)$$.

The final results for these calculations are found in [1], but seem worth deriving to exercise our commutator muscles.

### Heisenberg picture velocity operator

The first order of business is the Heisenberg picture velocity operator, but first note

\label{eqn:gaugeTx:60}
\begin{aligned}
\BPi \cdot \BPi
&= \lr{ \Bp – \frac{e}{c} \BA} \cdot \lr{ \Bp – \frac{e}{c} \BA} \\
&= \Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp + \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2.
\end{aligned}

The time evolution of the Heisenberg picture position operator is therefore

\label{eqn:gaugeTx:80}
\begin{aligned}
\ddt{\Bx}
&= \inv{i\Hbar} \antisymmetric{\Bx}{H} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\BPi^2} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp
+ \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2 } \\
&= \inv{i\Hbar 2 m}
\lr{
\antisymmetric{\Bx}{\Bp^2}
– \frac{e}{c} \antisymmetric{\Bx}{ \BA \cdot \Bp + \Bp \cdot \BA }
}
.
\end{aligned}

For the $$\Bp^2$$ commutator we have

\label{eqn:gaugeTx:100}
\antisymmetric{x_r}{\Bp^2}
=
i \Hbar \PD{p_r}{\Bp^2}
=
2 i \Hbar p_r,

or
\label{eqn:gaugeTx:120}
\antisymmetric{\Bx}{\Bp^2}
=
2 i \Hbar \Bp.

Computing the remaining commutator, we’ve got

\label{eqn:gaugeTx:140}
\begin{aligned}
\antisymmetric{x_r}{\Bp \cdot \BA + \BA \cdot \Bp}
&= x_r p_s A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s p_s x_r \\
&= \lr{ \antisymmetric{x_r}{p_s} + p_s x_r } A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s \lr{ \antisymmetric{p_s}{x_r} + x_r p_s } \\
&= \antisymmetric{x_r}{p_s} A_s + {p_s A_s x_r – p_s A_s x_r} \\
&\quad+ {x_r A_s p_s – x_r A_s p_s} + A_s \antisymmetric{x_r}{p_s} \\
&= 2 i \Hbar \delta_{r s} A_s \\
&= 2 i \Hbar A_r,
\end{aligned}

so

\label{eqn:gaugeTx:160}
\antisymmetric{\Bx}{\Bp \cdot \BA + \BA \cdot \Bp} = 2 i \Hbar \BA.

Assembling these results gives

\label{eqn:gaugeTx:180}
\boxed{
\ddt{\Bx} = \inv{m} \lr{ \Bp – \frac{e}{c} \BA } = \inv{m} \BPi,
}

as asserted in the text.

### Kinetic Momentum commutators

\label{eqn:gaugeTx:200}
\begin{aligned}
\antisymmetric{\Pi_r}{\Pi_s}
&=
\antisymmetric{p_r – e A_r/c}{p_s – e A_s/c} \\
&=
{\antisymmetric{p_r}{p_s}}
– \frac{e}{c} \lr{ \antisymmetric{p_r}{A_s} + \antisymmetric{A_r}{p_s}}
+ \frac{e^2}{c^2} {\antisymmetric{A_r}{A_s}} \\
&=
– \frac{e}{c} \lr{ (-i\Hbar) \PD{x_r}{A_s} + (i\Hbar) \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \lr{ -\PD{x_r}{A_s} + \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \epsilon_{t s r} B_t,
\end{aligned}

or
\label{eqn:gaugeTx:220}
\boxed{
\antisymmetric{\Pi_r}{\Pi_s}
=
\frac{i e \Hbar}{c} \epsilon_{r s t} B_t.
}

### Quantum Lorentz force

For the force equation we have

\label{eqn:gaugeTx:240}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&= \ddt{\BPi} \\
&= \inv{i \Hbar} \antisymmetric{\BPi}{H} \\
&= \inv{i \Hbar 2 m } \antisymmetric{\BPi}{\BPi^2}
+ \inv{i \Hbar } \antisymmetric{\BPi}{e \phi}.
\end{aligned}

For the $$\phi$$ commutator consider one component

\label{eqn:gaugeTx:260}
\begin{aligned}
\antisymmetric{\Pi_r}{e \phi}
&=
e \antisymmetric{p_r – \frac{e}{c} A_r}{\phi} \\
&=
e \antisymmetric{p_r}{\phi} \\
&=
e (-i\Hbar) \PD{x_r}{\phi},
\end{aligned}

or
\label{eqn:gaugeTx:280}
\inv{i \Hbar} \antisymmetric{\BPi}{e \phi}
=
=
e \BE.

For the $$\BPi^2$$ commutator I initially did this the hard way (it took four notebook pages, plus two for a false start.) Realizing that I didn’t use \ref{eqn:gaugeTx:220} for that expansion was the clue to doing this more expediently.

Considering a single component

\label{eqn:gaugeTx:300}
\begin{aligned}
\antisymmetric{\Pi_r}{\BPi^2}
&=
\antisymmetric{\Pi_r}{\Pi_s \Pi_s} \\
&=
\Pi_r \Pi_s \Pi_s – \Pi_s \Pi_s \Pi_r \\
&=
\lr{ \antisymmetric{\Pi_r}{\Pi_s} + {\Pi_s \Pi_r} }
\Pi_s
– \Pi_s
\lr{ \antisymmetric{\Pi_s}{\Pi_r} + {\Pi_r \Pi_s} } \\
&= i \Hbar \frac{e}{c} \epsilon_{r s t}
\lr{ B_t \Pi_s + \Pi_s B_t },
\end{aligned}

or

\label{eqn:gaugeTx:320}
\begin{aligned}
\inv{ i \Hbar 2 m} \antisymmetric{\BPi}{\BPi^2}
&= \frac{e}{2 m c } \epsilon_{r s t} \Be_r
\lr{ B_t \Pi_s + \Pi_s B_t } \\
&= \frac{e}{ 2 m c }
\lr{
\BPi \cross \BB
– \BB \cross \BPi
}.
\end{aligned}

Putting all the pieces together we’ve got the quantum equivalent of the Lorentz force equation

\label{eqn:gaugeTx:340}
\boxed{
m \frac{d^2 \Bx}{dt^2} = e \BE + \frac{e}{2 c} \lr{
\frac{d\Bx}{dt} \cross \BB
– \BB \cross \frac{d\Bx}{dt}
}.
}

While this looks equivalent to the classical result, all the vectors here are Heisenberg picture operators dependent on position.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.