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## errata for Feynman’s Quantum Electrodynamics (Addison-Wesley) ?

Posted by peeterjoot on May 28, 2010

# Motivation.

I got a nice present today which included one of Feynman’s QED books. I noticed some early mistakes, and since I can’t find an errata page anywhere, I’ll collect them here.

# Third Lecture

## Page 6 typos.

The electric field is given in terms of only the scalar potential

\begin{aligned}\mathbf{E} = -\boldsymbol{\nabla} \phi + \partial \phi/ \partial t,\end{aligned}

and should be

\begin{aligned}\mathbf{E} = -\boldsymbol{\nabla} \phi - \frac{1}{{c}} \partial \mathbf{A}/ \partial t.\end{aligned}

The invariant gauge transformation for the vector and scalar potentials are given as

\begin{aligned}\mathbf{A}' &= \mathbf{A} = \boldsymbol{\nabla} \chi \\ \phi' &= \phi + \partial \chi / \partial t\end{aligned}

But these should be

\begin{aligned}\mathbf{A}' &= \mathbf{A} + \boldsymbol{\nabla} \chi \\ \phi' &= \phi - \frac{1}{{c}} \partial \chi / \partial t\end{aligned}

The sign was crossed on the scalar potential transformation. Feynman is also probably used to using $c=1$, but he doesn’t do that explicitly at a different point on the page, so including it here is proper.

## Page 7 notes.

The units in the transformation for the wave function don’t look right. We want to transform the Pauli equation

\begin{aligned}i \hbar \frac{\partial {\Psi}}{\partial {t}} = \frac{1}{{2 m}} \left( \mathbf{p} - \frac{e}{c} \mathbf{A} \right)^2 \Psi + e \phi \Psi,\end{aligned}

with a transformation of the form

\begin{aligned}\mathbf{A}' &= \mathbf{A} + \boldsymbol{\nabla} \chi \\ \phi' &= \phi - \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \\ \Psi' &= e^{-i \mu} \Psi,\end{aligned}

Where $\mu \propto \chi$ is presumed, and we want to find the proportionality constant required for invariance. With $\mathbf{p} = - i \hbar \boldsymbol{\nabla}$ we have

\begin{aligned}\mathbf{p} \Psi' &=-i \hbar \boldsymbol{\nabla} e^{-i \mu} \Psi \\ &=-i \hbar \left( -i (\boldsymbol{\nabla} \mu) e^{-i \mu} \Psi + e^{-i \mu} \boldsymbol{\nabla} \Psi \right) \\ &=+ e^{-i \mu} \left( \mathbf{p} + \hbar \boldsymbol{\nabla} \mu \right) \Psi,\end{aligned}

so

\begin{aligned}(\mathbf{p} -\frac{e}{c} \mathbf{A}' )\Psi' &=e^{-i \mu} \left( \mathbf{p} - \frac{e}{c} \mathbf{A} - \boldsymbol{\nabla} (\hbar \mu + \frac{e}{c} \chi) \right) \Psi.\end{aligned}

For the time partial we have

\begin{aligned}\frac{\partial {\Psi'}}{\partial {t}} &= e^{-i \mu} \frac{\partial {\Psi}}{\partial {t}} -i \frac{\partial {\mu}}{\partial {t}} e^{-i \mu} \Psi,\end{aligned}

and the scalar potential term transforms as

\begin{aligned}e \phi' \Psi'&=e \left( \phi - \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \right) e^{-i \mu } \Psi\end{aligned}

Putting the pieces together we have

\begin{aligned}i \hbar e^{-i \mu}\left( \frac{\partial {}}{\partial {t}} -i \frac{\partial {\mu}}{\partial {t}} \right) \Psi &=\frac{1}{{2m}}\left(\mathbf{p} -\frac{e}{c} \mathbf{A} -\frac{e}{c} \boldsymbol{\nabla} \chi \right)e^{-i \mu} \left( \mathbf{p} - \frac{e}{c} \mathbf{A} - \boldsymbol{\nabla} (\hbar \mu + \frac{e}{c} \chi) \right) \Psi + e \left( \phi - \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \right) e^{-i \mu } \Psi\end{aligned}

We need one more intermediate result, that of

\begin{aligned}\mathbf{p} e^{-i \mu } \mathbf{D}&= - i \hbar e^{-i \mu} \left( -i (\boldsymbol{\nabla} \mu) + \boldsymbol{\nabla} \right) \mathbf{D} \\ &= e^{-i\mu} (\mathbf{p} - \hbar \boldsymbol{\nabla} \mu) \mathbf{D}.\end{aligned}

So we have

\begin{aligned}i \hbar \frac{\partial {\Psi}}{\partial {t}}+\hbar \frac{\partial {\mu}}{\partial {t}} \Psi &=\frac{1}{{2m}}\left(\mathbf{p} - \hbar \boldsymbol{\nabla} \mu -\frac{e}{c} \mathbf{A} -\frac{e}{c} \boldsymbol{\nabla} \chi \right)\left( \mathbf{p} - \frac{e}{c} \mathbf{A} - \boldsymbol{\nabla} (\hbar \mu + \frac{e}{c} \chi) \right) \Psi + e \left( \phi - \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \right) \Psi.\end{aligned}

To get rid of the $\mu$, and $\chi$ time partials we need

\begin{aligned}\hbar \frac{\partial {\mu}}{\partial {t}} = - \frac{e}{c} \frac{\partial {\chi}}{\partial {t}}\end{aligned}

Or

\begin{aligned}\mu = - \frac{e}{c\hbar} \chi\end{aligned}

This also kills off all the additional undesirable terms in the transformed $\mathbf{P}^2$ operator (with $\mathbf{P} = \mathbf{p} - e \mathbf{A}/c$), leaving the invariant transformation completely specified

\begin{aligned}\mathbf{A}' &= \mathbf{A} + \boldsymbol{\nabla} \chi \\ \phi' &= \phi - \frac{1}{{c}} \frac{\partial { \chi }}{\partial {t}} \\ \Psi' &= \exp\left( i \frac{e}{ \hbar c } \chi \right) \Psi,\end{aligned}

This is a fair bit different than Feynman’s result, but since he starts with the wrong electrodynamic guage transformation, that’s not too unexpected.

# Second Lecture

This isn’t errata, but I found the following required slight exploration. He gives (implicitly)

\begin{aligned}\overline{\sin^2(\omega t - \mathbf{K} \cdot \mathbf{x})} = \frac{1}{{2}}\end{aligned}

Is this an average over space and time? How would one do that? What do we get just integrating this over the volume? That dot product is $\mathbf{K} \cdot \mathbf{x} = 2 \pi \left(\frac{m}{\lambda_1} x + \frac{n}{\lambda_2} y + \frac{o}{\lambda_3} z \right)$. Our average over the volume, for $m \ne 0$, using wolfram alpha to do the dirty work

\begin{aligned}&\frac{1}{{\lambda_1 \lambda_2 \lambda_3}} \int_{z=0}^{\lambda_3} dz\int_{y=0}^{\lambda_2} dy\int_{x=0}^{\lambda_1}dx \sin^2 \left( -\frac{2 \pi m x}{\lambda_1} -\frac{2 \pi n y}{\lambda_2} -\frac{2 \pi o z}{\lambda_3} + \omega t \right) \\ &=\frac{1}{{\lambda_1 \lambda_2 \lambda_3}} \int_{z=0}^{\lambda_3} dz\int_{y=0}^{\lambda_2} dy{\left.\frac{-\lambda_1}{4 \pi m} \left( -\frac{2 \pi m }{\lambda_1} x -\frac{2 \pi n y}{\lambda_2} -\frac{2 \pi o z}{\lambda_3} + \omega t \right)\right\vert}_{x=0}^{\lambda_1} \\ &-\frac{1}{{\lambda_1 \lambda_2 \lambda_3}} \int_{z=0}^{\lambda_3} dz\int_{y=0}^{\lambda_2} dy{\left.\frac{-\lambda_1}{8 \pi m} \sin \left( 2 \left(-\frac{2 \pi m }{\lambda_1} x -\frac{2 \pi n y}{\lambda_2} -\frac{2 \pi o z}{\lambda_3} + \omega t \right) \right)\right\vert}_{x=0}^{\lambda_1}\end{aligned}

Since the sine integral vanishes, we have just $1/2$ as expected regardless of the angular frequency $\omega$. Okay, that makes sense now. Looks like $\omega$ is only relavent for the single $\mathbf{K} = 0$ Fourier component, but that likely doesn’t matter since I seem to recall that the $\mathbf{K} = 0$ fourier component of this oscillators in a box problem was entirely constant (and perhaps zero?).