Is this a typo? (Quantum Theory for Mathematicians by Brian C. Hall)

[danielristic]

TL;DR

Just started reading this book and wondering if this is a typo or if I'm already lost.

Hello,

After reading a few vulgarisation books, I'm looking into familiarising myself with the more mathematical aspects of quantum physics so I've started reading Quantum Theory for Mathematicians by Brian C. Hall.

I'm only 9 pages in but I've already spotted what I think is a typo. I checked online and the author is providing some corrections but this one is not part of the list.

Here's the relevant except:

As "e" was never introduced I'm assuming this is really "Q", what do you think?

 

[PeroK]

Summary:: Just started reading this book and wondering if this is a typo or if I'm already lost.

Hello,

After reading a few vulgarisation books, I'm looking into familiarising myself with the more mathematical aspects of quantum physics so I've started reading Quantum Theory for Mathematicians by Brian C. Hall.

I'm only 9 pages in but I've already spotted what I think is a typo. I checked online and the author is providing some corrections but this one is not part of the list.

Here's the relevant except:

View attachment 259386

As "e" was never introduced I'm assuming this is really "Q", what do you think?

Assuming you are a mathematician, can't you prove from those equations that $Q = \pm e$?

 

[danielristic]

From a mathematical standpoint, given the last two equations sure, you can infer that $Q^2 = e^2$ and that therefore $Q=e$ or $Q=-e$ but I'm not trying to solve a maths problem here. In physics, variables are attached to measurable quantities or constants and conventional notations tend to use the same symbols to designate the same things.

I don't have an extensive background in physics but I know that $e$ is commonly used for the charge of an electron but the previous except states "Q is the charge of the electron" so I was just trying to figure out if Q and e had a different conventional meaning for physicists or if it was simply a typographical error.

 

[PeroK]

From a mathematical standpoint, given the last two equations sure, you can infer that $Q^2 = e^2$ and that therefore $Q=e$ or $Q=-e$ but I'm not trying to solve a maths problem here. In physics, variables are attached to measurable quantities or constants and conventional notations tend to use the same symbols to designate the same things.

I don't have an extensive background in physics but I know that $e$ is commonly used for the charge of an electron but the previous except states "Q is the charge of the electron" so I was just trying to figure out if Q and e had a different conventional meaning for physicists or if it was simply a typographical error.

It certainly looks like he suddenly decided to use $e$ instead of $Q$ and then immediately changed his mind again!

From what I know, $Q$ is generally used as a generic symbol for the charge of anything. And, $e$ is often used specifically as the charge on an electron; or as minus the charge on an electron, i.e. the charge on a proton, depending on the author.

 

[danielristic]

I see, makes sense. Thanks!

 

[vanhees71]

From a mathematical standpoint, given the last two equations sure, you can infer that $Q^2 = e^2$ and that therefore $Q=e$ or $Q=-e$ but I'm not trying to solve a maths problem here. In physics, variables are attached to measurable quantities or constants and conventional notations tend to use the same symbols to designate the same things.

I don't have an extensive background in physics but I know that $e$ is commonly used for the charge of an electron but the previous except states "Q is the charge of the electron" so I was just trying to figure out if Q and e had a different conventional meaning for physicists or if it was simply a typographical error.

Be careful. Usually $e$ is the charge of the proton (positive). An electron then has charge $Q_{\text{e}}=-e$ (negative), but some textbooks/papers may use a different notation.

 

[forcefield]

Why is there no Coulomb's constant in (1.3)?

Btw, AFAIK this is all classical physics.

 

[Gaussian97]

I suppose that is because, "in appropriate units", $K=1$