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

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Summary:

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

Main Question or Discussion Point

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:

Hall_2013_pdf__page_26_of_566_.png


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

Answers and Replies

  • #2
PeroK
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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##?
 
  • #3
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.
 
  • #4
PeroK
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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.
 
  • #5
I see, makes sense. Thanks!
 
  • #6
vanhees71
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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.
 
  • #7
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Why is there no Coulomb's constant in (1.3)?

Btw, AFAIK this is all classical physics.
 
  • #8
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I suppose that is because, "in appropriate units", ##K=1##
 
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