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I am currently studying Laser Systems Engineering by Keith Kasunic. Chapter 1.2 Laser Engineering says the following:

If the energy of the incident electromagnetic field more-or-less matches that of the electron’s excited state energy $E_2$ compared with some lower-energy state $E_1$, then there is a high probability the electron will give up its energy in the form of a stimulated photon whose energy $E_p$ in Fig. 1.9(a) is $$E_p = E_2 - E_1 = h \nu = \dfrac{h c}{\lambda} \ \ \ \ \ \ \ [\mathrm{J}] \tag{1.2}$$ where $h = 6.626 \times 10^{-34}$ J-sec is Planck’s constant.

What is $[\mathrm{J}]$ supposed to denote in this case? I read this relevant question on square brackets in dimensional analysis, but I don't think it clarifies what it means in this particular case.

The Pointer
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    It means (1.2) is expressed in Joules. – ZeroTheHero Dec 07 '20 at 03:25
  • @ZeroTheHero That's all? There isn't some additional meaning given by the square brackets? Why wouldn't the author just leave it as $\mathrm{J}$, then? – The Pointer Dec 07 '20 at 03:25
  • depends on the authors I guess. It's sometimes indicated so that in particular the value of hbar is taken to be in correct numerical value in Joules. – ZeroTheHero Dec 07 '20 at 04:03
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    The [J] is pointless and confusing, since equations like 1.2 are true in any units. – G. Smith Dec 07 '20 at 04:22
  • @G.Smith What do you mean that such equations are true "in any units"? – The Pointer Dec 07 '20 at 04:23
  • Consider something simpler. The equation $x=vt$ does not need an [m] next to it because $x$ does not have to be measured in meters. You can measure $x$ in furlongs and $t$ in fortnights, or any other units that you find convenient, and $x=vt$ will still be true. – G. Smith Dec 07 '20 at 04:25
  • @G.Smith Oh, ok, I see what you mean. – The Pointer Dec 07 '20 at 04:27
  • I suppose that once the value of the Planck constant is specified (with units) as was done here, then it follows that the energy will be in Joules (if $\nu$ is in inverse seconds) but I agree with @G.Smith it's not terribly illuminating to include the [J]. – ZeroTheHero Dec 07 '20 at 04:39
  • And if $c$ is in meters per second. And if you don’t do any unit conversions. I think the [J] may be there as a “helpful reminder” because this is a book for engineers rather than physicists. Do they put the SI unit for the quantity being computed next to every equation, or just this one? – G. Smith Dec 07 '20 at 04:41
  • @ZeroTheHero This is awfully confusing. How does that follow? If $h$ is in J-sec, $\nu$ is in $s^{-1}$, and $c$ is in $m \cdot s^{-1}$, then don't we end up with final units of J-sec (by cancellation) instead of just J? – The Pointer Dec 07 '20 at 04:47
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    Clearly there's a typo somewhere as $h\nu \ne h c/\nu$. I think the last one should be $h c/\lambda$ since $\lambda \nu =c$. – ZeroTheHero Dec 07 '20 at 05:11
  • @ZeroTheHero Oh, you're right; that was my typo! Sorry about that. The units make sense now. – The Pointer Dec 07 '20 at 05:13

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$$E_p = E_2 - E_1 = h \nu = \dfrac{h c}{\lambda} \ \ \ \ \ \ \ [\mathrm{J}] \tag{1.2}$$ where $h = 6.626 \times 10^{-34}$ J-sec is Planck’s constant.

Another way of writing this is,

$$E_p/J = E_2 - E_1 = h \nu = \dfrac{h c}{\lambda} \ \ \ \ \ \ \ \tag{1.3}.$$

It means that $E_p$ is expressed in units of $Joules$, so $E_p/J$ is $E_p$ per unit $J$.

The Pointer
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Daddy Kropotkin
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