How to deduce the units of hw?

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Discussion Overview

The discussion revolves around deducing the units of angular frequency (\(\omega\)) and the reduced Planck's constant (\(\hbar\)) in the context of a calculation presented in Paul Harrison's book on quantum wells. Participants are trying to reconcile the numerical values provided with their respective units to ensure consistency with the energy value of 871.879 meV.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that \(\hbar\) is given as \(6.58214928 \times 10^{-13} \text{ meV}\) and infers \(\omega\) to be \(9.9879927\), questioning the units of \(\omega\) and \(\hbar\).
  • Another participant clarifies that in certain units, \(\hbar\) is \(6.5821 \times 10^{-13} \text{ meV s}\), suggesting that \(\omega\) could be \(1.325 \times 10^{15} \text{ radians/seconds}\).
  • Some participants express confusion about the units of \(\hbar\) and \(\omega\), with one stating that arbitrary units could be used for \(\hbar\) but questioning the validity of the numerical value of \(87.2827...\).
  • There is a claim that the initial calculation of \(\omega\) was incorrect, with a participant asserting that it should be \(1.32 \times 10^{15} \text{ radians/seconds}\) and challenging the use of units for \(m_0\).
  • Another participant points out that \(\hbar\) has units of meV s and that \(\omega\) has units of \(s^{-1}\), which would yield the correct units for the product \(\hbar \omega\) as meV.

Areas of Agreement / Disagreement

Participants express differing views on the units of \(\hbar\) and \(\omega\), with no consensus reached on the correct interpretation of the values and units involved in the calculations.

Contextual Notes

There are unresolved questions regarding the assumptions made in the calculations, particularly concerning the units used for mass and the numerical values of constants. The discussion highlights the complexity of unit conversions and the importance of clarity in definitions.

cashimingo
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Hi everyone, in the book of Paul Harrison (Quantum Wells, Wires and Dots, 2nd Ed.), it is said that [itex]\hbarω=871.879 meV[/itex] but it is known that [itex]\hbar=6.58214928\times10^{-13}meV[/itex] so i can inferred that [itex]\omega=9.9879927[/itex], but i don't know the units of [itex]\omega[/itex], or how can i deduce what units he used for [itex]\omega[/itex] and what units for [itex]\hbar[/itex], so that when i multiply both, it results 871.879 meV?.

Any suggestion would be really appreciated.
Thanks.
 
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Planck's constant, in those units, is actually [itex]\hbar = 6.5821... \times 10^{-13} meV s[/itex]. There is a 'seconds' in Planck's constant. Thus the angular frequency is [itex]1.325 \times 10^{15} {{radians}\over{seconds}}[/itex].
 
Ah ok, yeah, but it is also true that [itex]\hbar=87.2827...[/itex] but in what units?.
Because (87.2827..)(9.9878..)=871.879 meV, my problem is that i don't know what are the units of each number.

Thanks for the reply.
 
cashimingo said:
Ah ok, yeah, but it is also true that [itext]\hbar=87.2827...[\itext] but in what units?.
Because (87.2827..)(9.9878..)=871.879 meV, my problem is that i don't know what are the units of each number.

Thanks for the reply.

You can arbitrarily make up units to make Planck's constant whatever you want, but I've never seen Planck's constant with the numerical value of 87.2827...

The common values for Planck's constant is given at http://en.wikipedia.org/wiki/Planck's_constant .
 
Yes that's true, that is what I'm trying to figure out? in what units is the 87.2927...?,
 
cashimingo said:
Yes that's true, that is what I'm trying to figure out? in what units is the 87.2927...?,

You did your calculation incorrectly at the very start. You should have determined that [itex]\omega = 1.32\times 10^{15} {{radians}\over{seconds}}[/itex]. No SI units would allow you to get the 87.2927 number for Planck's constant.
 
Well, maybe i haven't explain the problem correctly. In Harrison's book, there is the following calculation

[itex]\omega=\sqrt{8\Delta V_{CB}(E_{g}^{AlAs}-E_{g}^{GaAs})(x_{max}-x_{min})} \frac{1}{a\sqrt{m}}[/itex]

where
[itex]V_{CB}=0.67,[/itex]
[itex]E_{g}^{AlAs}-E_{g}^{GaAs}=1247 meV,[/itex]
[itex]x_{max}=10,[/itex]
[itex]x_{min}=0,[/itex]
[itex]a=100angstroms[/itex]
[itex]m=0.067 m_{0}[/itex]

then (that's what he says) [itex]\hbar\omega=871.879 meV[/itex].
But when i perform the calculation, i get
[itex]\omega≈9.9879927[/itex], so i get [itex]\hbar≈87.2927[/itex].
That's why i asked in what units was the [itex]\hbar[/itex] and the [itex]\omega[/itex].

Any comment of the calculation or what is wrong, would be really appreciated.
Thanks.
By the way, how can i write the angstrom symbol?
 
Cashmingo pengwuino has answered your question.h bar has units of meVs and not meV as you wrote incorrectly in post 1.Omega has units of seconds to the minus one and so,therefore,h bar omega has units of meV as you wrote correctly in post 1.
 
cashimingo said:
But when i perform the calculation, i get
[itex]\omega≈9.9879927[/itex]

That is off by roughly 14 orders of magnitude. I crosschecked and got [itex]\omega≈1.325*10^{15}\frac{1}{s}[/itex] which matches the energy given in the book.

Are you sure you used the right units for m0? It is [itex]\frac{510998.928 eV}{c^2}[/itex].
 

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