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How to deduce the units of hw?

  1. Mar 16, 2012 #1
    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.
     
  2. jcsd
  3. Mar 16, 2012 #2

    Pengwuino

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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].
     
  4. Mar 16, 2012 #3
    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.
     
  5. Mar 16, 2012 #4

    Pengwuino

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    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 .
     
  6. Mar 16, 2012 #5
    Yes that's true, that is what i'm trying to figure out? in what units is the 87.2927...?,
     
  7. Mar 16, 2012 #6

    Pengwuino

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    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.
     
  8. Mar 17, 2012 #7
    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?
     
  9. Mar 17, 2012 #8
    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.
     
  10. Mar 17, 2012 #9

    Cthugha

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    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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