How to deduce the units of hw?

[cashimingo]

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

Any suggestion would be really appreciated.
Thanks.

 

[Pengwuino]

Planck's constant, in those units, is actually \hbar = 6.5821... \times 10^{-13} meV s. There is a 'seconds' in Planck's constant. Thus the angular frequency is 1.325 \times 10^{15} {{radians}\over{seconds}}.

 

[cashimingo]

Ah ok, yeah, but it is also true that \hbar=87.2827... 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.

 

[Pengwuino]

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 .

 

[cashimingo]

Yes that's true, that is what I'm trying to figure out? in what units is the 87.2927...?,

 

[Pengwuino]

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 \omega = 1.32\times 10^{15} {{radians}\over{seconds}}. No SI units would allow you to get the 87.2927 number for Planck's constant.

 

[cashimingo]

Well, maybe i haven't explain the problem correctly. In Harrison's book, there is the following calculation

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

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

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

Any comment of the calculation or what is wrong, would be really appreciated.
Thanks.
By the way, how can i write the angstrom symbol?

 

[Dadface]

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.

 

[Cthugha]

But when i perform the calculation, i get
\omega≈9.9879927

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

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