Anyone care to explain the TA's note?

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The discussion centers on the relationship between wavelength, frequency, and angular frequency in wave mechanics. The formula for angular frequency is established as ω = 2πf, which leads to the wavelength being expressed as λ = 2πc/ω. Participants clarify that while λ = c/f is valid, it does not account for the angular frequency representation. The confusion arises from the use of ħ (h-bar) in the context of energy differences, reinforcing the importance of distinguishing between linear and angular frequency in calculations.

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Shackleford
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Omega here is frequency.

I used lambda = c/f. Why is it (2*pi*c)/omega and not c/f or c/omega?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-04203213.jpg?t=1286245112

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-04203118.jpg?t=1286245065
 
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It looks like the formula you used was derived as

\omega_{ij} = \frac{E_j - E_i}{\hbar},

which is an angular frequency as can be seen from the presence of \hbar rather than h.
 
fzero said:
It looks like the formula you used was derived as

\omega_{ij} = \frac{E_j - E_i}{\hbar},

which is an angular frequency as can be seen from the presence of \hbar rather than h.

Yeah. But h-bar is already in the denominator in the formula I used to calculate omega_{ij}.
 
The wavelength is still 2\pi c/\omega.
 
Why not ask your TA for clarification?
 
fzero said:
The wavelength is still 2\pi c/\omega.

Why is lambda = c/f not valid?
 
cristo said:
Why not ask your TA for clarification?

I forgot. I already bugged him about two other problems. I did poorly on this damn homework.
 
Shackleford said:
Why is lambda = c/f not valid?

It is, but \omega = 2\pi f.
 
fzero said:
It is, but \omega = 2\pi f.

Well, that's my mistake. I wasn't taking omega to be an angular frequency. Why? I don't know. Son of a *****.
 

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