Fundamental Question

  • Thread starter Domnu
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  • #1
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Main Question or Discussion Point

Hi, I'm extremely new to quantum mechanics (my only knowledge of quantum mechanics is that taught in physics C, ergo none, and a bit of wave mechanics... however I have a pretty strong mathematics background... diff-eqs + linear alg. + vector calc.), and was wondering as to how the derivation of

[tex]

\[
E = \hbar \omega \iff p = \hbar k
\]

[/tex]

worked. Here's my derivation, which seems encouraging, but could someone tell me where my derivation messed up?

[tex]

\[
E = \hbar \omega \iff \frac{1}{2} pv = \hbar \cdot 2 \pi f \iff
\frac{1}{2} p= \hbar \cdot \frac{2\pi}{\lambda} \iff p = 2 \hbar k \neq \hbar k
\]

[/tex]
 

Answers and Replies

  • #2
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The problem lies in the difference between phase velocity and group velocity. And neither of these equations are really derived, although they are "consistent". de Broglie did it from analogies between classical mechanics and Fermat's ideas about light.
 
  • #3
dx
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[tex]

\[
E = \hbar \omega \iff \frac{1}{2} pv = \hbar \cdot 2 \pi f \iff
\frac{1}{2} p= \hbar \cdot \frac{2\pi}{\lambda} \iff p = 2 \hbar k \neq \hbar k
\]

[/tex]
You should use the relativistic expression for energy. For example, to verify the relation for a photon, use

[tex] E = pc [/tex].

This can be derived from electromagnetic theory, or from the relation

[tex] E^2 = p^2 c^2 + m^2 c^4 [/tex]

by setting m = 0. If we put this in [tex] E = \hbar \omega [/tex], we get

[tex] pc = \hbar \omega = \hbar 2\pi f[/tex].

Since [tex] f\lambda = c [/tex], we get

[tex] p = \hbar \frac{2\pi}{\lambda} = \hbar k [/tex].
 

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