I What is the significance of the dispersion relation?

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The dispersion relation is crucial as it connects energy and momentum, revealing how wave packets behave in a system. For free electrons in a vacuum, the relation E = (ħ²k²)/(2m) indicates that energy is quadratic in momentum, which informs about the particle's dynamics. The group velocity of a wave packet, given by v_g = ħk/m, reflects how the wave propagates through a medium. Understanding dispersion relations is essential for applications like semiconductor physics, particularly in analyzing energy and momentum conservation during light absorption and emission. This knowledge enhances insights into electron behavior within various materials.
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what is the significance of the dispersion relation?
hi guys
i would like to know what is the physical significance of the dispersion relation , i know that it relates the energy and momentum vector and correspondingly the energy and momentum with each other , but what does this tell me about the system , and why should i care that the dispersion relation for free electrons in vacuum is given by
$$E=\frac{\hbar^{2}k^{2}}{2m}$$
and for light just ##\omega=ck## , it seems that the electron's energy in vacuum is quadratic in the momentum but what do i gain by knowing that ??
i have too many questions , i will appreciate any help ,thanks .
 
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It tells you how the wave function behaves. Particularly it tells you that a wave packet, which is not too broad in momentum space moves with the group velocity ##\vec{v}_g=\vec{\nabla}_k \omega(\vec{k})##. What do you get for the dispersion relations you mentioned?
 
vanhees71 said:
It tells you how the wave function behaves. Particularly it tells you that a wave packet, which is not too broad in momentum space moves with the group velocity ##\vec{v}_g=\vec{\nabla}_k \omega(\vec{k})##. What do you get for the dispersion relations you mentioned?
since $$\omega(k) = \frac{\hbar*k^{2}}{2m} \;⇒ $$
$$
\left(\frac{\partial}{\partial\;k_{x}}\hat{k_{x}}+\frac{\partial}{\partial\;k_{y}}\hat{k_{y}}+\frac{\partial}{\partial\;k_{z}}\hat{k_{z}}\right)\left[\omega(\vec{k}) \right]= \left(\frac{\partial}{\partial\;k_{x}}\hat{k_{x}}+\frac{\partial}{\partial\;k_{y}}\hat{k_{y}}+\frac{\partial}{\partial\;k_{z}}\hat{k_{z}}\right)\left[\frac{\hbar*\vec{k}.\vec{k}}{2m}\right]
$$
$$\vec{v}_{g} = \frac{\hbar}{m}\left[k_{x}\hat{k_{x}}+k_{y}\hat{k_{y}}+k_{z}\hat{k_{z}}\right]$$
isn't that right
 
Yes, and that looks pretty familiar for the velocity of a free particle, because what you got is
$$\vec{v}_g=\hbar \vec{k}/m=\vec{p}/m.$$
 
so the dispersion relation is useful for determining the group velocity of the wave packet inside a medium , i am sorry i have another question , considering an electron moving inside a solid , after solving the Schrodinger equation ,say the following E,k diagram is obtained
gan_bands.jpg

is that complicated diagram just indicate that the electron wave packet is changing its velocity inside the lattice or is there more insight to it
 
patric44 said:
Summary:: what is the significance of the dispersion relation?

hi guys
i would like to know what is the physical significance of the dispersion relation , i know that it relates the energy and momentum vector and correspondingly the energy and momentum with each other , but what does this tell me about the system , and why should i care that the dispersion relation for free electrons in vacuum is given by
$$E=\frac{\hbar^{2}k^{2}}{2m}$$
and for light just ##\omega=ck## , it seems that the electron's energy in vacuum is quadratic in the momentum but what do i gain by knowing that ??
i have too many questions , i will appreciate any help ,thanks .

Well, one thing that could come to mind is "How can I obey both conservation of energy and conservation of momentum for the absorption and emission of light"? This question is highly relevant in semiconductor physics.
 
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