What is the significance of the wavevector in optics and k-space?

  • Context: Graduate 
  • Thread starter Thread starter iLIKEstuff
  • Start date Start date
Click For Summary

Discussion Overview

The discussion centers on the concept of the wavevector in optics and its significance in k-space, particularly in relation to E(k) vs k diagrams and optical absorption in semiconductors. Participants explore the mathematical and physical interpretations of the wavevector, its components, and its role in different contexts such as optics and solid state physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the wavevector and its relation to the wave number, seeking clarity on its significance in k-space and E(k) diagrams.
  • Another participant suggests that the wavevector is a convenient mathematical representation of both the direction of a wave and its wavelength.
  • A participant questions the need to break the wavevector into components, proposing that it is necessary for taking second derivatives in equations involving vector quantities.
  • It is noted that in dispersion relations, the wavevector is proportional to the momentum of particles or quasiparticles, linking it to energy versus momentum plots.
  • A distinction is made between the wavevector of electromagnetic waves and that of electron waves, highlighting the context-dependent nature of the concept.
  • One participant provides a trigonometric explanation of how to find the components of the wavevector in two and three dimensions.
  • Another participant emphasizes the importance of understanding the physical meaning behind the wavevector and its components.
  • A later reply connects the wavevector to momentum, stating that a smaller wavelength corresponds to greater momentum.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and interpretations of the wavevector, with no consensus reached on its significance or the best approach to comprehend it. Multiple competing views on its role in different contexts remain present.

Contextual Notes

Some participants highlight the complexity of the wavevector's role in different fields, such as optics versus solid state physics, and the necessity of breaking it into components for mathematical clarity. There are also unresolved aspects regarding the physical implications of the wavevector in relation to momentum and energy.

iLIKEstuff
Messages
24
Reaction score
0
What is wavevector?

So I've been studying optics on my own for about 2 months now and I am having a lot of trouble understanding what exactly is the wavevector

\overrightarrow{k} = (k_x , k_y , k_z)

where wavenumber k is

k = | \overrightarrow{k} | = \sqrt{{k_x}^2 + {k_y}^2 + {k_z}^2}

i was told when i first started that the meaning of wave vector would just come to me as i digged deeper into optics, but now I'm looking at E(k) vs k diagrams for semiconductors and this talk about k-space is confusing the heck out of me.

so i know k= \frac{2 \pi}{ \lambda}

and that the wave number k is the coefficient in the helmholtz equation

{\nabla}^{2} U + {k}^{2} U = 0 where U is the complex amplitude and a function of x,y,z,t

but how does this relate to the wave vector? i know it doesn't expand like
\frac{{\partial}^{2}}{{\partial x}^{2}} \left ( U (x,t) \right ) + {k_x}^{2} U(x,t) =0
\frac{{\partial}^{2}}{{\partial y}^{2}} \left ( U (y,t) \right ) + {k_y}^{2} U(y,t) =0
\frac{{\partial}^{2}}{{\partial z}^{2}} \left ( U (z,t) \right ) + {k_z}^{2} U(z,t) =0

because k is the wave number. I'm lost.

perhaps i am approaching this concept in the wrong way. i would gladly appreciate any pointers on how to understand this menacing construct

i'm looking for a way to understand the wave vector in the context of understanding k-space and E(k) vs k diagrams to then understand optical absorption by transitions across band gaps for different materials. but also i simply just don't understand why they break up the wavevector into components whose magnitude is the wave number which is clearly (and understandably) defined as inversely proportional to the wavelength

thanks guys.
 
Science news on Phys.org


k is just a convenient mathematical way of expression both the direction of a wave and it's wavelength. (lambda = 2 pi / |k|, as you know)
 


iLIKEstuff said:
i simply just don't understand why they break up the wavevector into components

If it is a vector then it has components, U looks like a 4D vector to me, and if you want the second gradiant of a vector your going to need to break the equation into it's components so that you can take the second partial derivative of each component and then "put it back together".
 


iLIKEstuff said:
i was told when i first started that the meaning of wave vector would just come to me as i digged deeper into optics, but now I'm looking at E(k) vs k diagrams for semiconductors and this talk about k-space is confusing the heck out of me.

Maybe one should add, that in dispersion relations the k is proportional to the momentum carried by some particle or quasiparticle (crystal electron, phonon, exciton or others) and therefore E(k) diagrams are more or less plotting energy versus carried momentum.
 


Are you studying optics or solid state physics? The reason I ask is because there are some subtle differences in talking about the wavevector of an EM wave and the wavevector of an electron wave.

For waves in general, Redbelly gave the best physical explanation (post #2). When we start to talk E-k plots, these come as the result of performing Fourier analysis on the crystal lattice. The crystal lattice obviously exists in a position space. The Fourier transform of position is momentum which is directly proportional to the wavevector. In quantum mechanics, it is convenient to use wavevector rather than momentum. But really, when you talk about one, you equivalently talk about the other.
 


The wavevector \vec{k} has magnitude equal to the wave number k = 2\pi / \lambda and a direction that is the same as the direction of propagation of the wave.

You can find the components of the wavevector using trigonometry. For example, in two dimensions the wavevector is:

\vec{k} = k_x \hat{i} + k_y \hat{j}

where

|\vec{k}| = \sqrt{k_x^2 +k_y^2}

and

k_x = |\vec{k}|cos \theta

where \theta is the angle between \vec{k} and the x axis, for example.

In three dimensions, things are a little more complicated, but you can still use trignometry.

Oh, and:

{\nabla}^{2} U + {k}^{2} U = 0

expands as

\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} + \frac{\partial^2 U}{\partial z^2} + (k_x^2 + k_y^2 + k_z^2) U = 0,

which is, of course, a scalar equation, not a vector equation.
 
Last edited:


hi ,
1ist of all try to understand physical meaning of physics that is what physics is
wave number :number of wavelengths(λ) present in 2pie radians
wavevector :it gives us direction of the wave in which it travels that,s why it is breakdown into components because u always cannot expect the wave to travel into one dimension .

please read intoduction to solid state physics by CHARLES KITTEL
 


how i can ask questions ?
 


Start with something that makes sense, like the fact that you have more momentum with a smaller wavelength. So basically, p=h/\lambda, where h is some constant.

Since \hbar=h/2\pi,
p=h/\lambda=2\pi\hbar/\lambda=\hbar k,
where k=2\pi/\lambda, ie, k is just a vector in the direction of the momentum.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K