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What does the wavevector \vec{k} mean?

  1. Mar 23, 2012 #1


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    Dear Forum
    The wavefunction for a wave traveling in direction r can be written as
    [tex]\psi ( \vec{r}, t ) = A \cos ( \vec{k} \cdot \vec{r} - \omega t + \phi ) [/tex], where [tex]\vec{k}[/tex] is the wave vector.
    In one dimension, [tex]k = 2*\pi/\lambda[/tex], so is it correct to write the vector components of [tex]\vec{k}= [k_1, k_2, k_3] = [2\pi/\lambda_1, 2\pi/\lambda_2, 2\pi/\lambda_3][/tex]?

    Thanks for any hints.
  2. jcsd
  3. Mar 26, 2012 #2
    Hi mzh,

    Yes, the vector specifies how many wavefronts one will measure in a unit length in each direction. (The geometric decomposition of a plane wave)

    Note that you can you create a scalar wavenumber which may also use k as a variable name. One convention when doing that is to not use italics (where you would use italics for the vector).

    Last edited: Mar 26, 2012
  4. Mar 26, 2012 #3


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    @phildsp, thanks. ok, if i get it right, this means the wavevector is a generalization of the wavenumber concept. ok. thanks.
  5. Mar 26, 2012 #4
    The first half is correct. However, you cannot write [tex][k_1, k_2, k_3] = [2\pi/\lambda_1, 2\pi/\lambda_2, 2\pi/\lambda_3][/tex] if [tex][\lambda_1,\lambda_2,\lambda_3]=\frac{2\pi}{k^2}[k_1,k_2,k_3][/tex] is a vector in the direction of propagation and with the length λ.

    A dot product between a position vector and a wave vector is well defined and easy to work with mathematically. That's why this notation is used. Dividing the position by a "wavelength vector" is not well defined, or rather the only sensible definition is via the wave vector. So rather than using some weird, indirect definition you just use the wave vector.

    Note that this is true for all kinds of waves: Wave functions in QM, electromagnetic waves, acoustics, ... As soon as you have to deal with diffraction and interference you are much better off with the wave vector. Condensed matter physics is done almost exclusively in reciprocal space, i.e. writing things as function of the wave vector instead of real space position.
  6. Mar 26, 2012 #5


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    thanks quack for your comprehensive reply. however, i'm not sure if i understand all. the following i see, the dot product [tex]\vec{k}\cdot\vec{r}[/tex] needs to be unitless. That means the unit of the components of [tex]\vec{k}[/tex] should be reciprocal distance. But if what I wrote is wrong, then what do the components of [tex]\vec{k}[/tex] look like?
  7. Mar 26, 2012 #6
    Units of lambda are meters (or mm, Anstrom or whatever distance units you use)

    Units of k are 1/m (1/mm 1/Angstrom ...), so each component of k has the same units.

    Just as [itex]\vec{r}[/itex]. Each component has units meters.
  8. Mar 26, 2012 #7
    It seems the definition of wavenumber that Wikipedia gives is a specific wavenumber scalar that could be called

    [tex]k_{mag} = \sqrt{k_x^2 + k_y^2 + k_z^2}[/tex]

    where [itex]k_x[/itex] is the spatial frequency in the x dimension such that

    [tex]k_x = \vec{k} \ \cdot <1, 0, 0>[/tex]
    [tex]k_y = \vec{k} \ \cdot <0, 1, 0>[/tex]
    [tex]k_z = \vec{k} \ \cdot <0, 0, 1>[/tex]
    Last edited: Mar 26, 2012
  9. Mar 27, 2012 #8


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    @Phildsp/quack: Ok, thanks. So I see it this way that the wavevector is a "generalization" of the wavenumber.
  10. Mar 27, 2012 #9
    Yes, it combines the wave number and the direction of propagation.

    The wave vector is an extremely useful concept in wave physics. Taking time to understand it properly will save you headaches later on :-)
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