What does the wavevector \vec{k} mean?

[mzh]

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

Thanks for any hints.

 

[PhilDSP]

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).

http://en.wikipedia.org/wiki/Wave_vector
http://en.wikipedia.org/wiki/Wavenumber

 

[mzh]

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).

http://en.wikipedia.org/wiki/Wave_vector
http://en.wikipedia.org/wiki/Wavenumber

@phildsp, thanks. ok, if i get it right, this means the wavevector is a generalization of the wavenumber concept. ok. thanks.

 

[M Quack]

In one dimension, $$k = 2*\pi/\lambda$$, so is it correct to write the vector components of $$\vec{k}= [k_1, k_2, k_3] = [2\pi/\lambda_1, 2\pi/\lambda_2, 2\pi/\lambda_3]$$?

The first half is correct. However, you cannot write $$[k_1, k_2, k_3] = [2\pi/\lambda_1, 2\pi/\lambda_2, 2\pi/\lambda_3]$$ if $$[\lambda_1,\lambda_2,\lambda_3]=\frac{2\pi}{k^2}[k_1,k_2,k_3]$$ 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.

 

[mzh]

The first half is correct. However, you cannot write $$[k_1, k_2, k_3] = [2\pi/\lambda_1, 2\pi/\lambda_2, 2\pi/\lambda_3]$$ if $$[\lambda_1,\lambda_2,\lambda_3]=\frac{2\pi}{k^2}[k_1,k_2,k_3]$$ 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.

thanks quack for your comprehensive reply. however, I'm not sure if i understand all. the following i see, the dot product $$\vec{k}\cdot\vec{r}$$ needs to be unitless. That means the unit of the components of $$\vec{k}$$ should be reciprocal distance. But if what I wrote is wrong, then what do the components of $$\vec{k}$$ look like?

 

[M Quack]

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 $\vec{r}$. Each component has units meters.

 

[PhilDSP]

It seems the definition of wavenumber that Wikipedia gives is a specific wavenumber scalar that could be called

$$k_{mag} = \sqrt{k_x^2 + k_y^2 + k_z^2}$$

where $k_x$ is the spatial frequency in the x dimension such that

$$k_x = \vec{k} \ \cdot <1, 0, 0>$$
$$k_y = \vec{k} \ \cdot <0, 1, 0>$$
$$k_z = \vec{k} \ \cdot <0, 0, 1>$$

 

[mzh]

It seems the definition of wavenumber that Wikipedia gives is a specific wavenumber scalar that could be called

$$k_{mag} = \sqrt{k_x^2 + k_y^2 + k_z^2}$$

where $k_x$ is the spatial frequency in the x dimension such that

$$k_x = \vec{k} \ \cdot <1, 0, 0>$$
$$k_y = \vec{k} \ \cdot <0, 1, 0>$$
$$k_z = \vec{k} \ \cdot <0, 0, 1>$$

@Phildsp/quack: Ok, thanks. So I see it this way that the wavevector is a "generalization" of the wavenumber.

 

[M Quack]

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 :-)