Plane Waves

As we have just seen, a wave of amplitude $A$, wavenumber $k$, angular frequency $\omega$, and phase angle $\varphi$, propagating in the positive $x$-direction, is represented by the following wavefunction:

$$\psi(x,t)=A \cos(k x-\omega t+\varphi).$$ (29)

Now, the type of wave represented above is conventionally termed a one-dimensional plane wave. It is one-dimensional because its associated wavefunction only depends on the single Cartesian coordinate $x$. Furthermore, it is a plane wave because the wave maxima, which are located at

$$k x-\omega t+\varphi = j 2\pi,$$ (30)

where $j$ is an integer, consist of a series of parallel planes, normal to the $x$-axis, which are equally spaced a distance $\lambda=2\pi/k$ apart, and propagate along the positive $x$-axis at the velocity $v=\omega/k$. These conclusions follow because Eq. (30) can be re-written in the form

$$x= d,$$ (31)

where $d=(j-\varphi/2\pi) \lambda + v t$. Moreover, as is well-known, (31) is the equation of a plane, normal to the $x$-axis, whose distance of closest approach to the origin is $d$.

Figure 1: The solution of ${\bf n}\cdot{\bf r} = d$ is a plane.

The previous equation can also be written in the coordinate-free form

$${\bf n}\cdot{\bf r} = d,$$ (32)

where ${\bf n} = (1, 0, 0)$ is a unit vector directed along the positive $x$-axis, and ${\bf r}=(x, y, z)$ represents the vector displacement of a general point from the origin. Since there is nothing special about the $x$-direction, it follows that if ${\bf n}$ is re-interpreted as a unit vector pointing in an arbitrary direction then (32) can be re-interpreted as the general equation of a plane. As before, the plane is normal to ${\bf n}$, and its distance of closest approach to the origin is $d$. See Fig. 1. This observation allows us to write the three-dimensional equivalent to the wavefunction (29) as

$$\psi(x,y,z,t)=A \cos({\bf k}\cdot{\bf r}-\omega t+\varphi),$$ (33)

where the constant vector ${\bf k} = (k_x, k_y, k_z)=k {\bf n}$ is called the wavevector. The wave represented above is conventionally termed a three-dimensional plane wave. It is three-dimensional because its wavefunction, $\psi(x,y,z,t)$, depends on all three Cartesian coordinates. Moreover, it is a plane wave because the wave maxima are located at

$${\bf k}\cdot{\bf r} -\omega t +\varphi= j 2\pi,$$ (34)

or

$${\bf n}\cdot{\bf r} = (j-\varphi/2\pi) \lambda + v t,$$ (35)

where $\lambda=2\pi/k$, and $v=\omega/k$. Note that the wavenumber, $k$, is the magnitude of the wavevector, ${\bf k}$: i.e., $k\equiv \vert{\bf k}\vert$. It follows, by comparison with Eq. (32), that the wave maxima consist of a series of parallel planes, normal to the wavevector, which are equally spaced a distance $\lambda$ apart, and which propagate in the ${\bf k}$-direction at the velocity $v$. See Fig. 2. Hence, the direction of the wavevector specifies the wave propagation direction, whereas its magnitude determines the wavenumber, $k$, and, thus, the wavelength, $\lambda=2\pi/k$.

Figure 2: Wave maxima associated with a three-dimensional plane wave.