5.1 Plane waves

Plane waves

We want to find the expression for a plane that is perpendicular to $\vec{k}$ , where $\vec{k}$ is a vector in the direction of propagation of the wave.The plane is the set of points that has the same projection onto the vector $\vec{k}$ That is any point $\vec{r}$ that satisfies $$\vec{k}\cdot \vec{r}=constant$$ is a point on the planeNow consider the function $$\psi (\vec{r})=A{e}^{i\vec{k}\cdot \vec{r}}$$ we see that the magnitude of $\psi (\vec{r})$ is the same over every plane that is defined by $\vec{k}\cdot \vec{r}=constant$ we want to construct harmonic waves, ie. they should repeat every wavelength along the direction of propagation so they should satisfy $$\psi (\vec{r})=\psi \left(\vec{r}+\frac{\lambda \vec{k}}{k}\right)$$ where $\lambda$ is the wavelengththen we must have $${\displaystyle \begin{array}{c}A{e}^{i\vec{k}\cdot \vec{r}}=A{e}^{i\vec{k}\cdot \left(\vec{r}+\frac{\lambda \vec{k}}{k}\right)}\\ =A{e}^{i\vec{k}\cdot \vec{r}}{e}^{i\vec{k}\cdot \vec{k}\lambda /k}\\ =A{e}^{i\vec{k}\cdot \vec{r}}{e}^{ik\lambda }\end{array}}$$ This is true if $$e^{i\lambda k}=1=e^{i2\pi }$$ or $$\lambda k=2\pi$$ $$k=\frac{2\pi }{\lambda }$$ This should have a familiar look to it! Finally we want these waves to propagate in time so you should be able to guess the answer from our work onmechanical waves $$\psi (r)=A{e}^{i\left(\vec{k}\cdot \vec{r}\mp \omega t\right)}$$