# Klein-Gordon equation: plane wave solutions

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problem 3.1.

The Klein-Gordon equation was one of the first attempts at producing a relativistic quantum theory. In natural units, the equation is

$\displaystyle \left(\partial_{\mu}\partial^{\mu}+m^{2}\right)\phi=0 \ \ \ \ \ (1)$

This equation also results from the Euler-Lagrange equation for a scalar field ${\phi}$with Lagrangian

$\displaystyle \mathcal{L}=\frac{1}{2}\left(\partial_{\mu}\phi\right)\left(\partial^{\mu}\phi\right)-\frac{1}{2}m^{2}\phi^{2} \ \ \ \ \ (2)$

This is the Lagrangian for zero potential ${V\left(\phi\right)=0}$.

To write solutions to equation 1, we can introduce some new notation. In natural units, the four-momentum is

 $\displaystyle p_{\mu}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} E & p_{i}\end{array}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} E & -p^{i}\end{array}\right] \ \ \ \ \ (4)$

The scalar product of four-momentum with a spacetime vector ${x^{\mu}}$ is therefore

$\displaystyle px\equiv p_{\mu}x^{\mu}=Et-\mathbf{p}\cdot\mathbf{x} \ \ \ \ \ (5)$

For a plane wave with angular frequency ${\omega}$, Planck’s relation is ${E=\hbar\omega=\omega}$, and the wave vector ${\mathbf{k}}$ has components in the three spatial directions of ${2\pi/\lambda_{i}}$, where ${\lambda_{i}}$ is the component of the wavelength in direction ${x_{i}}$. For example, a wave moving in the ${x_{1}}$ direction has ${\lambda_{2}=\lambda_{3}=\infty}$, so ${\mathbf{k}=\left[k_{1},0,0\right]}$. The four-vector ${k^{\mu}}$ is

$\displaystyle k^{\mu}=\left[\omega,\mathbf{k}\right] \ \ \ \ \ (6)$

and since ${\mathbf{p}=\mathbf{k}}$ in natural units, we have

$\displaystyle kx=k_{\mu}x^{\mu}=p_{\mu}x^{\mu}=px \ \ \ \ \ (7)$

A plane wave solution to 1 turns out to be

$\displaystyle \phi=\sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}e^{-ikx}+B_{\mathbf{k}}^{\dagger}e^{ikx}\right) \ \ \ \ \ (8)$

We can see this by direct substitution. Consider one term ${\phi_{\mathbf{k}}}$ from the sum. Then

 $\displaystyle \phi_{\mathbf{k}}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}e^{-ikx}+B_{\mathbf{k}}^{\dagger}e^{ikx}\right)\ \ \ \ \ (9)$ $\displaystyle \partial^{\mu}\phi_{\mathbf{k}}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(-ik_{\mu}A_{\mathbf{k}}e^{-ikx}+ik_{\mu}B_{\mathbf{k}}^{\dagger}e^{ikx}\right)\ \ \ \ \ (10)$ $\displaystyle \partial_{\mu}\partial^{\mu}\phi_{\mathbf{k}}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(-k_{\mu}k^{\mu}A_{\mathbf{k}}e^{-ikx}-k_{\mu}k^{\mu}B_{\mathbf{k}}^{\dagger}e^{ikx}\right) \ \ \ \ \ (11)$

However, using the invariant scalar ${p_{\mu}p^{\mu}}$ from relativity:

$\displaystyle k_{\mu}k^{\mu}=p_{\mu}p^{\mu}=E^{2}-p^{2}=m^{2} \ \ \ \ \ (12)$

Thus

$\displaystyle \partial_{\mu}\partial^{\mu}\phi_{\mathbf{k}}=-m^{2}\phi_{\mathbf{k}} \ \ \ \ \ (13)$

so 1 is true for a single component ${\phi_{\mathbf{k}}}$. Since the solution 8 is a linear combination of such solutions, and the original differential equation is linear, then 8 is also a solution. [The normalization factor ${1/\sqrt{2V\omega_{\mathbf{k}}}}$ is irrelevant in proving that 8 is a solution; it’s just there to make future calculations easier.]

Note that the first term (involving ${A_{\mathbf{k}}}$) is also a solution of the free-particle Schrödinger equation, which is

$\displaystyle i\frac{\partial\phi_{S}}{\partial t}=-\frac{1}{2m}\nabla^{2}\phi_{S} \ \ \ \ \ (14)$

If we take

$\displaystyle \phi_{S}=\sum_{\mathbf{k}}A_{\mathbf{k}}e^{-ikx} \ \ \ \ \ (15)$

then

 $\displaystyle i\frac{\partial\phi_{S}}{\partial t}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}E_{\mathbf{k}}A_{\mathbf{k}}e^{-ikx}\ \ \ \ \ (16)$ $\displaystyle -\frac{1}{2m}\nabla^{2}\phi_{S}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{k^{2}}{2m}A_{\mathbf{k}}e^{-ikx} \ \ \ \ \ (17)$

For a free particle, the energy ${E_{\mathbf{k}}=\frac{p^{2}}{2m}=\frac{k^{2}}{2m}}$, so the Schrödinger equation is satisfied. The second term (with ${B_{\mathbf{k}}^{\dagger}}$) does not satisfy the Schrödinger equation, since in that case we get

 $\displaystyle \phi_{S}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}B_{\mathbf{k}}^{\dagger}e^{ikx}\ \ \ \ \ (18)$ $\displaystyle i\frac{\partial\phi_{S}}{\partial t}$ $\displaystyle =$ $\displaystyle -\sum_{\mathbf{k}}E_{\mathbf{k}}B_{\mathbf{k}}^{\dagger}e^{ikx}\ \ \ \ \ (19)$ $\displaystyle -\frac{1}{2m}\nabla^{2}\phi_{S}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{k^{2}}{2m}B_{\mathbf{k}}^{\dagger}e^{ikx} \ \ \ \ \ (20)$

The extra minus sign means the two sides don’t match.