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.

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