Klein-Gordon Field – Why Can the Klein-Gordon Field Be Fourier Expanded in Terms of Ladder Operators?

Question

Using the plane wave ansatz
$$\phi(x) = e^{ik_\mu x^\mu}$$
the solution to the Klein-Gordon equation $(\Box + m^2) \phi(x) =0$ can be written as a sum of solutions, since the equation is linear and the superposition principle holds, as
$$\phi(x) = \sum_{{k}} \left( Ae^{ik_\mu x^\mu} + Be^{-ik_\mu x^\mu} \right).$$
How does one find the coefficients? More exactly, why does it turn out they are the annihilation and creation operators with the factor $1/\sqrt{2E}$?

The various books and sources I've checked just confused me even more. Peskin and Schroeder just plug in the integral equation (Fourier modes) by analogy with the harmonic oscillator solution. Schwartz gives a very strange reason that the energy factor is just for convenience. In Srednicki the author writes it as $f(k)$ without an explicit form. In Mandl and Shaw, they just state the equation without any justification.

My best guess is that those come from the quantization process, but how does one do it in this case explicitly?

Answer 1

You might feel more comfortable if we run the reasoning 'in reverse' a bit. I'm basically just recounting what all the standard texts do, but without jumping ahead in the interpretation.

• We begin with the canonically quantized Klein-Gordon field $\phi(\mathbf{x})$ and its conjugate momentum $\pi(\mathbf{x})$, so that $[\phi(\mathbf{x}), \pi(\mathbf{y})] \propto i \delta(\mathbf{x}-\mathbf{y})$.
• We know that generally physics is nicer in Fourier space, especially for free fields where plane waves are classical solutions, so we define the Fourier transforms $\phi(\mathbf{p})$ and $\pi(\mathbf{p})$.
• By analogy with the harmonic oscillator, we suspect that the quantity $$a_{\mathbf{p}} \propto \phi(\mathbf{p}) + i \pi(\mathbf{p})/\omega_p$$ will be nice to work with. At this stage, $a_{\mathbf{p}}$ is just some operator with no physical interpretation.
• When we compute the commutation relations, we find $$[a_{\mathbf{p}}^\dagger, a_{\mathbf{q}}] \propto \delta(\mathbf{p}-\mathbf{q})$$ with an annoying energy-dependent proportionality constant, and all other commutators zero.
• From undergraduate quantum mechanics, these commutation relations imply that applying $a_{\mathbf{p}}^\dagger$ gives a ladder of states for each value of $\mathbf{p}$. Then we physically interpret $(a_{\mathbf{p}}^\dagger)^n |0 \rangle$ to contain $n$ particles of momentum $\mathbf{p}$, so that $a_{\mathbf{p}}^\dagger$ really is a creation operator, i.e. it creates particles.
• The only issue is that these states are not normalized; we get an annoying energy-dependent normalization factor. (To see this, try calculating the norm of $a_{\mathbf{p}}^\dagger |0 \rangle$ using the commutation relation above.) Tracing backwards, we find that it goes away if we replace $a_{\mathbf{p}}$ with $\sqrt{\omega_{\mathbf{p}}} a_{\mathbf{p}}$, recovering the standard definition.

When Peskin does this whole thing in one step, you should view it as just a very inspired definition (where we've outlined where the 'inspiration' might come from above), not a statement that needs to be proved. It's a more advanced version of guessing a sinusoid to solve the harmonic oscillator.