In Peskin and Schroeder, they quantize the Klein-Gordon field in the following way. They write the Fourier transform of $\phi(x,t)$ $$ \phi(x,t)=\int \frac{d^3 p}{(2\pi)^3}e^{ipx}\phi(p,t) $$ after that they say that the Klein-Gordon equation reduces to $$ \left(\frac{\partial^2}{\partial t^2}+\omega_p\right)\phi(p,t)=0 $$ We have an harmonic oscillator for each $p$. They recall how to quantize an harmonic oscillator saying that for the following Hamiltonian $$ H=\frac{1}{2}p^2+\frac{1}{2}\omega^2 \phi^2 $$ we have $$ \phi=\frac{1}{\sqrt{2\omega}}(a+a^{\dagger}) $$ and that we have therefore $$ \phi(x)=\int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p e^{ipx}+a_p ^\dagger e^{-ipx})\, . $$

What I don't understand is that we should have $$ \phi(p)=\frac{1}{\sqrt{2\omega_p}}(a_p + a_p ^\dagger) $$ and then remplacing in the Fourier expansion $$ \phi(x)=\int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p+a_p ^\dagger)e^{ipx} $$ by this method if we are doing an analogy with a single harmonic oscillator. I understand that this can’t be correct but I don't see why, with this particular method we have this result. I understand that by solving the Klein-Gordon equation you get both types of plane wave but I’m having trouble understanding this particular method of quantizing the field.


1 Answer 1


To connect to the simple harmonic oscillator, note that the simplest Lorentz-invariant equation of motion a field can satisfy is
$\square \phi = (\partial_t^2 - \vec \nabla^2) \phi = 0$

Classical solutions are plane waves, e.g.
$\phi (t, \vec x) = a_p (t) e^{i \vec p \cdot \vec x}$
$(\partial_t^2 + \vec p \cdot \vec p) a_p (t) = 0$
which is the equation of motion of a harmonic oscillator,
but also
$\phi^* (t, \vec x) = a_p^* (t) e^{-i \vec p \cdot \vec x}$

A general solution is
$\phi (t, \vec x) = \int \frac{d^3p}{(2 \pi)^3} [a_p (t) e^{i \vec p \cdot \vec x} + a_p^* (t) e^{-i \vec p \cdot \vec x}]$
or explicitating $a_p (t) = a_p e^{-i \omega_p t}$
$\phi (x) = \int \frac{d^3p}{(2 \pi)^3} (a_p e^{-i p x} + a_p^* e^{i p x})$
$\eta_{\mu \nu} = diag(1, -1, -1, -1)$ metric tensor in Minkowski spacetime
$p^\mu = (\omega_p, \vec p)$
$\omega_p = |\vec p|$
$x^\mu = (t, \vec x)$


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