# Klein Gordon Field Quantization: why this is the correct way to express the field?

I'm reading a book in QFT and the first thing tackled is the quantization of the Klein Gordon Field. The classical Klein Gordon field satisfies the partial differential equation

$$(\partial^\mu\partial_\mu+m^2)\phi=0 \Longrightarrow (\partial_t^2-\nabla^2+m^2)\phi=0.$$

Taking the Fourier transform we get

$$(\partial_t^2+\omega_p^2)\hat{\phi}=0,$$

where now $\omega_p^2 = p^2+m^2$. In other words $\hat{\phi}(p,t)$ satisfies the simple harmonic oscilator equation of motion for each fixed $p$. In this case we have

$$\phi(x,t)=\int d^3p \dfrac{1}{(2\pi)^3}\hat{\phi}(p,t)e^{ix\cdot p}.$$

That is fine, and is all classical. Now, we want to quantize the field. As the book explains, quantizing the field means promoting $\phi(\mathbf{x},t)$ to an operator, such that

$$[\phi(\mathbf{x}),\phi(\mathbf{y})]=[\pi(\mathbf{x}),\pi(\mathbf{y})]=0,$$

$$[\phi(\mathbf{x}),\pi(\mathbf{y})]=(2\pi)^3\delta(\mathbf{y}-\mathbf{x}).$$

in the same way as we do with position $X$ and momentum $P$ in quantum mechanics.

Now, to do this the author uses the ladder operators from the quantum harmonic oscilator. In that case if $X$ and $P$ are position and momentum, the ladder oerators satisfies

$$X=\dfrac{1}{\sqrt{2\omega}}(a+a^\dagger), \quad P=-i\sqrt{\dfrac{\omega}{2}}(a-a^\dagger).$$

The author by analogy with this, then says that

$$\phi(x)=\int d^3p \dfrac{1}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_p}}(a_pe^{ix\cdot p}+a_p^\dagger e^{-ix\cdot p})$$ $$\pi(x)=\int d^3p \dfrac{1}{(2\pi)^3}(-i)\sqrt{\dfrac{\omega_p}{2}}(a_pe^{ix\cdot p}-a_p^\dagger e^{-ix\cdot p})$$

Now this is not at all clear to me. My main issues are:

1. First what leads to this expansion of the quantized field $\phi$? I mean, my guess is that the author considered $\hat{\phi}(p)$ behaves as the coordinate of a harmonic oscilator, so that we can write $\hat{\phi}(p)$ in terms of ladder operators $a_p$ and $a_p\dagger$. However if that is done we would get

$$\phi(x)=\int d^3p \dfrac{1}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_p}}(a_p+a_p^\dagger)e^{ix\cdot p}$$

This would be the analogy in my opinion, where we set $\hat{\phi}(p)$ as the position of the harmonic oscilator. Why do we get $(a_p e^{ix\cdot p}+a_p^\dagger e^{-ix\cdot p})$ instead?

1. If this analogy is being carried, why $\hat{\pi}(p)$ would be the momentum of the harmonic oscilator? I see no reason for this directly by the differential equation.

1. The $a_p$ and $a_p^\dagger$ are supposed to be Hermitian adjoints of each other, both classically and quantumly. Applying the complex conjugate to the usual expression, we see that it maps $a_p\mathrm{e}^{\mathrm{i}px}$ to $a_p^\dagger\mathrm{e}^{-\mathrm{i}px}$ and vice versa under the assumption that $a_p^\dagger$ is indeed the adjoint of $a_p$, so the expression is invariant under taking the adjoint, which is required by the the scalar field in question being real, i.e. $\phi(x)^\dagger = \phi(x)$. Your proposal, on the other hand, would require $(a_p)^\dagger = a_{-p}^\dagger$ - which is a possible choice for how you define the Fourier coefficients, but notationally rather confusing.
2. $\pi(p)$ is the analogue to the momentum variable of the harmonic oscillator simply because it fulfills the infinite-dimensional version of the correct commutation relation with the analogue of the position variable $\phi(p)$. It is not "the" momentum of "the" harmonic oscillator - the actual momentum operator in QFT is $$P^\mu = \int p^\mu a^\dagger_p a_p \frac{\mathrm{d}^3p}{(2\pi)^3}.$$