# A Naive Question of Canonical Quantization A Real Scalar Field

When we use canonical quantization method to quantize a free real scalar field in Schrodinger Picture. The free real scalar Lagrangian is $$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2$$ The EOM derived from this Lagrangian is \begin{align} (\partial_\mu\partial^{\mu} + m^2)\phi = 0 \end{align} If we do the Fourier transform from $$\phi(\vec{x}, t)$$ to $$\phi(\vec{p}, t)$$ as \begin{align} \phi(\vec{p}, t) = \int \mathrm{d}^3\vec{x} \phi(\vec{x}, t)e^{-i \vec{p}\cdot\vec{x}}\\ \phi(\vec{x}, t) = \int \frac{\mathrm{d}^3\vec{p}}{(2\pi)^3} \phi(\vec{p}, t)e^{i \vec{p}\cdot\vec{x}} \end{align} Then the EOM in momentum space is written as $$\left[ \frac{\partial^2}{\partial t^2} + (\vec{p}^2 + m^2) \right]\phi(\vec{p}, t) = 0$$ Thus, for each value of $$\vec{p}$$, $$\phi(\vec{p}, t)$$solves this equation simple harmonic oscillator vibrating at frequency: $$\omega_{\vec{p}} = \sqrt{\vec{p}^2 + m^2}$$ When I quantize a Simple Harmonic Oscillator in Quantum Mechanics as defining $$a = \sqrt{\frac{\omega}{2}}q + \sqrt{\frac{i}{2\omega}}p \quad ~, \quad a^{\dagger} = \sqrt{\frac{\omega}{2}}q - \sqrt{\frac{i}{2\omega}}p$$ And $$q = \sqrt{\frac{1}{2\omega}}(a + a^{\dagger}) \quad ~, \quad p = - i \sqrt{\frac{\omega}{2}}(a - a^{\dagger})$$ And if we do the similiar thing in QFT as writing $$\phi(\vec{p}) = \sqrt{\frac{1}{2\omega_{\vec{p}}}}(a_{\vec{p}} + a^{\dagger}_{\vec{p}})$$ It's manifestly wrong( Because we need a real $$\phi(\vec{x})$$, which needs $$\phi^{*}(\vec{p}) = \phi(-\vec{p})$$). I want to know why we need to write the $$\phi(\vec{p})$$ as $$\phi(\vec{p}) = \sqrt{\frac{1}{2\omega_{\vec{p}}}}(a_{\vec{p}} + a^{\dagger}_{-\vec{p}})$$

By the way, I know the $$\int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{\sqrt{2 \omega_{\vec{p}}}}\left[a_{\vec{p}}+a_{-\vec{p}}^{\dagger} \right] e^{i \vec{p} \cdot \vec{x}} = \int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{\sqrt{2 \omega_{\vec{p}}}}\left[a_{\vec{p}} e^{i \vec{p} \cdot \vec{x}}+a_{\vec{p}}^{\dagger} e^{-i \vec{p} \cdot \vec{x}}\right]$$

If the field $$\phi(x)$$ is assumed to be real as it is in this question a property of Fourier transforms (FT) requires, that the FT of a real function complies with the symmetry property:

$$\tilde{\phi^\ast}(p) = \tilde{\phi}(-p)$$

This can be checked be consulting properties of Fourier transforms. For complex functions it is no longer valid.

EDIT: If the field operator is now as chosen as:

$$\tilde{\phi}(\vec{p}) = \sqrt{\frac{1}{2\omega_\vec{p}}}( a_\vec{p} + a^\dagger_{-\vec{p}})$$

it can be easily seen that $$\tilde{\phi}^{\dagger}(\vec{p}) =\tilde{\phi}(-\vec{p})$$ (the above identity for fourier coefficients translates in the formalism of operators to a replacement of $$\ast \rightarrow \dagger$$) is fulfilled as required:

$$\tilde{\phi}^\dagger(\vec{p}) = \sqrt{\frac{1}{2\omega_\vec{p}}}( a^\dagger_\vec{p} + a_{-\vec{p}})$$

and

$$\tilde{\phi}(-\vec{p}) = \sqrt{\frac{1}{2\omega_\vec{p}}}( a_{-\vec{p}} + a^\dagger_\vec{p})$$

$$\omega_\vec{p}$$ does not change as $$\vec{p}$$ enters in its formula quadratically.