# How creation operator pops out while expanding field operator?

While doing QFT when we try to canonically quantize the Klein Gordon equation $$\Box \phi =0$$ we promote the $$\phi$$ to an operator field and impose the commutation rule $$[\phi(x,t),\pi (y,t)]=i\hbar\delta^3(x-y)$$.

So my doubt is following when we try to do expansion of $$\phi$$ in terms of $$f(x,t)$$ which are c-numbered scalar field and satisfy $$\Box f=0$$. All of sudden the creation and annihilation operator pops up i.e. $$\phi= \Sigma (af(x,t) + a^\dagger f(x,t))$$ I am right now going for box normalization so the $$\Sigma$$ sign is present instead of $$\int$$ sign. Since the $$\phi$$ field is a hermitian field therefore it has to be such that $$\phi ^\dagger = \phi$$ and since $$\phi$$ is also an operator it's expansion has to involve operator so if go for operators $$a$$ and $$a^\dagger$$ such that $$[a,a^\dagger]=\frac{1}{2}$$ then we retrieve the original $$\phi$$ and $$\pi$$ commutation relation.

So is this the reason that above expansion is the way it is i.e. we guessed the most simple and consistent expansion there possible and it worked out.

Also is there any text in the literature which tackles with the above idea of expanding an operator field in terms of its equivalent c-numbered field ?

You can think of the 'expansion' of $$\phi$$ as a Fourier transform. $$\phi(x) = \int d^4p \;\tilde{\phi}(p) \;e^{-i x_\mu p^\mu}$$
In order to satisfy the wave equation the Fourier variables need to satisfy $$p_\mu p^\mu = m^2$$. Thus we have two 'branches'
$$\phi(x) = \int d^4p \;\tilde{\phi}(p) \;\delta(p_\mu p^\mu -m^2)\;e^{-i x_\mu p^\mu} = \int \frac{d^3p}{2 E_p}\left( \tilde{\phi}(p) e^{-i x_\mu p^\mu}|_{p^0 = E_p} + \tilde{\phi}(p) e^{-i x_\mu p^\mu}|_{p^0 = -E_p} \right)$$ $$E_p = +\sqrt{\vec{p}^2+m^2}$$ Temporarily define $$\tilde{\phi}_1(p)\equiv\tilde{\phi}(p)|_{p^0 = E_p}$$ and $$\tilde{\phi}_2(p)\equiv\tilde{\phi}(p)|_{p^0 = E_p}$$. Because $$\phi$$ is Hermition, we must have $$\tilde{\phi}_1^\dagger = \tilde{\phi}_2$$. Now, the true moment of (canonical) 'quantisation' is when we promote the fields to operators and impose the commutation relations $$[\phi(\vec{x},t),\Pi(\vec{y},t)] = i \delta(\vec{x}-\vec{y}), \quad \quad \text{and all others equal zero.}$$ In terms of the expansion above it is of course the Fourier coefficients that inherit the new operator nature of $$\phi$$. One can write the Fourier coefficients in terms of the fields, e.g. $$\tilde{\phi}_1(p)= \int d^3x e^{ix^\mu p_\mu}(E_p \phi+i\Pi)$$ and now, using the commutators imposed on the fields we can derive the commutation relations for $$\tilde{\phi}_1(p)$$ and $$\tilde{\phi}_1(p)^\dagger$$ and we find that they are exactly those satisfied by the creation/annihilation operators in the Harmonic oscillator problem. Thus we can identify them as such $$a_p = \tilde{\phi}_1(p)$$
• @aitfel, Sorry for the possibly confusing use of word. I simply mean that the equation $p_\mu p^\mu = m^2$ has two distinct types of solutions ('branches'): $p^0 = \pm \sqrt{\vec{p}^2 + m^2}$. So it makes sense that in the Fourier transform of $\phi(x)$ we separate them out explicitly. In my equation this is done explicitly mathematically by the delta function integrated over $p^0$, which becomes a sum over the (two) zeros of $p_\mu p^\mu = m^2$ – Rudyard Sep 21 at 15:11