I have following question about creation and annihilation operators in QFT: The Klein-Gordon field is introduced as continuous interference of plane waves $\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}$ with positive energy (resp $\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})}$ with negative energy):
$$\varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[a(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + b^\dagger(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right].$$
Then, when we quantizing the coefficients $b_{\mathbf{p}}^{\dagger}$ and $a_{\mathbf{p}}$ we get
$$\varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[\hat{a}(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + \hat{b}^\dagger(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right].$$
with the annihilation operator $\hat{b}_{\mathbf{p}}^{\dagger}$ and creation operator $\hat{a}_{\mathbf{p}}$ ???
Previously I asked a question concerning distinguishing the annihilation and creation operators in this expression here: Creation and Annihilation Operators in QFT
and got indeed good answers.
But the point of this question is the following:
Earlier I asked the same question my prof and he gave seemingly a more easy/intuitive answer that $\hat{b}_{\mathbf{p}}^{\dagger}$ must correspond to the exponential with positive energy:
He used the argument that $\hat{b}_{\mathbf{p}}^{\dagger}$ must be annihilation since the inititial state must have positive energy.
Can anybody decrypt how to interpret this argument? I don't understand exactly this line of thought. What is here the initial state? The exponential with positive energy?
Does he mean an evaluation argument like $<f | \phi(x) | i>= e^{\text{blabla}}$ as neccessary condition?
Best Answer
Assuming that you are quantizing a free massive complex scalar field, whose Hamiltonian is:
$H=\int d^{D-1} x (\partial_{\mu}\varphi^{\dagger} \partial^{\mu}\varphi+m^2\varphi^{\dagger}\varphi) $
you can see easily that if you plugin for the expression of the field in terms of the creation/annihilation operators stated above and do the integration over $x$, you will obtain the expression:
$H=\int d^{D-1} k \hspace{0.1cm}\omega_{k}(a^{\dagger}_{\mathbf{k}}a_{\mathbf{k}}+b^{\dagger}_{\mathbf{k}}b_{\mathbf{k}}) $
(the cross terms should vanish upon integration). Now the particles created by $a^{\dagger}$, $b^{\dagger}$ carry different charge under the U(1) symmetry and that's why we distinguish between the operators a and b unlike the case of the real scalar, where the particle is it's own antiparticle. Say that you wanted to quantize slightly differently with
$$\varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[a^{\dagger}(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + b(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right].$$
$a\rightarrow a^{\dagger}$, $b^{\dagger}\rightarrow b$,
you would get a Hamiltonian which is not normal-ordered, and thus the energy of the ground state would be infinite because the constant that you gain from ordering the Hamiltonian the right way is infinite, as you can see below.
$H=\int d^{D-1} k \hspace{0.1cm}\omega_{k}(a_{\mathbf{k}}a^{\dagger}_{\mathbf{k}}+b_{\mathbf{k}}b^{\dagger}_{\mathbf{k}})=\int d^{D-1} k \hspace{0.1cm}\omega_{k}(a^{\dagger}_{\mathbf{k}}a_{\mathbf{k}}+b^{\dagger}_{\mathbf{k}}b_{\mathbf{k}})+\int d^{D-1} k \hspace{0.1cm}\omega_{k}$
I guess your professor insisted on this interpretation of "positive" energy to avoid introducing normal ordering (the operation that puts all creation the left of annihilation operators).