# Why is photon annihilation associated with the POSITIVE frequency component of the electric field?

I'm reading Glauber's paper "The quantum theory of optical coherence". In his work he does not introduce the annihilation and creation operators, but he refers instead to the positive and negative frequency components of the electric field (which probably are proportional to the former, I guess).

He says that "...the positive frequency part $E^{(+)}(\mathbb{r}t)$, may be shown to be a photon annihilation operator..." and then cites Dirac book "The principles of quantum mechanics", 3rd ed. pp 239-242, but even from the reference I couldn't understand why the positive frequency components is associated with the photon annihilation.

PS with the capitalization I don't mean to yell but to stress on the term "positive".

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If you look at a classical field, you have $2$ possibilities for a field "corresponding" to a positive energy $\omega_k$ : $\Phi_{\pm(x,t)} = e^{\pm i(\omega_kt-\vec k.\vec x)}$. Now, you have to choose a standard, and you may think to the Schrodinger equation where the energy operator is $i \frac{\partial}{\partial t}$ (in units $\hbar=1$). So the standard solution, for a positive energy $\omega_k$ is $e^{- i(\omega_kt-\vec k.\vec x)}$

Now, we want to retrieve some taste of these classical fields by looking at the action of the field operator $A_\mu(x)$ sandwiched between a one-particle state $|k, \lambda\rangle = a^\dagger(k,\lambda) |0\rangle$ and the vaccuum, so we want :

$\epsilon^\lambda_\mu(k) e^{- i(\omega_kt-\vec k.\vec x)} = \langle 0|A_\mu(x)|k, \lambda\rangle$

One sees, that, the decomposition :

$A_\mu(x) = \int d \tilde k \sum\limits_\lambda(\epsilon^\lambda_\mu(k) a(k,\lambda) e^{- i(\omega_kt-\vec k.\vec x)} + (\epsilon^\lambda_\mu(k))^* a^\dagger(k,\lambda) e^{+ i(\omega_kt-\vec k.\vec x) })$

does the job (using the properties of the creation/annihilation operators).

So, annihilation operators are associated to positive energies.

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