Ansatz of two-photon state

I am confused regarding the ansatz of the two-photon state. Generally, the two-photon state in the frequency domain is： $$|\psi\rangle=\int\int d\omega_1d\omega_2 f(\omega_1,\omega_2) a^+(\omega_1)a^+(\omega_2)|vac\rangle\,,$$ where $$f(\omega_1,\omega_2)$$ is the probability function, $$a$$ is the annihilation operator of photons.

I would like to know, if the frequencies of the two photons are the same, can we write the state in the following form? $$|\psi\rangle=\int d\omega f'(\omega) a^+(\omega)a^+(\omega)|vac\rangle\,,$$ where $$f'(\omega)$$ and $$f(\omega_1,\omega_2)$$ are different functions.

Any help would be appreciated.

• The second equation you wrote is a special case of the first. Let $f(\omega_1,\omega_2) = f(\omega_1)\delta(\omega_1 - \omega_2)$.
– hft
Commented Mar 27, 2023 at 18:08
• Also, your $f$ is a different function in the first vs second equations. I am using this same convention. (I.e., use the same symbol $f$ to denote the function, but differentiating between the two based on the number of arguments...).
– hft
Commented Mar 27, 2023 at 18:09
• Also, you probably want the RHSs of your equations to be operating on some ket... Like the vacuum state...
– hft
Commented Mar 27, 2023 at 18:10

I suspect what you meant to do was more like $$|\psi\rangle=\int d\omega_1d\omega_2 f(\omega_1)f(\omega_2)a^\dagger(\omega_1)a^\dagger(\omega_2)|\text{vac}\rangle$$ So both photons have the same frequency distribution - the two photons have the same wavefunction, but aren't entangled in any way.