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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.

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    $\begingroup$ The second equation you wrote is a special case of the first. Let $f(\omega_1,\omega_2) = f(\omega_1)\delta(\omega_1 - \omega_2)$. $\endgroup$
    – hft
    Commented Mar 27, 2023 at 18:08
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    $\begingroup$ 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...). $\endgroup$
    – hft
    Commented Mar 27, 2023 at 18:09
  • $\begingroup$ Also, you probably want the RHSs of your equations to be operating on some ket... Like the vacuum state... $\endgroup$
    – hft
    Commented Mar 27, 2023 at 18:10

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What you've done is maybe acceptable, but it's a bit odd. You've entangled the two photons rather than giving them both the same distribution of frequencies. The state you get can be described as "two photons either both have frequency A, or they both have frequency B, or they both have frequency C...". So if you measured one photon's frequency, the other's frequency would also collapse to the same frequency.

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.

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