# Decoherence of one-photon states

Let $$\rho_1$$ be the pure one-photon state described by the ket $$|\psi_1\rangle = \int dk\ A(k)a^\dagger(k)|0\rangle$$ for a complex amplitude function $$A(x)$$ and an empty ket $$|0\rangle$$. This state is obviously coherent, since non-diagonal terms with amplitudes $$A(k)A^*(k')$$ contribute to the density matrix.

The main tools to describe loss of coherence that I know of are the depolarizing and the dephasing channels. However, most treatments deal only with their application to a qubit state. How does one describe partial decoherence of such a state (preferably, in a calculation-friendly way)?

A dephasing channel acts on the state $$\rho$$ as $$\rho\to \frac{1}{2\pi}\int d\theta e^{i\theta a^\dagger a}\rho e^{-i\theta a^\dagger a}.$$ Here $$a^\dagger a$$ is the photon-number operator for a particular mode that gets dephased from the rest. You can see that the channel retains components that commute with $$a^\dagger a$$, because $$a^{\dagger l}|0\rangle\langle 0| a^m\to e^{i\theta(l-m)}a^{\dagger l}|0\rangle\langle 0| a^m$$. So you can dephase a particular mode from the rest by using $$\rho\to \frac{1}{2\pi}\int d\theta e^{i\theta a^\dagger(k) a(k)}\rho e^{-i\theta a^\dagger(k) a(k)}$$ to remove terms of the form $$|k\rangle\langle k^\prime|$$ with $$k^\prime\neq k$$ from $$\rho$$. To dephase all of the modes, we need an integral for each $$k$$: $$\rho=\int dk_1 dk_2 A(k_1)A^*(k_2)|k_1\rangle\langle k_2|\to \prod_k\frac{\int d\theta_k}{2\pi} e^{i\theta_k a^\dagger(k)a(k)}\rho e^{-i\theta_k a^\dagger(k)a(k)}=\int dk_1 |A(k_1)|^2 |k_1\rangle\langle k_1|.$$ There's probably a fancy way to write this integral over a continuous number of parameters but I don't have it offhand. A partially dephasing channel would mix this channel with an identity channel, such that with probability $$p$$ this channel is applied and with probability $$1-p$$ nothing happens to the state.
Completely depolarizing channels replace the state with the maximally mixed state. This is also a challenge with a continuous number of parameters, because that state is an identity matrix over all states and must be normalized: $$\rho\to \frac{\int d_k a^\dagger (k)|0\rangle\langle 0|a(k)}{\int d_k }=\frac{\mathbb{I}}{\mathrm{Tr}\mathbb{I}},$$ where the denominator goes to infinity. This channel has Kraus operators that take every possible state to every other possible state, so it will look like $$\rho\to\propto \int dk_1 dk_2 |k_1\rangle\langle k_2|\rho |k_2\rangle\langle k_1|.$$ Again, partially depolarizing channels apply this with some probability $$p$$.
• So, to be explicit, partial decoherence is given by the evolution $$\rho\to (1-p) \int dk_1dk_2A(k_1)A^*(k_2)|k_1\rangle\langle k_1|+p\int dk |A(k)|^2|k\rangle\langle k|$$ under the action of the dephasing channel, where $p\in[0,1]$ quantifies the amount of remaining coherence? Commented Jan 26 at 18:08
• @Bentanglement yes ish; that is explicitly partial dephasing where the dephasing acts equally on each mode. Yes that is a type of decoherence, but sometimes people mean different things when they say decoherence. You can see how each term $A(k_1)A^*(K_2)|k_1\rangle\langle k_2|$ gets multiplied by $(1-p)+p\delta(k-k^\prime)$; an even more general model that dephases different mode combinations by different amounts could multiply each term by $(1-p(k_1,k_2))+p(k_1,k_2)\delta(k-k^\prime)$ for some function $p(k_1,k_2)$ Commented Jan 26 at 19:53