# How does taking the modulus of the P-function of an optical state affect the trace norm?

I have two density matrices in the coherent-state basis:

$$\rho_i = \int_{\mathbb{C}}P_i(\alpha) \left|\alpha\right\rangle\!\left\langle\alpha\right| \textrm{d}^2\alpha$$

where $i=1,2$. I want to find the trace distance between these, so I look at the operator:

$$\sigma \equiv \rho_1-\rho_2 = \int_{\mathbb{C}}f(\alpha) \left|\alpha\right\rangle\!\left\langle\alpha\right| \textrm{d}^2\alpha$$

where $\!\!\quad f(\alpha) \equiv P_1(\alpha)-P_2(\alpha)$, and I want to find

$$\lVert \sigma \rVert_1 = \textrm{Tr}\left(\left|\sigma\right|\right)$$

where $\left|\sigma\right| = \sqrt{\sigma^\dagger\sigma}$.

Since this is difficult to directly calculate, I hypothesise that $\left|\sigma\right|$ may be equal to $\bar{\sigma}$ (or at least less than it, in the sense of the trace norm), where I have defined

$$\bar{\sigma} = \int_{\mathbb{C}}\left|f(\alpha)\right| \left|\alpha\right\rangle\!\left\langle\alpha\right| \textrm{d}^2\alpha$$

although I don't know how to go about showing this (or finding a counter example if it is not the case).

You can instantly get a counterexample by choosing two rank-1 states $\rho_1=|0\rangle\langle0|$ and $\rho_2=|\alpha\rangle\langle\alpha|$. Then, $\mathrm{tr}\,\bar\sigma=2$, while $\|\rho_1-\rho_2\|=2\sqrt{1-|\langle0|\alpha\rangle|^2}$. (The latter follows from the fact that the $\rho_1-\rho_2$ is supported in the 2-dimensional subspace spanned by $|0\rangle$ and $|\alpha\rangle$, and is fully determined by their overlap.)
On the other hand, as AccidentalFourierTransform pointed out, the inequality $$\mathrm{tr}\,\sigma\le \mathrm{tr}\,\bar\sigma$$ follows right away from the triangle inequality.